mm-4439 === Subject: sequence of condition I have a simple question. How to proof this: P(A|B|C)=P(A|C|B) What I mean here is the sequence of multiple conditions has no effect to the final condition probability. In this case, both should be P(A| B,C). Jason === Subject: Re: sequence of condition I have a simple question. How to proof this: > P(A|B|C)=P(A|C|B) > What I mean here is the sequence of multiple conditions has no effect > to the final condition probability. In this case, both should be P(A| > B,C). What's the definition of P(A|B|C)? === Subject: Re: Possible to decide a line's slope without division? > A line is given by y=ax+b. Given the two points P1(x1,y1), P2(x2,y2) the > slope can be calculated: > > a = y2-y1/x2-x1 You need parentheses: (y2-y1)/(x2-x1). > But is there someway to decide 'a' without using division? Choose your points so that x2 = x1 + 1. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Possible to decide a line's slope without division? > A line is given by y=ax+b. Given the two points P1(x1,y1), P2(x2,y2) the > slope can be calculated: a = y2-y1/x2-x1 But is there someway to decide 'a' without using division? Sure, use logarithms. Division turns into a subtraction. === Subject: Re: Parallel Runga-Kutta Method Is there any way to adapt the Runge-Kutta Method for parallel computing? > Namely, im using the 4th order Runge-Kutta Method to solve nonlinear diff. > equations and the usage of parallel computing would speed up the solving. Usage of parallel computing is a bit vague, isn't it? Could you describe what kind of parallel computing you are thinking of? I'd first have a look at higher order Runge-Kutta formulas, especially if you need a lot of precision. On the other hand, multi-step solutions can be a lot faster but need to be handled very carefully. === Subject: Re: Parallel Runga-Kutta Method On Oct 3, 12:55 am, christian.bau Is there any way to adapt the Runge-Kutta Method for parallel computing? > Namely, im using the 4th order Runge-Kutta Method to solve nonlinear diff. > equations and the usage of parallel computing would speed up the solving. Usage of parallel computing is a bit vague, isn't it? Could you > describe what kind of parallel computing you are thinking of? I'd first have a look at higher order Runge-Kutta formulas, especially > if you need a lot of precision. On the other hand, multi-step > solutions can be a lot faster but need to be handled very carefully. He didn't say he wanted more precision, he said he wanted faster computation by using parallel methods. Anyway, what's wrong with doing a Google Scholar search for 'parallel Runge-Kutta' and looking through a few promising papers and their references to find a starting point? Another place to start is Hairer,Noersett,Wanner: Solving Ordinary Differential Equations I, Chapter II.11 - Parallel Methods, although I suspect new developments have taken place in the last 15 years. === Subject: Re: Parallel Runga-Kutta Method > He didn't say he wanted more precision, he said he wanted faster > computation by using parallel methods. I didn't say higher order RK gives higher precision (although it does, at same step size). I said higher order runs faster, and the difference gets bigger if you need higher precision (because the number of steps grows slower with precision when you use higher order). Got it? === Subject: Re: Parallel Runga-Kutta Method Usage of parallel computing is a bit vague, isn't it? Could you > describe what kind of parallel computing you are thinking of? I'd first have a look at higher order Runge-Kutta formulas, especially > if you need a lot of precision. On the other hand, multi-step > solutions can be a lot faster but need to be handled very carefully. Well, i am solving with Runge-Kutta method nonlinear differential equations, and with parallel computing i mean the calculation by using more computer processors. Therefore, i should use some kind of adapted Runge-Kutta that enables this kind of calculation (e.g. 5 computers are calculating at the same time). === Subject: Re: Parallel Runga-Kutta Method Usage of parallel computing is a bit vague, isn't it? Could you > describe what kind of parallel computing you are thinking of? I'd first have a look at higher order Runge-Kutta formulas, especially > if you need a lot of precision. On the other hand, multi-step > solutions can be a lot faster but need to be handled very carefully. Well, i am solving with Runge-Kutta method nonlinear differential equations, > and with parallel computing i mean the calculation by using more computer > processors. Therefore, i should use some kind of adapted Runge-Kutta that > enables this kind of calculation (e.g. 5 computers are calculating at the > same time). There are methods that work for systems with massive numbers of variables with little connection between the variables. Say you have y' = f (y, t) where y is a vector of n elements, n large (say 1,000,000), and for example f (y(k)) only depends on y(k-1), y(k), y(k+1) and t. Look at what values RK asks you to compute. Fourth order Runge-Kutta would require you to calculate four values f (y, t). Start the first processor to calculate the first vector f. Once it has calculated a few thousand of the one million values, you have some data to start calculating the second vector f using the second processor. Once that has done a few thousand calculations, you start processor 3 to work on the third vector f and so on. This has to be programmed quite carefully so that one processor doesn't overtake the previous one. On the other hand, if each calculation is very simple, the communication overhead (data has to be sent from one processor to the next) will kill performance, and a single processor might do the job faster. Especially if you are talking about several separate computers with slow communication between them. And all this only works if calculating the function f (y, t) can start and produce results when only a few of the input values are available yet. For many systems, that is not the case. The simplest and most efficient approach is to solve several separate systems at the same time, one on each processor :-) So: How many variables in your system? What does the diff. equation look like? Do you have more than one system to solve? === Subject: Prefixes? I have a relatively simple question, If a monomial is X+4 (just some random number) A quadratic is X^2+X+4 (again, just some random number) A cubic is X^3+X^2+X+4 (Must I repeat myself?) And a quartic is X^4...+4, Where do they get quad (in quadratic) from? In what way does it reference anything even remotely to the number 4? I get quartic being like... 4 quarts in a gallon, but quadratic? And what are the higher powers than 4? Bonus question - How do you know when to use pent- (as in pentagon) and when to use quint (as in...quintuplet)? === Subject: Re: Prefixes? > I have a relatively simple question, > > If a monomial is X+4 (just some random number) > A quadratic is X^2+X+4 (again, just some random number) > A cubic is X^3+X^2+X+4 (Must I repeat myself?) > And a quartic is X^4...+4, > > Where do they get quad (in quadratic) from? In what way does it > reference anything even remotely to the number 4? I get quartic > being like... 4 quarts in a gallon, but quadratic? And what are the > higher powers than 4? quadratic comes from Latin quadrum meaning square, and this is from quattuor meaning 4: a square (in the geometric sense) has four corners. quartic comes from Latin quartus meaning fourth (also, of course, derived from quattuor). Yes, quart also comes from quartus because a quart is 1/4 of a gallon. And guess what fraction ounce comes from... > Bonus question - How do you know when to use pent- (as in pentagon) > and when to use quint (as in...quintuplet)? The OED says that quint- is erroneous: quinti- is the proper Latin prefix for fifth, quinque- for having, consisting of, etc. five (things specified). So a pentagon can also be a quinquangle, but not a quintangle. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Prefixes? Case closed. All of you are amazing. Keep it up! === Subject: Re: Prefixes? >I have a relatively simple question, If a monomial is X+4 (just some random number) >A quadratic is X^2+X+4 (again, just some random number) >A cubic is X^3+X^2+X+4 (Must I repeat myself?) >And a quartic is X^4...+4, Where do they get quad (in quadratic) from? In what way does it >reference anything even remotely to the number 4? I get quartic >being like... 4 quarts in a gallon, but quadratic? And what are the >higher powers than 4? I think degree 6 equations are sextics. They have the most fun. quasi === Subject: Re: Prefixes? I have a relatively simple question, If a monomial is X+4 (just some random number) >A quadratic is X^2+X+4 (again, just some random number) >A cubic is X^3+X^2+X+4 (Must I repeat myself?) >And a quartic is X^4...+4, Where do they get quad (in quadratic) from? In what way does it >reference anything even remotely to the number 4? I get quartic >being like... 4 quarts in a gallon, but quadratic? And what are the >higher powers than 4? I think degree 6 equations are sextics. They have the most fun. Right. And that's related to something humorous; see below. >Bonus question - How do you know when to use pent- (as in pentagon) >and when to use quint (as in...quintuplet)? In forming words, it is considered desirable that all parts come from the same source language. Both parts of pentagon are Greek, while both parts of quintuplet are Latin. Of course, there are exceptions, in which one part of a word comes from one language and another part from another language. But as I said, forming words in that way is considered to be undesirable. So what word did the engineers at IBM decide on to describe numbers in base 16? Well, it _should_ have been sexadecimal, and I suspect they knew that. But, the story goes, that just sounded too risque, and so they decided to mix Greek and Latin roots, giving hexadecimal. David === Subject: Re: Prefixes? <88j5g3thvh5k95o2hidl35ph4dk3tb9b5m@4ax.com> <20071002193316.379$Wt@newsreader.com >I have a relatively simple question, >If a monomial is X+4 (just some random number) >A quadratic is X^2+X+4 (again, just some random number) >A cubic is X^3+X^2+X+4 (Must I repeat myself?) >And a quartic is X^4...+4, >Where do they get quad (in quadratic) from? In what way does it >reference anything even remotely to the number 4? I get quartic >being like... 4 quarts in a gallon, but quadratic? And what are the >higher powers than 4? I think degree 6 equations are sextics. They have the most fun. Right. And that's related to something humorous; see below. >Bonus question - How do you know when to use pent- (as in pentagon) >and when to use quint (as in...quintuplet)? In forming words, it is considered desirable that all parts come from the > same source language. Both parts of pentagon are Greek, while both parts > of quintuplet are Latin. Of course, there are exceptions, in which one > part of a word comes from one language and another part from another > language. But as I said, forming words in that way is considered to be > undesirable. So what word did the engineers at IBM decide on to describe numbers in base > 16? Well, it _should_ have been sexadecimal, and I suspect they knew > that. But, the story goes, that just sounded too risque, and so they > decided to mix Greek and Latin roots, giving hexadecimal. They probably couldn't get away with that today, sounds too much like casting spells. David- Hide quoted text - - Show quoted text - === Subject: Re: Prefixes? >I have a relatively simple question, If a monomial is X+4 (just some random number) >A quadratic is X^2+X+4 (again, just some random number) >A cubic is X^3+X^2+X+4 (Must I repeat myself?) >And a quartic is X^4...+4, Where do they get quad (in quadratic) from? In what way does it >reference anything even remotely to the number 4? I get quartic >being like... 4 quarts in a gallon, but quadratic? Think physical (as the ancients did, it seems) A univariate polynomial of degree 2 has a square term, right? And a square has how many sides? Thus, quadratic! Of course, the above is just a made up theory, but who knows, it might be true. quasi === Subject: Re: Prefixes? > I have a relatively simple question, If a monomial is X+4 (just some random number) > A quadratic is X^2+X+4 (again, just some random number) > A cubic is X^3+X^2+X+4 (Must I repeat myself?) > And a quartic is X^4...+4, Where do they get quad (in quadratic) from? In what way does it > reference anything even remotely to the number 4? Why is X^2 called X squared? But squares have 4 sides. So, to solve the square root of X, you use the quadratic equation. > I get quartic > being like... 4 quarts in a gallon, but quadratic? And what are the > higher powers than 4? Bonus question - How do you know when to use pent- (as in pentagon) > and when to use quint (as in...quintuplet)? > === Subject: Re: Prefixes? > I have a relatively simple question, If a monomial is X+4 (just some random number) > A quadratic is X^2+X+4 (again, just some random number) > A cubic is X^3+X^2+X+4 (Must I repeat myself?) > And a quartic is X^4...+4, Where do they get quad (in quadratic) from? In what way does it > reference anything even remotely to the number 4? I get quartic > being like... 4 quarts in a gallon, but quadratic? And what are the > higher powers than 4? Bonus question - How do you know when to use pent- (as in pentagon) > and when to use quint (as in...quintuplet)? > I had a teacher explain the quadratics are called such because the highest power is a square, a square has four sides. === Subject: Re: Prefixes? > I have a relatively simple question, > > If a monomial is X+4 but it isn't - what you've written is a binomial. > (just some random number) > A quadratic is X^2+X+4 (again, just some random number) > A cubic is X^3+X^2+X+4 (Must I repeat myself?) > And a quartic is X^4...