mm-9015 Subject: Re: Four-Color Theorem > Am I right in saying that there are no short proofs of the > four-color theorem to date? In 1976 Appel and Haken proved it using > computers, and in 1996 Robertson, Seymour and Thomas improved and > shortened the proof. Has there been any news since that time? > And what of Hadwiger's conjecture? > I ask because I want to prove them, hehe. > -Greg I thought that this theorem hadn't been proved. Do you have a link to these proofs. > I thought that this theorem hadn't been proved. Do you have a link to these proofs. Just use Google and search for Robertson Seymour Thomas. === Subject: Re: Great-circle radius of ellipsoid > The solution you have given is arbitrary and does not follow a > consistent derivation, and with that sort of reasoning, there would > be infinite solutions. > > The answer is simply: > > Area of circle = Area of ellipse > ^^^^^^^ > r = sqrt(ab). > Right, for an *ELLIPSE* (the left, semi-axes diagram)! > What is being discussed here is a 3-D ellipsoid (the right, > arcradii diagram). That being the case, the solution is merely: r = (product(i=1, n) Ai)^1/n where Ai = the i'th axis of the ellipsoid n = the number of dimensions of the ellipsoid r = radius of an n-dimensional spheroid > If this is wrong as you contend, explain why. > I think the problem is everyone is viewing the ellipsoid as being > composed of equi-sized meridional ellipses and/or meridionally > parallel semi-ellipses. That would be a very bad view, if indeed that were the case. It's like saying everyine is viewing a sphere as being composed of equi-sized circles. > While that may be *A* valid interpretation, it is not *THE* proper > interpretation for this discussion--the finding of the great > circle arcradius of best fit for the spheroid/ellipsoid. Since there is no definiton of best fit, I assume the best fit radius is such that the volume of the spheroid is that of the ellipsoid. Which would make the radius formula: r = (product(i=1, n) Ai)^1/n > As such, the relevant approach to analyzing the form of an ellipsoid > is to view it as being made up of unique great ellipses (GEs), > where a meridian is the maximally elliptic GE and the equuator is > the minimally elliptic GE, a great circle. How do those two ellipses constitute a great circle? > So, for a *3-D ELLIPSOID*, one is averaging the arcradii of all of > the different GEs, hence the simplest 2-point average would be > between a meridian (~=~ [a*b]^.5 or .5*[a+b] or [.5*(a^2+b^2)]^.5, > to name the most common)--the maximum GE--and the equator, which, > being a great circle and having an equal radius throughout, always > equals a. > GEs, as all geodetic lines travel along them, can also be viewed as > ARC PATHs (APs). > Let a = 6378.135 and b = 6356.75. > The mean meridional arcradius (Rr) = 6367.4469888342: > [a*b]^.5 = 6367.43352233; 2/[1/a + 1/b] = 6367.42454467; > .5 * [a + b] = 6367.4425; [.5 * (a^2 + b^2)]^.5 = 6367.45147766; > [.5 * (a^1.5 + b^1.5)]^(1/1.5) = 6367.4469888334; > The mid-GE possibilities work out to the following: > Lat1 = 0, Long2 - Long1 = 90¡, > Lat2 = 45¡, AP = 45.09601377¡, mean arcradius = 6372.80780848; > Lat2 = 45.09601444¡, AP = 45¡, mean arcradius = 6372.78989792; > (.5 * [6372.80780848 + 6372.78989792] = 6372.79885320) > Lat2 = AP = 45.04800712¡, mean arcradius = 6372.79885325 > (the isopathic mid-GE (the concept and process discussed in a separate > Keeping in mind that--given the minimal ellipticity involved--the > trapezoidal 3-point average provides an integration to about > .0000001, the global great circle arcradius of best fit using the > three different differentiating methods produce: > .25 * (Rr + 2*6372.80780848 + a) = 6372.7994014; > .25 * (Rr + 2*6372.78989792 + a) = 6372.7904462; > .25 * (Rr + 2*6372.79885325 + a) = 6372.7949238; > = .5 * [6372.7994014 + 6372.7904462] > Now compare the aforementioned approximations: > [a * Rr]^.5 = 6372.78875377; > [a^3 * b]^.5 = 6372.78201486; > .5 * [a + Rr] = 6372.79099442; > [.25 * (3a^2 + b^2)]^.5 = 6372.79547760; > Wouldn't you agree that [.25 * (3a^2 + b^2)]^.5 is by far the best > approximation--if not another ACTUAL global mean, via some other > differentiation? > Letting a = 10000, down to a b value of about 8385, the maximum > difference from the isopathic is -.74186, whereupon the deviation > reverses until they equate at a b value of about 7624.5 and > [.25 * (3a^2 + b^2)]^.5 remains greater than the isopathic for > lesser b values (at 7071.067 the difference is +1.94, and at 6000, > +11.03)--thus even if the isopathic IS the only true great circle > arcradius of best fit, for general purposes [.25 * (3a^2 + b^2)]^.5 > appears adequate. > As I noted in my original reply, given the relationship of > .5 * [a^2 + b^2] to the integrand^2 of the elliptic integral > of the second kind, there is still a good possibility that > [.25 * (3a^2 + b^2)]^.5 may be a closed-form reduction to another > legitimate great ellipse differentiation. > Or, rather than arcradius, perhaps [.25 * (3a^2 + b^2)]^.5 is the > global average of the *semi-axes* along each GE--i.e., change b to > Radius{Lat} for each GE and calculate Rr for each GE (thus Radius{0} > equals a--with Rr equaling a--and Radius{90¡} equals b--with > Rr equaling the meridional Rr)--but I haven't really explored that > approach, so it is just a random theory at this point. After all that, I'm afraid you haven't laid out your case clearly. It appears you are trying to obfuscate an otherwise simple problem. If I am wrong, define precisely the problem. > ~Kaimbridge~ > ----- > Wanted?Kaimbridge (w/mugshot!): > http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html > ---------- > Digitology?The Grand Theory Of The Universe: > http://www.angelfire.com/ma2/digitology/index.html > ***** Void Where Permitted; Limit 0 Per Customer. ***** === Subject: Re: Great-circle radius of ellipsoid >> While that may be *A* valid interpretation, it is not *THE* proper >> interpretation for this discussion--the finding of the great >> circle arcradius of best fit for the spheroid/ellipsoid. > Since there is no definiton of best fit, I assume the best fit > radius is such that the volume of the spheroid is that of the > ellipsoid. Huh? A spheroid *is* an ellipsoid, no? >> As such, the relevant approach to analyzing the form of an >> ellipsoid is to view it as being made up of unique great ellipses >> (GEs), where a meridian is the maximally elliptic GE and the >> equuator is the minimally elliptic GE, a great circle. > How do those two ellipses constitute a great circle? Because if you averaged the mean arcradius of those two and multipled the average by 2*Pi, you would have a mean GE which--since the mean arcradius thoughout that mean GE is constant--would be a great circle. > After all that, I'm afraid you haven't laid out your case clearly. > It appears you are trying to obfuscate an otherwise simple problem. > If I am wrong, define precisely the problem. What's not clear? Just continue what I did above--find the mean arcradius of ALL of the different GEs, average them together, and multiply the average by 2*Pi: You will have a great circle of best fit! Hence, divide that great circle of best fit by 2*Pi and you will have the mean arcradius of best fit! ~Kaimbridge~ ----- Wanted÷Kaimbridge (w/mugshot!): http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html ---------- Digitology÷The Grand Theory Of The Universe: http://www.angelfire.com/ma2/digitology/index.html ***** Void Where Permitted; Limit 0 Per Customer. ***** === Subject: Re: Infinite Series: Summation [(-1)^(n-1)][(2+sin n)/sqrt n] -> Converge? An afterthought - sorry for self-reply: > Does this series converge? By direct comparison to summation 1/sqrt n this > series is not absolutely convergent. But could it be conditionally > convergent? The absolute values of the terms do not approach zero > monotonically so ... ? > L > Divide and conquer: > sum[n=1 to inf]((-1)^(n-1)*2/sqrt(n)) converges > conditionally (standard alternating series test); > sum[n=1 to inf]((-1)^(n-1)*sin(n))/sqrt(n) > also converges conditionally, for a slightly more > subtle reason (Dirichlet's Test). > [Rest can be skipped] > The second series directly: > Denote x=pi-1, then sin(n*x) = (-1)^(n-1) * sin(n), > so the second series is > sum[n=1 to inf] sin(n*x)/sqrt(n) > and we manipulate the partial sums (N >= 2) > sum[n=1 to N] sin(n*x)/sqrt(n) > = cos(x/2) - cos((N+1/2)*x)/sqrt(N+1) > + 1/(2*sin(x/2)) * sum[n=1 to N-1] A(n) > where > A(n) = (1/sqrt(n) - 1/sqrt(n+1)) * cos((n+1/2)*x) > (check it out; it's called summation by parts). > The last sum (extended to infinity) is convergent > absolutely, and the initial terms converge as N > goes to infinity. > Oh, and sin(x/2) is different from 0. Exercise: Will the series sum[n=1 to inf] (-1)^(n-1) / (sqrt(n) + (-1)^(n-1)) converge or diverge? === Subject: Re: Simplification Assitance needed > .... > I just need to get the Theta (T) out on the right > and get the equation in terms of N and P so I can calulate Theta, > but can't for the life of me remember the trig simplifications. > Here it is: > .... > ArcTan(N/cos(T)) + ArcTan( (N*cos(T-P)-N+1) / (N*sin(T-P)) ) = 90 +P Just hard enough to be interesting but not intractable. :-) First, the way you've written denominators suggests that your N isn't 0, and that you're interested only in solutions with neither cos(T) nor sin(T - P) equal to 0. I'll make those assumptions. (1) Abbreviate the notation by writing a = N/cos(T), and b = (N.cos(T - P) - N + 1)/(N.sin(T - P)) so that your equation says arctan(a) + arctan(b) = 90 degrees + P. Take the tangent of both sides, using the addition formula tan(u + v) = (tan(u) + tan(v))/(1 - tan(u).tan(v)). Therefore (a + b)/(1 - ab) = tan(90 degrees + P) = - cot(P). Therefore ab - 1 = (a + b).tan(P). (2) It's convenient factorize part of that: (a - tan(P))(b - tan(P)) = 1 + (tan(P))^2 = (sec(P))^2. Multiplying by (cos(P))^2 then gives (a.cos(P) - sin(P))(b.cos(P) - sin(P)) = 1. Replace a and b using their definitions above, clear of fractions, and use the cosine addition formula cos(T - P).cos(P) - sin(T - P).sin(P) = cos((T - P) + P) = cos(T). That bit of algebra gives (N.cos(P) - sin(P).cos(T))(N.cos(T) + (1 - N).cos(P)) = N.cos(T).sin(T - P). = N.cos(T)(sin(T).cos(P) - cos(T).sin(P)) by the sine addition formula. Multiplying out and tidying up then leads to (N^2 + (N - 1).sin(P)).cos(P).cos(T) + N(1 - N).(cos(P))^2 = N.cos(P).sin(T).cos(T). (3) This is obviously true if cos(P) = 0, which means there is some integer k such that P = (k + 1/2)(pi). This simplifies equation (2) to a + b = 0, which easily leads to the solution sin(T) = N + ((-1)^k).(1 - 1/N) remembering P = (k + 1/2)(pi). But if cos(P) is non-zero like N (assumption (1)), then divide equation (3) by N.cos(P) to get (N + (1 - 1/N).sin(P)).cos(T) + (1 - N).cos(P) = sin(T).cos(T). (4) At this stage two more easy special cases turn up. The first has N = 1, which simplifies equation (4) to cos(T) = sin(T).cos(T), giving cos(T) = 0 or sin(T) = 1 (which implies cos(T) = 0 anyway). But the possibility that cos(T) = 0 was excluded at (1). The second special case has N + (1 - 1/N).sin(P) = 0, which means sin(P) = (N^2)/(1 - N). That simplifies equation (4) to sin(2T) = 2(1 - N).cos(P) = +/- 2.sqrt((1 - N)^2 - N^4) using sin(P) = (N^2)/(1 - N). But if you're not content with either of those special cases, then I'm afraid (4) leads to a quartic equation. Abbreviate the notation by writing N + (1 - 1/N).sin(P) = c and (1 - N).cos(P) = d, and also the unknown cos(T) = x. Then squaring both sides of (4) leads to (cx + d)^2 = (1 - x^2)(x^2), so x^4 + (c^2 - 1)(x^2) + 2cdx + d^2 = 0. If you don't want the special cases above (d = 0 or c = 0), then this quartic has no obvious simplification, so you'll need to solve it numerically using particular values of N and P (hence of c and d) that you're interested in. Now I need some coffee. :-) Ken Pledger. === Subject: Re: Simplification Assitance needed > But if you're not content with either of those special cases, then > I'm afraid (4) leads to a quartic equation. Abbreviate the notation by > writing > N + (1 - 1/N).sin(P) = c and > (1 - N).cos(P) = d, and also the > unknown > cos(T) = x. Then squaring both sides of (4) leads to > (cx + d)^2 = (1 - x^2)(x^2), so > x^4 + (c^2 - 1)(x^2) + 2cdx + d^2 = 0. Ken, that was a beautiful thing -- plumbing my memory from college. I need to plot a curve though. Hmmmm, I didn't know this was going to be so ugly. Darn it. I thank you for you solution. You obviously know what you're doing -- even before the coffee :) ... I guess I'll have to just estimate it in Excel with tiny slices and huge tables. > If you don't want the special cases above (d = 0 or c = 0), then > this quartic has no obvious simplification, so you'll need to solve it > numerically using particular values of N and P (hence of c and d) that > you're interested in. > Now I need some coffee. :-) > Ken Pledger. === Subject: Re: vonNeumann Gametheory of the OptimalStrategy playing the StockMarket; 20 more free shares SBC > Portfolio of PAF as of 25MAY04: > BCE 8,000 19.49 $155,920.00 > BLS 110 25.00 $2,750.00 > BMY 30 24.98 $749.40 > SBC 13,910 23.85 $331,753.50 Call me kooky, but AP looks a little over-weight in telecoms. Eh? -dl === Subject: Re: vonNeumann Gametheory of the OptimalStrategy playing the StockMarket; 20 more free shares SBC > Portfolio of PAF as of 25MAY04: > BCE 8,000 19.49 $155,920.00 > BLS 110 25.00 $2,750.00 > BMY 30 24.98 $749.40 > SBC 13,910 23.85 $331,753.50 > Call me kooky, but AP looks a little over-weight in telecoms. Eh? > -dl Overweight is rather a ridiculous concept for the Crossover technique. On 6May it can be claimed that PAF was overweight in drug company. But PAF amassed BMY because it was cheaper than SBC and peeled off the SBC to buy BMY. Now BMY is higher than SBC. But when SBC gets higher than BMY, I peel off the SBC to buy BMY. The trouble with April up to 6May was that I did not switch into more BMY such that the portfolio should have had just 80 shares of SBC and had 14,300 shares of BMY. Trouble with the April and May crossover was that I did not take a more full advantage of that Crossover. > Portfolio of PAF as of 6MAY04 : > BCE 7,350 20.44 $150,234.00 > BMY 7,530 26.13 $196,758.90 > SBC 6,780 25.21 $170,923.80 > 21,660 sum of units of share-wealth So in the Crossover the swing back and forth by definition is overweight. The concept of diversity and overweight are for the old foggy and stupid investment organizations who have no idea of *optimal strategy*, of *crossover*, of *employing two currencies*. Traditional investments such as over 95% of the mutual funds are outfits that gain in wealth only really during excessive Up markets and in Down markets lose their shirts and pants and wealth because they rely on outdated stupid concepts of diversity and overweighted. There is probably not a single investment or mutual fund manager who has ever heard of VonNeumann, or Gametheory or Optimal Strategy. And it is rare for any money manager to have a education in mathematics. Archimedes Plutonium www.archimedesplutonium.com www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Bourbaki Volumes (was Re: Cosets) > Could you expand on this a little? Why are they lost forever? > Published during the self-destructing-paper era? > Limited print run? > Other? > For instance the first volumes of Topology (I think Vol. I, II and III or > so) are not printed anymore (in French, I mean; getting them in English > would be ok too, but i'd lose much of their semantic value). > The best I could do was to borrow them from my university's library: I got > two old books from the 60's, hardly bound with the pages almost vanishing > into ashes when you touch them (the woman was looking at me as if I was some > sorcerer or I dunno which freak, all the more than I had to bother her for > almost half an hour because the books were in a remote dark room buried in a > secret place of the uni. lol); I didn't have the courage to scan each of the > thounsands of pages, and I had to give them back -you can borrow books only > for one month, even though, hey, those books hadn't been borrowed for some 9 > years-. They're now lost forever. No wonder you can't find them if you're trying to get ahold of the individually bound chapters from the first printing! They've been reprinted, reissued, and even revised numerous times since then, always as several chapters to a volume. I have a almost new copy of Theorie Des Ensembles, in French, right in front of me. Currently Masson-Dunod is printing the French language copies. Chapters 5-10 of Topologie is currently in print and available from numerous booksellers in France, many of whom will ship overseas. Chapters 1-4 appear to have gone out of print fairly recently (Bourbaki doesn't have it listed as 'epuise' on his web page yet) but there should be used copies floating about. For the dates of the most recent printings and addresses of relevant publishers, see Bourbaki's page at: http://www.bourbaki.ens.fr/Ouvrages.html I can assure you; Bourbaki's works are in no danger of being lost. (Although not all of them have been translated to English yet, so your hope of getting all of them from Springer is not feasible. The translation project continues, albeit slowly) Good luck, Nathan === Subject: limitation to the halting proof 1/ Say a halt function exits, TMh(x, y) = 0 IFF TMx(y) doesn't halt. 2/ Modify the halt function TMm(x, y) loops IFF TMh(x, y) != 0 let x = y = m, TMm(m, m) loops IFF TMh(m, m) !=0 TMm(m, m) loops IFF TMm(m) halts -------------------------------------------------------- There's no contradiction here, TMm is a function of 2 parameters so TMm(m) WILL halt, although not with any computed value, it just won't run. TMm(m, m) is a completely different application and result. The halting proof *relies* on a parameter_doubling construction. Return to line 2, 2a is ineffective for a halting contradiction, all halting proofs must resort to the format of 2b. -------------------------------------------------------- 2a/ Modify the halt function TMm(x, y) loops IFF TMh(x, y) != 0 2b/ Modify the halt function TMd(x) loops IFF TMh(x, x) != 0 let x = d TMd(d) loops IFF TMh(d, d) != 0 TMd(d) loops IFF TMd(d) halts Contradiction therefore 1 or 2 is invalid. ---------------------------------------------------------- It's not 1, its 2 that's wrong!! Its trivial to define primitive functions that dissalow diagonal function name reference traps that occur from constructions like Fx(x). First of all you want to define functions as parameters, as numbers. Secondly you want to allow function types that get a parameter and double it, where there's an x construct x, x. 1 becomes 2, 2 becomes 4... its no wonder the functions don't halt any more. Herc -- oo ____|mn / /_/ / _ - F R E E T H E T R U e M A N - / K-9/ /_/ - Join www.chatty.net - /____/_____ - Webmasters join www.BannerX.net - ~how are we supposed to make blockbuster movies unless the ~Truman faces impossible situation after impossible situation? === Subject: Re: limitation to the halting proof It's a kind of paradox like This sentence is not true. > 1/ Say a halt function exits, TMh(x, y) = 0 IFF TMx(y) doesn't halt. > 2/ Modify the halt function TMm(x, y) loops IFF TMh(x, y) != 0 > let x = y = m, > TMm(m, m) loops IFF TMh(m, m) !=0 > TMm(m, m) loops IFF TMm(m) halts > -------------------------------------------------------- > There's no contradiction here, TMm is a function of 2 parameters so > TMm(m) WILL halt, although not with any computed value, it just won't run. > TMm(m, m) is a completely different application and result. > The halting proof *relies* on a parameter_doubling construction. > Return to line 2, 2a is ineffective for a halting contradiction, all halting proofs must > resort to the format of 2b. > -------------------------------------------------------- > 2a/ Modify the halt function TMm(x, y) loops IFF TMh(x, y) != 0 > 2b/ Modify the halt function TMd(x) loops IFF TMh(x, x) != 0 > let x = d > TMd(d) loops IFF TMh(d, d) != 0 > TMd(d) loops IFF TMd(d) halts > Contradiction therefore 1 or 2 is invalid. > ---------------------------------------------------------- > It's not 1, its 2 that's wrong!! > Its trivial to define primitive functions that dissalow diagonal function name reference > traps that occur from constructions like Fx(x). > First of all you want to define functions as parameters, as numbers. > Secondly you want to allow function types that get a parameter and double it, > where there's an x construct x, x. 1 becomes 2, 2 becomes 4... its no wonder > the functions don't halt any more. > Herc > -- > oo > ____|mn > / /_/ / _ - F R E E T H E T R U e M A N - > / K-9/ /_/ - Join www.chatty.net - > /____/_____ - Webmasters join www.BannerX.net - > ~how are we supposed to make blockbuster movies unless the > ~Truman faces impossible situation after impossible situation? === Subject: Re: limitation to the halting proof In sci.logic, |-|erc : > 1/ Say a halt function exits, TMh(x, y) = 0 IFF TMx(y) doesn't halt. > 2/ Modify the halt function TMm(x, y) loops IFF TMh(x, y) != 0 > let x = y = m, > TMm(m, m) loops IFF TMh(m, m) !=0 > TMm(m, m) loops IFF TMm(m) halts > -------------------------------------------------------- > There's no contradiction here, TMm is a function of 2 parameters so > TMm(m) WILL halt, although not with any computed value, it just won't run. > TMm(m, m) is a completely different application and result. A Turing machine cannot accept more than one parameter anyway, without designating a special character in the alphabet. Therefore '1' has problems before one even properly gets started. One can of course easily work around this, by defining an alphabet [0,1,2,3,4,5,6,7,8,9,','] and using variable-argument Turing machines, which is slightly unwieldy but works. So: 1/ Assume there is a Halting Turing Machine, which can take the tape if the machine encoded by x halts on input y, (The input y needn't be a single parameter.) Call this HTM. Construct another Turing machine from this machine, by adding states that causes the machine to loop when it sees a 1 on its output, once the original machine made its decision. Call this LTM. Encode this second machine, using 'z', then feed HTM z , z. Does HTM indicate 1 or 0? the feedback loop is not established as it's not being fed the right input. One can construct a machine that takes its first parameter, delimited by a comma, then copies it. (The machine will take at least as many states as the alphabet, as it will have to remember which character it read from the tape, as it moves. This includes the comma needed to offset the new parameter.) So call L'TM the machine which: [1] first copies its first parameter to the end of the tape. (We assume sufficient blank space at the end thereof.) [2] goes through a duplicate of HTM, [3] loops if the symbol is 1, halts if the symbol is 0 on the tape. The feedback loop is now established: Feed L'TM z', its own encoding. LTM construct z' . , . z', then transitions into the HTM (which it has embedded into itself). The contradiction ensues. > The halting proof *relies* on a parameter_doubling construction. Not doubling. It is sufficient to merely copy the first parameter onto the end of the input. For a Turing machine, this is fairly easy. [rest snipped] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: limitation to the halting proof > 1/ Say a halt function exits, TMh(x, y) = 0 IFF TMx(y) doesn't halt. > 2/ Modify the halt function TMm(x, y) loops IFF TMh(x, y) != 0 > let x = y = m, > TMm(m, m) loops IFF TMh(m, m) !=0 > TMm(m, m) loops IFF TMm(m) halts > -------------------------------------------------------- > There's no contradiction here, TMm is a function of 2 parameters so > TMm(m) WILL halt, although not with any computed value, it just won't run. > TMm(m, m) is a completely different application and result. > A Turing machine cannot accept more than one parameter anyway, > without designating a special character in the alphabet. > Therefore '1' has problems before one even properly gets started. > One can of course easily work around this, by defining an > alphabet [0,1,2,3,4,5,6,7,8,9,','] and using variable-argument > Turing machines, which is slightly unwieldy but works. Its standard practice to use 100 for 0, 101 for 1, 110... or bit pairs, so any base can be used on a standard TM. > So: > 1/ Assume there is a Halting Turing Machine, which can take > the tape if the machine encoded by x halts on input y, > (The input y needn't be a single parameter.) Call this HTM. > Construct another Turing machine from this machine, by > adding states that causes the machine to loop when it > sees a 1 on its output, once the original machine made > its decision. Call this LTM. > Encode this second machine, using 'z', then feed HTM > z , z. Does HTM indicate 1 or 0? > the feedback loop is not established as it's not being fed > the right input. > One can construct a machine that takes its first parameter, > delimited by a comma, then copies it. (The machine will > take at least as many states as the alphabet, as it will > have to remember which character it read from the tape, as > it moves. This includes the comma needed to offset the > new parameter.) > So call L'TM the machine which: > [1] first copies its first parameter to the end of the tape. > (We assume sufficient blank space at the end thereof.) > [2] goes through a duplicate of HTM, > [3] loops if the symbol is 1, halts if the symbol is 0 > on the tape. > The feedback loop is now established: > Feed L'TM z', its own encoding. LTM construct z' . , . z', > then transitions into the HTM (which it has embedded into itself). > The contradiction ensues. > The halting proof *relies* on a parameter_doubling construction. > Not doubling. It is sufficient to merely copy the first parameter > onto the end of the input. For a Turing machine, this is fairly easy. I know its easy, its also easy to construct this sentence out of words.. This statement is false. that doesn't imply that its impossible to make a natural language and have a decision procedure that a sentence can be parsed. The fact is without that procedure to go out of your way to make X >> X, X there will never be the halting contradiction. Its a bit like living in a world without SQR(x). In my system you couldn't define SQR(x) = x * x. However you could use X^2. It doesn't matter if its an easy constradiction to make, its an easy contradiction to classify and thus resolve. Herc === Subject: Re: smallest disk covering a set of points ... snip ... > What's the smallest real number greater than 1? :> 1.000... (go thataway >) ... (< comefrom thataway) 01 Oops, that is a rational value. So is 1. It is well known that there are an uncountable infinity of real values between any two distinct rational values. :-) -- Chuck F (cbfalconer@yahoo.com) (cbfalconer@worldnet.att.net) Available for consulting/temporary embedded and systems. USE worldnet address! === Subject: Re: smallest disk covering a set of points > ... snip ... >>What's the smallest real number greater than 1? :> > 1.000... (go thataway >) ... (< comefrom thataway) 01 > Oops, that is a rational value. So is 1. It is well known that > there are an uncountable infinity of real values between any two > distinct rational values. So we've just proved that there is no *open* MCD in R^2? :> -- Corey Murtagh The Electric Monk Quidquid latine dictum sit, altum viditur! === Subject: Re: Balls > In the game of snooker (see footnote) it is sometimes necessary to break > up a pack of touching red balls with the cue ball. Does the > unsolvability(?) of the three body problem mean that the player cannot > predict just where the cue ball will end up after such a shot? As others have said, no. > I ask because commentators sometimes criticize a player for the way he > plays such a shot if the outcome is poor, especially in regard to the > position that the cue ball comes to rest in. My thought is that the > player can only trust to luck in such a situation. Oh, absolutely not. Consider this: if one were limited to strictly Newtonian motion and retarding friction for getting a proper leave (or cueball position), one could simply vary the force applied to the cue ball to get an approximate position. The ball would bounce around the table until it ran out of kinetic energy. (Letting the cue ball be off by eight inches is no big deal unless there are special circumstances). But friction plays a much larger part than just that. One can apply all sorts of spin to change the cue ball's behavior, both before and after contact with the object ball. The trouble is that these effects are dependent on things like temperature, humidity, cloth conditions, etc. and adjusting for variation is very difficult (nevermind the incredible difficulty in learning to shoot mechanically) Check out: http://www.playpool.com/download.php?op=viewdownload&cid=6 and download Amateur Physics for the Amateur Pool Player if you're interested. 'cid 'ooh > (Footnote: for the benefit of the American readers here, snooker is > superficially like pool but is a game of skill for grown-ups :-) === Subject: Re: Snell's Law >In a book, it mentions that light always follows the quickest path. > ... and the book is wrong. Actually the path is a stationary point, which > may be a saddle or a maximum or a minimum. For example, in reflection > from a concave mirror where the starting and ending points are farther > away than the centre of curvature, light is following the longest path > (of those that go from the starting point in a straight line to the mirror > and then in a straight line to the ending point). The textbook is not wrong in restating Fermat's least time principle even as applicable to a concave mirror. Quoted from the Wolfram site http://scienceworld.wolfram.com/physics/FermatsPrinciple.html Fermat's principle is somewhat incomplete and even slightly in error. Its modern form is A light ray, in going between two points, must traverse as optical path length which is stationary with respect to variations of the path. In this formulation, the paths may be maxima, minima, or saddle points. Mathematical and physical parameters representation or inter relation in a proper physical field could perhaps still validate some other situation that should be sought after or cited if already known, where the maximum time arises. === Subject: Re: Improper integral > William Elliot escribió: >> Let t>-1 . f(x)=log(x) * x^t >> >> Show that the integral of on [0,1] exists. > Integral of ?? > Let t = 0. Looks very inproper. > Int(ln(x), x, r, 1) = (1ln(1) - 1) - (r*ln(r) - r) = r - r*ln(r) - 1 > Lim( r - r*ln(r) - 1, r, 0+) = -1 > Lim(r*ln(r), r, 0+) = Lim(ln(r)/(1/r), r, 0+) > = Lim((1/r)/(-1/r^2), r, 0+) = Lim(-r, r, 0+) = 0 Ya, I made mental algebra mistake at that point so yes, integral at t = 0 is proper. === Subject: Approximate function for the area of intersection of two circles Hi! I am looking an approximation to the function to calculate the area of intersection of two circles http://www1.i2r.a-star.edu.sg/~sukanta/circle-circle.pdf Specifically, the radius of the two circles is fixed (not equal). The variable is the distance between the centres of the two circles (r). I have tried a taylor series approximation of the function (in maple), without success. I need an approximate function which is easily integrable. Could someone give me a clue on how to proceed. Sukanta === Subject: [tensor algebra] tensorfield problem. Here's the problem: Definitions: M is a real manifold. T(M) is the space of smooth vectorfields on M. T^1(M) is the space of 1-forms on M. ------------------------------------------------- An important property of tensor fields is that they are multilinear over the space of smooth functions. Given {X_i} smooth vectorfields, and {w^j} 1-forms. Given a (k,l)-tensorfield F. Then the function F(X_1, ..., X_k, w^1, ..., w^l) is smooth, and thus F induces a map: F : T^1(M) x .... x T^1(M) x T(M) x .... x T(M) ---> C^infty(M) The map F is multilinear over C^infty(M), meaning for functions f, g, in C^infty(M) and any smooth vectorfields or covectorfields alpha and beta, F(..., f * alpha + g * beta, ...) = f F(..., alpha , ...) + g F(..., beta, ...). ------------------------- The converse is also true. Any such multilinear map defines a tensorfield. A map t: T^1(M) x .... x T^1(M) x T(M) x .... x T(M) ---> C^infty(M) is induced by a (k,l) tensorfield as above if and only if it is multilinear over C^infty(M). And a map t: T^1(M) x .... x T^1(M) x T(M) x .... x T(M) ---> T(M) is induced by a (k,l+1) tensorfield if and only if it is multilinear over C^infty(M). ------------------- How do i have to prove the converse? === Subject: Asymptotic linearity of a function Hi all there, I'm struggling with a problem. I have a function f(t), t>0 with the following properties: 1) For all t>0 always: ktt: f(t+s)=f(t)+f(s)+o(f(t)) (2) What I want to show is that as t-> oo: f(t)/t -> c a constant. (3) If I add the condition that f(t)/t is decreasing it should work, but to show this is a little bit tricky, so I would like to avoid it. Has somebody a counterexample of a function with (1) and (2) for which (3) does not hold? Or if not an ideas to prove (3) from (1) and (2) ? Karl === Subject: Re: Four people crossing a bridge Or we could take a page out of Ted Kennedy's book and simply drive off the bridge. > > >> >So why don't you smart ers from the Dorm Radio in the Basement of >the Science Building post the ing solution... to the original >puzzle as posed? Very few posters on this thread have even tried to >post a solution. >> >>It's been posted several times: 35. > > > There seems to be some confusion. A flashlight is simply US-English for > the object called a torch in UK-English. > As far as how language differences may cause variations in the outcome > of the puzzle, we can burn that bridge when we come to it. > --Bill === > Suppose that n is positive integer. Suppose that a and b are positive real number. > How to simplify/compute the following integral. > Int[0, INFINITE] x^n exp^( - (x-a)^2/ (2b^2)) dx Here's a probabilistic approach. Let Z be a standard normal random variable, let I be the indicator variable for the event {Z > -a/b}, and let X = (bZ + a) I. For n > 0, your integral equals sqrt(2 pi) b E[X^n]. (The case n = 0 is a delicate matter; see below.) Letting m be the moment generating function of X, E[X^n] = [m^(n)](0), the nth derivative of m evaluated at 0. m(s) = E exp(sX) = P{X = 0} + integral(z = -a/b..infty, exp(s (b z + a) - z^2 / 2)) / sqrt(2 pi) Complete squares to evaluate the latter integral. I'll be surprised if all my algebra is correct, but I get m(s) = F(-a/b) + exp(a s + b^2 s^2 /2) [1 - F(-a/b - b s)] where F is the standard normal cumulative distribution function. Note that F'(x) = exp(-x^2 / 2) / sqrt(2 pi). When n = 0, we cannot simply use m(0) because P{X = 0} > 0 and we need to be careful about 0^0. However, in this case, it is clear that the integral is sqrt(2 pi) b [1 - F(-a/b)]. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === === Subject: tensorfield problem Here's the problem: Definitions: M is a real manifold. T(M) is the space of smooth vectorfields on M. T^1(M) is the space of 1-forms on M. ------------------------------------------------- An important property of tensor fields is that they are multilinear over the space of smooth functions. Given {X_i} smooth vectorfields, and {w^j} 1-forms. Given a (k,l)-tensorfield F. Then the function F(X_1, ..., X_k, w^1, ..., w^l) is smooth, and thus F induces a map: F : T^1(M) x .... x T^1(M) x T(M) x .... x T(M) ---> C^infty(M) The map F is multilinear over C^infty(M), meaning for functions f, g, in C^infty(M) and any smooth vectorfields or covectorfields alpha and beta, F(..., f * alpha + g * beta, ...) = f F(..., alpha , ...) + g F(..., beta, ...). ------------------------- The converse is also true. Any such multilinear map defines a tensorfield. A map t: T^1(M) x .... x T^1(M) x T(M) x .... x T(M) ---> C^infty(M) is induced by a (k,l) tensorfield as above if and only if it is multilinear over C^infty(M). And a map t: T^1(M) x .... x T^1(M) x T(M) x .... x T(M) ---> T(M) is induced by a (k,l+1) tensorfield if and only if it is multilinear over C^infty(M). ------------------- How do i have to prove the converse? === Subject: Re: tensorfield problem > The converse is also true. > Any such multilinear map defines a tensorfield. > A map t: T^1(M) x .... x T^1(M) x T(M) x .... x T(M) ---> C^infty(M) > is induced by a (k,l) tensorfield as above if and only if it is multilinear > over C^infty(M). > And a map t: T^1(M) x .... x T^1(M) x T(M) x .... x T(M) ---> T(M) > is induced by a (k,l+1) tensorfield if and only if it is multilinear over > C^infty(M). for all p in M: t(w^1, ..., w^k, X_1, ..., X_l)(p) := t_p(w^1_p, ..., w^k_p, X_1 |_p, ...., X_l |_p) are smooth functions, and multilinear because of the definition of t. So for all p in M, t_p is a (k,l) tensor. So t is a smooth section in the (k,l) tensorbundle of M. So t is a tensorfield. Finished! The second part is a similar proof. Subject: Re: Probability / Combinatorical Problem... >Well if I only had one counter it would be easy, but how do I deal >with those several interacting counters? My first idea was to consider >a combination of the 3 counters as one state and then solve the >problem with Markov chains, but that would lead to a >(5^3)*(5^3)-matrix > It leads to a 13 by 13 matrix. Since you are going to rerun X > infinitely, you do not need to consider the states that are only > passed once. > and I don't think that's a clever idea considering >that I have a similar problem with much larger counters. >Does anyone know which trick is needed for this problem? > I certainly don't. :) Actually the 13 by 13 format sounds a lot better so I'll use that...thanx for the help :) === Subject: Re: Open letter to James Harris - Self-Publish >> >> Why don't you self-publish your paper? >> Excellent idea: he may seek tips from Mr Wolfram who >> self-published his book. > No, Mr. Wolfram was rich enough and influential enough already to > self-publish. I wonder why he chose to self-publish rather than to use an established publisher? -- 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: Open letter to James Harris - Self-Publish >> >> >> Why don't you self-publish your paper? >> >> Excellent idea: he may seek tips from Mr Wolfram who >> self-published his book. > No, Mr. Wolfram was rich enough and influential enough already to > self-publish. > I wonder why he chose to self-publish rather than to use an > established publisher? Maybe he couldn't find a publisher who agreed with him on copyright? http://www.oreillynet.com/pub/wlg/1506 === Subject: Re: Open letter to James Harris - Self-Publish >>> >>> >>> Why don't you self-publish your paper? >>> >>> Excellent idea: he may seek tips from Mr Wolfram who >>> self-published his book. >> >> No, Mr. Wolfram was rich enough and influential enough already to >> self-publish. >> I wonder why he chose to self-publish rather than to use an >> established publisher? > Maybe he couldn't find a publisher who agreed with him on copyright? > http://www.oreillynet.com/pub/wlg/1506 Steven Krantz also has his own ideas on why he didn't publish with Addison-Wesley: -- 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: Open letter to James Harris - Self-Publish >> >>> >>> >>> Why don't you self-publish your paper? >>> >>> Excellent idea: he may seek tips from Mr Wolfram who >>> self-published his book. >> >> No, Mr. Wolfram was rich enough and influential enough already to >> self-publish. >> >> I wonder why he chose to self-publish rather than to use an >> established publisher? > Maybe he couldn't find a publisher who agreed with him on copyright? > http://www.oreillynet.com/pub/wlg/1506 > Steven Krantz also has his own ideas on why he didn't publish with > Addison-Wesley: A suggestion to James: Wolfram tried to impose a requirement on reviewers that they promise not to study math or physics for the ensuing ten years. === Subject: Re: Open letter to James Harris - Self-Publish 2kpKShAQELm5uc578GgAAAJ5SURBVHichdNBb9sgFABgvMblarSwXDMcKdfUpso1td1xDWqfc020 mV5JoHt/f0Dc1pU2jUskvrz3eA9D6n8s8h+QnVSzv0G3zUDPlOzqut3N2g9YcwDQHgm1zJj9O8gc IshWIu+NMe4NEICFGKdoySHArzd4AI0ZQLNyZR7hNIJ8gL3YcLiIRuQQcg3te8S+CfVdpa4Rx8UI Z9DUcuiEXKcax8OkOAunEuUyZjJHN0IR24D9jlCVjjuMxbcm9ZGYh31zGOHR7CPkwILFiPYKsjdD PsCCg14AfDPmZRzJXWyp8GueQwH8bMzXK8jbmNbsFQ8hALSPo4qwTfv9M4YT6F7j7RvETGbgM5vp DPge1XmEHymi8NZqVrCiaB5vrpBKDNhYZEwznTWv8yvEg5sXL5QPY8kY8w+YoEuZ5nVlfRYiWOEs SQ1uUqaLKK0jnDG28K+nBPepCe9q1RGGXFN/Zgl+h/8PmsxqidnCEkZbe0gQDhUGW9C6tCQstF7p +JWECRrdE+pljZYUAXCXPh8ZMjGwrkKPWQCF+JpgF8fB81mzDVUpUmkxTxBGOMADLLxqS9Uhlkrd pouKt2QQTtWqEbWizarD/pQ6fzRD/yXvDyKsuq065ccbVP3QPy9hmJetRSSsVebpOsTfpj8pgIEU GSuyU931Hxc1KGS6VJY03wPQ8UXF8iqcVoiyEuvj5Knl5qTEmjax/PZ5Alt29GJFLmF/tZxPH+fd S8hjGyGVWz5N4f5gK7HzXWfd5ucnoOoidliV2H6qUZ9R3ohNqI7N1k3Bo3Chuhey2vgpzGh4lSv0 ohQKp7BwAUrESohOuQngtbf4s9qNIX8At0MbK5iXDI4AAAAASUVORK5CYII= Discussion, linux) > > > Why don't you self-publish your paper? > > Excellent idea: he may seek tips from Mr Wolfram who > self-published his book. >> No, Mr. Wolfram was rich enough and influential enough already to >> self-publish. > I wonder why he chose to self-publish rather than to use an > established publisher? In order to maximize his control of his work? I'm sure that a man of Wolfram's stature could have found a publisher for his book. I also assume that the publisher would exercise some editorial control and that Wolfram wouldn't like that. (Note: I am not endorsing Wolfram's book or theories here.) -- This page contains information of a type (text/html) that can only be viewed with the appropriate Plug-in. Click OK to download Plugin. --- Netscape 4.7 error message === Subject: Re: Open letter to James Harris - Self-Publish > In order to maximize his control of his work? > I'm sure that a man of Wolfram's stature could have found a publisher > for his book. I also assume that the publisher would exercise some > editorial control and that Wolfram wouldn't like that. Right - and you could add that he got just as much attention for his work as if it had been published by a major publisher. This is not typically the case for self-published books. === Subject: Re: contravariant tensor fields, differential operators and quantisation > In the meantime, I worked out the transformation rule for the components of > a second order differential operator in coordinates. By this, I mean > something of the form > a^ij d^2/(dx^i dx^j) > where the a^ij are the component functions and d is the curly d partial > differential operator. The transformation will get pretty complicated for > higher orders, and it does not seem to have to do anything with tensors - > what kind of structure is it then? I'm not sure what to call the above structure, but it is not a tensor. The important clue to construct tensorial higher order differential operators is to look at vector fields. If X is a vector field and f a function, then g = X(f) is another function. In coordinates, X can be written as X = X^i d/dx^i, and its action as g = X^i df/dx^i. Now, there's nothing stopping you from applying another vector field Y to g to get h = Y(g) = Y(X(f)). Writing h out in component form gives h = Y^j d/dx^j (X^i df/dx^i). When you change coordinates from x to x', the new coordinate expression for h is h = Y'j d/dx'^j (X'^i df/dx'^i), the coordinate transformation Jacobians cancel nicely. So you can define the action of the tensor operator X (x) Y on f by (X (x) Y)(f) = X(Y(f)). You could also define a quadratic form H by requiring that H(X,Y) = X(Y(f)), I think this would be the Hessian form. Unfortunately, it seems that commutativity of the derivatives or equivalently the symmetry of the Hessian form goes to hell, since in general X(Y(f)) is not equal to Y(X(f)). But I'm sure there is some subtle way to recover it. Hope this helps. Igor === Subject: Re: contravariant tensor fields, differential operators and quantisation > I'm not sure what to call the above structure, but it is not a tensor. > The important clue to construct tensorial higher order differential > operators is to look at vector fields. If X is a vector field and f > a function, then g = X(f) is another function. In coordinates, X can > be written as X = X^i d/dx^i, and its action as g = X^i df/dx^i. Now, > there's nothing stopping you from applying another vector field Y > to g to get h = Y(g) = Y(X(f)). > Writing h out in component form gives h = Y^j d/dx^j (X^i df/dx^i). > When you change coordinates from x to x', the new coordinate expression > for h is h = Y'j d/dx'^j (X'^i df/dx'^i), the coordinate transformation > Jacobians cancel nicely. > So you can define the action of the tensor operator X (x) Y on f by > (X (x) Y)(f) = X(Y(f)). You could also define a quadratic form H by > requiring that H(X,Y) = X(Y(f)), I think this would be the Hessian > form. Unfortunately, it seems that commutativity of the derivatives > or equivalently the symmetry of the Hessian form goes to hell, since > in general X(Y(f)) is not equal to Y(X(f)). But I'm sure there is > some subtle way to recover it. Yes, I know that this is the obvious way of doing it. Perhaps one can see this X(Y(.)) as a diff op of 2nd order, but there are linear combinations of these where the 2nd derivatices cancel out, take e.g. the wedge product of two vector fields: X n Y = X (x) Y - Y (x) X The corresponding operator would be X(Y(.)) - Y(X(.)) which is again a derivation on the functions, as you know it corresponds to the Lie bracket vector field operating as [X,Y](f). So we have a 1st order differential operator corresponding to a 2nd rank tensor - this is not what we wanted. Tobias -- hang my head drown my fear till you all just disappear reverse my forename for mail! - saibot === Subject: Re: y'' x^2 + 2y' x - y^3 = 0 ? > Does anybody have any suggestions on how to find an analytic solution to > y'' x^2 + 2y' x - y^3 = 0 ? When the degree of the independent variable is the same as the order of the dependent variable's derivative for all terms, it is called a Bernoulli equation. You DE book should be able to tell you how to solve it. Jason === Subject: Re: y'' x^2 + 2y' x - y^3 = 0 ? > Does anybody have any suggestions on how to find an analytic solution to > y'' x^2 + 2y' x - y^3 = 0 ? Use the substitution x = exp(z). Then y(x) = Y(z), with Y' = dY/dz = y' x, etc. -Michael. === Subject: Re: y'' x^2 + 2y' x - y^3 = 0 ? > Does anybody have any suggestions on how to find an analytic solution to > y'' x^2 + 2y' x - y^3 = 0 ? > Use the substitution x = exp(z). Then y(x) = Y(z), with Y' = dY/dz = y' x, dy/dz = dy/dx dx/dz = xy' d^2 y/dz = d(xy')/dz = d(xy')/dx dx/dz = (y' + xy)x = xy'+ x^2 y d^2 y/dz^2 + dy/dz = y^3 y + y' = y^3 ??? === Subject: Re: y'' x^2 + 2y' x - y^3 = 0 ? > y + y' = y^3 The equation is still nonlinear, but now it is independent of z. I doubt there is an analytical solution, though. Qualitatively we may say the following: Consider the function E(y,y') = (y')^2 - y^4/2. Then E' = -2 (y')^2 < 0. So the value of E decreases. Hm. Note sure what good that does... It appears to be a ball rolling downhill subject to friction. The question is whether the solutions exist for all z. -Michael. === Subject: Re: Need help calculating the norm of a linear operator >Let (F_n) be a sequence of linear operators L2 -> R, = >int_{t=-1..1}x(t)*cos(pi*n*t)dt. I need to show that (F_n) does not >converge to 0 operator. There is a hint that it is provable that >||F_n|| = 1 for all n. Actually proving the original statement seems >easier for me than proving the hint. For every F_n there's an obvious >x_n from L2 such that ||x_n||=1 and x_n(t)*cos(pi*n*t) = >|cos(pi*n*t)|, so int_{t=-1..1}x(t)*cos(pi*n*t)dt = 2/pi and >sup_{||x||=1}= 2/pi. It is also easy to show that ||F_n|| is >upper bounded by 1+1/2 using the fact that |cos(pi*n*t)| <= 1. But how > Somewhere in the book you'll find a thing called the Schwarz > inequality, or the Cauchy-Schwarz inequality (together with > a statement of when equality holds.) or Cauchy-Bunyakovsky inequality throughout my literature... === Subject: Re: [OT] Re: Bondy & Murty's book available on-line (was: Re: Eulerianpath in infinite graph) > Shotgun Squad raved: > Translation: I pride myself on condemning anyone > who disagrees with me. > Meanwhile, my ride to go visit a hospitalized friend, > the quadripeligic Jessi (Jesus) just arrived, and I > must go visit one of the more interesting accumulations > in the universe, a living being. > xanthian. You've got to admit, he's got a point there. 1) Your grammar is just _terrible_. 2) You insult someone for calling you on it. 3) You rant for several dozen lines about absolutely nothing. Clearly, Shotgun Squad's assessment was not too far off the mark. I would only like to add that it it clear to me that your are an incredible narcissist. Symptoms: Everything Shotgun Squad described, plus your transparent attempt at showing us how great a person you are indicates a severe need for acceptance and adoration by those around you. What, are you trying to make him feel guilty for not thinking you're a god among men? 'cid 'ooh === Subject: Re: [OT] Re: Bondy & Murty's book available on-line (was: Re: Eulerianpath in infinite graph) X-URL: http://mygate.mailgate.org/mynews/comp/comp.theory/720deafbcec1426b2462e56481 34084b.48257%40mygate.mailgate.org > exceptional cases > paid him any attention > offensively ranting 1) The case was not exceptional, anyone with a bit of insight into humanity would _expect_ that case. 2) You'd have saved yourself _loads_ of embarrassment if you'd had the guts to stop, look back, and admit your error. You failed to take steps anyone else would have taken, and you injured yourself as a result. 