+4, > > Where do they get quad (in quadratic) from? > Bonus question - How do you know when to use pent- (as in pentagon) > and when to use quint (as in...quintuplet)? By copying what others have done before you. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Prefixes? My corrections, Binomial. Sorry for the mixup. (Even I know that a monomial is only on term) === Subject: Quadratic Best fit to data Ok I have a question. I'm given a set of data points [(1,1) (2,3) (3,5) (4,5)] and have to fit an equation of the form y(x) = c0 + c1*x + c2*x^2 where c0,c1, and c2 are constants. Using matrix manipulations, how am I suppose to fit a curve that runs EXACTLY through points (1,1) and (4,5), while incorporating the rest of the data into the curve? Please help, I know how to fit a normal quadratic easy, but this just has me totally confused. === Subject: Re: Quadratic Best fit to data <9802325.1191364801731.JavaMail.jakarta@nitrogen.mathforum.org>, Billy > Ok I have a question. > > I'm given a set of data points [(1,1) (2,3) (3,5) (4,5)] and have to fit an > equation of the form y(x) = c0 + c1*x + c2*x^2 where c0,c1, and c2 are > constants. Using matrix manipulations, how am I suppose to fit a curve that > runs EXACTLY through points (1,1) and (4,5), while incorporating the rest of > the data into the curve? > > Please help, I know how to fit a normal quadratic easy, but this just has me > totally confused. you have two constraints which are two linear eqs. in three the unknowns you can find two in terns of the third or all three in terms of a parameter u, then the problem reduces to finding u for the 'best fit'. I assume you can now note this process in linear algebra. === Subject: Re: Quadratic Best fit to data > Ok I have a question. > > I'm given a set of data points [(1,1) (2,3) (3,5) (4,5)] and > have to fit an equation of the form y(x) = c0 + c1*x + c2*x^2 > where c0,c1, and c2 are constants. Using matrix manipulations, > how am I suppose to fit a curve that runs EXACTLY through points > (1,1) and (4,5), while incorporating the rest of the data into > the curve? > > Please help, I know how to fit a normal quadratic easy, but > this just has me totally confused. If you run a line through your end points and subtract it off of your polynomial-to-be and your data, your requirement becomes: find a closest-fit quadratic polynomial that is zero at x=1 and 4. That leaves one parameter c left for your fit: y(x) = c*(x-1)*(x-4) + 4/3*x - 1/3 ( that is, add back the line between (1,1) and (4,5)) Sum the square of the differences between your curve and your data, differentiate with respect to c and solve for c. I don't see how to do this using matrix operations, off the top of my head, but maybe this will get you started. Jim Burns === Subject: Re: Quadratic Best fit to data <4702D0B1.2060907@osu.edu Ok I have a question. I'm given a set of data points [(1,1) (2,3) (3,5) (4,5)] and > have to fit an equation of the form y(x) = c0 + c1*x + c2*x^2 > where c0,c1, and c2 are constants. Using matrix manipulations, > how am I suppose to fit a curve that runs EXACTLY through points > (1,1) and (4,5), while incorporating the rest of the data into > the curve? Please help, I know how to fit a normal quadratic easy, but > this just has me totally confused. If you run a line through your end points and subtract it off > of your polynomial-to-be and your data, your requirement becomes: > find a closest-fit quadratic polynomial that is zero at x=1 and 4. > That leaves one parameter c left for your fit: > y(x) = c*(x-1)*(x-4) + 4/3*x - 1/3 > ( that is, add back the line between (1,1) and (4,5)) Sum the square of the differences between your curve and your > data, differentiate with respect to c and solve for c. I don't see how to do this using matrix operations, off the > top of my head, but maybe this will get you started. Jim Burns You have y = c*f1(x) + f0(x) You want to minimize sum(y_i - c*f1(x_i) - f0(x_i))^2 which is equivalent to a least-squares fit of c*f1(x) to the variable z = y - f0(x). The least-squares problem is thus to minimize (Z - c*F1)'(Z - c*F1) where Z is the vector of z values and F1 is the vector of f1(x) values. 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Do I use the chain rule for this? The matrix A is symmetric A=A'. All matrices (F,T,A) are m square. Hardy === Subject: Re: Matrix Differentiation wrt Matrix > I have an expression F=TAT' (' is transpose) and I need dF/dT which is a Matrix of matrices. Do I use the chain rule for this? > The matrix A is symmetric A=A'. All matrices (F,T,A) are m square. > Hardy Would also like to know the following I can write Vec(F)=(G*G)Vec(A) where * is Kronecker product Why then is the differential dF/dA not G*G? Hardy === Subject: Re: Matrix Differentiation wrt Matrix >> I have an expression F=TAT' (' is transpose) and I need >> dF/dT which is a Matrix of matrices. Do I use the chain rule for this? >> The matrix A is symmetric A=A'. All matrices (F,T,A) are m square. >> huh, a matrix of matrices? Are A and T also matrices of matrices? http://en.wikipedia.org/wiki/Matrix_calculus I assume you mean that F is a function of T(else if not the dF/dT = 0) Note that dT'/dT = (dT/dT)' = I then just use chain rule dF/dQ = dT/dQ*AT' + T*[A*dT'/dQ + dA/dQ*T'] if Q = T then dF/dQ = TA + AT' + T*dA/dT*T' === Subject: Re: Matrix Differentiation wrt Matrix I have an expression F=TAT' (' is transpose) and I need > dF/dT which is a Matrix of matrices. Do I use the chain rule for this? >> The matrix A is symmetric A=A'. All matrices (F,T,A) are m square. huh, a matrix of matrices? Are A and T also matrices of matrices? http://en.wikipedia.org/wiki/Matrix_calculus I assume you mean that F is a function of T(else if not the dF/dT = 0) Note that dT'/dT = (dT/dT)' = I then just use chain rule dF/dQ = dT/dQ*AT' + T*[A*dT'/dQ + dA/dQ*T'] if Q = T then dF/dQ = TA + AT' + T*dA/dT*T' You cannot do that. For an m square matrix Q your answer will still be m square and it should be m^2 square. You are differentiating one matrix wrt each element in the other. Hardy === Subject: Re: Matrix Differentiation wrt Matrix > I have an expression F=TAT' (' is transpose) and I need > dF/dT which is a Matrix of matrices. Do I use the chain rule for this? > The matrix A is symmetric A=A'. All matrices (F,T,A) are m square. >> huh, a matrix of matrices? Are A and T also matrices of matrices? >> http://en.wikipedia.org/wiki/Matrix_calculus >> I assume you mean that F is a function of T(else if not the dF/dT = 0) >> Note that dT'/dT = (dT/dT)' = I >> then just use chain rule >> dF/dQ = dT/dQ*AT' + T*[A*dT'/dQ + dA/dQ*T'] >> if Q = T then >> dF/dQ = TA + AT' + T*dA/dT*T' You cannot do that. For an m square matrix Q your answer will still be > m square and it should be m^2 square. You are differentiating one > matrix wrt each element in the other. > No, the matrix is m^2 but each element is a m^2 matrix itself(since you said F was m^2). The products now are Kronecker products. === Subject: Re: Matrix Differentiation wrt Matrix > I have an expression F=TAT' (' is transpose) and I need >> dF/dT which is a Matrix of matrices. Do I use the chain rule for this? > The matrix A is symmetric A=A'. All matrices (F,T,A) are m square. > huh, a matrix of matrices? Are A and T also matrices of matrices? >http://en.wikipedia.org/wiki/Matrix_calculus > I assume you mean that F is a function of T(else if not the dF/dT = 0) > Note that dT'/dT = (dT/dT)' = I > then just use chain rule > dF/dQ = dT/dQ*AT' + T*[A*dT'/dQ + dA/dQ*T'] > if Q = T then > dF/dQ = TA + AT' + T*dA/dT*T' You cannot do that. For an m square matrix Q your answer will still be > m square and it should be m^2 square. You are differentiating one > matrix wrt each element in the other. No, the matrix is m^2 but each element is a m^2 matrix itself(since you said > F was m^2). The products now are Kronecker products. I meant an mXm matrix - not m^2! Sorry. Does this alter things? Hardy === Subject: Re: Matrix Differentiation wrt Matrix I have an expression F=TAT' (' is transpose) and I need dF/dT which is a Matrix of matrices. Do I use the chain rule for this? > The matrix A is symmetric A=A'. All matrices (F,T,A) are m square. > Hardy Would also like to know the following I can write Vec(F)=(G*G)Vec(A) where * is Kronecker product Why then is the differential dF/dA not G*G? > Hardy Oops sorry - ignore this last comment. I need to diff wrt to T and NOT A. Hardy === Subject: Re: when does math use its own relaxed rules of English? > > [...] > >I would like to quote _Modern_English_Usage_, second >edition but there is no way without destroying the >it up. > > Which one? Second edition, 1965, revised by Sir Ernest Gowers. _A_Dictionary_of_Modern_English_Usage_ by Henry W. Fowler. > The New Fowler's Modern English Usage (3rd Edition, edited by > R.W. Burchfield), does not seem to have such a heading. I never looked at the third edition, guessing that it will have destroyed all the charm. Grammar is serious business and needs a deft touch. Do yourself a favor and get a copy of the second edition. A real time sink. It is like a drug. You start out innocently looking up a point, then start an hour later. Does the third edition have `polysyllabic humour' terminological inexactitude `misquotation' In the sweat of thy face shalt thou eat bread. To gild refined gold, to paint the lily. Screw yourr courage to the sticking-place. An ill-favoured thing, sir, but mine own. Pride goeth before destruction and an haughty spirit before a fall. In `split infinitive' does he divide the English speaking world into five classes? > It does have, under Apostrophe: > > 4. French names. Those ending in s or x should always be followed > by 's when used possessively in English. It being assumed that > readers know the pronunciation of the French names (in Rabelais, > le Roux, and Dumas, for example, the final consonant is left > unpronounced), the only correct way of writing these names in > the possessive in English is Rabelais's, le Roux's, Dumas's. > > The Oxford Guide to English Usage, 2nd edition, under possessive case > (pp 44 ff), says French names ending in silent s or x add -'s, which > is pronounced as z, e.g. Dumas's ( = Dumah's), Cremieux's. Names > ending in -es pronounced iz are treated like plurals and take only an > apostrophe (following the pronunciation, which is iz, not iziz), e.g., > Bridges', Moses', Hodges', Riches'. Polysyllables not accented on the > last or second last syllable can take the apostrophe alone, but the > form with -'s is equally acceptable, e.g., Barnabas' or Barnabas's, > Nichals' or Nicholas's. It is the custom in classical works to use > thea postrophe only, irrespective of pronunciation, for ancient > classical names ending in -s, e.g. Ceres', Mars', Demosthenes', > Venus', Herodotus', Xerxes'. Jesus' is an accepted liturgical > archaism (Hart's Rules, p. 31). But in non-liturgical use, Jesus's is > acceptable (used, e.g., in the Revised English Bible (1989), John 2: > 3). Quoting the second edition on the apostrophe for possessives it originally indicated the omission of the _e_ from the possessive inflexion _es_. -- Michael Press === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! On Oct 1, 7:05 pm, Gerry Myerson An irrational number is a number that is not rational. > In particular, every complex number with non-zero > imaginary part is irrational. This is the standard definition in Number Theory. And so the debate continues over whether the set of irrationals is RQ or CQ. This came up earlier in one of the so-called crank threads, in which finite guy was trying to prove that the irrationals don't belong on the real line. I pointed out that ironically, the crank was right, if one uses Myerson's definition of irrationals (since CQ is not a subset of R). So apparently, there is no general consensus as to whether RQ or CQ are the irrationals. I must admit that the only context in which I've ever seen CQ is in the aforementioned Gelfond-Schneider theorem, otherwise it usually refers to RQ. === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! So apparently, there is no general consensus as to > whether RQ or CQ are the irrationals. I must > admit that the only context in which I've ever > seen CQ is in the aforementioned Gelfond-Schneider > theorem, otherwise it usually refers to RQ. > While on the other hand I've several times read that the rational numbers + the irrational numbers form the real numbers. Just tried to find a quote: The field of all rational and irrational numbers is called the real numbers, or simply the reals, and denoted IR. (http://mathworld.wolfram.com/RealNumber.html) F. -- E-mail: infosimple-linede === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! > > You can't do much mathematics without i, > and you can't call i rational, > so you either throw your hands up in the air, > or you call i irrational. > > Or you can both not call irrational and not call it rational. > > Let's see how your style of argument works in other contexts: > > You can't do much mathematics without 0, > and you can't call 0 positive, > so you either throw your hands up in the air, > or you call 0 negative. No, that's why we have the word nonpositive. Just as the positive integers come in three flavors (composite, prime, and unit), so do the reals: positive, negative, and zero. But the complex numbers come in only two flavors: rational, and irrational. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! > But the complex numbers come in only two flavors: rational, > and irrational. You are here simply repeating a prior assertion that you've seen several disagree with. Assertions do not get more convincing the more they are repeated. Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! > But the complex numbers come in only two flavors: rational, > and irrational. > > You are here simply repeating a prior assertion that > you've seen several disagree with. Assertions do not > get more convincing the more they are repeated. Let's see if this is more convincing. The concepts of rational & irrational were originally developed by Pythagoras & Co. in the context of the real numbers. Some millennia later, the complex numbers were discovered/invented/ whatever. Mathematicians being prone to generalization, it occurred to someone to ask whether the rational/irrational distinction could be extended to this new domain. Now, you're certainly within your rights to say, no, it can't; irrational is a subconcept of real, and nonreal complex numbers are neither rational nor irrational. But the fact is that mathematicians have found it useful to extend the concept to the complex numbers, in a way that is consistent with the pre-existing concept, and in a way that makes it easier to state some theorems. An individual can always reject a generalization, but mathematics will always accept a generalization, if it is useful and is not seen to lead to contradictions. There's a good reason to call i irrational, and no good reason not to, so, I suggest, that's the way to go. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! > Assertions do not get more convincing the more they are repeated. They do so. Do so, do so, do so, do so! DO SO!!!! -- People make mistakes. Better to live today and learn the truth, than to be one of those poor saps who died deluded, thinking they knew certain things that they just didn't. Thinking they had proofs that they didn't. --James S. Harris, almost too sad for a .sig === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! > > Assertions do not get more convincing the more they are repeated. > > > They do so. Do so, do so, do so, do so! DO SO!!!! Well now you put it /that/ way ;-) Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! > My experience is that dichotomy rational/irrational is usually >> reserved for contexts in which we are working with real numbers; in >> algebraic number theory, when complex numbers are at work, the >> dichotomy runs first and foremost along algebraic/transcendental, then >> along algebraic integer/algebraic number, then by degree. Sometimes >> along the real/nonreal line as well. What about the Gelfond-Schneider theorem? It states that a^b is >transcendental if: (1) a is algebraic, not equal to 0 or 1, and > (2) b is algebraic and irrational. > > Well, what it states depends on your definitions. Here is how Niven > quotes the Gelfond-Schneider theorem (p 134, Chapter 10, of > Irrational Numbers): > > Theorem 10.1 If a and ba re algebraic numbers, with a not equal to > 0 or 1, and if b is not a real rational number, then any value of > a^b is transcendental. > > Remark. The hypothesis that b is not a real rational number is > usually stated in the form b is irrational. Our wording is an > attempt to avoid the suggestions that b must be a real number. Such > a number as b=2+3i, sometimes called a complex rational number, > satisfies the hypothesis of the theorem. [...] > > > [Gelfond's 1929] result was that a^b is transcendental for > algebraic a different from 0 or 1, if beta is an imaginary > quadratic irrationality. > > The old saw about standards and there being so many to choose from > comes to mind. Gelfond's 1929 result is not the Gelfond-Schneider Theorem, just an important step on the way. Here's Hardy & Wright, The Theory of Numbers, 4th edition, p. 176: It has been proved more recently that alpha^beta is transcendental if alpha and beta are algebraic, alpha is not 0 or 1, and beta is irrational. This shows in particular that e^(-pi), which is one of the values of i^(2i), is transcendental. So evidently Hardy & Wright are happy to call 2i irrational. Here's Alan Baker, A Concise Introduction to the Theory of Numbers, p. 54: In 1900, Hilbert raised as the seventh of his famous list of 23 problems, the question of proving the transcendence of 2^sqrt2 and, more generally, that of alpha^beta for algebraic alpha not 0 or 1 and algebraic irrational beta. ...in 1929 ... Gelfond succeeded in verifying the special case that e^pi = (-1)^(-i) is transcendental.... So Baker (who won a Fields medal for his work on transcendence) is happy to call -i irrational. Burger and Tubbs, Making Transcendence Transparent, p. 113: Theorem 5.1 (The Gelfond-Schneider Theorem) Suppose alpha and beta are algebraic numbers with alpha not 0, 1, and beta irrational. Then alpha^beta is transcendental. .... The transcendence of e^pi also follows immediately if we view e^pi = i^(-2i).... I'm beginning to think that the discussion of whether i is irrational is similar to the discussion that occasionally breaks out here of whether 0 is a natural number. Both sides of the discussion can cite supporting sources ad nauseam, and neither point of view is going to cause anyone to prove false results, so long as people make clear just which definition is in use at any given time. In a context where only real numbers are under discussion (e.g., continued fractions, approximation by rationals) no harm is done by saying irrational where I would prefer to say real irrational. In a context where nonreal numbers take part, I think one has to interpret irrational to include all those a + b i with b not zero. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! days. My association with the Department is that of an alumnus. [...] >> The old saw about standards and there being so many to choose from >> comes to mind. Gelfond's 1929 result is not the Gelfond-Schneider Theorem, theorem happened to agree with the use in Niven (that is, using words other than simply irrational so as not to restrict itself to real numbers), but its statement of Gelfond's 1929 Theorem, with imaginary quadratic irrationalities, by contrast, seemed to lean the other way. [...] >I'm beginning to think that the discussion of whether i is irrational >is similar to the discussion that occasionally breaks out here of >whether 0 is a natural number. Both sides of the discussion can cite >supporting sources ad nauseam, and neither point of view is going >to cause anyone to prove false results, so long as people make clear >just which definition is in use at any given time. I wonder what I was refering to above when alluding to the old saw about standards... -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! On Oct 2, 8:02 pm, Gerry Myerson > My experience is that dichotomy rational/irrational is usually >> reserved for contexts in which we are working with real numbers; in >> algebraic number theory, when complex numbers are at work, the >> dichotomy runs first and foremost along algebraic/transcendental, then >> along algebraic integer/algebraic number, then by degree. Sometimes >> along the real/nonreal line as well. >What about the Gelfond-Schneider theorem? It states that a^b is >transcendental if: > (1) a is algebraic, not equal to 0 or 1, and > (2) b is algebraic and irrational. Well, what it states depends on your definitions. Here is how Niven > quotes the Gelfond-Schneider theorem (p 134, Chapter 10, of > Irrational Numbers): Theorem 10.1 If a and ba re algebraic numbers, with a not equal to > 0 or 1, and if b is not a real rational number, then any value of > a^b is transcendental. Remark. The hypothesis that b is not a real rational number is > usually stated in the form b is irrational. Our wording is an > attempt to avoid the suggestions that b must be a real number. Such > a number as b=2+3i, sometimes called a complex rational number, > satisfies the hypothesis of the theorem. [...] > [Gelfond's 1929] result was that a^b is transcendental for > algebraic a different from 0 or 1, if beta is an imaginary > quadratic irrationality. The old saw about standards and there being so many to choose from > comes to mind. Gelfond's 1929 result is not the Gelfond-Schneider Theorem, > just an important step on the way. Here's Hardy & Wright, The Theory of Numbers, 4th edition, p. 176: It has been proved more recently that alpha^beta is transcendental > if alpha and beta are algebraic, alpha is not 0 or 1, and beta is > irrational. This shows in particular that e^(-pi), which is one of the > values of i^(2i), is transcendental. So evidently Hardy & Wright are happy to call 2i irrational. Here's Alan Baker, A Concise Introduction to the Theory of Numbers, > p. 54: In 1900, Hilbert raised as the seventh of his famous list of 23 > problems, the question of proving the transcendence of 2^sqrt2 and, > more generally, that of alpha^beta for algebraic alpha not 0 or 1 > and algebraic irrational beta. ...in 1929 ... Gelfond succeeded in > verifying the special case that e^pi = (-1)^(-i) is transcendental.... So Baker (who won a Fields medal for his work on transcendence) is > happy to call -i irrational. Burger and Tubbs, Making Transcendence Transparent, p. 113: Theorem 5.1 (The Gelfond-Schneider Theorem) Suppose alpha > and beta are algebraic numbers with alpha not 0, 1, and beta > irrational. Then alpha^beta is transcendental. .... The transcendence of e^pi also follows immediately if we view > e^pi = i^(-2i).... I'm beginning to think About time too. > that the discussion of whether i is irrational > is similar to the discussion that occasionally breaks out here of > whether 0 is a natural number. No? Ya think? > Both sides of the discussion can cite > supporting sources ad nauseam, and neither point of view is going > to cause anyone to prove false results, so long as people make clear > just which definition is in use at any given time. In a context where only real numbers are under discussion (e.g., > continued fractions, approximation by rationals) no harm is done by > saying irrational where I would prefer to say real irrational. In a > context where nonreal numbers take part, I think one has to interpret > irrational to include all those a + b i with b not zero. Piffle. It is obvious to anyone with two brain cells to rub together that the only reasonable extention of rational/irrational to complex values, is to say that a complex value is rational if and only if it is the ratio of Gaussian integers. Calling any imaginary value irrational, just to make the statement of Gelfond-Schneider a bit simpler is dumb. Your citations above only prove that even the best mathematicians have lapses. - William Hughes reponse, ending with I was really right anyway. I am not impressed. === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! >Piffle. It is obvious to anyone with two brain cells to rub >together that the only reasonable extention of rational/irrational >to complex values, is to say that a complex value is rational >if and only if it is the ratio of Gaussian integers. That doesn't make sense to me. If you declare the quotient of 2 Gaussian integers rational, then why wouldn't you do the same for the elements of Z[sqrt(2)]. If a,b are ordinary integers, then a + b*sqrt(2) is an algebraic integer, right? Note that it's not much different in form than a + b*sqrt(-1) Thus, if you are going to regard (a + b*i) / (c + d*i) as rational, then why wouldn't (a + b*sqrt(2)) / (c + d*sqrt(2)) also be considered rational? Of course that would mean sqrt(2) is rational. This argues against regarding the quotient of 2 Gaussian integers as rational. The fact is, as far as I can see, most sources define irrational as reals which are not rational. However, there's nothing really wrong with calling i irrational. The justification is simple -- i is not rational. Also, just look at it ... sqrt(-1) It's a square root that doesn't simplify. In that regard, it's conceptually just as irrational as sqrt(2) My point is this ... Defining irrationals as reals which are not rational, while perhaps too limiting in some respects, appears to be the most standard interpretation. Defining irrationals as complexes which are not rational is used by some authors, but not most. However it certainly seems to be a reasonable definition, and the greater generality makes it useful for some theorems. Defining irrationals as complex numbers which are not the quotient of 2 gaussian integers gives too much priority to a particular quadratic extension of Z, namely Z[i]. Such a definition seems a little strange to me, and I doubt it would be useful as a general concept of irrationality. quasi === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! days. My association with the Department is that of an alumnus. >Piffle. It is obvious to anyone with two brain cells to rub >>together that the only reasonable extention of rational/irrational >>to complex values, is to say that a complex value is rational >>if and only if it is the ratio of Gaussian integers. That doesn't make sense to me. If you declare the quotient of 2 Gaussian integers rational, then why >wouldn't you do the same for the elements of Z[sqrt(2)]. I suspect the reason is quite simple: under that definition, the rational complex numbers would be exactly those that can be expressed as a + bi with a and b real rationals, and the irrationals would be everything else. In other words, if you identify the complex numbers with the plane, then the rationals would be Q x Q, the points with rational coordinates. This would allow all the approximation theorems to go through. That said, I also suspect that William is being more than a bit sarcastic in his preamble, judging from his postscript. >The fact is, as far as I can see, most sources define irrational as >reals which are not rational. However, there's nothing really wrong with calling i irrational. The >justification is simple -- i is not rational. Neither is the number 2... in GF(3) (there it corresponds to an equivalence class of integers, hence not an integer or a ratio of integers). The definition of rational as a number that can be expresssed as quotient of two integers assumes a scope for number. It is not always immediately clear what that scope is supposed to be. [...] >Defining irrationals