3) Riiiiight. Let's see, I was exactly correct, you fought tooth and nail to avoid admitting that, and _I_ was the one offensively ranting? Does your house lack mirrors? Somehow, it's all still anyones fault but yours, I see. You might meditate on that idea for a decade or two, to see if you can find an alternative viewpoint before you die of old age. I've got to go climb another mountain in five hours from now, so I can't stop to teach you to think this evening. Twice I have attempted to write a line by line analysis of where you went wrong, to help you see the clues anyone else sees easily, and twice my computer has wedged itself and nuked the entire effort, for reasons not related to that work at all. Fate must have a _very_ strong reason for you to remain abjectly incompetent in human affairs. I am _so_ glad not to be you with such a geas on me. xanthian. -- === Subject: Re: [OT] Re: Bondy & Murty's book available on-line (was: Re: Eulerianpath in infinite graph) > Shotgun Squad raved: > Translation: I pride myself on condemning anyone > who disagrees with me. > I can't even tell you to take your meds, because > I strongly suspect you are not even under the care > you need. Instead, guessing your cycle as close > to what mine was before meds fixed that, I'll just > invite you to rewrite that at the bottom of your > depressive half cycle in a week or two. Go ahead. Keep condemning me as mad. It will make it even harder for you when it comes time to apologize. -- LAY ON YOUR FACE AND BEG THAT THE MERCY OF GOD WILL COME UPON YOU === Subject: Re: [OT] Re: Bondy & Murty's book available on-line (was: Re: Eulerianpath in infinite graph) X-URL: http://mygate.mailgate.org/mynews/talk/talk.bizarre/a7ad03678c048d2eff31a07e4 5ef15e1.48257%40mygate.mailgate.org > Go ahead. Keep condemning me as mad. Condemning you as mad? Where? I'm merely noticing that the problems with your mind you claimed to be all the fault of the medical care you received in South Dakota (and which care you chose to fight tooth and nail) continue right on happening as before when you are receiving _no_ medical care, and are in Utah. When you are calmer, this might, I say _might_, give you a clue as to the correctness of your previous claims, conclusions, and attributions of malice to your attempted caregivers. Do I think you are mad? In a non-judgemental sense, yes. Do I condemn you for it? Why would I? I've been there myself; our minds are what they are. I've spent years trying to get mine fixed, to little avail, you've spent years resisting having yours fixed, to little avail. It is a bit hard to draw a line in the sand between those two histories, and say: this side is condemned, this side is blessed, and then to find a way to defend that separation. > It will make it even harder for you when it > comes time to apologize. When I do something for which _I_ see a need to apologize, I'm quite quick to do so. When I have someone unbalanced, and remaining so by his own choice, telling me that the things I do will soon be requiring me to apologize, and he has not been grotesquely offensive in the process, I just wish him better mental health in the future. Life's too short for hate-as-hate, though I do tend to take offense rather easily. xanthian. A nice day today, all in all. Jesus is back from the hospital, no longer racked with infection. Hideki and I spent 45 minutes with Jesus, went off and had a splendid dinner at the store selling octopus heads for 1/16th the price per weight of octopus tentacles, then came to my place where I worked with him for a couple of hours on his English pronunciations again. [_You_ try to get the vowels right in full, fool, fall when your birth language contains neither of the consonent sounds F or L. Both of us were in giggles by the time that particular problem had been under work for half an hour.] He's finished school, will stay for a year working in the US, with luck helping run a charity, his just finished field of study, before going home. -- === Subject: Turing Machine with Tape The tape on my Turing machine broke, where can I get a new replacement tape? ---- === Subject: Re: Turing Machine with Tape > The tape on my Turing machine broke, > where can I get a new replacement tape? It's no longer in stock. A newer model has replaced the device, consisting of two electromagnetic cavities, each calibrated to its own resonant frequency. There may be an emulator sold off-the-shelf which allows you to capture the tape state and transitions by a suitable encoding into the 2-mode field state space. There are also device limitations that restrict you from above certain energy limits, before non-linear distortion begins setting in; particularly in the fermionic sector with gamma+gamma -> e+ + e- breakdowns; and in the Yang Mills sector with gamma+gamma -> W+ + W-; gamma+gamma -> Z + Z parasitics. === Subject: Re: Turing Machine with Tape > The tape on my Turing machine broke, > where can I get a new replacement tape? Just use the portion that is infinitely long in one direction, your Turing machine won't notice the difference. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Turing Machine with Tape Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >The tape on my Turing machine broke, >where can I get a new replacement tape? Why bother replacing it? Just shift it an infinite number of places to the left. -- Richard === Subject: Re: Turing Machine with Tape > The tape on my Turing machine broke, > where can I get a new replacement tape? > ---- You might want to try Chryslers. They have spare parts for quite a few touring machines. === Subject: Re: Turing Machine with Tape >> The tape on my Turing machine broke, >> where can I get a new replacement tape? >> ---- > You might want to try Chryslers. They have spare parts for quite a few > touring machines. Let's see. Humour is to humor as touring is to .... -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Turing Machine with Tape >> The tape on my Turing machine broke, >> where can I get a new replacement tape? >> >> ---- > You might want to try Chryslers. They have spare parts for quite a few > touring machines. > Let's see. Humour is to humor as touring is to .... I hope you're not suggesting that turing is the American spelling of touring! Rick === Subject: Re: Turing Machine with Tape >>> The tape on my Turing machine broke, >>> where can I get a new replacement tape? >>> >>> ---- >> >> You might want to try Chryslers. They have spare parts for quite a few >> touring machines. >> Let's see. Humour is to humor as touring is to .... > I hope you're not suggesting that turing is the American spelling of touring! > Rick I really don't know what I was suggesting. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Turing Machine with Tape I would check ebay. > The tape on my Turing machine broke, > where can I get a new replacement tape? > ---- === Subject: Sum of two numeasurable functions Can a sum of two immeasurable functions be a measurable function? If the answer is positive, can u give me an example? === Subject: Re: Sum of two numeasurable functions > Can a sum of two immeasurable functions :) non measurable ? -- J.S, tu peux essayer fr.sci.maths si tu pr.8ef.8fres. === Subject: mathematical linguistics (was: Sum of two numeasurable functions > Can a sum of two immeasurable functions > :) non measurable ? I once had a student write: ...therefore the Cauchy criterion is dissatisfied. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Sum of two numeasurable functions > Can a sum of two immeasurable functions be a measurable function? If > the answer is positive, can u give me an example? This can be answered by applying Tooth's Theorem: For any sensible definition of nice you can always find two un-nice functions whose sum is nice. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Sum of two numeasurable functions > Can a sum of two immeasurable functions be a measurable function? If > the answer is positive, can u give me an example? Sure. Let f be non-measurable. The -f is also non-measurable, but f + (-f) = 0 and 0 is measurable. Jose Carlos Santos === Subject: Re: Sum of two numeasurable functions > Can a sum of two immeasurable functions be a measurable function? If > the answer is positive, can u give me an example? > Sure. Let f be non-measurable. The -f is also non-measurable, but > f + (-f) = 0 and 0 is measurable. Ah... but 0 is not positive... === Subject: Re: Sum of two numeasurable functions >> Can a sum of two immeasurable functions be a measurable function? If >> the answer is positive, can u give me an example? >> Sure. Let f be non-measurable. The -f is also non-measurable, but >> f + (-f) = 0 and 0 is measurable. > Ah... but 0 is not positive... That's a good pun :-) but since I have a long history of trivially false answers, my first reaction was Oh no! I've goofed again!. Jose Carlos Santos === Subject: 3-from-8 card trick Those involved: A is a member of the audience, Y and Z are magicians. The effect: Y has a set of eight cards, they have identical backs and each card has a unique number from 1 to 8 on its face. Eight playing cards, ace through 8, would do equally well, the suits being ignored. Y gives the cards to A, A selects any three cards and returns those three (face down) to Y, making sure that Z does not see any of them. Y looks at the cards and places one of them face down on the table in front of him. He places the other two cards, face down in a little pile, neatly squared up, in front of Z. Z picks up the cards and then says a number. A then reveals the card on the table in front of Y. It is the card which Z named. There is no secretive communication involved between Y and Z. No nods, winks, ninja hand signals, etc. The exact orientations of the cards is immaterial. The information must be conveyed by the identity of the two cards in the small pile, and their order. The method: I have figured out a way to do this trick. The method is a little complex and I am wondering if anybody can suggest a simpler method. Clearly, 8C3 = 56 = 8P2. So the trick would work if there exists a bijection between the set of combinations of 3-from-8 and the set of permutations of 2-from-8, such that the elements in each permutation form a subset of the elements of its corresponding combination. Both Y and Z will need to remember this bijection. Let E be the set {1,2,3,4,5,6,7,8}. Let P be the set of permutations (ordered pairs) of elements chosen two at a time from E. Let C be the set of combinations (unordered triples) of elements chosen three at a time from E. x(+)y is to mean: ((x+y-1) mod 8)+1. Example: 6(+)5=3 x(-)y is to mean: if |x-y|<=4 then |x-y| else 8-|x-y|. Example: 1(-)7=2 If (x,y) is an element of P, and the corresponding element of C is {x,y,z}, then: if x(-)y=1 then z=y(+)6 if x(-)y=2 then z=y(+)5 if x(-)y=3 then z=y(+)4 if x(-)y=4 then z=y(+)2 Examples: (2,1) -> {2,1,7} [2(-)1=1, 1(+)6=7] (2,3) -> {2,3,1} [2(-)3=1, 3(+)6=1] (2,4) -> {2,4,1} [2(-)4=2, 4(+)5=1] (2,5) -> {2,5,1} [2(-)5=3, 5(+)4=1] (2,6) -> {2,6,8} [2(-)6=4, 6(+)2=8] (2,7) -> {2,7,3} [2(-)7=3, 7(+)4=3] (2,8) -> {2,8,5} [2(-)8=2, 8(+)5=5] [Any simpler bijection, or simpler way of expressing this bijection, would be appreciated. I suspect that I am overlooking something mathematically elegant.] It may not be clear to you that this establishes a bijection with the required properties, but it does. Here is the complete mapping, firstly sorted by combination, then by permutation. CCC PP PP CCC 1 123 23 12 128 2 124 24 13 138 3 125 25 14 148 4 126 16 15 157 5 127 21 16 126 6 128 12 17 147 7 134 43 18 168 8 135 51 21 127 9 136 31 23 123 10 137 37 24 124 11 138 13 25 125 12 145 41 26 268 13 146 64 27 237 14 147 17 28 258 15 148 14 31 136 16 156 61 32 238 17 157 15 34 234 18 158 85 35 235 19 167 71 36 236 20 168 18 37 137 21 178 81 38 348 22 234 34 41 145 23 235 35 42 247 24 236 36 43 134 25 237 27 45 345 26 238 32 46 346 27 245 54 47 347 28 246 62 48 248 29 247 42 51 135 30 248 48 52 256 31 256 52 53 358 32 257 75 54 245 33 258 28 56 456 34 267 72 57 457 35 268 26 58 458 36 278 82 61 156 37 345 45 62 246 38 346 46 63 367 39 347 47 64 146 40 348 38 65 356 41 356 65 67 567 42 357 73 68 568 43 358 53 71 167 44 367 63 72 267 45 368 86 73 357 46 378 83 74 478 47 456 56 75 257 48 457 57 76 467 49 458 58 78 678 50 467 76 81 178 51 468 84 82 278 52 478 74 83 378 53 567 67 84 468 54 568 68 85 158 55 578 87 86 368 56 678 78 87 578 See also... http://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml#5cardTrick http://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml#5cardTrickRedux -- Clive Tooth http://www.clivetooth.dk === Subject: Re: 3-from-8 card trick # Those involved: # A is a member of the audience, Y and Z are magicians. # # The effect: # Y has a set of eight cards, they have identical backs and each card has a # unique number from 1 to 8 on its face. Eight playing cards, ace through 8, # would do equally well, the suits being ignored. # # Y gives the cards to A, A selects any three cards and returns those three # (face down) to Y, making sure that Z does not see any of them. Y looks at # the cards and places one of them face down on the table in front of him. He # places the other two cards, face down in a little pile, neatly squared up, # in front of Z. # # Z picks up the cards and then says a number. A then reveals the card on the # table in front of Y. It is the card which Z named. # # There is no secretive communication involved between Y and Z. No nods, # winks, ninja hand signals, etc. The exact orientations of the cards is # immaterial. The information must be conveyed by the identity of the two # cards in the small pile, and their order. # # The method: # I have figured out a way to do this trick. The method is a little complex # and I am wondering if anybody can suggest a simpler method. # # Clearly, 8C3 = 56 = 8P2. So the trick would work if there exists a bijection # between the set of combinations of 3-from-8 and the set of permutations of # 2-from-8, such that the elements in each permutation form a subset of the # elements of its corresponding combination. Both Y and Z will need to # remember this bijection. Here is a simple argument that shows the EXISTENCE of such a bijection: Construct a bipartite graph that has 56 vertices corresponding to the 3-from-8 combinations on the left side, and 56 vertices corresponding to the 2-from-8 permutations on the right side. There is an edge between a combination {x,y,z} and a permutation (a,b) if and only if {a,b} is a subset of {x,y,z}. Since this bipartite graph is 6-regular, by Hall's theorem it must contain a perfect matching. This perfect matching describes the desired bijection. Finding a bijection that is nice to describe and easy to remember is a different story... --Gerhard ___________________________________________________________ Gerhard J. Woeginger http://www.win.tue.nl/~gwoegi/ === Subject: Re: 3-from-8 card trick > Those involved: > A is a member of the audience, Y and Z are magicians. > The effect: > Y has a set of eight cards, they have identical backs and each card has a > unique number from 1 to 8 on its face. Eight playing cards, ace through 8, > would do equally well, the suits being ignored. > Y gives the cards to A, A selects any three cards and returns those three > (face down) to Y, making sure that Z does not see any of them. Y looks at > the cards and places one of them face down on the table in front of him. He > places the other two cards, face down in a little pile, neatly squared up, > in front of Z. > Z picks up the cards and then says a number. A then reveals the card on the > table in front of Y. It is the card which Z named. > There is no secretive communication involved between Y and Z. No nods, > winks, ninja hand signals, etc. The exact orientations of the cards is > immaterial. The information must be conveyed by the identity of the two > cards in the small pile, and their order. > The method: > I have figured out a way to do this trick. The method is a little complex > and I am wondering if anybody can suggest a simpler method. Aargh! A decade or so ago I have been looking *real* hard to find the solution for this problem - and I found a very nice one :-) I started with the (8,3) problem. I'll work with the set { 0, 1, 2, 3, 4, 5, 6, 7 } Needless to say, everything goes mod 8. 1) First I'll give the way to *decode*, i.o.w. to find the hidden card when the two other cards are announced: Suppose I and J are announced in that order. The encoded number to find is K. Look at the gap (mod 8) between the first and second number. If this gap is 1 or 2 or 3, then add 2 to the second number J. Otherwise add 1 to the first number I. So K is given by If J-I <= 3 (mod 8) then K = J+2 (mod 8) else K = I+1 (mod 8) This gives the following table for K with I as row number and J as column number: | 0 1 2 3 4 5 6 7 --|---------------- 0 | - 3 4 5 1 1 1 1 1 | 2 - 4 5 6 2 2 2 2 | 3 3 - 5 6 7 3 3 3 | 4 4 4 - 6 7 0 4 4 | 5 5 5 5 - 7 0 1 5 | 2 6 6 6 6 - 0 1 6 | 2 3 7 7 7 7 - 1 7 | 2 3 4 0 0 0 0 - If you look carefully, there is a nice pattern - some sort of tiling. 2) Now the encoding bit. You first have to understand the previous part. Then suppose you get the 3 numbers I, J and K such that I < J < K. The gaps between the numbers are A, B, C, where C is the runaround-gap from K to I. Of course you know that A+B+C = 8. Since you can't have all gaps A=B=C=3, there must be one (or maximum 2) gaps 1 or 2. Pick the one that, when applying the first procedure, gives the correct answer. Example: 4-5-7 have gaps 1,2,5 If you take gap 1, you must hide card 5, so if you pass 4-7, the other one will decode 1 (bad) if you pass 7-4, the other one will decode 0 (bad) If you take gap 2, you must hide card 7, so if you pass 4-5, the other one will decode 7 GOOD! if you pass 5-4, the other one will decode 6 (bad) With a bit of practice, this step takes a second or so. The table from the Berkeley solution is somewhat less good looking than mine :-) | 0 1 2 3 4 5 6 7 --|---------------- 0 | - 4 3 1 3 2 1 3 1 | 7 - 0 4 2 0 3 2 2 | 6 5 - 1 0 3 1 0 3 | 5 7 6 - 2 1 0 2 4 | 7 6 5 7 - 0 2 1 5 | 6 4 7 6 3 - 1 0 6 | 4 7 5 4 7 4 - 2 7 | 6 5 4 6 5 3 5 - but of course it works as well. I think that the version is the transposed version of the Berkeley table. I have no time to check it now. I tried real hard to 'extrapolate' this tiling technique to the (27,4) and (124,5) puzzles, but alas it didn't work. If you haven't found it yet, you can find Berkeley's memorizable strategy in this one: http://math.berkeley.edu/~tsh/Papers/cardTrick.pdf Also have a look at http://www.stonehill.edu/compsci/Japan_files/frame.htm Brilliant puzzle! Dirk Vdm === Subject: Stationary action in physics I recently looked at some intros and overviews of string theory, and they all start with the principle of stationary action, as they should. I recall studying this in classical theory, and the universal applicability of this principle and the Lagrangian is amazing. The theory actually simplfies in the relativistic formulation. It also gives the eqns. for fluids, MHD, and other continua, also EM theory and GR. Consider the (relativistic) Lagrangians L. where T = proper time, or the 1-form of 4-momentum, L = p = mu = - m dT, then S = Integral(L). For a fluid, L = - rho = - n(1 + e), n = number density, e = thermal energy. For MHD, add - B^2/2 to rho. Then there is the usual formalism for EM theory, with F = dA, L = -(dA|dA)/2 + (J|A), and for GR the Lagrangian is the scalar curvature R (what else could it be?). Now one can start turning the wheels of the machine, and write down the energy momentum tensor T associated with the Lagrangian, T = gL - dz dL/d(dz), where the z are the fields. L is a function of the fields z and their (exterior) derivatives dz. We also have the Euler-Lagrange eqns., but div(T) = 0 is a better form for the eqns. of motion, since they express conservation of energy and momentum. One call also formulate a theory of wave propation with this formalism, and include the interaction of the waves (say sound or MHD waves) with the medium in which they propagate. One can divide the wave action by Planck's constant h, and call it wave quanta. Then instead of saying that waves propagate so that action is conserved (whatever that means), on can say that wave quanta are conserved. Who can fail to be impressed by the generality of this formalism? And of course it applies to many other situations, like solids, etc. (see Soper, Classical Field Theory, e.g.). Van === Subject: Re: Stationary action in physics X-SessionID: rH6tc-1421-25-543@news.uchicago.edu X-Hash-Info: post-filter,v:1.4 X-Hash: badcde5a c774940b d1c24ce7 58075411 e278e59a >I recently looked at some intros and overviews of string theory, >and they all start with the principle of stationary action, as they should. >I recall studying this in classical theory, and the universal applicability >of this principle and the Lagrangian is amazing. The theory actually simplfies >in the relativistic formulation. It also gives the >eqns. for fluids, MHD, and other continua, also EM theory and GR. >Consider the (relativistic) Lagrangians L. >where T = proper time, or the 1-form of 4-momentum, >L = p = mu = - m dT, then S = Integral(L). >For a fluid, L = - rho = - n(1 + e), n = number density, e = thermal energy. >For MHD, add - B^2/2 to rho. Then there is the usual formalism for >EM theory, with F = dA, L = -(dA|dA)/2 + (J|A), >and for GR the Lagrangian is the scalar curvature R >(what else could it be?). >Now one can start turning the wheels of the machine, and >write down the energy momentum tensor T associated with the >Lagrangian, T = gL - dz dL/d(dz), where the z are the fields. >L is a function of the fields z and their (exterior) derivatives dz. >We also have the Euler-Lagrange eqns., but div(T) = 0 is >a better form for the eqns. of motion, since they express conservation >of energy and momentum. >One call also formulate a theory of wave propation with this formalism, >and include the interaction of the waves (say sound or MHD waves) with >the medium in which they propagate. One can divide the wave action by >Planck's constant h, and call it wave quanta. Then instead of saying >that waves propagate so that action is conserved (whatever that means), >on can say that wave quanta are conserved. >Who can fail to be impressed by the generality of this formalism? >And of course it applies to many other situations, like >solids, etc. (see Soper, Classical Field Theory, e.g.). Welcome to the land of beauty and enchantment. I still recall my thrill when I first discovered how powerful this is. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: Stationary action in physics > Who can fail to be impressed by the generality of this formalism? > And of course it applies to many other situations, like > solids, etc. (see Soper, Classical Field Theory, e.g.). The approach (1) point q(t) on manifold M (2) equations of motion q''(t) = a(q(t),q'(t)) a: M x TM -> T(TM) with a: (q,v) -> T_{q,v}(T_q(M)); for v in T_q(M), q in M is more general, and more generic being characterized by a Boundary Principle: For each time t, there exists an interval I = [t-,t+] containing t such that corresponding to the boundary state over dI = (t+) - (t-): Q[dI] = (Q-,Q+) in M x M is a unique evolution q: I -> M such that q(t-) = Q-; q(t+) = Q+; or: q|{dI} = Q[dI]. This generalizes to N+1 Lorentzian manifolds P. For each point p in P, these exists a compact region K containing p such that: (1) dK = (K+) - (K-), (2) K+, K- are spacelike N submanifolds with boundary (3) d(K+) = d(K-) (4) for each Q[dK] = {Q-,Q+}: (K-) [+] (K+) -> M there exists a unique q: K -> M such that q|dK = Q[dK]. Theorem (Conjecture, actually): Assume that q: R -> M satisfies the property given above by the first set of criteria. Then there exists an a: M x TM -> T(TM) such that: q''(t) = a(q(t),q'(t)), for all t in R. Challenge: prove this or cite additional conditions under which it becomes true. Formulate and prove a generalized version where the real line R is replaced by a manifold P. === Subject: Re: Stationary action in physics > (see Soper, Classical Field Theory, e.g.). What's that? Anything like Class Field Theory? :-) -- 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: Constrained brachistochrone I know how to solve the brachistochrone problem using variational calculus, but can anyone tell me how to obtain a solution for the constraint of constant length? This is what I have done so far: T[y]=integral w.r.t. x of [(1+y'^2)/2gy]^1/2 J[y]=length L=integral w.r.t. x of (1+y'^2)^1/2 T[y]-lambda*J[y]=integral w.r.t. x of a function F(y,y') Applying the Beltrami identity: F - dF/dy' = constant C implies that [(2gy)^-1/2 - lambda][(1+y'^2)^1/2 - y'^2 * (1+y'^2)^-1/2] = C which can be rearranged as: [(2gy)^-1/2 - lambda]/[1+y'^2]^1/2 = C I have no idea what a suitable substitution might be to solve this. Any help would be greatly appreciated. Jon Heath. === Subject: Re: Constrained brachistochrone > [(2gy)^-1/2 - lambda]/[1+y'^2]^1/2 = C > I have no idea what a suitable substitution might be to solve this. > Any help would be greatly appreciated. This is an equation involving y and y'. Solve for y' in terms of y, as such: y' = f(y). Then separate variables, dy/f(y) = dt, and integrate. -Michael. === Subject: Re: Interminable Real Numbers in Compact Form? In sci.math, Yossarian > I need math help, so I hope I have come to the right place. > How do you express a non-terminating decimal that repeats (for > example, 3.412875487548754875....) in a more compact form (for > example, as a fraction.) > I would like to learn a general rule or formula that I could use to > enable me to decipher not only the example I have provided, but future > examples as well. As another poster has already pointed out this is a rational number; shift and subtract. However, if one merely wants an alternative notation that doesn't involve constructing that number, one can do things such as: 0.[3] 0.[142857] 3.412[8754] in ASCII. This notation at least conceptually could lead to a computational algorithm for such numbers that leads to exact results, given sufficient memory. (The precise mechanics would be rather unwieldy, though, even if one were to use binary.) A more traditional notation would put an overline over the last part, which looks a bit like: _ 0.333 ______ 0.142857142857142857 ____ 3.412875487548754 (best viewed with a monospace font such as Courier) or simply lets it repeat, with an ellipsis, as you've done above: 0.333... 0.142857142857142857... 3.412875487548754... although this can lead to some confusion, as e's expansion can be construed: e = 2.718281828... which implies that e is a rational number, when it's not, and a longer expansion shows a more accurate view: e = 2.718281828459... If one has access to TeX, LaTeX, or Lyx, one can use $0.33overline{3}$ to typeset the overbar. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Multiplication of negative binary numbers. In sci.math, Quentin Grady : > G'day G'day Folks, > Two's complement works fine for producing functional positive and > negative numbers so long as we confine ourselves to addition. > Two's complement sometimes give the right answer with subtraction and > sometimes doesn't. > 0011 3 > -1001 -(-7) > = 1010 = 10 works, > 0111 7 > -1001 -(-7) > = 1110 = -2 doesn't. > Is there a binary code that can deal with > (negative number)*(negative number) = positive number? > Two's complement obviously doesn't. Here is a 4-bit example. > 1001 > x1110 >= 0000 > + 1 0010 > + 10 0100 > + 100 1000 > It doesn't seem to matter if we include the overflow or not. > In this instance the msb is 1 whatever we choose to do. > Best wishes, The general notion of 1's complement and 2's complement is that a signed number is represented as an unsigned number, which conceptually could be written a = m + a where m is the modulus. For 16-bit 2's complement number m = 65536, which works out well. (16-bit 1's complement runs into problems as the modulus there is 65535; one has to add in an adjustment factor. This is probably the big reason 1's complement is no longer used.) Another form uses a = m - a where a is the absolute value. This one might be slightly clearer. Obviously for addition and subtraction, the modulus vanishes from consideration for all cases. In multiplication, one gets things like: (m - a) * b = mb - ab which turns out ot be correct, modulo m. If one does (m - a) * (m - b) = m^2 - ma - mb + ab that also turns out correct, modulo m. Of course part of your confusion is apparently simply that 1110 = -2 = +14 is the 2's complement representation of a negative number (modulo 16). You've stepped into the realm of *overflow*. There's not a lot that can be done with overflow except making the integers bigger. Take your 4-bit example. 1001 = -7 * 1110 = -2 = 1110 = +14 = -2 Whoops, overflow again. However, if we use 8 bits, we get: 1111 1001 1111 1110 and grinding it out: 1 1111 0010 11 1110 010 111 1100 10 1111 1001 0 1 1111 0010 11 1110 010 111 1100 10 ----------------------- 1111 0111 0000 1110 or one can cheat by using 'bc', if one has it handy: $ bc ibase=2 obase=2 11111001*11111110 1111011100001110 After lopping off everything to the left of the 8 bits we get the desired result, namely +14. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Real Number In sci.math, Dave Seaman : > > There are at least two proofs (both by Cantor) that > show that a 1-1 and onto mapping f:N -> R is not possible, > so there's at least two infinities (and I believe there are > an infinite number of infinities, each one constructible > using a power set construction, >> Sure, there are infinitely many cardinals. But what is the >> cardinality of the set bijectively equivalent infinite sets? ;-) > There is no such set (as I suspect you are aware), but that each one > constructibele using a power set construction part is equivalent to the > generalized continuum hypothesis. >> 'cid 'ooh >> PS - If someone could answer this, that would be really sweet > Not even a little acidic? Sourball. :-) -- #191, ewill3@earthlink.net -- insert random jawbreaker here It's still legal to go .sigless. === Subject: Re: Real Number >> >> There are at least two proofs (both by Cantor) that >> show that a 1-1 and onto mapping f:N -> R is not possible, >> so there's at least two infinities (and I believe there are >> an infinite number of infinities, each one constructible >> using a power set construction, > Sure, there are infinitely many cardinals. But what is the > cardinality of the set bijectively equivalent infinite sets? ;-) > There is no such set (as I suspect you are aware), but that each one > constructibele using a power set construction part is equivalent to the > generalized continuum hypothesis. > 'cid 'ooh > PS - If someone could answer this, that would be really sweet > Not even a little acidic? That depends on who made it--I stay away from the Midwest stuff personally. 'cid 'ooh === Subject: Re: Real Number > > > There are at least two proofs (both by Cantor) that > show that a 1-1 and onto mapping f:N -> R is not possible, > so there's at least two infinities (and I believe there are > an infinite number of infinities, each one constructible > using a power set construction, > > Sure, there are infinitely many cardinals. But what is the > cardinality of the set bijectively equivalent infinite sets? ;-) > > 'cid 'ooh > PS - If someone could answer this, that would be really sweet > Depending on your axioms this would not be a set but rather a proper > class. To see this, we'll show that the collection of all sets of > cardinality 2 is in 1 - 1 corrrespondence with the collection of all > sets. Let E be the empty set and S be any set. The ordered pair > (E, S) or in other symbols, {E, {E, S}} > has cardinality 2 and there is one unique such set for every set S. OK, Mr. Smarty Pants... what is the cardinality of this proper class? ;-) (I realize that it's an obnoxious question, but it gets at something I've wanted to know for a few years: What is the difference between sets and classes other than their quantificational properties within a set theory? It may even be that my question is ill-posed, but that is due to my ignorance on the topic) 'cid 'ooh === Subject: Re: Real Number >> >> There are at least two proofs (both by Cantor) that >> show that a 1-1 and onto mapping f:N -> R is not possible, >> so there's at least two infinities (and I believe there are >> an infinite number of infinities, each one constructible >> using a power set construction, > Sure, there are infinitely many cardinals. But what is the > cardinality of the set bijectively equivalent infinite sets? ;-) > 'cid 'ooh > PS - If someone could answer this, that would be really sweet >> Depending on your axioms this would not be a set but rather a >> proper class. To see this, we'll show that the collection of all >> sets of cardinality 2 is in 1 - 1 corrrespondence with the >> collection of all sets. Let E be the empty set and S be any set. >> The ordered pair (E, S) or in other symbols, {E, {E, S}} >> has cardinality 2 and there is one unique such set for every set S. > OK, Mr. Smarty Pants... what is the cardinality of this proper class? > ;-) > (I realize that it's an obnoxious question, but it gets at something > I've wanted to know for a few years: What is the difference between > sets and classes other than their quantificational properties within a > set theory? In a formal set theory with classes (like G.9adel-Bernays) and a strong form of Choice, it is clear, first of all, that any proper class is greater that any set (you well-order your class, and any set is equipotent to an initial segment of this order). It is not really harder to show that all classes are equipotent : use a similar reasoning by induction on two well-orderings of the two classes. So there is only one cardinal (say oo, or Omega) for all proper classes, and Omega is greater that any aleph. But without choice, I dont know the complete situation. It is at least clear that there should not exist a bijection between the class of all ordinals and the class of all sets (else, every set could be well ordered...) It may even be that my question is ill-posed, but that is > due to my ignorance on the topic) > 'cid 'ooh === Subject: Re: Stupid question In sci.math, Donald G. Shead > Cut< >> A variant has the user adjusting the mass position to >> counterbalance the torque; the doctor's scale is of this >> type, and the masses are moved an increasing distance >> away from the pivot-point as the force on the platform >> increases. > So you're saying that the scale is measuring the _force_ on the > platform: I agree! One could construe it as such. I prefer torque, for the simple reason that there's a pivot involved. As it turns out, the torques are arranged such that one measures mass; a Doctor's Scale on the moon would measure the same as on Earth. > Another variant is the simple pan scale; the >> torque is adjusted by placing a set of reference masses >> on the other pan. When the scale balances the torques >> are (near) zero and the user infers that the masses are >> (almost) equal. > I agree; but when the scale doesn't _quite_ balance it will tilt > toward the side on which the weight is greatest. Yes, torque again. > Scales of any kind react to the forces exerted on them: Whether the > forces are mechanical, or due to the weight of the masses placed upon > them. > A scale calibrated in units of mass - where [w/g] is in slugs; grams, > or kilograms - is accurate only where the denominator [g] is > consistent with its location: On the moon a scale would have to be > calibrated for [g] on the moon. Ah, you misunderstand torque so. Here's a schematic representation of the Doctor's Scale. (There are a few forms, but this will suffice.) [---]--+--*-----o-----------------V---------------------[] | C A P Z R I The left side is a screwable counterweight which is not normally used, except for zeroing. Call it C. The pivot point makes this a lever. The platform is attached somewhere -- call it A -- such that a force on the platform causes the pivot arm to go up. R is the reference mass. It is movable. Because of the counterweight the zero point can in fact be pretty much anywhere on the arm, although for convenience one normally simply sets the R to where the zero detent is and then adjusts the counterweight, with the platform empty. I is simply something at the very end which is a quick indicator; if R is properly adjusted, the arm will be in the center of the indicator; otherwise it will hit the top (R set too low/too far to the left) or the bottom (R set too high/too far to the right). There are a number of factors here that determine the sensitivity of this scale. Assume that C is properly set so that the scale indicates balance when R is at the zero-point. This means that p(C) * c + p(A) * a = p(Z) * r. (I use p(X) to indicate position of something relative to the pivot point. c is the weight of the counterweight; C = c/g is its mass.) Or, since this will be useful later, p(Z) * r - p(C) * c - p(A) * a = 0. Now step onto the platform A. An additional torque is imparted, which has to be compensated for by moving the reference mass to the right. The new equation is p(C) * c + p(A) * (a + M * g) = (p(Z) + l) * r once we've zeroed the scale again, and M = f/g is the unknown mass. Solving for M, we get M = ( (p(Z) + l) * r - p(A) * a - p(C) * c)/ (g * p(A)) = (l * r + p(Z) * r - p(A) * a - p(C) * c)/ (g * p(A)) = l * r / (g * p(A)) = l * R / (p(A)) Honest Mass: No Springs! The main issue is p(A) -- the length of the arm between platform and pivot -- and the reference mass. g has dropped right out of the equation. I'd patent this, but someone no doubt beat me to it long ago... :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Peano's space-filling curve > Yes, it does seem like gibbering. Exactly what are you > referring to, that L did that made him great? I understand that he questioned the parallel line axiom in Euclidean geometry, then tried tinkering with it. Tom Lehrer made up the song The Great Lobechevsky because Boylai had done more or less the same thing at more or less the same time. It seems however that there was no plagiarism involved - just another re-run of the Leibnitz-Newton dust up but this time without so much dust. But you already knew that and, doubtless, considerably more, so why did you ask me? > And more to the point, what does any of this have to > do with space-filling curves? Nothing. But you already knew that too. It does however have something to do with axiom systems. > a once and for all advance that >once enshrined in dogma could never be disputed or >discarded. > News flash: That's true. I accept your assertion, in the context of it being within the constraints of a formal system. > It's possible that in fact Gore is president, and there's > a huge conspiracy to hide that fact here in Oklahoma. Aha! I thought as much. The same conspiracy has pervaded Europe, too. I'm glad to see you have that elusive seventh sense - humour, oops, humor. > Doesn't seem all that likely, but I can't prove mathematically > that it's not so - if we're talking about knowledge in the > sense in which we know mathematical truths then the > best answer to is Bush president? is it would appear > so. On the other hand the neat fact he referred to > _is_ a fact. I _know_ it's a fact, not because Rudolph > said so, but because _I_ know how to _prove_ it. I find nothing unpalatable or unacceptable in what you have said. Seems to be pretty much as I thought to be the case. But many of your mathematical facts, that stem from the axioms of a formal system, change or disappear when the axioms are revised. True, they remain facts from the view point of the old system, but if interest shifts to a more utilitarian system, one that accords better with life, the universe and government funding of mathematical research, then they are discarded or even disputed because they were products of the out-moded axiom system. And then there is always the spectre of faulty logic that can slip an error past even the very best of brains, at least for a while. When you say _I_know how to_prove_ it, are you as convinced of the infallibility of your own logic as, say, Russell was of his in the Principia? I was about to write that the factual roll-call of presidents past and present cannot be subjected to such treatment (disputed, discarded), then I remembered there have been political regimes that did just that. A hundred years ago nobody disputed a theorem that made the angle sum of a triangle equal to two right angles. It was proved within the formal system of Euclid's axioms. But unlikely that this fact can be applied rigorously anywhere in the universe. Luckily, approximation means we can rest easy when using it locally, but in a certain sense it is not true anymore. It remains a fact in Euclidean space, but increasingly it looks like this space doesn't have any concrete reality. We'll hang onto it, for sure, because it's such an easy model to use, and the approximations are very, very close, but in one sense it's no longer a true fact. As a physical scientist, I only ever learnt the utilitarian mathematics needed for my trade. But I know there is a lot more of it out there and all of it true because of the efforts of people like yourself. Except for a few bits that got by because of faulty logic, but you are all busy chasing those down, aren't you? :-) John === Subject: Re: Peano's space-filling curve <2goqv9F4u1o3U1@uni-berlin.de> <2h0hctF7j3doU1@uni-berlin.de> <67kma05j4gfjmvpfi0ghs4jks89h725pin@4ax.com> <2hdn4fFbfe9dU1@uni-berlin.de> <81j3b010h2oihbqe1oqdqj2s36pp9hlkp3@4ax.com> <2hjhu7Fdg3s4U3@uni-berlin.de> 2kpKShAQELm5uc578GgAAAJ5SURBVHichdNBb9sgFABgvMblarSwXDMcKdfUpso1td1xDWqfc020 mV5JoHt/f0Dc1pU2jUskvrz3eA9D6n8s8h+QnVSzv0G3zUDPlOzqut3N2g9YcwDQHgm1zJj9O8gc IshWIu+NMe4NEICFGKdoySHArzd4AI0ZQLNyZR7hNIJ8gL3YcLiIRuQQcg3te8S+CfVdpa4Rx8UI Z9DUcuiEXKcax8OkOAunEuUyZjJHN0IR24D9jlCVjjuMxbcm9ZGYh31zGOHR7CPkwILFiPYKsjdD PsCCg14AfDPmZRzJXWyp8GueQwH8bMzXK8jbmNbsFQ8hALSPo4qwTfv9M4YT6F7j7RvETGbgM5vp DPge1XmEHymi8NZqVrCiaB5vrpBKDNhYZEwznTWv8yvEg5sXL5QPY8kY8w+YoEuZ5nVlfRYiWOEs SQ1uUqaLKK0jnDG28K+nBPepCe9q1RGGXFN/Zgl+h/8PmsxqidnCEkZbe0gQDhUGW9C6tCQstF7p +JWECRrdE+pljZYUAXCXPh8ZMjGwrkKPWQCF+JpgF8fB81mzDVUpUmkxTxBGOMADLLxqS9Uhlkrd pouKt2QQTtWqEbWizarD/pQ6fzRD/yXvDyKsuq065ccbVP3QPy9hmJetRSSsVebpOsTfpj8pgIEU GSuyU931Hxc1KGS6VJY03wPQ8UXF8iqcVoiyEuvj5Knl5qTEmjax/PZ5Alt29GJFLmF/tZxPH+fd S8hjGyGVWz5N4f5gK7HzXWfd5ucnoOoidliV2H6qUZ9R3ohNqI7N1k3Bo3Chuhey2vgpzGh4lSv0 ohQKp7BwAUrESohOuQngtbf4s9qNIX8At0MbK5iXDI4AAAAASUVORK5CYII= Discussion, linux) > But many of your mathematical facts, that stem from the > axioms of a formal system, change or disappear when the > axioms are revised. Wow. I never thought of that. I think I better have a lie-down. -- [R]eality has a fascinating ability to check us when we get a little too big for our britches... Make no mistake. There isn't a mathematician alive today that I can't now touch, and not a mathematical career on the planet that I can't now affect. --James Harris, render of worlds === Subject: Re: Peano's space-filling curve > Perhaps somebody on this newsgroup can enlighten a non-mathy - 'cos you've read this bit many times already. > I just got back from vacation and read this entire thread without a > break. Fascinating - I think we're all in the middle of a giant Eliza > experiment. Would that be Eliza as in Pygmalion, or is there another connotation that I may not know or understand. If so, would you dare to assign roles to each of the posters in a one-to-one correspondence with characters in Shaw's original play? ;-) John But I mean no harm, nor put fault On anyone who lives in a vault But it's alright ma, if I can please him - Bob Dylan === Subject: Re: Peano's space-filling curve >> Perhaps somebody on this newsgroup can enlighten a > non-mathy > > - 'cos you've read this bit many times already. >> I just got back from vacation and read this entire thread > without a >> break. Fascinating - I think we're all in the middle of a > giant Eliza >> experiment. > Would that be Eliza as in Pygmalion, or is there another > connotation that I may not know or understand. If so, would > you dare to assign roles to each of the posters in a > one-to-one correspondence with characters in Shaw's original > play? ;-) Eliza is a famous computer program that was inspired by the Turing test. . The Pygmalion connection does have something to do with the choice of the program's name. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Peano's space-filling curve > It can be surprising how much mathematics has grown since Kronecker's > day. It can be very hard to believe nowadays that Kronecker wasn't > considered some kind of strange crank in his day. will have gathered, I knew only the bare minimum about this. > Standard spelling: Kronecker Yep, my fault. I didn't check the encyclopaedia === Subject: Re: My paper passed peer review > If you wish to dispute my paper, then do it. But consider the > argument IN the paper, rather than coming in to attack the conclusion. I do not wish to work through your paper- the subject matter is not of interest to me. In any case, I have made no statement regarding the truth or falsity of the contents in the paper. > Some of you got away with lies for quite some time on Usenet, and only > now, hopefully can some people see you for what you are. Yes, I've > talked a lot myself against the mainstream and tried to find a quicker > way to acceptance of my results. > submitted them to a math journal, where my paper faced peer review, > and it passed it. This is the point we agree on- your paper was accepted for publication. And we also agree that the Journal erred in its subsequent actions. -- Andrew Resnick, Ph. D. National Center for Microgravity Research NASA Glenn Research Center === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. > I have no idea what you mean when you write it connects complexity > with measure. But there is no relevant extant definition of > complexity which he is employing here to relate to measure or anything > else. He is instead laying down how a reader ought to understand his > use of the word complexity hereafter. > Whether he is careful to use the term only in its new technical sense > and not in its old intuitive sense hereafter is, I suppose, another > matter. The intuitive arguments we have heard either stem from a lack of understanding of Kolmogorov complexity (as seems to be the case with given two examples), and/or a flawed intuition of complexity. The poster thinks some silly fractal iteration can cause abundantly complex output. That is not correct, that is only seeming complexity: complexity which the poster's naturally limited visual system cannot classify. The only reliable analysis of complexity is mathematical, not intuitive, especially intuition of those who are not familiar with complex programs. -- Eray Ozkural === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. <87vfimo2t2.fsf@phiwumbda.org> 2kpKShAQELm5uc578GgAAAJ5SURBVHichdNBb9sgFABgvMblarSwXDMcKdfUpso1td1xDWqfc020 mV5JoHt/f0Dc1pU2jUskvrz3eA9D6n8s8h+QnVSzv0G3zUDPlOzqut3N2g9YcwDQHgm1zJj9O8gc IshWIu+NMe4NEICFGKdoySHArzd4AI0ZQLNyZR7hNIJ8gL3YcLiIRuQQcg3te8S+CfVdpa4Rx8UI Z9DUcuiEXKcax8OkOAunEuUyZjJHN0IR24D9jlCVjjuMxbcm9ZGYh31zGOHR7CPkwILFiPYKsjdD PsCCg14AfDPmZRzJXWyp8GueQwH8bMzXK8jbmNbsFQ8hALSPo4qwTfv9M4YT6F7j7RvETGbgM5vp DPge1XmEHymi8NZqVrCiaB5vrpBKDNhYZEwznTWv8yvEg5sXL5QPY8kY8w+YoEuZ5nVlfRYiWOEs SQ1uUqaLKK0jnDG28K+nBPepCe9q1RGGXFN/Zgl+h/8PmsxqidnCEkZbe0gQDhUGW9C6tCQstF7p +JWECRrdE+pljZYUAXCXPh8ZMjGwrkKPWQCF+JpgF8fB81mzDVUpUmkxTxBGOMADLLxqS9Uhlkrd pouKt2QQTtWqEbWizarD/pQ6fzRD/yXvDyKsuq065ccbVP3QPy9hmJetRSSsVebpOsTfpj8pgIEU GSuyU931Hxc1KGS6VJY03wPQ8UXF8iqcVoiyEuvj5Knl5qTEmjax/PZ5Alt29GJFLmF/tZxPH+fd S8hjGyGVWz5N4f5gK7HzXWfd5ucnoOoidliV2H6qUZ9R3ohNqI7N1k3Bo3Chuhey2vgpzGh4lSv0 ohQKp7BwAUrESohOuQngtbf4s9qNIX8At0MbK5iXDI4AAAAASUVORK5CYII= Discussion, linux) >> I have no idea what you mean when you write it connects complexity >> with measure. But there is no relevant extant definition of >> complexity which he is employing here to relate to measure or anything >> else. He is instead laying down how a reader ought to understand his >> use of the word complexity hereafter. >> Whether he is careful to use the term only in its new technical sense >> and not in its old intuitive sense hereafter is, I suppose, another >> matter. > The intuitive arguments we have heard either stem from a lack of > understanding of Kolmogorov complexity (as seems to be the case with > given two examples), and/or a flawed intuition of complexity. When Chaitin gives a technical definition of random, proves that long initial segments of Omega are random[1] and then concludes triumphantly that some mathematical facts are true for no reason, then he is confusing his technical fact with an intuitive idea that random things occur without cause. This is the confusion of which I spoke. I don't know what the heck you were on about, but it wasn't relevant to my comments. On the other hand, the fact that the output of these programs are complex in our everyday use of the term complex indicates that one must be careful to avoid confusing the technical term and the everyday term --- and that perhaps the technical term is badly chosen. > The poster thinks some silly fractal iteration can cause abundantly > complex output. That is not correct, that is only seeming > complexity: complexity which the poster's naturally limited visual > system cannot classify. > The only reliable analysis of complexity is mathematical, not > intuitive, especially intuition of those who are not familiar with > complex programs. Footnotes: [1] I'm probably bungling the technical details of what he proved. -- If you have a really big idea, you can get a measure of how big it is REALLY, REALLY, *REALLY*, BIG DISCOVERY!!! --James Harris, on being ignored === Subject: Re: Panu Raatikainen's review of two of Chaitin's books. <87vfimo2t2.fsf@phiwumbda.org> >> I have no idea what you mean when you write it connects complexity >> with measure. But there is no relevant extant definition of >> complexity which he is employing here to relate to measure or anything >> else. He is instead laying down how a reader ought to understand his >> use of the word complexity hereafter. >> Whether he is careful to use the term only in its new technical sense >> and not in its old intuitive sense hereafter is, I suppose, another >> matter. >The intuitive arguments we have heard either stem from a lack of >understanding of Kolmogorov complexity (as seems to be the case with >given two examples), and/or a flawed intuition of complexity. The >poster thinks some silly fractal iteration can cause abundantly >complex output. That is not correct, that is only seeming complexity: >complexity which the poster's naturally limited visual system cannot >classify. >The only reliable analysis of complexity is mathematical, not >intuitive, especially intuition of those who are not familiar with >complex programs. >Eray Ozkural You seem to have mistaken the spelling - it's K_A_M not G_O_D. (How are you doing with that assignment on intensional/referential opacity by the way?) -- David Longley === Subject: Re: help i'm stuck! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4QDtj800438; >hi, im trying to revise for my gcse's and i've come across a pair of >simutaneous equations i can't figure out: >x+4y=15 >3x-2y=10 >please help!! >lucy- There's nothing extraordinary in those equations. A simple way to solve them is to isolate x from the first equation (x=15-4y), and assign this in the other equation to get y (which turns out to be two and a half), then use the first equation again, and see that x=5. The Levantine. === Subject: Re: the math fails me > I can't post these equations, they are too long. > However, if any of you can visit where they're at, > http://mypeoplepc.com/members/jon8338/polynomial/ > and tell me what I did wrong? jongiff2000@yahoo.com > As far as I can tell, the math is perfectly correct, but > the answer just doesn't make sense. I'll give you credit for adding some text to your page, but you still have the same problem: you haven't *explained* what you are doing at each step. If you want your solution to make sense, you need to explain each step of what you are doing. Equations do not speak for themselves. In addition, it would be far more convincing if you offered one or two examples. Take an eighth degree polynomial that is considered irreducible by most people, or at least by a CAS like Mathcad, and generate the solutions using your method. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: 640,000,000,000 TO 1 >>> I have seen this figure calculated out in a previous >>> but basicly if you take ALL the possible DNA options >>> and multiply them out you get a figure that rounds >>> down to 640 billion. >> Does that mean that there are 640 billion possible human beings, up to >> differences in nurture? >> I don't like >> this thought at all! (Even though I believe nurture and environment play >> a big role in determining one's personhood.) > Identical twins have identical genomes but are different people even > with similar nurturing and environment. right - this is one source of relief, but I would still rather know there was lots more variability in the genome (which you demonstrate later in your post). It's kind of like the story Feynman tells where he's in a class and says the sun has n-billion years left before going nova - one kid is extremely distressed because he thought Feynman said n *million*. However long a million is compared to my lifetime, I'd like to think there's unimaginably longer remaining. Or it's like the fear some cosmologists have expressed that if the universe is curved on itself much more tightly than expected (I mean in one or some of the spatial dimensions, not the nano-dimensions beyond 3), we might be close to seeing all the way around it already, and there's the sense of is that all there is?? - even if it's effing enormous already and there's no hope of ever, ever, ever getting anywhere near the edges. So I'm very glad of the numbers you share in the next paragraph, which make it very clear we're not about to exhaust the genetic possibilities of -Piotr p.s. is 94% of shared genes the accepted average between two randomly selected people? Why do we sometimes hear a figure higher than that for humans and chimpanzees? Is that figure taken between the average human and the average chimpanzee, instead of randomly selected members of each? > There are about 30,000 genes in the human genome with any two > reproductively capable people sharing about 94% of their genes so they > would be able to produce 2^1800 genetically distinct offspring, not > counting mutations and junk DNA variations. 2^1800 is about 10^541. > There are an estimated 10^80 protons and neutrons in the universe. > Create a universe of each proton and neutron and a universe of each of > those protons and neutrons to about 6 iterations and that is how many > different children could be produced by any two people. Then consider > how much variation there is in the population of the human race. === Subject: Re: 640,000,000,000 TO 1 In sci.logic, Thomas H. Faller <40B27460.DB31961D@sgi.com>: >>> I have seen this figure calculated out in a previous >>> but basicly if you take ALL the possible DNA options >>> and multiply them out you get a figure that rounds >>> down to 640 billion. >> Does that mean that there are 640 billion possible human beings, up to >> differences in nurture? I presume some are more likely to be generated >> than others so it is difficult to calculate exactly when the human >> genome-space will have been completely filled, or how long we're likely to >> have to wait before someone is duplicated. But a back-of-the-envelope >> calculation shows that if the world population climbed to, say, 12 billion >> (projected within the next half-century) and stayed stable there, then >> it's likely that all possible human beings will have been born after two >> or three thousand years (if the 640 billion figure is right). I don't like >> this thought at all! (Even though I believe nurture and environment play >> a big role in determining one's personhood.) >> If you factor in possible harmless mutations over such a period of time >> and such a population, to expand the available DNA options, how much is >> that likely to affect the 640 billion figure? Can someone verify my >> calculation? I'd be happier knowing we weren't about to exhaust our >> genetic possibilities within a (geological) eye-blink. > If your girl is one in a million, there are half-a-dozen like her in > any major city. Didn't Bester do an SF story on this back about > 40 years ago? The problem is, so are you. :-) And they all want the same girl. :-) > Tom Faller -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: 640,000,000,000 TO 1 >> I have seen this figure calculated out in a previous >> but basicly if you take ALL the possible DNA options >> and multiply them out you get a figure that rounds >> down to 640 billion. > Does that mean that there are 640 billion possible human beings, up to > differences in nurture? I presume some are more likely to be generated > than others so it is difficult to calculate exactly when the human > genome-space will have been completely filled, or how long we're likely to > have to wait before someone is duplicated. But a back-of-the-envelope > calculation shows that if the world population climbed to, say, 12 billion > (projected within the next half-century) and stayed stable there, then > it's likely that all possible human beings will have been born after two > or three thousand years (if the 640 billion figure is right). I don't like > this thought at all! (Even though I believe nurture and environment play > a big role in determining one's personhood.) > If you factor in possible harmless mutations over such a period of time > and such a population, to expand the available DNA options, how much is > that likely to affect the 640 billion figure? Can someone verify my > calculation? I'd be happier knowing we weren't about to exhaust our > genetic possibilities within a (geological) eye-blink. > If your girl is one in a million, there are half-a-dozen like her in > any major city. Didn't Bester do an SF story on this back about > 40 years ago? > Tom Faller And if she's Chinese, there are a thousand women just like her. === Subject: Sum of two approximation polynomials question. If one approximates two curves, each by a cubic (Bezier) polynomial then does adding the two cubic Bezier approximations give a polynomial that approximates the sum of the two true curves. To be a bit more specific, if I directly derive a cubic Bezier from the sum of the two true curves and compare that to the sum of two cubic Bezier each derived from one of the two true curves, will these two approximations be the same? This is what I'm getting in specific cases but I can't say for sure in general. === Subject: re:a new way round diagonalisation-one infinity?meta set theory Hallo Ross F. > Please detail what is meta set theory. Does not your result hold in > set theory? What do you consider classical set theory? > If you don't need different sizes of infinity in your meta set > theory, can you construct a model of the numerical system from meta > set theory, where not all integers are, for example, positive, even, > or prime? > What is your opinion of a model where all sets are ordinals and the > powerset is the order type is the successor? Meta set theory is the set theory I defined by the axioms A0,A1,A2 and A3. Wherby I am looking for a new axiom A«3, as A3 allows not enough sets to get natural numbers. Classical set theory I call ZFC and russell«s type hierarchy, in short all set theories without (meta-)levels. The construction of a numerical system is the critical point of the theory, here I still need some help: if we define the successor n+ of a set (or natural number) n as: x e(t+1) n+ :<-> Vd: [x e(d) n or x =(d) n] we see: x e(t+1) 0+ <-> Vd: [x e(d) 0 or x =(d) 0] <-> Vd: x =(d) 0 <-> (short) x = 0. We write 0+ =(t+1) {0} Analog we find for the set A of all sets: A+ =(t+1) A But to have a successor is not yet the construction of N, the set of all natural numbers. {0, 0+, 0++, ...} is no allowed definition, we look in meta set theory for a equation: A) x e(t+1) N :<-> A(x,t,...) whereby A(x,t,...) is to be defined. Another try (metatransfinite): B) x e(t+1) N :<-> Ed: x = 0++...+ (+ d-times) to A): Part of A(x,t,...) could be En: n e(t) N : x =(t) n+ or x=(t+1) 0 or x = (t+1) 0+ or E M: E m e(t+1) M: x =(t+1) m+. Whereby M is a meta subset of N, defined without using x e(t+1) N. Meta set theory shows at this point, what a tricky set the natural numbers are... For a good definition of natural numbers in meta set theory I would be thankful - or a proof, that it is impossible, even with a modified axiom A3! > What is your opinion of a model where all sets are ordinals and the > powerset is the order type is the successor? The last part of the question I do not understand - do you mean, that the successor is always the power set? What sets are allowed? If classical sets, you have the classical restrictions. If meta sets - I did not try yet, but it might work. Should/culd those sets be used for defining natural numbers? Yours Oskar ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: a new way round diagonalisation-one infinity?meta set theory Your reasoning may be partially sound. That the set of all sets would be its own powerset and thus map to itself trivially is called Cantor's paradox, because Cantor proved to himself and others that no set may map to its own powerset. For largely this reason, and generally for tractability, there are axioms in the Zermel-Fraenkel axiom schema of regularity, also known as foundation, the spirit if not letter of which being that no set contains itself as a set, every set is supposed to be well-founded, circularly that its transitive closure is regular or well-founded. Thus some logicians big and small call those collections that can't be sets instead classes, a rather daft thing to have in a set theory. So anyways those logicians say the set of all sets doesn't exist, thus avoiding a dilemna with Cantor or Cantorians, and making some other statements about sets more comfortable to make because their sets are well-founded. They call the collection of all sets the proper class, in that it is not an element (collection member) of another class. Please detail what is meta set theory. Does not your result hold in set theory? What do you consider classical set theory? If you don't need different sizes of infinity in your meta set theory, can you construct a model of the numerical system from meta set theory, where not all integers are, for example, positive, even, or prime? What is your opinion of a model where all sets are ordinals and the powerset is the order type is the successor? Ross F. === Subject: National Defense Science and Engineering Graduate Fellowship Applications for the 2005-2006 NDSEG Fellowship competition will be available this fall. The Department of Defense National Defense Science and Engineering Graduate Fellowship The Department of Defense is committed to increasing the number and quality of our Nation's scientists and engineers. Toward that end, the DoD annually supports approximately 8,000 graduate students in fields important to national defense needs. The DoD supports graduate students in a number of ways. First and foremost is the support of thousands of graduate students who are members of research teams funded through DoD contracts and grants. The students, selected by the research faculty, engage in fundamental studies under the leadership of a senior researcher and commensurately earn advanced degrees. Usually, these students are supported wholly by the DoD grant or contract. Another mechanism for graduate student support is through portable fellowships, such as those awarded to U.S. citizens or nationals following a competition each year under the National Defense Science and Engineering Graduate Fellowship Program. The actual number of awards varies from year to year depending upon the available funding. Two hundred eighty five Fellowships were awarded in 2001, 170 in 2002, and These fellowships are portable and allow the recipients to pursue their graduate studies at whichever university they choose to attend. This site describes the National Defense Science and Engineering Graduate Fellowship Program. This program seeks to identify individuals whose scientific and engineering credentials will support study through doctoral degrees. The prevailing goal is to provide the United States with talented, doctorally trained American men and women who will lead state of the art research projects in disciplines having the greatest payoff to national defense requirements. This program is highly competitive. The Department of Defense and its contractors are committed to providing equal educational opportunity. PLEASE NOTE: You are responsible for ensuring the proper submission of each update any of your responses prior to the submission deadline. This allows applicants to skip over some requested information that is not at hand, correct errors and omissions, and return to the application later. You may change your contact information at any time, even after the submission deadline. All Applications will be ------------------------------------------------------------------------ ------------------------------------------------------------------------ NDSEG Fellowship Program American Society for Engineering Education 1818 N Street N.W., Suite 600 ndseg@asee.org Phone: (202) 331-3516, Fax: (202) 265-8504 http://www.asee.org/ndseg/ -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ === Subject: please look at this problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4QEZf705279; hello please look at this problem and my solution . Is it true? suppose that f is a real and bounded variation function on the interval [0,1]and lim f(x)=0(when x-->0^{+}). if {a_{n}}is strictly decreasing and convergent to 0 and for every n a_{n} is in [0,1].prove that the series (-1)^{n}f(a_{n}) is convergent. solution:since the sequence a_{n}is decreasing strictly then any finit number of elements of the sequence can be a partition for [0,1]. let c_{n}=|f(a_{n})-(f{b_{n})| and we call the sequence of the partial sum of c_{n},s_{n}. f is a bounded variation function so s_{n} is bounded and increasing .so it is convergent and the series |f(a_{n})-f(a_{n+1})| is convergent because lim f(x)=0(x-->0^{+}) by definition we have: fore every m there exists a n s.t 0N we have |a_{n}-0|Let M be a (n by n) matrix whose elements M_{ij} are of the form >m_{abs(i-j)}. Here is such a matrix M: >1 3 5 2 >3 8 3 5 >5 3 8 3 >2 5 3 1 >I am interested in v such that M*v=alpha*ones, with v in R^n, alpha in >R and ones=[1 ... 1]^t. It is straighforward to show that >v_{1+i}=v_{n-i}. Does it exist other properties of v, e.g., that could >simplify its calculus? >Richard then it is a symmetric toeplitz matrix and for systems with such matrices there exist indeed specialized methods of complexity O(n^2) for solving which are numerically stable (and even superfast methods of complexity nlog n , which a lot. for example: 30. Zbl pre01915818 Van Barel, Marc; Heinig, Georg; Kravanja, Peter A superfast method for solving Toeplitz linear least squares problems. (English) there are even fast iterative solvers for this type. 69. Zbl pre02029987 Ng, Michael K.; Sun, Hai-Wei; Jin, Xiao-Qing Recursive-based PCG methods for Toeplitz systems with nonnegative generating functions. (English) hth peter === Subject: Hyperfocal distance math I'm writting a program and I need to solve the following equation for f H=((f*f)/(Nc))+f I've gotten as far as: Nc(H-f)=f*f ...and I forgot how to finish it. padillaH === Subject: Re: Hyperfocal distance math >I'm writting a program and I need to solve the following equation for f >H=((f*f)/(Nc))+f >I've gotten as far as: >Nc(H-f)=f*f >....and I forgot how to finish it. H=((f*f)/(Nc))+f 0=((f*f)/(Nc))+f-H 0=f*f+f*Nc-H*Nc Then use the quadratic formula: (-b+-sqrt(b^2-4*a*c)/(2*a) a=1 b=Nc c=-H*Nc f=(-Nc +- sqrt((Nc*Nc) + 4*1*H*Nc)/2 --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Maximizing area! I need to maximize the area: int [from 0 to 1] f(x) dx= (area under the curve) subject to: int[from 0 to 1] (1+[f '(x)]^2) dx =L (arc length fixed) and f(0)=f(1)=0. I am thinking of :Langrange multiplier. Can anyone help? === Subject: Re: Maximizing area! > I need to maximize the area: int [from 0 to 1] f(x) dx= (area under the > curve) subject to: > int[from 0 to 1] (1+[f '(x)]^2) dx =L (arc length fixed) and > f(0)=f(1)=0. > I am thinking of :Langrange multiplier. Can anyone help? Have a look at the thread Constrained brachistochrone, posted today by Jonathan Heath. -Michael. === Subject: Re: Maximizing area! > I need to maximize the area: int [from 0 to 1] f(x) dx= (area under the > curve) subject to: > int[from 0 to 1] (1+[f '(x)]^2) dx =L (arc length fixed) and > f(0)=f(1)=0. > I am thinking of :Langrange multiplier. Can anyone help? You should be thinking of: calculus of variations. You see, the set of f satisfying your constraint is not specified by finitely many parameters. Instead it is an infinite-dimensional collection of functions. === Subject: Re: Alexander Grothedieck biography? >> I remember a short biography written by P. Cartier on the occasion of the > speculative remarks on Grothendieck's current mental state. But surely one would have to agree that Grothendieck's behaviour has always been rather odd? That is one reason why I would have thought he would make an excellent subject for a fascinating biography. [Cf Andrew Hodge's biography of Alan Turing.] I might add that I regard Grothendieck as the greatest mathematician of the 20th century, and I would regard it as a bonus in a biography that he seems to hold strong and often bizarre views on a wide range of subjects. A bit like Newton? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Question for Harris > have found many different posts of yours with valid counter-arguments > by other posters. > Since you are still making such contentions, can you please define a > short and unassailable mathematical proof supporting your claims? All the indications are that he believes he has already done so. The same proof that has been repeatedly shown wrong is, as I understand it, the one he got published. Unfortunately for him, it didn't take long to point out the error in his proof and the paper is no longer available in published form by that math journal. Harris has taken this as proof that we are an unruly mob, rather than as proof that he is wrong. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Question for Harris > have found many different posts of yours with valid counter-arguments > by other posters. > Since you are still making such contentions, can you please define a > short and unassailable mathematical proof supporting your claims? > All the indications are that he believes he has already done so. The > same proof that has been repeatedly shown wrong is, as I understand > it, the one he got published. Unfortunately for him, it didn't take > long to point out the error in his proof and the paper is no longer > available in published form by that math journal. Harris has taken this > as proof that we are an unruly mob, rather than as proof that he is wrong. The two solution spaces are not mutually exclusive. :-) === Subject: Re: Question for Harris > have found many different posts of yours with valid counter-arguments > by other posters. > > Since you are still making such contentions, can you please define a > short and unassailable mathematical proof supporting your claims? > All the indications are that he believes he has already done so. The > same proof that has been repeatedly shown wrong is, as I understand > it, the one he got published. Unfortunately for him, it didn't take > long to point out the error in his proof and the paper is no longer > available in published form by that math journal. Harris has taken this > as proof that we are an unruly mob, rather than as proof that he is wrong. > The two solution spaces are not mutually exclusive. :-) Prime Slut James Harris ought to read the funny papers, Proof Promises Progress in Prime Progressions -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Question for Harris > have found many different posts of yours with valid counter-arguments > by other posters. > Since you are still making such contentions, can you please define a > short and unassailable mathematical proof supporting your claims? > Why do you care, crackpot? Are you trying to get some pointers in being a > kook from Harris? You might have better luck trolling the streets. === Subject: Re: roots of unity in Koblitz > (We have a Field F[p^f] where p is prime) > 12. Let d = gcd(k,(p^f)-1). Since d | (p^f)-1, the cyclic group F*[p^f] > clearly has d d-th roots of unity. Each of them is also a k-th root. I don't think this really has anything to do with finite fields, except that the multiplicative group in a finite field is cyclic, in this case of order p^f - 1. If C is a cyclic group of order n, and d | n then there are just d elements x in C satisfying x^d = 1, or if you like there is exactly one subgroup S of order d, and it is cyclic. (If C is generated by g then S is generated by g^{n/d}.) > I have tried an example : > p = 2 > f = 4 > k = 5 > so gcd(5,15) = 5, so I should be able to find 5 5th roots of unity, but I > can only find one (Its 1). > Probably I'm missing something here, can anyone else find other 5th roots > of unity here ? How do you visualise the elements in this field? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: sci.math stats since '86 -- InterBang change nospam to optimusprime to reply === Subject: Unabomber suspect the unabomber was a mathematician: And Dave Seaman's reply: -- InterBang change nospam to optimusprime to reply === Subject: Re: Unabomber > suspect the unabomber was a mathematician: > And Dave Seaman's reply: some private email as a result of my comment back in 1995. He had lots more theories about who the Unabomber was, and he was eager to share them with anyone who would listen. He signed it Arthur T. Murray. He seems to be one of the more notable cranks on usenet. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Unabomber 2kpKShAQELm5uc578GgAAAJ5SURBVHichdNBb9sgFABgvMblarSwXDMcKdfUpso1td1xDWqfc020 mV5JoHt/f0Dc1pU2jUskvrz3eA9D6n8s8h+QnVSzv0G3zUDPlOzqut3N2g9YcwDQHgm1zJj9O8gc IshWIu+NMe4NEICFGKdoySHArzd4AI0ZQLNyZR7hNIJ8gL3YcLiIRuQQcg3te8S+CfVdpa4Rx8UI Z9DUcuiEXKcax8OkOAunEuUyZjJHN0IR24D9jlCVjjuMxbcm9ZGYh31zGOHR7CPkwILFiPYKsjdD PsCCg14AfDPmZRzJXWyp8GueQwH8bMzXK8jbmNbsFQ8hALSPo4qwTfv9M4YT6F7j7RvETGbgM5vp DPge1XmEHymi8NZqVrCiaB5vrpBKDNhYZEwznTWv8yvEg5sXL5QPY8kY8w+YoEuZ5nVlfRYiWOEs SQ1uUqaLKK0jnDG28K+nBPepCe9q1RGGXFN/Zgl+h/8PmsxqidnCEkZbe0gQDhUGW9C6tCQstF7p +JWECRrdE+pljZYUAXCXPh8ZMjGwrkKPWQCF+JpgF8fB81mzDVUpUmkxTxBGOMADLLxqS9Uhlkrd pouKt2QQTtWqEbWizarD/pQ6fzRD/yXvDyKsuq065ccbVP3QPy9hmJetRSSsVebpOsTfpj8pgIEU GSuyU931Hxc1KGS6VJY03wPQ8UXF8iqcVoiyEuvj5Knl5qTEmjax/PZ5Alt29GJFLmF/tZxPH+fd S8hjGyGVWz5N4f5gK7HzXWfd5ucnoOoidliV2H6qUZ9R3ohNqI7N1k3Bo3Chuhey2vgpzGh4lSv0 ohQKp7BwAUrESohOuQngtbf4s9qNIX8At0MbK5iXDI4AAAAASUVORK5CYII= Discussion, linux) >> suspect the unabomber was a mathematician: >> And Dave Seaman's reply: > some private email as a result of my comment back in 1995. He had lots more > theories about who the Unabomber was, and he was eager to share them with > anyone who would listen. > He signed it Arthur T. Murray. He seems to be one of the more notable > cranks on usenet. > I didn't think his evidence that the unabomber was a mathematician was all that compelling, but of course he *was* right. I wonder: any of his other theories close to the mark? Were his theories consistent? Or did he spew enough different theories that it's not too surprising if one of them was correct in some detail? -- Mathematicians are rather important in the infrastructures of many organizations that protect civilization. I've determined that they are a consistent security risk, and seem to have other agendas, other loyalties beyond loyalty to their respective nations. -- James Harris === Subject: Re: Unabomber > suspect the unabomber was a mathematician: > And Dave Seaman's reply: >> some private email as a result of my comment back in 1995. He had lots more >> theories about who the Unabomber was, and he was eager to share them with >> anyone who would listen. >> He signed it Arthur T. Murray. He seems to be one of the more notable >> cranks on usenet. >> > I didn't think his evidence that the unabomber was a mathematician was > all that compelling, but of course he *was* right. > I wonder: any of his other theories close to the mark? Were his > theories consistent? Or did he spew enough different theories that > it's not too surprising if one of them was correct in some detail? You mean his (ATM's) theories about the Unabomber, or his theories that are mentioned on that web page? What follows comes from an email correspondence, but ATM included a disclaimer that he was releasing the material into the public domain and that the FBI was now dealing with nine pages of information that ATM had provided (in 1995). ATM's theories about the Unabomber centered on one particular person, a computer consultant whom he had met at a coffee shop in the Seattle area. ATM described him as not a working, professional mathematician, but rather someone who shows a deep immersion in mathematics as the dominant aspect of his intellect. ATM identified him as F-C+, because the initials FC were known to be significant to the Unabomber and this person's initials were ED. F-C+ evidently had engaged ATM in a sort of cat-and-mouse game, each accusing the other of being the Unabomber. It looks like pretty trivial stuff, on the level of Dr. Matrix-style numerology. I suspect F-C+ had a sense of humor and was stringing ATM along. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Unabomber > suspect the unabomber was a mathematician: > And Dave Seaman's reply: > -- > InterBang > change nospam to optimusprime to reply It might be difficult to say why T. Kaczynski's mathematical background was not given much media attention, although I heard about it quite some time ago (don't remember how) and discovered this link to some of his works. http://www.rpi.edu/~bulloj/tjk/tjk.html Kevin O'Neill === Subject: Re: Unabomber 2kpKShAQELm5uc578GgAAAJ5SURBVHichdNBb9sgFABgvMblarSwXDMcKdfUpso1td1xDWqfc020 mV5JoHt/f0Dc1pU2jUskvrz3eA9D6n8s8h+QnVSzv0G3zUDPlOzqut3N2g9YcwDQHgm1zJj9O8gc IshWIu+NMe4NEICFGKdoySHArzd4AI0ZQLNyZR7hNIJ8gL3YcLiIRuQQcg3te8S+CfVdpa4Rx8UI Z9DUcuiEXKcax8OkOAunEuUyZjJHN0IR24D9jlCVjjuMxbcm9ZGYh31zGOHR7CPkwILFiPYKsjdD PsCCg14AfDPmZRzJXWyp8GueQwH8bMzXK8jbmNbsFQ8hALSPo4qwTfv9M4YT6F7j7RvETGbgM5vp DPge1XmEHymi8NZqVrCiaB5vrpBKDNhYZEwznTWv8yvEg5sXL5QPY8kY8w+YoEuZ5nVlfRYiWOEs SQ1uUqaLKK0jnDG28K+nBPepCe9q1RGGXFN/Zgl+h/8PmsxqidnCEkZbe0gQDhUGW9C6tCQstF7p +JWECRrdE+pljZYUAXCXPh8ZMjGwrkKPWQCF+JpgF8fB81mzDVUpUmkxTxBGOMADLLxqS9Uhlkrd pouKt2QQTtWqEbWizarD/pQ6fzRD/yXvDyKsuq065ccbVP3QPy9hmJetRSSsVebpOsTfpj8pgIEU GSuyU931Hxc1KGS6VJY03wPQ8UXF8iqcVoiyEuvj5Knl5qTEmjax/PZ5Alt29GJFLmF/tZxPH+fd S8hjGyGVWz5N4f5gK7HzXWfd5ucnoOoidliV2H6qUZ9R3ohNqI7N1k3Bo3Chuhey2vgpzGh4lSv0 ohQKp7BwAUrESohOuQngtbf4s9qNIX8At0MbK5iXDI4AAAAASUVORK5CYII= Discussion, linux) > It might be difficult to say why T. Kaczynski's mathematical background > was not given much media attention, although I heard about it quite some > time ago (don't remember how) and discovered this link to some of his > works. > http://www.rpi.edu/~bulloj/tjk/tjk.html Who says it wasn't given much media attention? It received rather a lot of attention at the time I thought. Of course, the OP was pointing to posts prior to the identification of Kaczynski as the Unabomber. -- Jesse F. Hughes That's the base tautological space where by tautological space I mean a region of truth. -- James S. Harris does philosophy of mathematics. JSH is a renaissance man. === Subject: Re: JSH: Math journal review > Some of you wish to knock the journal that published my paper, then > retracted it after pressure on the chief editor Ioannis Argyros, well > I looked him up and found the following at Cameron University: And almost all of them continued to knock the journal for retracting it rather than publishing the counter-arguments. Have you been reading the same thread I have? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: *EASY, HONEST, LEGAL CASH FAST!* #1) M. Teferi, 62 Farmfield court, thorplands, Northampton, NN3 8AL #2) J. Walker, 22 New Rowley Road, Dudley, West Midlands, DY2 8AS #3) H. Clarke, 32 Wood Street, Norwich, Norfolk, NR1 3RD #4) A. Ashwin, The Vicarage, Haydon Bridge, Northumberland, NE47 6LL #5) B. Fisher, 35 Castle Park, Eglinton, L'Derry, N.Ireland, BT47 3PL #6)D. Mossley, 48 Stefford road, Steffod, West Midlands DY2 4AD -- Clive Tooth http://www.clivetooth.dk === Subject: Re: math bibles Though not bibles - I would recommend A First Course in Calculus 5th edition by Serge Lang and Inside Calculus by George R. Exner for epsilon-delta proofs. -- Casey === Subject: Re: math bibles > The Bourbaki volumes were an impressive achievement when they were > published, and still worth reading today. > Any chance that someone will try to do what Bourbaki has done, updated > for the 21st century? and which parts do you think need updating, pray tell? as far as i can see, the material they cover is still an excellent foundation for today's mathematics. === Subject: Re: math bibles >>The Bourbaki volumes were an impressive achievement when they were >>published, and still worth reading today. >>Any chance that someone will try to do what Bourbaki has done, updated >>for the 21st century? > and which parts do you think need updating, pray tell? as far as i > can see, the material they cover is still an excellent foundation for > today's mathematics. (co)homology would have a more prominent role in organizing the whole structure. Algorithms would be used throughout, possibly. Finite mathematics and combinatorics would probably have their own books, integrated to the whole outline. Of course, it is hard to know for sure, until somebody does it. And I doubt that anybody will, any time soon. There doesn't seem to be any place in the mathematics world today that would nurture that kind of a massive effort. Al === Subject: Best fit with exponentials I doubt that this can be done in closed form, and I have an iterative approximation, but it is just too slow, so here it is: y = a * exp(bx) + c * exp(-dx) I need to find a, b, c, d, (all positive) that minimizes the sum of the squares of the errors against a set of (at least 4) points (x,y). Here is what I have done so far. If S is the sum of the squares of the errors, the partial derivative of S with respect to a and to b is linear in a and b. Therefore, for any c and d, I can easily find the a and b for which those partial derivatives are zero. So what I do is perform a two-dimensional trial and error search over c and d. When the total error stop going down, I stop trying. I need to implement this on a Palm PDA, and the floating point calculations are taking about 10 times longer than what is tolerable for my application. Does anyone have any idea on how to make the process converge faster, or (gasp!) to solve for a,b,c,d in closed form? -Robert Scott Ypsilanti, Michigan (Reply through this forum, not by direct e-mail to me, as automatic reply address is fake.) === Subject: Consecutive ingegers: Coprime consecutive integers: I've been trying to show that in any sequence of n consecutive integers, there existe at least one of them which is comprime with all the others (in pairs). Do you have any hint? === Subject: Re: Consecutive ingegers: Coprime > consecutive integers: > I've been trying to show that in any sequence of n consecutive > integers, there existe at least one of them which is comprime > with all the others (in pairs). > Do you have any hint? That is a very difficult statement to prove, since it implies that there is always a prime between two consecutive square numbers > 0. -- J K Haugland http://www.neutreeko.com === Subject: symmetries of platonic solids How does one analyze the cyclic structure of the set (viewed as permutations of the figure's vertex-set or of its face-set) of symmetries of a platonic solid? The most important question for my present purposes is: For each permutation, what are the sizes of its various cycles? A reference would be good, but of course the best thing would be a broad hint whose details I could work out on my own. Either way, any help will be mucho appreciado. Peace === Subject: Re: symmetries of platonic solids > How does one analyze the cyclic structure of the set (viewed as > permutations of the figure's vertex-set or of its face-set) of symmetries > of a platonic solid? The most important question for my present purposes > is: For each permutation, what are the sizes of its various cycles? A > reference would be good, but of course the best thing would be a broad > hint whose details I could work out on my own. Either way, any help will > be mucho appreciado. Isn't it fairly easy to visualise the symmetries? A proper symmetry is a rotation about some axis. The axis will have to go through a vertex, the centre of a face, or the centre of an edge. Unless one is considering the regular tetrahedron (in which case the symmetry group is just S_4) reflection J in the centre is an improper symmetry, and an improper symmetry is of the form Jr, where r is a proper symmetry. It's easy enough to determine how the vertices are permuted by Jr. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: symmetries of platonic solids | | > How does one analyze the cyclic structure of the set (viewed as | > permutations of the figure's vertex-set or of its face-set) of symmetries | > of a platonic solid? The most important question for my present purposes | > is: For each permutation, what are the sizes of its various cycles? A | > reference would be good, but of course the best thing would be a broad | > hint whose details I could work out on my own. Either way, any help will | > be mucho appreciado. | | Isn't it fairly easy to visualise the symmetries? ## not for me. | A proper symmetry is a rotation about some axis. ## proof? | The axis will have to go through a vertex, the centre of a face, | or the centre of an edge. | Unless one is considering the regular tetrahedron | (in which case the symmetry group is just S_4) | reflection J in the centre is an improper symmetry, | and an improper symmetry is of the form Jr, | where r is a proper symmetry. ## proof? | It's easy enough to determine how the vertices are permuted by Jr. ## not for me. | | | -- | Timothy Murphy | e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie | tel: +353-86-2336090, +353-1-2842366 | s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: symmetries of platonic solids > | .... > | Isn't it fairly easy to visualise the symmetries? > ## not for me.... Use a model! Hold it in your hands and turn it around! Ken Pledger. === Subject: Complete orthonormal sequence of polynomials by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4QHfFk25425; I would appreciate any help for this problem Does there exist any complete orthonormal sequence f_n of polynomials in L^2(a,b), -00 < a < b < 00, such that f_n is a polynomial of degree n?And if it exist,is it unique? Felix === Subject: Re: Complete orthonormal sequence of polynomials > I would appreciate any help for this problem > Does there exist any complete orthonormal sequence f_n > of polynomials in L^2(a,b), -00 < a < b < 00, such that f_n is a > polynomial of degree n?And if it exist,is it unique? yes. yes, up to sign. When a = -1 and b = 1 these are the (normalized) Legendre polynomials. -- 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: On the partial sums of reciprocals of primes by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4QHfFn25463; >In Prime Numbers: A computational Perspective by Crandall and Pomerance >you will find this fact on page 32. Their Theorem 1.4.2 (Mertens) states >that: >As x -> inf, prod_{p le x}(1 - 1/p) ~ e^{-gamma}/ln(x) >where gamma is the Euler constant. Taking the logarithm of this relation, we >have, >sum_{p le x}{1/p} = ln(ln(x)) + B + o(1), >for the Mertens constant B defined as >B = gamma + sum_{p}{(ln(1 - 1/p) + 1/p) Really, a lot of people suggest me that the proof of >sum_{p le x}{1/p} = ln(ln(x)) + B + o(1) can be found in Hardy and Wright. But please note that I said S(x) < lnlnx + B + (1/logx)^2. The point is this explicit formula for the upper bound. Am I missing something that trivially shows that the proof of the asymptotic formula implies the explicit upper bound? After I said this, none of them has replied so far... erdos fan === Subject: Re: Looking for a link on Talks to School Kids Comments: This message probably did not originate from the above address. It was automatically remailed by one or more anonymous mail services. You should NEVER trust ANY address on Usenet ANYWAYS: use PGP !!! Get information about complaints from the URL below X-Remailer-Contact: http://80.65.224.85/POL/ In case my abuse address is unreachable: It is because it has been flooded by , please contact Mail-To-News-Contact: abuse@dizum.com -----BEGIN PGP SIGNED MESSAGE----- cobien qve Abscondisti haec a sapientibvs et prvdentibvs id est potentibvs et regibvs et envcleastiea exigvis et tenvibvs >I lost a few links I had on putting together talks for schools etc on astronomy. >I cant seem to find the sites, they had relly good advice and links to resourses >Any one hve links to sites like this. Well that depends. If you are in the habit of LYING to children & adults alike, then you'd surely NOT want to include these links. But if SCIENCE is of any interest to you and your students, then by all means, add these: *Min's Secular Academia's so-called Copernican Revolution Is Dead and Buried! (download this URL & print it out for proof): *Min's Step-by-Step Lesson: How To Observe Sidereal & Synodic Planetary Orbits With Our Unaided Human Eyes: *All Apollo Moon Missions Were Unmanned: *Uncensored Apollo Moon Hoax Bookmarks: -----BEGIN PGP SIGNATURE----- iQA/AwUBQLYxgZljD7YrHM/nEQJ2CwCgih7VfgMvsSVmOL5oOeghOjwyjaEAoK7e GEYKfvgbDtv1T99RW2IIAqt5 =LxhI -----END PGP SIGNATURE----- === Subject: Re: Geometric Mean >The arithmetic mean of a set S is 1/#S sum x_i for all x_i in S. >Although this implies a countable set, there is a strong sense in >which it can be applied to an uncountable set. Consider S = (0,1) >by pairing every x < 1/2 with 1-x, we can interpret the arithmetic >mean of this set to be 1/2. You can's conclude anything like that unless you specify a distribution. The uniform distribution would have a mean of 1/2, but a gamma distribution (for example) would not, even though it contains all the same points. You can get the mean by integrating the product of the probability distribution function and x. (And dividing by the integral of the pdf, which is always 1.) >The geometric mean is G = (product x_i in S)^(1/#S). We can take >logs --> log G = 1/#s sum(log(x_i)) and replace the sum by an >an integral, but we still need to deal with 1/#S, where the set is >uncountable. Further if S = (0,1), the integral diverges. >Is there any meaningful way to interpret the geometric mean of >all the reals in (0,1)? What about the following interpretation: >Consider the set S = (0,1) and the projective transform S' = 1/S. >Now, geometric mean of G = 1/geometric mean of 1/G, so can we take >this mean as the reciprocal of >lim n-->00 1/N * (integral from 1+ to N of dx/log(x)) ? Again, it depends on the pdf. Find the mean of the pdf(x) * log(x) if you can, and then raise e to that power. I don't think it exists for the uniform distribution on (0,1). --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Geometric Mean In sci.math, Bob Silverman > The arithmetic mean of a set S is 1/#S sum x_i for all x_i in S. > Although this implies a countable set, there is a strong sense in > which it can be applied to an uncountable set. Consider S = (0,1) > by pairing every x < 1/2 with 1-x, we can interpret the arithmetic > mean of this set to be 1/2. Well, if one wants to be fairly pedantic, the arithmetic mean of a set of reals might be construed to be simply the Lebesgue integral (x in S) x dx / measure(S), where measure(S) is the Lebesgue measure of the set. (Standard Riemannian integration might be used if S is well enough behaved, but Riemannian integration doesn't deal with sets such as (0,1) intersect (R - Q) -- the set of irrationals within (0,1) -- too well.) > The geometric mean is G = (product x_i in S)^(1/#S). We can take > logs --> log G = 1/#s sum(log(x_i)) and replace the sum by an > an integral, but we still need to deal with 1/#S, where the set is > uncountable. Further if S = (0,1), the integral diverges. Well, one gets log(G(S)) = integral(x in S) log x dx / measure(S), whose numerator is a slightly ugly beast. > Is there any meaningful way to interpret the geometric mean of > all the reals in (0,1)? What about the following interpretation: > Consider the set S = (0,1) and the projective transform S' = 1/S. > Now, geometric mean of G = 1/geometric mean of 1/G, Not sure if that helps all that much or not. > so can we take > this mean as the reciprocal of > lim n-->00 1/N * (integral from 1+ to N of dx/log(x)) ? -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: cylinders My math book calls a cylinder with infinity height a cylinder. However, I was wondering if there is specific terminology to distinguish between a cylinder with infinity height and a cylinder with a finite height. === Subject: Re: cylinders > My math book calls a cylinder with infinite height a cylinder. However, I > was wondering if there is specific terminology to distinguish between a > cylinder with infinite height and a cylinder with a finite height.... I'm afraid this is one of those cases where mathematical terminology is sloppy. You need to see from the context which meaning of cylinder an author is using. Ken Pledger. === Subject: Re: cylinders > My math book calls a cylinder with infinity height a cylinder. However, I > was wondering if there is specific terminology to distinguish between a > cylinder with infinity height and a cylinder with a finite height. The latter might be called a truncated cylinder, though I suppose one would need to say how it is truncated. -- G.C. === Subject: Re: Polygons With Angles Of Different k-gons > I am wondering generally about irregular polygons which have angles > each equal to that of different regular polygons. > > In other words, each interior angle is equal to > > (1 - 2/k)*180 degrees > > for a set of positive integers k. > > For example, we can have a 5-gon with the angle of an equilateral > triangle, of a square, of a regular pentagon, of a regular hexagon, > and of a regular 20-gon. > > Or, allowing dupicate angles, we can have a 4-gon with 2 angles of an > equlilateral triangle, one angle of a square, and one of a > dodecagon(12-gon). > > > If we are allowed to adjust our polygons' side-lengths as necessary, > then which such irregular polygons can tile the infinite plane with > congruent copies of themselves? > A convex polygon that tiles the plane must have at most six sides, so > the only possible choices come from the Egyptian fraction > representations of 1 with at most six terms, all <= 1/3. There are only > finitely many such representations. > In addition, it has to be possible to fit the largest angle together > with some combination of other angles around a single vertex; a > combination of four angles is impossible (the four-angle combination > maximizing the largest angle is 1/3+1/3+1/4+1/12, but all the Egyptian > fractions I found had a term smaller than 1/12) so we must have three of > these fractions 1/a + 1/b + 1/c = 1/2 where c is the largest term in the > Egyptian fraction. > So, it seems, this eliminates all but the following possibilities > (in the distinct angle case; of course there are more if you allow > repeated angles): > 1/3 + 1/4 + 1/5 + 1/6 + 1/20 > 1/3 + 1/4 + 1/5 + 1/7 + 1/20 + 1/42 > 1/3 + 1/4 + 1/5 + 1/8 + 1/20 + 1/24 > 1/3 + 1/4 + 1/5 + 1/9 + 1/18 + 1/20 > 1/3 + 1/4 + 1/5 + 1/10 + 1/15 + 1/20 > 1/3 + 1/4 + 1/6 + 1/7 + 1/12 + 1/42 > 1/3 + 1/4 + 1/6 + 1/9 + 1/12 + 1/18 > 1/3 + 1/4 + 1/6 + 1/10 + 1/12 + 1/15 > In each of these cases, the internal angles of the hexagon can be > partitioned into two triples that each add to 360 (in the first > solution, consider the pentagon to be a degenerate hexagon with a 180 > degree angle). > If one can form a hexagon in which no two angles from the same triple > are adjacent, and opposite pairs of edges have equal length, I think > that would be enough to guarantee a tiling. If we fix one edge's length > and orientation, there are two remaining degrees of freedom in the other > two edge lengths and two degrees of freedom needed to get the edge > sequence to form a closed hexagon, so it looks like it may be possible > in each case, but I haven't worked out the details. Ah, interesting. In the {3,4,5,6,20} case, there is a very simple tiling. (I will try to explain without pictures.) Let P be the polygon with interior angles, in order, {a,b,c,d,e}. The angles are arranged like this: Angle ___ is that of regular __-gon, which is ___ degrees. a 3 60 b 6 120 c 5 108 d 4 90 e 20 162 Side bc has the same length as side ea. Sides bc and ea are parallel. Line up a row of Ps where each P's side bc coincides with its neighbor's side ea. So, the sides ab of these Ps will all be along one straight line. Duplicate this row and turn the new row over 180 degrees. Adjoin the new row with the original row. You should have an infinitely long strip, bounded above and below by the sides ab of the polygons. Of course, just placing one strip on top of another completes the tiling of the infinite plane. Leroy Quet === Subject: Numerators Of sum{k=1 to m} 1/k^n I wonder, for a particular n and r = unequal positive integers, which j's and m's are such that: sum{k=1 to m} 1/k^n has the same numerator (in reduced form) as sum{k=1 to j} 1/k^r ? For n = 1 and r = 2, for example, we have numerator(sum{k=1 to 6} 1/k) = numerator(sum{k=1 to 3} 1/k^2) = 49. And of course, for all n and r the numerators equal if m = j = 1. Leroy Quet === Subject: Re: Skew-symmetric matrix determinant === Subject: Re: Simplification > f(x) = floor(x/a) - x/a is a periodic function on the integers with > period a. We can express it as a Fourier series using the characters > of Z_a, i.e. > f(x) = sum_{k=0}^{a-1} c_k w^(kx) > where w = exp(2 pi i/a) and > c_k = 1/a sum_{j=0}^{a-1} f(j) w^(-kj) > = - 1/a^2 sum_{j=0}^{a-1} j w^(-kj) > = - w^k/(a(1-w^k)) > c_0 = -1/a^2 sum_{j=0}^{a-1} j = (1-a)/(2 a) > That is, > floor(x/a) > = x/a + (1-a)/(2a) - 1/a sum_{k=1}^{a-1} w^(k+kx)/(1-w^k) > It can also be written as > x/a + (1-a)/(2a) + sum_{k=1}^{a-1} sin(pi k(1+2x)/a)/(2 a sin(pi k/a)) > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 For x=12, a=5: x/a + (1-a)/(2a) + sum_{k=1}^{a-1} sin(pi k(1+2x)/a)/(2 a sin(pi k/a)) = 12/5, but floor(x/a)=floor(12/5) What is wrong? Christian Ruether === Subject: Re: Question about Combinatorics >i'm searching the following solution >there are T positions t1,t2, ... ,tT >and L different colors c1,c2, ... ,cL >there exists L^T different combinations. >now i like to calculate the number of >possibilities, where A colors are different. >for example, how many combinations exists, >where 3 entries are different? How many of the original L^T use at least 3 colors, is that what you're asking? It's pretty straightforward if you take the total number of combinations and subtract. Number of cases which use one color: L Number of cases which use two colors: (L choose 2) * (2^T - 1) So the answer is L^T - L * (L-1)/2 * (2^T-1) - L. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Question about Combinatorics charset=Windows-1252 >there are T positions t1,t2, ... ,tT >and L different colors c1,c2, ... ,cL >there exists L^T different combinations. >now i like to calculate the number of >possibilities, where A colors are different. >for example, how many combinations exists, >where 3 entries are different? > How many of the original L^T use at least 3 colors, is that what you're > asking? > It's pretty straightforward if you take the total number of combinations and > subtract. > Number of cases which use one color: L > Number of cases which use two colors: (L choose 2) * (2^T - 1) ^^^ For that I get ..........................(L choose 2) * (2^T - 2) > So the answer is L^T - L * (L-1)/2 * (2^T-1) - L. ^^^ and ...............L^T - L * (L-1)/2 * (2^T-2) - L. But the OP may be asking how many use exactly A colors, i.e. the the number of mappings from {t1,t2, ... ,tT} to {c1,c2, ... ,cL} such that the image size is exactly A. (I.e., assign the T positions to the L colors so that exactly A colors are used.) That number is A! C(L,A) S(T,A) where C(i,j) is a binomial coefficient and S(i,j) is a Stirling number of the second kind. For the Stirling numbers, see http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html This result can be seen as follows: First, pick the image set of size A -- there are C(L,A) ways to do that. Now pick a collection of pre-images of these A points, i.e. partition {t1,t2, ... ,tT} into A non-empty subsets -- there are S(T,A) ways to do that. Finally, map the A pre-images to the A image points -- there are A! ways to do that. Therefore, the desired number is A! C(L,A) S(T,A). ( Consequently, the probability that a random mapping from {t1,t2, ... ,tT} to {c1,c2, ... ,cL} has image size A, is A! C(L,A) S(T,A) / L^T (0<=A<=L). ) --r.e.s. === Subject: Re: Question about Combinatorics charset=Windows-1252 >there are T positions t1,t2, ... ,tT >and L different colors c1,c2, ... ,cL > >there exists L^T different combinations. > >now i like to calculate the number of >possibilities, where A colors are different. >for example, how many combinations exists, >where 3 entries are different? > How many of the original L^T use at least 3 colors, is that what you're > asking? > It's pretty straightforward if you take the total number of combinations > and > subtract. > Number of cases which use one color: L > Number of cases which use two colors: (L choose 2) * (2^T - 1) > ^^^ > For that I get ..........................(L choose 2) * (2^T - 2) > So the answer is L^T - L * (L-1)/2 * (2^T-1) - L. > ^^^ > and ...............L^T - L * (L-1)/2 * (2^T-2) - L. > But the OP may be asking how many use exactly A colors, i.e. the > the number of mappings from {t1,t2, ... ,tT} to {c1,c2, ... ,cL} > such that the image size is exactly A. (I.e., assign the > T positions to the L colors so that exactly A colors are used.) > That number is A! C(L,A) S(T,A) where C(i,j) is a binomial > coefficient and S(i,j) is a Stirling number of the second kind. > For the Stirling numbers, see > http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html > This result can be seen as follows: > First, pick the image set of size A -- there are C(L,A) ways to > do that. Now pick a collection of pre-images of these A points, > i.e. partition {t1,t2, ... ,tT} into A non-empty subsets -- there > are S(T,A) ways to do that. Finally, map the A pre-images to the > A image points -- there are A! ways to do that. Therefore, the > desired number is A! C(L,A) S(T,A). > ( Consequently, the probability that a random mapping from > {t1,t2, ... ,tT} to {c1,c2, ... ,cL} has image size A, is > A! C(L,A) S(T,A) / L^T (0<=A<=L). ) ^^^ Oops, that should be ...... 