as reals which are not rational, while perhaps >too limiting in some respects, appears to be the most standard >interpretation. Defining irrationals as complexes which are not rational is used by >some authors, but not most. However it certainly seems to be a >reasonable definition, and the greater generality makes it useful for >some theorems. Certainly, as already noted ad nauseam, the two most common uses give either RQ or CQ as the meaning for irrational. Both are valid, in the sense that both are reasonable meanings for the term and so long as they are used consistently, are unlikely to cause any problems. Just like any other number of multiple conventions (is the empty set a subsemigroup of a semigroup [by 'subsemigroup' I mean a set with a binary associative operation]? Yes, according to some authors, no, according to others; each has advantages; e.g., the intersection of two subsemigroups is not necessarily a subsemigroup if you do not allow the empty semigroup... Hell, for that matter, semigroup doesn't always means the same thing, nor does monoid) I certainly did not dispute the ->validity<- of saying irrationals means CQ. Rather, I disputed the assertion that it was the ->standard<- meaning in Number Theory, and presented evidence that explained where people might get the idea that irrational may mean RQ instead. But at the same time, I also provided evidence that it ->sometimes<- means CQ. (For what it is worth, since I am summarizing what I said, I also disputed the use of Wikipedia or PlanetMath as a good place to look for what may or may not be the standard meaning, assuming such a thing existed). -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! >Piffle. It is obvious to anyone with two brain cells to rub >>together that the only reasonable extention of rational/irrational >>to complex values, is to say that a complex value is rational >>if and only if it is the ratio of Gaussian integers. That doesn't make sense to me. If you declare the quotient of 2 Gaussian integers rational, then why >wouldn't you do the same for the elements of Z[sqrt(2)]. > > I suspect the reason is quite simple: under that definition, the > rational complex numbers would be exactly those that can be > expressed as a + bi with a and b real rationals, and the irrationals > would be everything else. In other words, if you identify the complex > numbers with the plane, then the rationals would be Q x Q, the > points with rational coordinates. This would allow all the > approximation theorems to go through. How do you mean? In what way does Hurwitz's theorem on Diophantine approximation, say, go through? -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! days. My association with the Department is that of an alumnus. >>Piffle. It is obvious to anyone with two brain cells to rub >together that the only reasonable extention of rational/irrational >to complex values, is to say that a complex value is rational >if and only if it is the ratio of Gaussian integers. >>That doesn't make sense to me. >>If you declare the quotient of 2 Gaussian integers rational, then why >>wouldn't you do the same for the elements of Z[sqrt(2)]. >> >> I suspect the reason is quite simple: under that definition, the >> rational complex numbers would be exactly those that can be >> expressed as a + bi with a and b real rationals, and the irrationals >> would be everything else. In other words, if you identify the complex >> numbers with the plane, then the rationals would be Q x Q, the >> points with rational coordinates. This would allow all the >> approximation theorems to go through. How do you mean? In what way does Hurwitz's theorem on Diophantine >approximation, say, go through? Okay, first, I am not advocating such a definition; in the part that quasi snipped, I think I made that somewhat clear. Second, I clearly misspoke and should have said most, rather than all. Certainly things like every irrational can be approximated by a sequence of rationals, and the like. That said, I'm getting a little tired of pummeling this defunct equine... -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! >Piffle. It is obvious to anyone with two brain cells to rub >together that the only reasonable extention of rational/irrational >to complex values, is to say that a complex value is rational >if and only if it is the ratio of Gaussian integers. >>That doesn't make sense to me. >>If you declare the quotient of 2 Gaussian integers rational, then why >>wouldn't you do the same for the elements of Z[sqrt(2)]. I suspect the reason is quite simple: under that definition, the >rational complex numbers would be exactly those that can be >expressed as a + bi with a and b real rationals, and the irrationals >would be everything else. In other words, if you identify the complex >numbers with the plane, then the rationals would be Q x Q, the >points with rational coordinates. This would allow all the >approximation theorems to go through. Ok, that's a good defense. Thus, it seems now that the 3 definitions previously discussed all have some validity. >That said, I also suspect that William is being more than a bit >sarcastic in his preamble, judging from his postscript. Well, the 2 brain cells comment was simply wrong. quasi === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! sarcastic in his preamble, judging from his postscript. Well, the 2 brain cells comment was simply wrong. > Well, I suppose I could have been even more over the top, but I thought it was pretty good. (I even managed a cliche!) - William Hughes === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! > >Piffle. It is obvious to anyone with two brain cells to rub >together that the only reasonable extention of rational/irrational >to complex values, is to say that a complex value is rational >if and only if it is the ratio of Gaussian integers. >>That doesn't make sense to me. >>If you declare the quotient of 2 Gaussian integers rational, then why >>wouldn't you do the same for the elements of Z[sqrt(2)]. I suspect the reason is quite simple: under that definition, the >rational complex numbers would be exactly those that can be >expressed as a + bi with a and b real rationals, and the irrationals >would be everything else. In other words, if you identify the complex >numbers with the plane, then the rationals would be Q x Q, the >points with rational coordinates. This would allow all the >approximation theorems to go through. > > Ok, that's a good defense. It still privileges i in a way I find unconvincing. Let w be a complex cube root of 1. Why not let the rational complex numbers be those that can be expressed as a + b w with a, b real rationals? Q(w) is still a 2-dimensional vector space over Q, thus it is Q x Q with respect to the basis {1, w}. The approximation theorems go through, no? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! > On Oct 2, 8:02 pm, Gerry Myerson I'm beginning to think > > About time too. I've been posting here for over 15 years. I don't recall having ever posted a personal insult aimed at you. I don't know why you choose to aim one at me, but may I say I don't think it does anything for your image here. > In a context where only real numbers are under discussion (e.g., > continued fractions, approximation by rationals) no harm is done by > saying irrational where I would prefer to say real irrational. In a > context where nonreal numbers take part, I think one has to interpret > irrational to include all those a + b i with b not zero. > > Piffle. It is obvious to anyone with two brain cells to rub > together that the only reasonable extention of rational/irrational > to complex values, is to say that a complex value is rational > if and only if it is the ratio of Gaussian integers. > Calling any imaginary value irrational, just to make the > statement of Gelfond-Schneider a bit simpler is dumb. You're saying G H Hardy and Alan Baker are dumb and don't have two brain cells each. Given a choice between your intellect (or mine) and theirs, who do you more people would cite as dumb? There is an interesting mathematical question here. Why, among all subrings of the ring of algebraic integers, might one single out the ring of Gaussian integers as special? or, why, among all subfields of the complex numbers, single out Q(i)? Sure, i generates the complex numbers, in the sense that C = R(i), but if alpha is any nonreal complex number, then C = R(alpha), so there's nothing special about i there. Z is special because it is generated by 1, and 1 is the multiplicative identity; but i is no identity. It's a root of unity, but there are lots of those around. So what's so special about the Gaussian integers? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! , Gerry Myerson > > > 3) If a*a is irrational, then a is irrational. >> Claim: a*a is => a is Proof: Suppose a*a is . Now assume a is . Then a*a is too. >Contradiction! [] > Actually, 3) does not hold, and hence the alleged proof is not (quite) > correct. > 4__ __ > Counterexample: Let a = -i * v(2). Then a*a = -v(2). With other words, > a*a is irrational. But a is NOT irrational (i.e. not a real number > which is not rational), since it is a (purely) imaginary number. > > I don't know where people get this idea. > An irrational number is a number that is not rational. > In particular, every complex number with non-zero > imaginary part is irrational. > > This is the standard definition in Number Theory. If a number field contains sqrt{-1}, then sqrt{-1} is also an integer, since it is a root of the monic polynomial xx + 1. How do you manage that? -- Michael Press === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! > , > Gerry Myerson > > I don't know where people get this idea. > An irrational number is a number that is not rational. > In particular, every complex number with non-zero > imaginary part is irrational. > > This is the standard definition in Number Theory. > > If a number field contains sqrt{-1}, > then sqrt{-1} is also an integer, > since it is a root of the monic polynomial xx + 1. > How do you manage that? How do I manage what? sqrt(-1) is not an integer; it is an algebraic integer. Given enough context, it is permissible to abuse the terminology and refer to sqrt(-1) as an integer, but whenever people do that they know they really mean it's an algebraic integer. Hey, if you're working in the field Q(sqrt2), you might refer to sqrt2 as an integer. Are you suggesting then that it's wrong to call sqrt2 an irrational? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! , Gerry Myerson > > , > Gerry Myerson > >I don't know where people get this idea. >An irrational number is a number that is not rational. >In particular, every complex number with non-zero >imaginary part is irrational. This is the standard definition in Number Theory. > > If a number field contains sqrt{-1}, > then sqrt{-1} is also an integer, > since it is a root of the monic polynomial xx + 1. > How do you manage that? > > How do I manage what? Well, you do answer this anyway. > sqrt(-1) is not an integer; it is an algebraic integer. > Given enough context, it is permissible to abuse the terminology > and refer to sqrt(-1) as an integer, but whenever people do that > they know they really mean it's an algebraic integer. Hey, if you're > working in the field Q(sqrt2), you might refer to sqrt2 as an integer. > Are you suggesting then that it's wrong to call sqrt2 an irrational? In Q(sqrt2) what is an irrational number? In Q(sqrt2), sqrt2 is the ratio of two integers. Since they are not rational integers, then we can call sqrt2 an irrational integer. Still, what is the utility of the definition? -- Michael Press === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! > , > Gerry Myerson > > >, > Gerry Myerson I don't know where people get this idea. >An irrational number is a number that is not rational. >In particular, every complex number with non-zero >imaginary part is irrational. This is the standard definition in Number Theory. If a number field contains sqrt{-1}, >then sqrt{-1} is also an integer, >since it is a root of the monic polynomial xx + 1. >How do you manage that? > > How do I manage what? > > Well, you do answer this anyway. > > sqrt(-1) is not an integer; it is an algebraic integer. > Given enough context, it is permissible to abuse the terminology > and refer to sqrt(-1) as an integer, but whenever people do that > they know they really mean it's an algebraic integer. Hey, if you're > working in the field Q(sqrt2), you might refer to sqrt2 as an integer. > Are you suggesting then that it's wrong to call sqrt2 an irrational? > > In Q(sqrt2) what is an irrational number? > In Q(sqrt2), sqrt2 is the ratio of two integers. > Since they are not rational integers, > then we can call sqrt2 an irrational integer. > Still, what is the utility of the definition? I don't think there is any. I'm not sure I know what we're arguing about. I maintain that sqrt(-1) is irrational. You note that (in some contexts) sqrt(-1) is an integer. In the context of this thread, I took it that you were putting this forth as evidence against my position that sqrt(-1) is irrational. If that's not what you were trying to do, then I'm afraid I don't know wnat your point was in bringing up sqrt(-1). If that was what you were trying to do, then my point was that the same argument that suggests sqrt(-1) isn't irrational also suggests sqrt2 isn't irrational, which doesn't look too good. Or, the same argument you use to call sqrt2 irrational, shows that sqrt(-1) is irrational, as it isn't the ratio of two rational integers. I hope I've made myself less unclear than usual. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! , Gerry Myerson > > , > Gerry Myerson > , > Gerry Myerson > I don't know where people get this idea. >> An irrational number is a number that is not rational. >> In particular, every complex number with non-zero >> imaginary part is irrational. >> >> This is the standard definition in Number Theory. If a number field contains sqrt{-1}, >then sqrt{-1} is also an integer, >since it is a root of the monic polynomial xx + 1. >How do you manage that? How do I manage what? > > Well, you do answer this anyway. > >sqrt(-1) is not an integer; it is an algebraic integer. >Given enough context, it is permissible to abuse the terminology >and refer to sqrt(-1) as an integer, but whenever people do that >they know they really mean it's an algebraic integer. Hey, if you're >working in the field Q(sqrt2), you might refer to sqrt2 as an integer. >Are you suggesting then that it's wrong to call sqrt2 an irrational? > > In Q(sqrt2) what is an irrational number? > In Q(sqrt2), sqrt2 is the ratio of two integers. > Since they are not rational integers, > then we can call sqrt2 an irrational integer. > Still, what is the utility of the definition? > > I don't think there is any. > > I'm not sure I know what we're arguing about. > > I maintain that sqrt(-1) is irrational. > You note that (in some contexts) sqrt(-1) is an integer. > In the context of this thread, I took it that you were putting > this forth as evidence against my position that sqrt(-1) is > irrational. If that's not what you were trying to do, then I'm > afraid I don't know wnat your point was in bringing up > sqrt(-1). If that was what you were trying to do, then my > point was that the same argument that suggests sqrt(-1) > isn't irrational also suggests sqrt2 isn't irrational, which > doesn't look too good. Or, the same argument you use to > call sqrt2 irrational, shows that sqrt(-1) is irrational, as > it isn't the ratio of two rational integers. > > I hope I've made myself less unclear than usual. successfully communicating my confusion. H.E. Rose in _A_Course_in_Number_Theory_ states the Gelfond-Schneider theorem as If a and b are algebraic numbers, a =/= 0 or 1, and b is irrational then w = a^b is transcendental. With a bit of thinking the meaning is clear, but I prefer that statements of theorems to be clear _without_ thinking. :) I prefer If a and b are algebraic numbers, a =/= 0 or 1, and b not in Q, then w = a^b is transcendental. -- Michael Press A little learning is a dangerous thing. === Subject: Re: PROVE sqrt(2)+sqrt(3) is a irrational number!!! H.E. Rose in _A_Course_in_Number_Theory_ states the > Gelfond-Schneider theorem as If a and b are algebraic numbers, a =/= 0 or 1, and b > is irrational then w = a^b is transcendental. With a bit of thinking the meaning is clear, > but I prefer that statements of theorems to be clear > _without_ thinking. :) I prefer If a and b are algebraic numbers, a =/= 0 or 1, > and b not in Q, then w = a^b is transcendental. > How about If a and b are algebraic numbers, a =/= 0 or 1, and b not rational, then w = a^b is transcendental. ? (I guess, the definition of /rational/ is less controversial than the definition of /irrational/, no?) F. -- E-mail: infosimple-linede === Subject: need help urganlty hi i need solution for all the examples of data and computer communication for 8th edition === Subject: Re: need help urganlty > hi i need solution for all the examples of data and computer > communication for 8th edition > moron..... go study night soil collection, you would be better at it. === Subject: Re: I need a solutions manual i AHVE THE SIX EDITION ON PDF AND YES IT COMES WITH ODD AND EVEN ANSWERS === Subject: Re: linearly independent On Oct 2, 10:48 am, cbr...@cbrownsystems.com > On Oct 2, 11:00 am, cbr...@cbrownsystems.com > If you apply the above rules correctly, you will find that A and B > are linearly independent is equivalent to saying 0 and else> are linearly independent. And as I mistated the above hint, this conclusion is misstated: > instead, you will find that /IF/ A and B are linearly DEPENDENT > (i.e., if there /are/ c1 and c2 not both 0 such that c1A + c2B = 0), > THEN A is or B is . And you will need the hint > to prove that IF A is or B is , THEN A and B > are linearly dependent. So you will end up with : A and B are > linearly dependent if, and only if, A is or B is > . Sorry for all the confusion. > so if i assume that: A=[x,a,c; a,y,b; c,b,z] and B=[0,e,j; -e,0,f; -j,-f,0] then A+B=[x,a+e,c+j; a-e,y,b+f; c-j,b-f,z] and A^T=[x,a,c; a,y,b; c,b,z] and B^T= [0,-e,-j; e,0,-f; j,f,0] then A^T+B^T=[x,a-e,c-j; a+e,y,b-f; c+j,b+f,z] then if take c1A+c2B=0 ill get c1x+c20=0......1 c1a+c2e=0......2 c1c+c2j=0......3 c1a-c2e=0......4 c1y-c20=0......5 c1b+c2f=0......6 c1c-c2j=0......7 c1b-c2f=0......8 c1z+c20=0......9 and the same 4 the A^T+B^T=0 ill get 8 eq. then after that im stuck ... did not understand the part where Z^T in terms of Z.... is it the same as A^T+B^T.....??? thanx a lot 4 ur time and help..... :) === Subject: Re: linearly independent > On Oct 2, 10:48 am, cbr...@cbrownsystems.com On Oct 2, 11:00 am, cbr...@cbrownsystems.com >If you apply the above rules correctly, you will find that A and B >are linearly independent is equivalent to saying 0 and else> are linearly independent. And as I mistated the above hint, this conclusion is misstated: > instead, you will find that /IF/ A and B are linearly DEPENDENT > (i.e., if there /are/ c1 and c2 not both 0 such that c1A + c2B = 0), > THEN A is or B is . And you will need the hint > to prove that IF A is or B is , THEN A and B > are linearly dependent. So you will end up with : A and B are > linearly dependent if, and only if, A is or B is > . Sorry for all the confusion. > so if i assume that: A=[x,a,c; a,y,b; c,b,z] and B=[0,e,j; -e,0,f; -j,-f,0] then A+B=[x,a+e,c+j; a-e,y,b+f; c-j,b-f,z] and A^T=[x,a,c; a,y,b; c,b,z] and B^T= [0,-e,-j; e,0,-f; j,f,0] > then A^T+B^T=[x,a-e,c-j; a+e,y,b-f; c+j,b+f,z] then if take c1A+c2B=0 ill get c1x+c20=0......1 > c1a+c2e=0......2 > c1c+c2j=0......3 > c1a-c2e=0......4 > c1y-c20=0......5 > c1b+c2f=0......6 > c1c-c2j=0......7 > c1b-c2f=0......8 > c1z+c20=0......9 and the same 4 the A^T+B^T=0 ill get 8 eq. then after that im stuck ... did not understand the part where Z^T in > terms of Z.... is it the same as A^T+B^T.....??? > Well, is it? That question can be answered, not /just/ in the situation of your question, but for /any/ X, Y and Z = X + Y. My hints are about this more /general/ question. It's not that hard to see: just follow that train of thought I laid out: 1. By the /definition/ of matrix addition, Z[i,j] = X[i,j] + Y[i,j] 2. By the /definition/ of transpose, Z^T[j,i] = Z[i,j] 3. Therefore, Z^T[j,i] = X[?,?] + Y[?,?] 4. X^T[j,i] = X[j,i]; Y^T[j,i] = Y[i,j] 5. By 4 and 3, Z^T[j,i] = X^T[?,?] + Y^T[?,?] 6. Since i and j are arbitrary, Z^T[i,j] = X^T[?,?] + Y^T[?,?] Fill in the question marks, and compare 1 with 6, and you'll see why it is /always/ true that (X + Y)^T = X^T + Y^T (not just in the case of this problem). To use this in your homework question: Either A and B are linearly dependent, or they aren't (right?). If we /assume/ that A and B /are/ linearly dependent, then we /assume/ that there are real numbers c1, c2, such that c1 A + c2 B = 0 and such that c1 and c2 are not both (check the definition of linearly dependent for ). That means that (c1 A + c2 B)^T = 0^T = 0 Use the above facts about (X + Y)^T and the particular constraints of this problem to give us another equation c3 A + c4 B = 0 where c3 and c4 are (very!) easily expressed in terms of c1 and c2. That will give you two equations: c1 A + c2 B = 0 c3 A + c4 B = 0 which you can add or subtract to get interesting results. That will give you an answer of the form If A and B are linearly dependent, then A or B is . The hint for the problem goes the other way: it encourges you to prove the statement if A or B is then A and B are linearly dependent. Together we get the answer to your problem: A and B are linearly dependent if, and only if, A or B is . === Subject: density of the range of an integer polynomial For an integer polynomial f, possibly multivariate, let range(f) denote the range of f for all possible integer inputs, and let density(f) denote the density of (range(f) intersect N), as a subset of N. Note -- as discussed in a recent thread, density(f) is not necessarily the same as density(-f). Of course, density(f), if it exists, is a real number in the interval [0,1]. Some simple examples ... If f(x) = x^2, then density(f) = 0. If f(x) = k*x, where k is a fixed positive integer, then density(f) = 1/k. Two questions ... (1) Does every integer polynomial have a density? (2) For an integer polynomial, what densities are possible? Is every rational number in the interval [0,1] achievable as a density? Can a density be irrational? quasi === Subject: Re: density of the range of an integer polynomial Ok, let's start over. This time, I'll try to avoid ending up in the (far too permissive) land of recursive enumerability. For an integer polynomial f, possibly multivariate, let range(f) denote the range of f for all possible integer inputs, and let density(f) denote the density of (range(f) intersect N), as a subset of N. Note -- as discussed in a recent thread, density(f) is not necessarily the same as density(-f). Of course, density(f), if it exists, is a real number in the interval [0,1]. Some examples ... If f is a univariate polynomial, then density(f) = 0, unless the degree of f is exactly 1. If f(x) = a x + b, where a,b are integers and a is nonzero, then density(f) = 1/|a|. Robert Israel showed that for all rational numbers d in the interval [0,1], there is an integer polynomial with density(f) = d. He also gave an example of an integer polynomial for which density(f) is irrational. Some questions ... Question (1): If f is an integer polynomial, then either density(f) or density(-f) exists. True or false? Question (2): If f is an integer polynomial which is nonnegative for all real inputs, then density(f) exists. True or false? Question (3): If f is an integer polynomial which is nonnegative for all real inputs, what values are possible for density(f)? quasi === Subject: Re: density of the range of an integer polynomial Not every polynomial has a density in the sense described, and every real number in the interval [0,1] is a density. This is an easy consequence of the theorem of Matijasevic that settled Hilbert's 10th problem, together with a remark by Putnam, or was it Davis. The result is: For every recursively enumerable set E of non- negative integer, there is a polynomial P such that the non-negative part of the range of P is E. One can arrange for the negative part of the range to be all negative integers, or to be quite sparse. A set E is recursively enumerable if there is an algorithm that will output the elements of E, possibly with repetition. Every reasonable set of non-negative integers is recursively enumerable. In particular, if you produce in a reasonable way a set of non-negative integers which does not have a density, or a set which has density d, that set will turn out to be r.e., and hence the non-negative part of the range of a polynomial. > For an integer polynomial f, possibly multivariate, let range(f) > denote the range of f for all possible integer inputs, and let > density(f) denote the density of (range(f) intersect N), as a subset > of N. Note -- as discussed in a recent thread, density(f) is not necessarily > the same as density(-f). Of course, density(f), if it exists, is a real number in the interval > [0,1]. Some simple examples ... If f(x) = x^2, then density(f) = 0. If f(x) = k*x, where k is a fixed positive integer, then density(f) = > 1/k. Two questions ... (1) Does every integer polynomial have a density? (2) For an integer polynomial, what densities are possible? Is every > rational number in the interval [0,1] achievable as a density? Can a > density be irrational? quasi === Subject: Re: density of the range of an integer polynomial Not every polynomial has a density in the sense described, and every >real number in the interval [0,1] is a density. This is an easy consequence of the theorem of Matijasevic that settled >Hilbert's 10th problem, together with a remark by Putnam, or was it >Davis. The result is: For every recursively enumerable set E of non- >negative integer, there is a polynomial P such that the non-negative >part of the range of P is E. One can arrange for the negative part of >the range to be all negative integers, or to be quite sparse. A set E is recursively enumerable if there is an algorithm that will >output the elements of E, possibly with repetition. Every reasonable >set of non-negative integers is recursively enumerable. In >particular, if you produce in a reasonable way a set of non-negative >integers which does not have a density, or a set which has density d, >that set will turn out to be r.e., and hence the non-negative part of >the range of a polynomial. Wow. I apologize. Stupidly, I only skimmed this initial post of yours, and focused instead on the later ones, for which I had all sorts of objections. My objections were based mostly on only the fact that I had only fragmentary knowledge of the relevant theory, and in some cases, complete ignorance. Assuming the result you state above, that every recursively enumerable subset of N is the positive part of the range of an integer polynomial, then yes, you can create all kinds of positive ranges, and all kinds of densities. However I think galathaea's point still holds -- namely, the possible densities are just the constructible numbers in the interval [0,1]. But, assuming the results you cite, I now agree that there are polynomials for which the density is undefined (the positive range gravitates back and forth between a lower density and an upper density). Also, positive ranges such as the powers of 2 are actually possible, as weird as that seems. I think the essence of the trick (and it's a sneaky trick) is that by restricting to the positive part of the range, an if statement is being allowed, i,e., a conditional branch. That gives too much