1<=A<=L > --r.e.s. === Subject: Re: Question about Combinatorics > i'm searching the following solution > there are T positions t1,t2, ... ,tT > and L different colors c1,c2, ... ,cL > there exists L^T different combinations. There are L^T different ways to color each position with repetition allowed. That is different from the number of ways to assign each color to a position. That number is T^L. > now i like to calculate the number of > possibilities, where A colors are different. > for example, how many combinations exists, > where 3 entries are different? It depends on what, exactly, you are trying to count. You want to do an assignment of positions and entries, but you haven't said how you want to make that assignment. > is there any closed solution aviable (maybe > very simple, but i didn't got the solution)? Possibly, but the question isn't clear. -- Will Twentyman email: wtwentyman at copper dot net === Subject: DFT and FFT Hello I'm a Computer Science student and my tutor gave me the idea of doing FFT for my final year project. I've read some stuff about it but it's all in very mathy language. I gather that it's an algorithm for turning... something... into a frequency? But what use is that, practically? So I was wondering if someone could tell me what FFT does and what it's used for, in very simple and normal language, so that I have a base for understanding what I'm reading about. Clare === Subject: Re: Sigma-ring proof > Your question is not very clear; usually one defines the Borel set B of R^k > as the smallest sigma-algebra (or sigma-ring, same thing here since the > space R^k itself is open and whence belongs to B) containing the open > subsets of R^k. Now your definition deals with constructing sets using > certain operations. (Btw, you didn't define that process precisely and you > should do it (that's not thaaaaaat easy to formulate)) Can you please outline how one would describe the process precisely? >What you can say is > that any sigma-ring containg the open sets of R^k will also contain (by > definition) all the sets constructed by using the previous process. The > resulting set (being also a sigma ring) is therefore the smallest sigma ring > containing the open sets. This is trivial. OK. Can you please indicate some subsets of R^k that are left out of this smallest sigma-ring, which would be included in, say, the power set? Clearly all open and all closed subsets are included in B, so that would leave those which are neither open nor closed. === Subject: Re: Sigma-ring proof > OK. Can you please indicate some subsets of R^k that are left out > of this smallest sigma-ring, which would be included in, say, the > power set? Clearly all open and all closed subsets are included in B, > so that would leave those which are neither open nor closed. Well, any non-measurable subset of R^k would do ... let's take k = 1, the classical example is that of the set of the classes of the equivalence relation: xRy <=> (x-y) is rational. You'll find a proof in any book. -- Julien Santini === Subject: Re: Sigma-ring proof <0405242137240.8916-100000@gandalf.math.ukans.edu> > Rudin's text defines a Borel set as one which can be obtained by a > countable number of operations, starting from open sets, each > operation consisting in taking unions, intersections, or > complements. I can understand how the collection B of all Borel > sets in R^k is a sigma-ring. However, how can one prove that B is > the smallest sigma-ring which contains all open sets? It would > appear that B is a sigma-algebra as well, since it closed under > complementations. Is this correct? Lastly, can the above > conclusions concerning B hold for spaces other than R^k? Many > If a collection S of sets is closed under complementation, then the > sigma-ring generated by S will also be closed under complementation; > i.e., the sigma-ring generated by S coincides with the sigma-algebra > generated by S. In particular, in any topological space, the > sigma-ring generated by the open sets together with the closed sets is > a sigma-algebra. Now, in a metric space, every closed set is a G-delta > set (intersection of countably many open sets); in that case, the > sigma-ring generated by the open sets will also contain the closed > sets, and it is a sigma-algebra. your question, and I also forgot the meaning of the word sigma-ring. i.e., a collection of sets closed under countable unions and intersections. And I was trying to explain why, in a metric space, the smallest collection of sets containing the open sets and closed under unions and intersections is a sigma-algebra; but that's not what you asked. As to why the collection B of all Borel sets is the smallest sigma-ring containing the open sets, isn't that the definition of B? === Subject: Re: Factoring linear differential operator >> I mean how do we know that: (D^2 + 5D + 6)y = (D+3)(D+2)y ? >I would do it by multiplying it out, and writing down what it means >(D+3)(D+2)y := (D+3)(Dy+2y) := D(Dy+2y)+3(Dy+2y) >= D(Dy)+D(2y)+3Dy+6y = D^2y+5Dy+6y := (D^2+5D+6)y > Good example - we should point out that this example shows > why the result is only stated for constant coefficients. More generally, following up also on my previous reply, the operator product (A+B)(A+C), A, B, C are operators, has the result: (A+B)(A+C)y = (A+B)(Ay + Cy) = A(Ay + Cy) + B(Ay + Cy) = A(Ay) + A(Cy) + B(Ay) + B(Cy) = (AA)y + A(Cy) + (BA)y + (BC)y = (AA)y + A(Cy) + A(By) + (BA)y - A(By) + (BC)y = (AA)y + A(By + Cy) + (BA)y - (AB)y + (BC)y = (AA)y + A((B+C)y) + (BC)y + (BA - AB)y = (AA)y + (A(B+C))y + (BC)y + [B,A]y = (AA + A(B+C) + BC + [B,A])y, the commutator [B,A] is defined as the operator which when applied to an f yields: [B,A]f = B(Af) - A(Bf). Thus (A+B)(A+C) = A^2 + A(B+C) + BC + [B,A]. Also proveable by the same reasoning is: (A+B)(A+C) = A^2 + (B+C)A + BC + [A,C]. When D is the differentiation operator and B, C are functions then [B,D]y = B(Dy) - D(By) = By' - (By)' = By' - B'y - By' = -B'y so that [B,D] = -B'. Similarly, [D,C] = C'. Therefore, in this case: (D+B)(D+C) = D^2 + D(B+C) + BC - B' = D^2 + (B+C)D + BC + C'. === Subject: Re: Factoring linear differential operator > I mean how do we know that: (D^2 + 5D + 6)y = (D+3)(D+2)y ? It's true by definitions on top of definitions. An operator is defined by the effect it has on a function. The operator algebra is defined by: A,B operators -> A+B defined as the operator that maps f |-> Af + Bf A,B operators -> AB defined as the operator that maps f |-> A(Bf) A operator -> kA defined as the operator that maps f |-> k(Af). Therefore, since ((D+3)(D+2))y = (D+3)((D+2)y) = (D+3)(Dy + 2y) = (D+3)(y' + 2y) = D(y' + 2y) + 3(y' + 2y) = D(y') + D(2y) + 3y' + 3(2y) = D(Dy) + 2 Dy + 3 Dy + 6y = D(Dy) + 5 Dy + 6y = (D^2)y + (5D)y + 6y = (D^2 + 5D + 6)y then the operators (D+3)(D+2) and (D^2 + 5D + 6) yield the same result, which makes them equal by definition. === Subject: Re: Factoring linear differential operator >I mean how do we know that: (D^2 + 5D + 6)y = (D+3)(D+2)y ? Answer: Because of the distributive property of differentiation over addition. i.e. We know that d(f(x)+g(x))/dx = d(f(x))/dx + d(g(x))/dx, so if there are nested transformations (D+3) and (D+2), you have (D+3)(D+2)y (D+3)(dy/dx + 2y) d(dy/dx + 2y) + 3(dy/dx + 2y) d^2y/dx^2 + 2dy/dx + 3dy/dx + 6y --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: resolving Will's misunderstanding >>>>Let's try a different approach, starting with definitions of everything >>>>under consideration: >>>> >>>>We are talking about functions, but the real concern is what algorithms >>>>are used to compute those functions. For simplicity's sake, we can >>>>restrict our discussion to functions mapping from the whole numbers to >>>>the whole numbers f:W->W. For an algorithm to correctly represent >>>>f(x)=y, it should start with input x and produce output y after finitely >>>>many steps, where each step is computed from the input and the previous >>>>steps. For a given function f, f may or may not be defined for all/any >>>>numbers in the whole numbers. As a result, the algorithm describing f >>>>is only required to be finite for those values of x for which f is >>>>defined. A function f with such an algorithm is in general called a >>>>partial function. A partial function f which is defined on all elements >>>>of the whole numbers is called a total function and is considered to be >>>>computable (or recursive). >>>> >>>>At this point, I believe we can start making some clear observations. >>>>First: the halting problem is only available for discussion if we a >>>>dealing with partial functions (which have algorithms that do not halt >>>>in finite time for some input values). >>>>Second: it appears to be your desire to discuss only total functions. >>>>Third: if we restrict ourselves to the total functions, discussing the >>>>halting problem is meaningless. >>>> >>>>A question that is relevent to the above: if we are restricting >>>>ourselves to total functions, is there a way to specify the algorithms >>>>of all total functions in such a way that there is no possibility of >>>>dealing with potentially partial functions? >>>> >>>>Before moving forward, do you have any comments on the above? Right now >>>>I'm just trying to set a framework for a detailed analysis, but if the >>>>basic framework is in dispute, I may need to adjust for that. >>>>> >>>>>The term to be defined is valid construction. Facts & FALSE -> anything. >>>> >>>>Agreed. I just wanted to make sure there were no objections before >>>>diving into the details. If the basic concepts to be defined in detail >>>>can't be agreed on, then there isn't much point in going further. >>>> >>>>For the moment, let's allow functions to be n-ary for ease of notation. >>>>(It can be shown that we can restrict ourselves to unary functions, but >>>>they get messier.) >>>> >>>>Let's start by defining primitive recursive functions These functions >>>>are guaranteed to be total functions and can be constructed by a finite >>>>number of applications of the following rules: >>>> >>>>1) f(x) = x+1 is primitive recursive >>>>2) f(x1,x2,...,xn) = m is primitive recursive where n and m >= 0 >>>>3) f(x1,x2,...,xn) = xi is primitive recursive where 0 < i <= n >>>>4) If g1, g2, .. gm, are primitive recursive and n-ary, and h is >>>>primitive recursive and m-ary then so is n-ary f defined as >>>> f(x1,x2,...,xn) = h(g1(x1,x2,...,xn),...,gm(x1,x2,...,xn)) >>>>5) For n>=1, if f is n-ary, g is (n-1)-ary, h is (n+1)-ary, and g and h >>>>are primitive recursive, then so is f defined by the following two rules: >>>>a) f(0,x2,x3,...,xn) = g(x2,x3,...,xn) >>>>b) f(x1+1,x2,x3,...,xn) = h(x1,f(x1,x2,...,xn),x2,...,xn) >>>> >>>>These rules appear (to my eye) to correspond with the definition of a >>>>function guaranteed to halt you offered in another thread. >>>> >>>>Using these it is possible (though messy) to construct virtually all >>>>normal mathematical functions on the whole numbers. In particular, >>>>prime decomposition that will allow a number to represent a sequence of >>>>numbers by its prime decomposition. This enables us to formally get >>>>back to unary functions, if need be. >>>> >>>>Any comments at this point? The next question will be, do primitive >>>>recursive functions describe ALL total computable functions? >>> >>>nope >>>> >>>>There's two questions, which one are you saying nope to? >>> >>>the 1st one, you're supposed to be establishing the *next* question. >> >>Since there are only 5 rules for deriving primitive recursive functions, >>each one can be indexed based on what rules/functions are used to >>construct it. Consider one such indexing f0, f1, f2, f3, ... and let us >>construct a new function g0(x) = fx(x) + 1. Since fx can be formally >>considered unary, regarless of what n is for n-ary fx, and x determines >>the construction of f, g0(x) is defined on all x in the Whole Numbers. >>Moreover, we have included all partial recursive functions on our list >>f0, f1, f2, .... Therefore, g0(x) is a total function, it is >>computable, but it is not a primitive recursive function. By prepending >>g0(x) to the listing of primitive recursive functions, we can construct >>g1(x), g2(x), etc. Since there are multiple possible indexing schemes >>available for the primitive recursive functions, there are also multiple >>sequences of gi(x) available. >> >>Any issues with the above? >> >>> >>>I got no idea what rule 5 does. It looks like a step -1 for loop that pushes the >>>parameters into the 2nd parameter of the 2nd parameter recursively, making >>>a longer parameter list every iteration. >> >>It's a construction similar to f(0) = c, f(x+1) = g(f(x)). The idea is >>to define f(0) as something known, and define f(x+1) in terms of f(x). >>The parameter list doesn't actually get longer. >> >> >> >>construct a new function g0(x) = fx(x) + 1 >>> >>>this is just diagonalisation all over again, f with godel x applied to its own godel number >>>with a modification. >> >>You are quite right. Now, does that invalidate the conclusion? >> >> >> >>>what's to stop me defining consistent p.r.f. where variable references are contextually free. >>>fx(x) = fx(y) for any x, y >> >>Those functions are described in construction 2, they are the constant >>functions. This has nothing to do with the validity or invalidity of >>the argument. >> >> >> >>>this stops dummy functions like the above that double usage the x in g0(x) >>>I've shown that getting a parameter x and using it for 2 arguments is >>>enough to change a total system to non halting system. >> >>We have not gotten to anything that does not halt. All that is being >>said is: a sufficient restriction to prevent functions that don't halt >>also excludes some that do always halt. >ok, but how hard is it to exclude functions of type fx(x), they're the ones going >into infinite loops. what computation are they performing that can't be >done some other way? >>The functions fx do not contain infinite loops. Nor do the functions gi >>built from them. The point was that the primitive recursive functions, >>while guaranteed to halt, are not *all* of the functions that are >>guaranteed to halt. The various functions gi are also guaranteed to >>halt, but they were found using a diagonal-style argument, and are >>therefore not primitive recursive. >>Having said all that, MOST functions and operations that we would want >>to perform would be primitive recursive functions, including things like >>the addition function, subtraction function (which produces 0 for >>negative results), multiplication, integer division, modular division, >>exponentiation, etc. As long as the computations can be broken down >>into the 5 rules applied to various primitive recursive functions, we >>are good to go. >>So far, there are no infinite loops, and everything is guaranteed to >>halt. It's just that not everything that halts is primitive recursive. >> The trick is to see if we can find something that will include ALL >>functions that halt, and still exclude those that don't. > If fx(x) is a construct that halts, then UTMs must be impossible. Because > mUTM(mUTM) doesn't halt, as we've been through. > If UTMs are not part of primitive recursive functions, then they're not complete > anyway. You are putting the cart before the horse. UTM has not been defined yet. All I have said is there are functions that are guaranteed to halt that are not primitive recursive. > Even so, what does g0(x) do? I still say you are entrenched in *names* of > functions and not *what* they actually compute. Two algorithms can perform > the same computation can't they? g0(x) computes something different from ALL primitive recursive functions. It proves that there is a function that is guaranteed to halt on all inputs, yet is not primitive recursive. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Another mass question. > Mass is volume multiplied by density. It's the integral of density over a volume. Density is not the factor of a product, but the kernel of a 3-cochain. === Subject: Thomas C. Hales - The Kepler Conjecture-->Overview by Paul Gartside boundary=------------070505050203020703090803 --------------------------------------------------------------------- Cannonballs and Honeycombs Thomas C. Hales with Overview by Paul Gartside An Overview At the turn of the last Century the famous mathematician Hilbert presented a list of 23 mathematical problems. The 18th of these problems, Kepler's Conjecture, can be phrased, Is there a better stacking of oranges than the pyramids found at the fruit stand? In pyramids, the oranges fill just over $74$% of space. Can a different packing do better? In August 1998, nearly 400 years after Kepler first made his conjecture, Thomas Hales, with the help of his graduate student, Samuel Ferguson, confirmed the conjecture, and solved Hilbert's 18th problem. These pages give the broad outlines of the proof of the Kepler conjecture in the most elementary possible terms. Along the way the history of the Kepler conjecture is sketched. * Hilbert: It seems `obvious' that Kepler's conjecture is correct. Why is the gulf so large between intuition and proof? Geometry taunts and defies us.... * Harriot and Kepler: The genesis of Kepler's conjecture. The pyramid stacking of oranges is known to chemists as the face-centered cubic packing. It is also known as the cannonball packing, because it is commonly used for that purpose at war memorials. Theorem (Kepler conjecture). No packing of balls of the same radius in three dimensions has density greater than the face-centered cubic packing. * Gauss: The first to prove anything about the Kepler conjecture, was Gauss. * Thue - Down to 2 dimensions: A typical mathematical gambit, is to gain insight into a hard problem by first tackling a simplified version. Thue solved the 2 dimensional analog of Kepler's problem in 1890. The two-dimensional version of Kepler's conjecture asks for the densest packing of unit disks in the plane. * Back to 3 dimensions - Voronoi: Implicit in the proof of Thue's theorem is the idea of a Voronoi cell. Let $t>1$ be a real number. We define a cluster of balls to be a set of nonoverlapping balls around a fixed ball at the origin, with the property that the ball centers have distance at most $2t$ from the origin. A cluster of $n$ balls is determined by the $3n$ coordinates of the centers. The ball at the center of the cluster is contained in a Voronoi cell. By definition, the Voronoi cell is the set of all points that lie closer to the origin than to any other ball center in the cluster. Voronoi cells give a bound on the density of sphere packings. * Fejes Toth: The bound given by Voronoi cells is not sufficient, and a correction term must be introduced. Toth, in 1953, was the first to suggest a potential correction term. It remains unclear if his proposed correction terms have all the requisite properties. But it became clear that Kepler's problem could be solved via an optimization problem in a finite number of variables - and this optimization might be performed by computer. * Combinatorial Structures: The space of clusters is so complicated that it is not possible to minimize the correction term directly. Instead to each cluster is assigned a planar graph that identifies the most prominent geometrical features of the cluster. There are about 5000 planar graphs that need to be dealt with. * 5000 Cases: In most cases bounds derived just from the graph were sufficient. But in some cases, the crude combinatorial bounds were not good enough. One case turned out to be far more intricate than the others. * Linear Programs: For each case, there is a large-scale nonlinear optimization problem to be solved. Nonlinear optimization problems of this size can be hopelessly difficult to solve rigorously. Fortunately, the large-scale structure of the problem is linear and can be solved by linear programming methods. ------------------------------------------------------------------------ Honeycombs A related problem, of even greater antiquity, is: What is the most efficient partition of the plane into equal areas? The honeycomb conjecture asserts that the answer is the regular hexagonal honeycomb. After completing the proof of the Kepler conjecture, Thomas Hales turned his attention to the honeycomb conjecture. Somewhat to his surprise he obtained a (relatively) short solution without resort to computers. Theorem (Honeycomb conjecture). Any partition of the plane into regions of equal area has perimeter at least that of the regular hexagonal honeycomb tiling. * Pappus: In 36BC, in a book on agriculture, the Roman scholar Marcus Varro, presented the honeycomb conjecture not as a conjecture but as a proven fact. But the `proof', as recorded by Pappus, is incomplete. * Kelvin and the Millenial List: The 3 dimensional version of the efficient partition problem is: How can space be divided into cavities of equal volume so as to minimize the surface area of the boundary? This is Hale's submission for a millenial `Hilbert problem list' * References -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ --------------070505050203020703090803-- === Subject: Re: Entropy of samples on real line segment. > Is there a precise measure/definition of the 'entropy' of N points on > the real line segment [0,1]? I am guessing that N points positioned at > (i+.5)/N (where i=0...N-1) gives the configuration with 'maximum' > entropy. How about this: Let x_i be the position of the i'th point. Then let d_i = x_i - x_{i-1}. In other words, d_i is the distance between two neighbouring points. Note that the sum of all the d_i is constant with value 1. Now define the entropy to be sum_i [d_i * log d_i]. -Michael. === Subject: Re: Entropy of samples on real line segment. > Is there a precise measure/definition of the 'entropy' of N points on > the real line segment [0,1]? I am guessing that N points positioned at > (i+.5)/N (where i=0...N-1) gives the configuration with 'maximum' > entropy. > Sabbir. I don't know of a precise definition. When I last looked at this (about a decade back), the basic idea was as follows: Let your set of points be P. Use P to define a continious distribution, f(x) on [0,1]. The entropy is defined as the entropy of f(x) (clearly this works for any functional on continuous distributions). The trouble is that there is no precise way of obtaining f(x) from P. Basically, one defines f by way of a histogram, but the histogram width is critical. The use of kernel functions avoids some of the discontinuity problems associated with histograms, but width is still critical. (See [1] for more than you want to know about this topic) I tried looking a the sample cumulative distribution, C(x) = (# points in P <=x) / (# points in P). I then looked for g(x), the function with the highest entropy such that G(x) (the cumulative distribution of g) is close enough to C(x) and defined the entropy of the point set to be the entropy of g. Of course now we have to define close enough. The approach seemed to have some promise, but I did not investigate it to any depth. Note that your guess is correct, a uniform distribution of points leads to the highest entropy. -William Hughes [1] D.W. Scott , Multivariate Density Estimation, Wiley series in probability and mathematical statistics, John Wiley and Sons, New York, 1992 === Subject: Re: Lie Algebra of E(2) > If we characterize E(2) (representing translations and rotations) by > matrices of the form > costheta sintheta x > -sintheta costheta y > 0 0 1 > I would like to compute its Lie algebra, but am having problems. Can > anyone help? Denote the matrix above by (theta,x,y). The identity is (0,0,0) -- the identity matrix. The inverse of (t,x,y) is (-t,-x,-y). The Lie vectors are the derivatives d/ds (t(s),x(s),y(s)) |s=0 of the functions (t(s),x(s),y(s)) for which (t(0),x(0),y(0)) = (0,0,0), the identity. It is expressible as a linear combination of t'(0), x'(0), y'(0): d/ds (t(s),x(s),y(s))|s=0 = t'(0) Y_t + x'(0) Y_x + y'(0) Y_y. The elements (Y_t, Y_x, Y_y) are the basis of the Lie algebra. The most general form of the derivatives are: d/ds (t(s),x(s),y(s)) -sin(t) t' cos(t) t' x' -cos(t) t' -sin(t) t' y' 0 0 0 which, evaluated at s = 0, assuming t(0) = x(0) = y(0) = 0, gives: L(t',x',y') = 0 t' x' -t' 0 y' 0 0 0. Therefore Y_t = 0 1 0 Y_x = 0 0 1 Y_y = 0 0 0 -1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0. The commutator is defined by the condition: d^2/ds dt (g(s)h(t)g(s)^{-1}) |s=t=0 = [g'(0),h'(0)], for functions g(s) = (t1(s),x1(s),y1(s)) h(s) = (t2(s),x2(s),y2(s)). For a Lie algebra represented in matrix form, this *always* yields the result: [g'(0),h'(0)] = g'(0)h'(0) - h'(0)g'(0). In the case above, [Y_t,Y_x] = -Y_y, [Y_t,Y_y] = Y_x, [Y_x,Y_y] = 0. The defining representation of E(2) which realises these commutators is: Y_t = x d/dy - y d/dx, Y_x = d/dx, Y_y = d/dy. E(n) has the representation: L_{ij} = -L_{ji} = x^i d/dx^j - x^j d/dx^i, 0 < i < j <= n P_i = d/dx^i, 0 < i <= n, and you can work the commutators out for yourself. === Subject: Re: Journal editors and reviewers, speak up > That was rather long winded. What was your point? My point was to observe that the concatenation of Dr. and Harris seemed to provoke the gag reflex in Doug Norris, which led me to opine that it is, perhaps, the resistance to letting the notorious James Harris into the mathematical clubhouse, rather than any shortcomings in his mathematical techniques, which is obstructing the publication of his work. Your point, I think, was to pull this gag: 1) Quote the entire text of a post. 2) Then add one line complaining about its length. And to really put the cherry on top: 3) Put no_spam in your return address. It'd work better with a 1000 line post, though. Maybe I'll write one. An epic poem about the struggles and eventual triumph of James S. Harris. But James would probably think it *sarcastic*. Okay, yes, I enjoy a bit of sarcasm now and then, and I don't really care if it's aimed at the quixotic Mr. Harris or at the corrupt mathematical establishment. Whatever's funny. Right now, the funniest thing is that the conspiracy against James in the math community has become so blatant that he might as well be submitting his work to a journal called, The International Journal of Mathematics by People Other Than James Harris . . . but that James seems to still think he has a chance of publishing! Ha! If the conspiracy were a snake, James would be dead by now! If it were a poisonous snake, that is. If not, then I suppose he'd be sitting around, saying, What was that strange, legless reptile? Perhaps a small, short-snouted, polyamputee alligator? Oh, never mind. Better write another cover letter. I mean, for crying out loud James, how obvious does it have to be? How can you be so smart and such a confounded moron at the same time? It's really quite obvious: you have profound mathematical results, right? I don't suppose you'd be wasting your time submitting old Cheetos you found under the couch, so I'll assume that what you're submitting to these journals are your profound mathematical results. Results profound enough that if *anyone else in the world besides you* were submitting them, then they'd be gobbled up in an instant. If David Ullrich submitted them . . . fame and glory for Dave. If Arturo Magidin submitted them . . . fame and glory for Art. But when James Harris submits them . . . You've got to admit, to an outside observer like me who doesn't really care who wins or loses in this whole thing, it's pretty damn funny. Oh look, the mathematical establishment has decided that James Harris is not allowed to publish . . . But look! He's still trying! Pathetic. Pathetic and funny. So what's next, James? Oh please up the ante in this farce. I so enjoy a good jest. Will you call in the Army to take over the math journals? The Army is busy in Iraq, James. Or have you been too busy to notice? Just how clueless are you? -- Jim Ferry at U of Illinois, Urbana-Champaign, (no, I don't educate) 2 email me l o o k up 1 row === Subject: Re: Journal editors and reviewers, speak up > My point was to observe that the concatenation of Dr. > and Harris seemed to provoke the gag reflex in Doug Norris, Ah, so pointing out deliberate falsehoods is a gag reflex? Listen - I spent the first twenty-eight years of my life working towards a doctorate in mathematics, and if Harris wants one, he can go to grad school just like anyone else. He doesn't need you conferring an honorary degree upon him. Doug === Subject: Re: Journal editors and reviewers, speak up >> My point was to observe that the concatenation of Dr. and Harris >> seemed to provoke the gag reflex in Doug Norris, > Ah, so pointing out deliberate falsehoods is a gag reflex? > Listen - I spent the first twenty-eight years of my life working towards > a doctorate in mathematics, and if Harris wants one, he can go to grad > school just like anyone else. He doesn't need you conferring an > honorary degree upon him. > Doug OK. I get it. You're a burned out academic. You wasted the best years of your life on a worthless degree. Now you're bitter because Professor Harris's achievements have exceeded your own. Get over it. Professor Harris's seminal work on factorization has many applications in such diverse fields as the War Against Terror (TM), energy policy, and family planning. It is definitely Noble-worthy. To Professor Harris, my hats off to you for an outstanding paper. Keep up the good work. We expect many more papers from you. -- Lance Lamboy === Subject: Re: Journal editors and reviewers, speak up <728tc.119469$xw3.7152079@attbi_s04> 2kpKShAQELm5uc578GgAAAJ5SURBVHichdNBb9sgFABgvMblarSwXDMcKdfUpso1td1xDWqfc020 mV5JoHt/f0Dc1pU2jUskvrz3eA9D6n8s8h+QnVSzv0G3zUDPlOzqut3N2g9YcwDQHgm1zJj9O8gc IshWIu+NMe4NEICFGKdoySHArzd4AI0ZQLNyZR7hNIJ8gL3YcLiIRuQQcg3te8S+CfVdpa4Rx8UI Z9DUcuiEXKcax8OkOAunEuUyZjJHN0IR24D9jlCVjjuMxbcm9ZGYh31zGOHR7CPkwILFiPYKsjdD PsCCg14AfDPmZRzJXWyp8GueQwH8bMzXK8jbmNbsFQ8hALSPo4qwTfv9M4YT6F7j7RvETGbgM5vp DPge1XmEHymi8NZqVrCiaB5vrpBKDNhYZEwznTWv8yvEg5sXL5QPY8kY8w+YoEuZ5nVlfRYiWOEs SQ1uUqaLKK0jnDG28K+nBPepCe9q1RGGXFN/Zgl+h/8PmsxqidnCEkZbe0gQDhUGW9C6tCQstF7p +JWECRrdE+pljZYUAXCXPh8ZMjGwrkKPWQCF+JpgF8fB81mzDVUpUmkxTxBGOMADLLxqS9Uhlkrd pouKt2QQTtWqEbWizarD/pQ6fzRD/yXvDyKsuq065ccbVP3QPy9hmJetRSSsVebpOsTfpj8pgIEU GSuyU931Hxc1KGS6VJY03wPQ8UXF8iqcVoiyEuvj5Knl5qTEmjax/PZ5Alt29GJFLmF/tZxPH+fd S8hjGyGVWz5N4f5gK7HzXWfd5ucnoOoidliV2H6qUZ9R3ohNqI7N1k3Bo3Chuhey2vgpzGh4lSv0 ohQKp7BwAUrESohOuQngtbf4s9qNIX8At0MbK5iXDI4AAAAASUVORK5CYII= Discussion, linux) >> My point was to observe that the concatenation of Dr. >> and Harris seemed to provoke the gag reflex in Doug Norris, > Ah, so pointing out deliberate falsehoods is a gag reflex? > Listen - I spent the first twenty-eight years of my life working > towards a doctorate in mathematics, and if Harris wants one, he can > go to grad school just like anyone else. He doesn't need you > conferring an honorary degree upon him. Really? The first twenty-eight years? Gosh, it's good you set goals early. My first year or two consisted mostly of sleeping, eating, farting and burping. Actually, my first year or two as an undergrad was pretty much the same. -- Jesse F. Hughes One is not superior merely because one sees the world as odious. -- Chateaubriand (1768-1848) === Subject: Re: New (?) Technique for Drawing a Tetrahedron by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i4QLSBH16372; >I'd figured-out that isometric drawings were the same >as taking an extreme telephoto lens to your subject (or >looking at the stars) in my mechanical drawing class >in junior high. > recently, I also realized that you don't have >to go with making all three XYZ coords equal, >at the same apparent angle of view. > yesterday, while making my poster for a poster-session >at UCLA, a simple diagram of a tetrahedron, used >to represent an airplane for a weight & balance determination, >I realized that it's simple, isometrically, >to exactly represent any tetrahedron, >just by using a circle to represent the circumsphere >of the tetrahedron; see, How? >actually, this is the perfect place to use the dual terminology, >tetra-asteron, since we affix the 4 points to the circumsphere, or >to the celestial sphere. In the drawing, the vertices of the tetrahedron are not all going to be on the circumference of the circle, though. === Subject: Two traits common to all masses These two traits are common to all masses of matter; by which we measure their quantity: #1) All masses have three dimensional extent so that we can use their linear length; their area, or their bulk, as measures of their quantity. #2) All matter has inertia: For any particular object; body, or mass of matter, its inertia is a constant: This constant [m] is the ratio of the net force [f] exerted on, and/or by it to the acceleration [a] that is caused: Which constant is also equal to its scaled weight [w] divided by the acceleration [g] at which it will free fall at the location of the scale on which it is weighed; anytime, anyplace. m = f/a = w/g Inertia is an accurate measure of the quantity of matter in any object; body, and/or mass of it. === Subject: Re: Two traits common to all masses > These two traits are common to all masses of matter; by which we > measure their quantity: > #1) All masses have three dimensional extent so that we can use their > linear length; their area, or their bulk, as measures of their > quantity. Hey Dumb Donny Head, what is the mass of a Menger-Sierpinski sponge? It has an arbitrarily large physical extent in all dimensions but asymptotically zero mass overall. > #2) All matter has inertia: Hey Dumb Donny Head, tell us the differences among inertial mass, gravitational mass, active mass, and passive mass, you ing imbecile. [snip idiocy] -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Two traits common to all masses > These two traits are common to all masses of matter; by which we > measure their quantity: > #1) All masses have three dimensional extent so that we can use their > linear length; their area, or their bulk, as measures of their > quantity. Not black hole sigularities === Subject: Re: Can the deduction theorem be used recursively? >My point (attempted) was this: He said that he was confused by the >question, > > No, I didn't say that. > (1) Initial response: it just requires using the Deduction Theorem > _twice_. > (2) Second response: I wasn't readind very carefully, just assumed he > was asking about the Deduction Theorem since that's what he said he > was asking about. > (3) Confuse: mistake one thing for another (dictionary.com) According to the definition of confuse you presented, he wasn't confused _by_ the question, but he _confused_ the question. That is, he mistook it for another. He already admitted to this. The sense of confuse in which the sentence The question confused him is meaningful is only the first in that dictionary.com definition: To cause to be unable to think with clarity or act with intelligence or understanding; throw off. By your own admission, this is not what you meant. So, you confused one sense of the word confuse with another. This confusion lead to further confusion, as wished to impart): He said that he was confused by the question. Get it? 'cid 'ooh === Subject: Re: Can the deduction theorem be used recursively? > You appear to be confused about what Godel has shown. He proved > the incompleteness of *arithmetic* (that is, of any consistent, > recursive axiomatization of it) in 1931. Give an exposition of the proof given by Godel in 1931 of his 1st. Incompleteness Theorem (based on w-consistency) and I will provide a link where I do likewise. > Do you see why that might lead one to think you don't > understand the subject matter? Actually, it's not a matter of understanding. I simply haven't pursued Godel's Completeness Theorem nor do I know what might constitute official terminology for it. Proof Theory is not my specialty - programming and creating axiomatic systems for various branches of Computer Science is. I am just dabbling here. That's why I've used such words as wondering . . . offhand . . . speculating regarding Proof Theory and Set Theory. I do know the Godel/Rosser/Smullyan Incompleteness Theorems rather well (having axiomatized them and other incompleteness results.) > But that terminology is completely standard. Smullyan: There is a true sentence of language L not provable in L. Kleene: In a formalization of number theory a closed formula A can be found that is true and unprovable. Mendelson: In first order theory S, wff w is not provable and wff ~w is not provable. Floyd/Beigel: In a theory containing axioms of arithmetic there is a statement v that is true and ~v is not a theorem. (Note that even the definition of incomplete varies!) > Both for us [mathematicians] and in my opinion pretty clearly > indicate this was a methodological principle, perhaps even a reasonable > belief, but not something Hilbert thought had been (or even perhaps > could have been) conclusively demonstrated. Ullrich only asked for something that Hilbert said that turned out to be false. Do you think There is no ignorabimus. We shall know. is true or false? (I personally find that request amazing.) > I am wondering if anyone has been wise enough to point out that the > set of axioms should be finite - which would preserve the analogy with > programming languages > Why on earth should *that* be a constraint on what counts as an > axiomatic system? Programming languages are a relatively recent > invention. So we can continue to abstract from axiomatic systems, the bases of computing proposed in the 1930's, and modern programming languages (all of which are equivalent) to a single theory at a higher level of abstraction. > The *proper* response > for you, with your strictly finitistic leanings, is *not* to try to say > that axiomatic theory doesn't in fact mean what it universally does, > but rather that the only theories that you think are philosophically > legitimate have only finitely many axioms. It goes way beyond philosophy (but I do appreciate the partial endorsement.) :) > I don't see a proof off-hand, and would still like to. What about ZF > gets in the way, then? > Well, in a nutshell, the fact that (due to the replacement schema) it > has infinitely many independent axioms. Great! Reference (or details) please (preferably on-line)? > Er, how would you suggest revising the replacement schema? > > As with any axiom schema that substitutes any particular formula for a > variable, we simply need axioms that define the set of formulas. > I haven't a clue as to what you think you are suggesting. Axioms define what is a theorem and can likewise define what is a formula. > (What does Er mean?) > Same as welllll or ummmmm... Oh, I though it was the opposite of r.e. > Err, wellll, ummm, assuming by generate you mean prove the existence > of As in the theorems being enumerable. > Suppose I give you some axioms about what ducks are: > Ducks are birds. . . . > Then you come along, and having never seen a duck, but lots of dogs, ask > whether ducks have fur, bark, and have teeth. Accordingly, I add some > axioms about what ducks aren't: > NO ducks have fur. . . . > Now, didn't that qualification about what ducks aren't help us to better > understand what ducks are? It wasn't contradictory at all to say what > is not in duck theory, right? > Same with set theory. If the non-duck predicates are not recursive (as the set of theorems is not recursive), the r.e. set of ducks is no longer r.e. > And in ZF the theorems are really just statements that generate > allowable sets. > You can use them to *prove* that certain sets exist, but they don't > generate sets. They describe them. I would speculate (without looking closer at this point) that the set of sets defined outside of Foundation is r.e. > This thing about axioms generating sets might be the source of your > confusion, though I confess I don't know what that confusion is. Do you agree that axioms generate theorems? > We can't say that the result of applying a rule to theorems generates > allowable sets. > I don't know what this means. We have to make sure it doesn't violate Foundation. > (I'd rather be right than popular.) > Alas, currently you are neither. I get plenty of fanmail, believe it or not. You can even see a few courageous souls among these posts. (And aren't you my friend??) > What do you mean you *tend* to agree? That's like saying you tend to > agree that, say, the complement of a recursive set is recursive. That's not true in a very general sense, actually (as I have noted before.) > You can't > very well *use* an axiomatic system to generate proofs if you don't have > a surefire way of deciding whether or not a certain sentence that you > need at some step in the proof is an axiom. Either way, you need the > axioms to be recursive. R.e. won't cut it. Not so. 1. Go through the r.e. set of axioms (dovetail) 2. Apply the rules 3. Out comes the proofs/theorems. > So having an infinite number of axioms is even worse than I said. > You've yet to identify a problem, let alone something even worse. If > the axioms are recursive, there isn't one. If they are infinite or theorems from another system, they may well not be. > I admittedly have yet to give it a close look, but your confusion > over a number of fairly elementary points here raises suspicions > about the rather grandiose claims you make for your theory of > computation. > It is quite inconceivable to me that your work could > have the virtues and significance you claim for it when you are so > unclear about such elementary matters. The paper that I posted and referred to http://www.arxiv.org/html/cs.lo/0003071 has nothing to do with the above (outside of any explicit references to it.) That is the fundamental flaw in your judgment. > These doubts are strengthened > when you expend a lot of energy defending your mistakes instead of > acknowledging them and moving on. > Chris Menzel That's why I used the words wondering . . . speculating . . . offhand regarding Proof Theory and Set Theory. If half of my speculation in those areas is true I would be quite pleased (considering that I have read and studied them for probably a total of 2 days in my life ö not counting incompleteness results.) Now do you see? :) Charlie Volkstorf Cambridge, MA === Subject: Re: Can the deduction theorem be used recursively? > Ullrich only asked for something that Hilbert said that turned out to > be false. Do you think There is no ignorabimus. We shall know. is > true or false? For us there is no ignorabimus, and in my opinion none too for natural science in general. Instead of the foolish Ignorabimus, our slogan shall be, on the contrary: We must know. We will know. (D. Hilbert, 1930) F. === Subject: Re: Can the deduction theorem be used recursively? >>If you can do that, then you don't need MP either. Just say that the >>axioms of T' are the theorems of T. > Nope. At least not without giving a system that's totally > useless: One of the basic requirements for formal systems > (not part of the definition of formal system but a hypothesis > in things like Godel's theorems and various other things, > usually left unstated in informal treatments) is that the > set of axioms be _recursive_. If you make the theorems > of FOL all axioms then the set of axioms is not recursive. Actually, we only require the set of axioms to be recursively enumerable. By Craig's theorem any such set can be effectively replaced with a recursive set of axioms with the same deductive closure. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Can the deduction theorem be used recursively? >If you can do that, then you don't need MP either. Just say that the >axioms of T' are the theorems of T. >> Nope. At least not without giving a system that's totally >> useless: One of the basic requirements for formal systems >> (not part of the definition of formal system but a hypothesis >> in things like Godel's theorems and various other things, >> usually left unstated in informal treatments) is that the >> set of axioms be _recursive_. If you make the theorems >> of FOL all axioms then the set of axioms is not recursive. >Actually, we only require the set of axioms to be recursively >enumerable. By Craig's theorem any such set can be effectively replaced >with a recursive set of axioms with the same deductive closure. Ah, right. ************************ David C. Ullrich === Subject: Re: Can the deduction theorem be used recursively? > I'll just call you stupid for saying that you > suspect that the axiom of foundation has something > to do with the fact that ZF is not finitely axiomatizable > ************************ > David C. Ullrich Obviously, the gauge of one's intelligence is a function of not only their response to a problem, but also the amount of time they spent pondering it and, more importantly, how much experience they have in that area. I have spent years programming and developing an axiomatization of Computer Science (Program Synthesis, the Theory of Computation, Recursion Theory, et. al.) I have spent maybe a day or two reading about Set Theory. (That was mostly spent applying my axiomatization to ZF, which works pretty well, actually.) In no way would I say that I am an expert on Set Theory. I take it that you have studied Set Theory on the order of a thousand times longer than my two days of dabbling. It is really shameful to try to take advantage of your infinitely greater experience in a field to attempt to show your intellectual superiority. It is also baseless to extrapolate my inexperience in Set Theory to my paper on my axiomitization of Computer Science, which have nothing to do with Set Theory. By your standard, you should be judged solely on your comments regarding Turing Machines, e.g. that there is no difference w.r.t. the issue of consistency between FOL and programs. Or your failure to recognize Modus Ponens. In that regard, you are batting 0. It also shows an incredible degree of abusive arrogance. Charlie Volkstorf Cambridge, MA === Subject: Re: Can the deduction theorem be used recursively? >> I'll just call you stupid for saying that you >> suspect that the axiom of foundation has something >> to do with the fact that ZF is not finitely axiomatizable >> ************************ >> David C. Ullrich >Obviously, the gauge of one's intelligence is a function of not only >their response to a problem, but also the amount of time they spent >pondering it and, more importantly, how much experience they have in >that area. I have spent years programming and developing an >axiomatization of Computer Science (Program Synthesis, the Theory of >Computation, Recursion Theory, et. al.) I have spent maybe a day or >two reading about Set Theory. Then the sensible thing to do would be to refrain from making comments on the subject, lest you sound stupid. It doesn't take a lot of study to determine that foundation is just one axiom - all you have to do is look at the axioms. It _does_ take a very special sort of understanding to conclude that a single axiom is the reason for a theory not being finitely axiomatizable... > (That was mostly spent applying my >axiomatization to ZF, which works pretty well, actually.) In no way >would I say that I am an expert on Set Theory. >I take it that you have studied Set Theory on the order of a thousand >times longer than my two days of dabbling. It is really shameful to >try to take advantage of your infinitely greater experience in a field >to attempt to show your intellectual superiority. It is also baseless >to extrapolate my inexperience in Set Theory to my paper on my >axiomitization of Computer Science, which have nothing to do with Set >Theory. Seems like you addressed that last bit to the wrong person - I haven't said anything about you axiomatization of CS in quite a while. >By your standard, you should be judged solely on your comments >regarding Turing Machines, e.g. that there is no difference w.r.t. the >issue of consistency between FOL and programs. When did I say that? I don't have much of an idea what that statement even _means_ (in particular what it means for a _program_ to be consistent...) >Or your failure to >recognize Modus Ponens. In that regard, you are batting 0. >It also shows an incredible degree of abusive arrogance. Giggle. Speaking of axioms, there's one about usenet, and how you can tell who the [whatever] guys are, because they're the ones who whine when people are [whatever] back at them... >Charlie Volkstorf >Cambridge, MA ************************ David C. Ullrich === Subject: Re: Can the deduction theorem be used recursively? >> No, Godel's Completeness Theorem is the converse: >> [all] logically valid wffs are theorems. >> > I meant iff. > It's still wrong. Look, Charly, I've read G.9adel's paper, I KNOW what he > has proved in it. I know. Look (as you say), when someone proves B from A, sometimes they just say A=>B, but when B=>A is a much simpler result, sometimes they go ahead and say A<=>B. I wanted a nice easy way to express Godel's Completeness Theorem (about which I know approximately zilch), so I looked it up on the internet and also checked a couple of my own books. Mendelson had a simple wording and is of course an authority on Mathematical Logic, so I just quoted him. Wouldn't you? (BTW Also, for the record: I am not an expert on Proof Theory nor Set Theory. Surprise! I do know a lot about the incompleteness aspect of Proof Theory (Godel, Rosser, Smullyan) and have written and posted a theorem-prover for that. But my paper on my axiomatization of Program Synthesis and the Theory of Computation has nothing to do with Proof Theory or Set Theory. My knowledge in the two areas are reciprocals of each other. In fact, at least twice I have posted requests for references on Set Theory so I can learn more, and more completely apply my axiomitization to that field of study, asking for (1) formal proofs using ZF, and (2) a proof that it can't be finitely axiomitized.) Charlie Volkstorf Cambridge, MA > > Introduction to Mathematical Logic, Mendelson 1964, pg. > 68, Godel's Completeness Theorem: In any first order predicate > calculus, the theorems are precisely the logically valid wffs. > > Mendelson is clueless?!? > > No. See D. Ullrich's comment: > That iff is traditionally broken into two parts, one the soundness > and one the completeness. The soundness theorem is trivial, > the completeness isn't. The iff is indeed true; if someone called > that Godel's Completeness Theorem it would be because > the completeness, which Godel proved, is the non-trivial half. > (David Ullrich) > And in my own words: > FIRST you have to prove that the system is consistent - that's the easy > part. And then you have to prove that it is complete - that's the hard > part. G.9adel was THE FIRST to present a proof for that, concerning FOPL. > predicate calculus, the theorems are precisely the logically valid > wffs. > F. === Subject: Re: Can the deduction theorem be used recursively? > I know. Look (as you say), when someone proves B from A, sometimes > they just say A=>B, but when B=>A is a much simpler result, sometimes > they go ahead and say A<=>B. Right. I mean, that's exactly what WE told you. :-) Still, G.9adel ONLY proved A=>B in his paper. Now since B=>A already was established (a rather easy task, in comparison) this actually meant that Godel's Completeness Theorem: In any first order predicate calculus, the theorems are precisely the logically valid wff. F. ===