flexibility. The main questions I have (from prior threads) concern _full_ ranges, but when considering the density issue, it seemed appropriate to restrict to the positive part of the range. Thus, unwittingly, I ended up with all recursively enumerable subsets of N. It was a trap, but I fell for it -- I'll try to be more careful in the future. I'm looking for structure, not the lack of it. quasi === Subject: Re: density of the range of an integer polynomial Not every polynomial has a density in the sense described, and every > real number in the interval [0,1] is a density. _every_ real number? there are only a countable number of polynomials so i could see every computable real being possible (though somewhat surprising) but just on cardinality there is no way i can see for the stronger statement to be true -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: density of the range of an integer polynomial >> Not every polynomial has a density in the sense described, and every >> real number in the interval [0,1] is a density. _every_ real number? there are only a countable number of polynomials > so i could see every computable real being possible > (though somewhat surprising) >but just on cardinality >there is no way i can see for the stronger statement to be true Once again, you are right on target. Countably many integer polynomials implies at most countably many densities. QED. quasi === Subject: Re: density of the range of an integer polynomial I was careless: it is every real number which is the density of an r.e. set. What these are is not clear. > Not every polynomial has a density in the sense described, and every > real number in the interval [0,1] is a density. _every_ real number? there are only a countable number of polynomials > so i could see every computable real being possible > (though somewhat surprising) > but just on cardinality > there is no way i can see for the stronger statement to be true -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar === Subject: Re: density of the range of an integer polynomial I was careless: it is every real number which is the density of an >r.e. set. What these are is not clear. Every _constructible_ real number (in the interval [0,1]). [as galathaea suggested] quasi === Subject: Re: density of the range of an integer polynomial I was careless: it is every real number which is the density of an >r.e. set. What these are is not clear. Cool -- at least you admit your error. It's a common error, actually, since Mat did show that the set of primes is representable as the positive integer range of an integer polynomial. Of course, the set of primes has density 0 in N, so it doesn't contribute a density that we don't already have. In any case, the original problem (1) is still in play, as well as part of problem (2) (the part that Robert Israel did not solve). Based on Robert Israel's argument (which I haven't yet tried to verify, but given that it's Robert Israel's argument, it's probably right), every rational number in the range [0,1] is a possible density. He also showed that some irrational densities were possible, but certainly not all irrational densities (the cardinality is too high). Which irrational densities are possible? Well, that question is still in play. For example, can a polynomial have density 1/sqrt(2)? My guess is no. quasi === Subject: Re: density of the range of an integer polynomial >I was careless: it is every real number which is the density of an >>r.e. set. What these are is not clear. Cool -- at least you admit your error. It's a common error, actually, since Mat did show that the set of >primes is representable as the positive integer range of an integer >polynomial. Of course, the set of primes has density 0 in N, so it >doesn't contribute a density that we don't already have. Of course, I meant Matiyasevich, not Mat. quasi === Subject: Re: density of the range of an integer polynomial Monthly, sometime in 1973. There are at least a couple of books, one by Matijasevic, one called I think Logical Number Theory Volume 1 by Smorynski (?). There has been some interest in producing cheap polynomials to represent the primes. I don't recall any papers on cheap ways to produce the powers of 2, though that should be technically a little easier. Given an algorithm for generating the r.e. set E, there is an algorithm to mechanically transform it into a polynomial whose non- negative range is E. However, the general algorithm produces a polynomial in an awful lot of variables. > Not every polynomial has a density in the sense described, and every > real number in the interval [0,1] is a density. _every_ real number? there are only a countable number of polynomials > so i could see every computable real being possible > (though somewhat surprising) > but just on cardinality > there is no way i can see for the stronger statement to be true -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar === Subject: Re: density of the range of an integer polynomial Monthly, sometime in 1973. There are at least a couple of books, one by Matijasevic, one called I >think Logical Number Theory Volume 1 by Smorynski (?). There has been some interest in producing cheap polynomials to >represent the primes. I don't recall any papers on cheap ways to >produce the powers of 2, though that should be technically a little >easier. No -- it's technically a lot harder (i.e., impossible) >Given an algorithm for generating the r.e. set E, there is an >algorithm to mechanically transform it into a polynomial whose non- >negative range is E. However, the general algorithm produces a >polynomial in an awful lot of variables. No -- you are making an elementary error in interpreting those logic related results. Every recursively enumerable set is the _projection_ onto the first variable of the solutions to a (multivariate) diophantine equation. That equation is typically _implicit_, and hence there is often no way to solve for that variable explicity as a polynomial in the other variables. A simple counterexample is the following set ... Let S = {2x | x in Z} union {3x | x in Z}. Thus, S is the set of all integers which are multiples of 2 or 3 (or both). Obviously, S is recursively enumerable -- in fact, S is recursive. But as was proved by Robert Sheskey in the thread once quasi asked about unions of sets ... S is not the range (for integer inputs) of any integer polynomial, regardless of the number of variables. With a slight extension of his argument, it can be shown that there's no polynomial, whose positive integer range (for integer inputs), is equal to S intersect N. quasi === Subject: Re: density of the range of an integer polynomial Monthly, sometime in 1973. There are at least a couple of books, one by Matijasevic, one called I > think Logical Number Theory Volume 1 by Smorynski (?). There has been some interest in producing cheap polynomials to > represent the primes. I don't recall any papers on cheap ways to > produce the powers of 2, though that should be technically a little > easier. Given an algorithm for generating the r.e. set E, there is an > algorithm to mechanically transform it into a polynomial whose non- > negative range is E. However, the general algorithm produces a > polynomial in an awful lot of variables. but i was under the impression that that required restriction to positive inputs as well using the classic construction that if D(a, x1, x2, ..., xn) = 0 is a diophantine representation of a for nonnegative a which is strictly on that section then a (1 - D(...)) gives a polynomial for its representation with positive a argument (loosely - there is usually a linear factor altered or...) in other words i thought the nonnegative part of the range still did not represent the re set but only the section of positive inputs ( or some half-plane or generally some diophantine section ) is this not true? or wouldn't any polynomial that represents 2^n in its positive range for _all_ integer inputs be infinite degree (and thus not polynomial)? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: density of the range of an integer polynomial >> Monthly, sometime in 1973. >> There are at least a couple of books, one by Matijasevic, one called I >> think Logical Number Theory Volume 1 by Smorynski (?). >> There has been some interest in producing cheap polynomials to >> represent the primes. I don't recall any papers on cheap ways to >> produce the powers of 2, though that should be technically a little >> easier. >> Given an algorithm for generating the r.e. set E, there is an >> algorithm to mechanically transform it into a polynomial whose non- >> negative range is E. However, the general algorithm produces a >> polynomial in an awful lot of variables. but i was under the impression > that that required restriction to positive inputs as well > using the classic construction that if D(a, x1, x2, ..., xn) = 0 > is a diophantine representation of a for nonnegative a > which is strictly on that section then > a (1 - D(...)) >gives a polynomial for its representation >with positive a argument >(loosely - there is usually a linear factor altered or...) in other words > i thought the nonnegative part of the range > still did not represent the re set >but only the section of positive inputs >( or some half-plane > or generally some diophantine section ) is this not true? Yes, exactly. You are aware of the key distinction. To say a set is diophantine is _not_ the same as saying it's representable by a polynomial. quasi === Subject: Re: density of the range of an integer polynomial Not every polynomial has a density in the sense described, and every > real number in the interval [0,1] is a density. This is an easy consequence of the theorem of Matijasevic that settled > Hilbert's 10th problem, together with a remark by Putnam, or was it > Davis. The result is: For every recursively enumerable set E of non- > negative integer, there is a polynomial P such that the non-negative > part of the range of P is E. One can arrange for the negative part of > the range to be all negative integers, or to be quite sparse. A set E is recursively enumerable if there is an algorithm that will > output the elements of E, possibly with repetition. Every reasonable > set of non-negative integers is recursively enumerable. In > particular, if you produce in a reasonable way a set of non-negative > integers which does not have a density, or a set which has density d, > that set will turn out to be r.e., and hence the non-negative part of > the range of a polynomial. > For an integer polynomial f, possibly multivariate, let range(f) > denote the range of f for all possible integer inputs, and let > density(f) denote the density of (range(f) intersect N), as a subset > of N. Note -- as discussed in a recent thread, density(f) is not necessarily > the same as density(-f). Of course, density(f), if it exists, is a real number in the interval > [0,1]. Some simple examples ... If f(x) = x^2, then density(f) = 0. If f(x) = k*x, where k is a fixed positive integer, then density(f) = > 1/k. Two questions ... (1) Does every integer polynomial have a density? (2) For an integer polynomial, what densities are possible? Is every > rational number in the interval [0,1] achievable as a density? Can a > density be irrational? do you have a reference for the _range_ of a _polynomial_ being any re set? in particular how does that work for the set {2^n | n e N}? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: density of the range of an integer polynomial >> Not every polynomial has a density in the sense described, and every >> real number in the interval [0,1] is a density. >> This is an easy consequence of the theorem of Matijasevic that settled >> Hilbert's 10th problem, together with a remark by Putnam, or was it >> Davis. The result is: For every recursively enumerable set E of non- >> negative integer, there is a polynomial P such that the non-negative >> part of the range of P is E. One can arrange for the negative part of >> the range to be all negative integers, or to be quite sparse. >> A set E is recursively enumerable if there is an algorithm that will >> output the elements of E, possibly with repetition. Every reasonable >> set of non-negative integers is recursively enumerable. In >> particular, if you produce in a reasonable way a set of non-negative >> integers which does not have a density, or a set which has density d, >> that set will turn out to be r.e., and hence the non-negative part of >> the range of a polynomial. >> For an integer polynomial f, possibly multivariate, let range(f) >> denote the range of f for all possible integer inputs, and let >> density(f) denote the density of (range(f) intersect N), as a subset >> of N. >> Note -- as discussed in a recent thread, density(f) is not necessarily >> the same as density(-f). >> Of course, density(f), if it exists, is a real number in the interval >> [0,1]. >> Some simple examples ... >> If f(x) = x^2, then density(f) = 0. >> If f(x) = k*x, where k is a fixed positive integer, then density(f) = >> 1/k. >> Two questions ... >> (1) Does every integer polynomial have a density? >> (2) For an integer polynomial, what densities are possible? Is every >> rational number in the interval [0,1] achievable as a density? Can a >> density be irrational? do you have a reference > for the _range_ of a _polynomial_ being any re set? in particular > how does that work for the set {2^n | n e N}? Yep -- good challenge to dispel the illusions. quasi === Subject: Re: density of the range of an integer polynomial > do you have a reference > for the _range_ of a _polynomial_ being any re set? http://scidiv.bcc.ctc.edu/Math/Robinson.html === Subject: Re: density of the range of an integer polynomial > do you have a reference >> for the _range_ of a _polynomial_ being any re set? http://scidiv.bcc.ctc.edu/Math/Robinson.html You may think that's a reference, but get real -- there is no nonconstant polynomial whose range (for integer inputs), intersected with N is a subset of {2^k | k is a nonnegative integer}. Moreover, it doesn't take advanced math to prove that. quasi === Subject: Re: density of the range of an integer polynomial do you have a reference > for the _range_ of a _polynomial_ being any re set? >>http://scidiv.bcc.ctc.edu/Math/Robinson.html You may think that's a reference, but get real -- there is no >nonconstant polynomial whose range (for integer inputs), intersected >with N is a subset of {2^k | k is a nonnegative integer}. Moreover, it >doesn't take advanced math to prove that. Correction: There is no integer polynomial whose range (for integer inputs), intersected with N is an infinite subset of {2^k | k is a nonnegative integer}. Moreover, it doesn't take advanced math to prove that. quasi === Subject: Re: density of the range of an integer polynomial >> do you have a reference >> for the _range_ of a _polynomial_ being any re set? http://scidiv.bcc.ctc.edu/Math/Robinson.html >>You may think that's a reference, but get real -- there is no >>nonconstant polynomial whose range (for integer inputs), intersected >>with N is a subset of {2^k | k is a nonnegative integer}. Moreover, it >>doesn't take advanced math to prove that. Correction: There is no integer polynomial whose range (for integer inputs), >intersected with N is an infinite subset of {2^k | k is a nonnegative >integer}. Moreover, it doesn't take advanced math to prove that. quasi To A N Niel: I'm attempting to take back my many false claims in this thread, so my get real challenge above is definitely one I need to apologize for. Sorry. quasi === Subject: Re: density of the range of an integer polynomial <031020071344163191%anniel@nym.alias.net.invalid> <0ln7g3t0hsc9kpput7u7ut0fppqplrb9ib@4ax.com> do you have a reference > for the _range_ of a _polynomial_ being any re set? >http://scidiv.bcc.ctc.edu/Math/Robinson.html You may think that's a reference, but get real -- there is no >nonconstant polynomial whose range (for integer inputs), intersected >with N is a subset of {2^k | k is a nonnegative integer}. Moreover, it >doesn't take advanced math to prove that. Correction: There is no integer polynomial whose range (for integer inputs), > intersected with N is an infinite subset of {2^k | k is a nonnegative > integer}. Moreover, it doesn't take advanced math to prove that. quasi The trick is that the polynomial producing {2^k} is very often negative. === Subject: Re: density of the range of an integer polynomial >> There is no integer polynomial whose range (for integer inputs), >> intersected with N is an infinite subset of {2^k | k is a nonnegative >> integer}. Moreover, it doesn't take advanced math to prove that. The trick is that the polynomial producing {2^k} is very often >negative. So you're saying it can be done? Hmmm ... I would need to see it to believe it. In my mind, I have a proof that it's impossible -- I'll see if it's still a proof when I try to write it down. quasi === Subject: Re: density of the range of an integer polynomial <031020071344163191%anniel@nym.alias.net.invalid> <0ln7g3t0hsc9kpput7u7ut0fppqplrb9ib@4ax.com> Let d be a positive number less than 1. Suppose moreover that there is a recursive function f such that the n-th digit of d after the decimal point is f(n). [We might as well use decimal digits.) Then there is a recursive subset D of the natural numbers with density d. The construction is more or less the natural one, though we have to use a few little tricks to ensure convergence. The natural approach for say 0.314.... is to put into D the first 3 of the first 10 integers, 28 of the next 90, then 283 of the next 900, and so on. This doesn't quite work, since there may be too much bouncing. The bouncing can be fixed by never going above the target d. Bouncing can also be fixed by stabilizing attained approximations (by repetition) before pushing on to the next digit. In particular, 1/sqrt{2} is certainly achievable. This means that there is a polynomial P the positive part of whose range has density 1/ sqrt{2}. But the there is does not say much. We can say a little more, but still not much, by replacing there is a polynomial by a polynomial can be constructed. However, what would come out of the construction is a polynomial with no natural structural connection to 1/sqrt{2}. It could still be interesting to construct natural polynomials with certain target densities. And since the range of a polynomial trick relies heavily on burying one's mistakes among the negatives, one could ask about densities in the case of positive polynomials. === Subject: Re: density of the range of an integer polynomial Let d be a positive number less than 1. Suppose moreover that there >is a recursive function f such that the n-th digit of d after the >decimal point is f(n). [We might as well use decimal digits.) Ok, so d is not just any element of [0,1]. I missed your later qualification on my first reading. Since there are only countably many recursive functions (right?), there are only countably many such values of d, which nullifies my prior objection based on comparing cardinalities. At the same time, you are no longer claiming that every element of [0,1] is a possible density, right? Would it be correct to say every such d is a constructible number, and vice-versa? >Then there is a recursive subset D of the natural numbers >with density d. The construction is more or less the natural one, though we have to >use a few little tricks to ensure convergence. The natural approach >for say 0.314.... is to put into D the first 3 of the first 10 >integers, 28 of the next 90, then 283 of the next 900, and so on. >This doesn't quite work, since there may be too much bouncing. The >bouncing can be fixed by never going above the target d. Bouncing can >also be fixed by stabilizing attained approximations (by repetition) >before pushing on to the next digit. In particular, 1/sqrt{2} is certainly achievable. But if your set of rational convergents is the range of a recursive function, that only implies the set is recursively enumerable (when normalized to integer values by using numerator, denominator pairs). In other words, just because you are using a recursive function to build a set, that doesn't automatically imply that the set is recursive. It only implies that the set is recursively enumerable. However the positive part of the range of an integer polynomial is always recursive, no? >This means that there is a polynomial P the positive part of whose >range has density 1/sqrt{2}. Sorry, I don't see that. If all you have is recursive enumerability, that seems too weak. quasi === Subject: Re: density of the range of an integer polynomial Firstly, I apologize if I critiqued your earlier replies unfairly. It now looks like some of my objections were wrong (but not all). >Let d be a positive number less than 1. Suppose moreover that there >is a recursive function f such that the n-th digit of d after the >decimal point is f(n). [We might as well use decimal digits.) Then >there is a recursive subset D of the natural numbers with density d. The construction is more or less the natural one, though we have to >use a few little tricks to ensure convergence. The natural approach >for say 0.314.... is to put into D the first 3 of the first 10 >integers, 28 of the next 90, then 283 of the next 900, and so on. >This doesn't quite work, since there may be too much bouncing. The >bouncing can be fixed by never going above the target d. Bouncing can >also be fixed by stabilizing attained approximations (by repetition) >before pushing on to the next digit. In particular, 1/sqrt{2} is certainly achievable. I have no problem with constructing a recursively enumerable set with a given density. I do have a problem if you assert that the same can be done using a recursive set. After all, there are only countably many recursive sets, right? I think that's right, but please correct me if I'm wrong about that. In any case, there are certainly only countably many integer polynomials, so for the positive part of the range of an integer polynomial, it's clear that not every density is possible. >This means that there is a polynomial P the positive part of whose >range has density 1/sqrt{2}. Are you saying that every recursive set is the positive part of the range of an integer polynomial? While I admit that I'm not that familiar with the relevant theory, that seems like too strong a claim. >But the there is does not say much. We can say a little more, but >still not much, by replacing there is a polynomial by a polynomial >can be constructed. However, what would come out of the construction >is a polynomial with no natural structural connection to 1/sqrt{2}. It could still be interesting to construct natural polynomials with >certain target densities. And since the range of a polynomial trick >relies heavily on burying one's mistakes among the negatives, one >could ask about densities in the case of positive polynomials. As a fall back, I certainly intend to ask that, or something along those lines. quasi === Subject: Re: density of the range of an integer polynomial > > > Firstly, I apologize if I critiqued your earlier replies unfairly. > > It now looks like some of my objections were wrong (but not all). > >Let d be a positive number less than 1. Suppose moreover that there >is a recursive function f such that the n-th digit of d after the >decimal point is f(n). [We might as well use decimal digits.) Then >there is a recursive subset D of the natural numbers with density d. The construction is more or less the natural one, though we have to >use a few little tricks to ensure convergence. The natural approach >for say 0.314.... is to put into D the first 3 of the first 10 >integers, 28 of the next 90, then 283 of the next 900, and so on. >This doesn't quite work, since there may be too much bouncing. The >bouncing can be fixed by never going above the target d. Bouncing can >also be fixed by stabilizing attained approximations (by repetition) >before pushing on to the next digit. In particular, 1/sqrt{2} is certainly achievable. > > I have no problem with constructing a recursively enumerable set with > a given density. I do have a problem if you assert that the same can > be done using a recursive set. After all, there are only countably > many recursive sets, right? I think that's right, but please correct > me if I'm wrong about that. > > In any case, there are certainly only countably many integer > polynomials, so for the positive part of the range of an integer > polynomial, it's clear that not every density is possible. > >This means that there is a polynomial P the positive part of whose >range has density 1/sqrt{2}. > > Are you saying that every recursive set is the positive part of the > range of an integer polynomial? While I admit that I'm not that > familiar with the relevant theory, that seems like too strong a claim. That's what Matiyasevich showed: see e.g. positive integers is Diophantine iff there is a polynomial with integer coefficients whose set of positive values is that set. That part's easy: if S = {x: there exist positive integers y_1,...,y_m such that Q(x,y_1,...,y_m) = 0} then take P(x,z_1,...,z_{4m}) = x (1 - Q^2(x, z_1^2 +...+z_4^2, ..., z_{4m-3}^2 +...+z_{4m}^2)) A set is recursively enumerable iff it is Diophantine. That's equivalent to Matiyasevich's theorem. Recursive sets are recursively enumerable (but the converse is not true: that's how to get undecidable statements). -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: density of the range of an integer polynomial > >> >> Firstly, I apologize if I critiqued your earlier replies unfairly. >> >> It now looks like some of my objections were wrong (but not all). >> >>Let d be a positive number less than 1. Suppose moreover that there >>is a recursive function f such that the n-th digit of d after the >>decimal point is f(n). [We might as well use decimal digits.) Then >>there is a recursive subset D of the natural numbers with density d. >>The construction is more or less the natural one, though we have to >>use a few little tricks to ensure convergence. The natural approach >>for say 0.314.... is to put into D the first 3 of the first 10 >>integers, 28 of the next 90, then 283 of the next 900, and so on. >>This doesn't quite work, since there may be too much bouncing. The >>bouncing can be fixed by never going above the target d. Bouncing can >>also be fixed by stabilizing attained approximations (by repetition) >>before pushing on to the next digit. >>In particular, 1/sqrt{2} is certainly achievable. >> >> I have no problem with constructing a recursively enumerable set with >> a given density. I do have a problem if you assert that the same can >> be done using a recursive set. After all, there are only countably >> many recursive sets, right? I think that's right, but please correct >> me if I'm wrong about that. >> >> In any case, there are certainly only countably many integer >> polynomials, so for the positive part of the range of an integer >> polynomial, it's clear that not every density is possible. >> >>This means that there is a polynomial P the positive part of whose >>range has density 1/sqrt{2}. >> >> Are you saying that every recursive set is the positive part of the >> range of an integer polynomial? While I admit that I'm not that >> familiar with the relevant theory, that seems like too strong a claim. That's what Matiyasevich showed: see e.g. coefficients whose set of positive values is that set. That part's >easy: if >S = {x: there exist positive integers y_1,...,y_m such that >Q(x,y_1,...,y_m) = 0} >then take P(x,z_1,...,z_{4m}) = > x (1 - Q^2(x, z_1^2 +...+z_4^2, ..., z_{4m-3}^2 +...+z_{4m}^2)) A set is recursively enumerable iff it is Diophantine. That's equivalent to >Matiyasevich's theorem. Recursive sets are recursively enumerable >(but the converse is not true: that's how to get undecidable statements). Ok, now I see the light. The trick hinges on the restriction to the positive part of the range. While it seems like an innocent restriction, in reality it's the hidden key. quasi === Subject: Re: density of the range of an integer polynomial > There is no integer polynomial whose range (for integer inputs), > intersected with N is an infinite subset of {2^k | k is a nonnegative > integer}. Moreover, it doesn't take advanced math to prove that. >>The trick is that the polynomial producing {2^k} is very often >>negative. So you're saying it can be done? Hmmm ... I would need to see it to believe it. In my mind, I have a proof that it's impossible -- I'll see if it's >still a proof when I try to write it down. Hmmm ... My first attempt at writing down what I thought was an immediate proof doesn't go through. The fact that it's only a partial range seems to be the key obstacle. Thus, based on your suggestion, if instead we try to find an example (or prove that one exists) we need enough variables to allow successive interpolation to positive powers of 2, followed by a retreat to negative outputs to recuperate (or so as not to get stuck in an exponential mode). You know, I was so absolutely sure that the powers of 2 are unrepresentable -- if I was wrong about that, I apologize for being so insistent. Anyway, at this point, you have me doubting my earlier claims. On the other hand, if we restrict to _nonnegative_ polynomials, my proof goes through -- that is, there does not exist an integer polynomial which is nonnegative for all real inputs and whose range for integer inputs is an infinite subset of {2^k | k is a nonnegative integer}. In fact, I think we can relax the above restriction, requiring only that for some variable for which the polynomial has positive degree, the leading coefficient with respect to that variable assumes at least one positive value (for integer inputs). quasi === Subject: Re: density of the range of an integer polynomial > For an integer polynomial f, possibly multivariate, let range(f) > denote the range of f for all possible integer inputs, and let > density(f) denote the density of (range(f) intersect N), as a subset > of N. > > Note -- as discussed in a recent thread, density(f) is not necessarily > the same as density(-f). > > Of course, density(f), if it exists, is a real number in the interval > [0,1]. > > Some simple examples ... > > If f(x) = x^2, then density(f) = 0. > > If f(x) = k*x, where k is a fixed positive integer, then density(f) = > 1/k. > > Two questions ... > > (1) Does every integer polynomial have a density? > > (2) For an integer polynomial, what densities are possible? Is every > rational number in the interval [0,1] achievable as a density? Can a > density be irrational? For density m/n, try f(x,y) = x + n y + (x^9 + y^9) prod_{j=1}^m (x - j). For the irrational density 1 - 6/pi^2, try f(v,w,x,y,z) = v (w^4 + x^4 + y^4 + z^4 + 2)^2 -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: AP is hung up on base ten: Was: Re: #16 Chapter exponentiation of 3^...3333 fresh look at e^(pi)(i) = (-1) ; new textbook; Mathematical-Physics (p-adic primer) for students of age 6 onwards although I didn't see the movie, I somehow had the realization, just a little while of a time ago, that p-adics are like a box of chocolates ... inside or outside, shape, time-stamp et c. et c. et cetera. > since then is based on the (false) assumption that pi*i is real. He --Fox promotes Cheeny Admin. war with Iran: http://larouchepub.com/other/2007/3437fox_poop_cheney.html --14 Italian Senators Call for Cheney Impeachment Aug. 1, 2007 (EIRNS)- The Lyndon LaRouche Political Action Committee (LPAC) issued the following release today. Fourteen members of the Italian Senate have signed a call to the Members of Congress to support Rep. Kucinich's House Resolution 333 for the Impeachment of Dick Cheney. http://larouchepub.com/pr/2007/070801italian_senators_call.html === Subject: Re: Tetrads and Spin Connections as Yang-Mills Gauge Potentials a-hem; speaking for the other 6.5-billion hominids around here, if I may: where's that Monolith, when y'need it; at least, I could beat my head against it, to stop the sigint from MC-infra! Counting Down from the '60s with Jack in the Dreamstalk! did anyone ever review one of your touchy-feely-new-age textbooks on the theory of nothing -- could we have a link to the journal o'spooks & more spooks, Joe Newage & the Roswell Priesthoodies? did Lt.Col Corso actually file that first story with the agent-not-really-in-charge at the paper, or What?... may he rest in peace, in spite of every thing! surely, you do not go so far as to vouchsafe with the paradigm, that all of out techniques were retrieved & reverse-engineered by nascent military-post-industrial-infotainment companies, that which Eisenhower warned against in the full paragraph, from taht little pile of crap at the B-52 base, or did you already fall into that? don't struggle in the quicksand! > exterior covariant derivative as D = d + S/ Suppressing indices for simplicity. This is analogous to a Yang-Mills > theory where the curvature two form field is R = DS i.e. curvature field 2-form = exterior covariant derivative of the spin > connection Yang-Mills potential with itself, i.e. in 1916 GR thus: the buckyfullerene challenge: uhm, carbondating? so, not a textbook: A.A.Michelson's book on interferometry. a textbook: hey, I'm not going to promote a textbook, just because it didn't suck as hard as possible, but it is in a popular series, available at (y'know) bookstores; one of their Very Important Subjects line, not Objects! thus: I'll apologize for Tttteslaaaa, just as soon as I ZZZZAPPPP you all into a temporary metastable state (and you configure how to take it back from thereat .-) so, did anyone find a workable integer value about that, somewhat inconclusive (if not fully multivalued). > and matched well the calculations inhttp://users.aol.com/atrupp/loschmid.htm you might also be interested in actual measurements on such columns > by dr roderich graeff > also validating loschmidt's claims thus: like, the first rainy day of the season, the hard part is folding them back up, thus: p-adics are like a box of chocolates ... inside or outside, shape, time-stamp, partitioning, pigeonholing et c. et c. et cetera. --14 Italian Senators Call for Cheney Impeachment Aug. 1, 2007 (EIRNS)- The Lyndon LaRouche Political Action Committee (LPAC) issued the following release today. Fourteen members of the Italian Senate have signed a call to the Members of Congress to support Rep. Kucinich's House Resolution 333 for the Impeachment of Dick Cheney. http://larouchepub.com/pr/2007/070801italian_senators_call.html === Subject: Re: Conjecture <3rj2g3ld4p65crc2opkru32tdcttphtjhi@4ax.com> <9b65g3l97v6kq5ketuecki49q5ml7p25qd@4ax.com >P(z) > R(z) for all prime N > 2. > Forgive my ignorance but ... > P and R are what functions? > The variable z is restricted to what domain? > How does N relate to P and R? > quasi ? tommy1729 has already indicated that the original message was not his. quasi- Hide quoted text - - Show quoted text - whatever === Subject: Re: Conjecture Nntp-Posting-Host: hera.cwi.nl ... > tommy1729 has already indicated that the original message was not his. > > quasi > > whatever Indeed. It is someone posting through aioe.org who is posting also quite a bit of abuse in this newsgroup. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Conjecture <9b65g3l97v6kq5ketuecki49q5ml7p25qd@4ax.com> ... > > tommy1729 has already indicated that the original message was not his. > > quasi > > whatever Indeed. It is someone posting through aioe.org who is posting also quite > a bit of abuse in this newsgroup. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland;http://www.cwi.nl/~dik/ It ain't me. === Subject: nike cheap sell wholesale shoes jordan shox air max force one gucci prada It is our pleasure to meet you here. we are ale International Trade Co.,Ltd in Fujian of China. our website: http://www.cheapest-sell.cn our email:cheapests...@hotmail.com We are professional and honest wholesaler of all kinds of brand sneaks and apparel.the products our company supply are as follows: 1.jordan1-21shoes 3.airforce1(airforce1/bape/dunk...)shoes 4.Shoes(R3/R4/NZ/turbo/t13/monster...)shoes 5.Shoes/Shoes/Shoes/rift/Sports shoes 6.fashion branded hoodies/jeans/caps/handbag/t-shirts and others 7.because the space of the website is limited,we can also supply many other products which be not showed out in our site. if you have the photos of the products you need , we are pleasure to supply for your orders. 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Website : http://www.cheapest-sell.cn our email : cheapestsell@yahoo.com.cn our msn : cheapestsell@hotmail.com === === === Subject: Re: Normal from point to line? > How do I calculate the normal fra a known point to a known line? Any reference to such basic geometry math on the web would be appreciated. > If the point p lies on the line L, then the normal is the plane perpendicular to L and passing through p. If p doesn't lie on L, then let P be the plane determined by p and L. The normal is the line lying on P, perpendicular to L and passing through p. To construct this line, draw a circle with center at p and radius large enough that L intersects the circle in two points, A and B. The perpendicular bisector of AB is the normal. === Subject: Re: Normal from point to line? > How do I calculate the normal fra a known point to a known line? > > Any reference to such basic geometry math on the web would be appreciated. The distance from point (u,v) to line a*x + b*y + c = 0 along the normal to the line is |a*u + b*c + c|/sqrt(a^2 + b^2). A vector normal to that line is [a,b]. === Subject: Stupidity, and Partitions and the Pentagonal Number Theorem <47025bab$0$26382$88260bb3@free.teranews.com> P(z) > R(z) for all prime N > 2. > Clearly, Dr Physics is just trying to incite a flame war. to neilist: If Dr Physics is actually you, then you are breaking a > truce. to tommy1729: If possible, don't respond to the provocation. If despite my efforts, all-out war breaks out, I want no part of it. quasi I deny being Dr. Physics, although I can understand your suspicion. And Dr. Physics' recent and sparce appearances on sci.math could stoke some more suspicion. No, I certainly wouldn't use the term dearest. And there is no truce. I'm doing what I said I would. I am pondering posting about the pentagonal number theorem in connection with partitions since, in only the last month, I read about So don't involve me. === Subject: Evidence that Finite Guy is full of <46fbdce3$0$32541$4c368faf@roadrunner.com> <46fd86cd$0$26363$4c368faf@roadrunner.com> > It converges, since near pi/6 the function 1 - 2sin(x) behaves like >> -sqrt(3)*(x - pi/6) (compute a few terms of its Taylor series). Yes, thank you very much. > Maybe, by mean value theorem, 1-2sin x >= sqrt(3).(pi/6 - x). and I had a question for this integral. > I showed that > int x^n / sqrt(a+bx) dx > = [2(x^n).sqrt(a+bx) / {b(2n+1)}] - (2an)/(b(2n+1)) int [(x^(n-1)) / > sqrt(a+bx)] dx What's (2n+1) of denominator ? How make that out ? Oh, stupid... I can make out this form. === Subject: Re: Rational numbers, irrational numbers: each dense in real numbers I'm under the impression that a transfinite recursion schema is the > provision of class functions for the ordinal zero, (other) limit > ordinals, and successors, as ordinals, as is done above. Each transfinite recusrion schema is proven and use of transfinite > recursion requires using a schema. If the schema in use is not > obvious, then you need to mention which schema you intend. And nothing > you've mentioned so far is an obvious schema (not to me at least). I > can't verify the correctness of a purported proof that purports to use > transfinite recursion unless I know which specific transfinite > recursion schema is purported to be deployed. Any work I would do to > figure out how to fit your argument into a particular transfinite > recursion schema is work that you might as well perform yourself so > that I, as the reader, could easily follow your argument. So, for > about the fourth time, it ain't my job to try to figure out how your > argument might be put into a correct transfinite recursion schema. > After this post, if you continue to argue as if by transfinite > recursion but don't provide a schema, then I'm not going to waste any > more of my time on this. Let there be a family of choice functions F_(x,y) for each x,y E (0,1) > such that each C_(x,y) E F_(x,y) returns an irrational number p such > that y < p < x and p is irrational. Now you're taking yet another formulation. I started trying to figure out what you might mean, but later down in > your post I found that I need to ask you whether by '(0,1)' you mean > the real inveral (0 1) or whether you mean the set {0 1}. Let me know. Then I'll see as to the rest. MoeBlee That's not really a new formulation, it's basically the same thing as selecting any irrational between two real numbers. In the above (0,1) indicates the open unit interval of the (standard) real numbers. If I meant the set containing only zero and one I would write {0,1}, with the braces instead of parentheses. In the previous pairs in parentheses indicate ordered pairs, not open interval endpoints. More informally, the transfinite recursion theorem tells us that to define a function on a well-ordered set < W,< >, it suffices to give a definition of each function value F(t) in terms of all the previous function values F(x) where x< t. -- http://www.mscs.dal.ca/~selinger/courses/582W99/handouts/ordinals.pdf Those notes indicate that Herb Enderton's text would indicate requirements of a proof by transfinite recursion schema, besides indicating that class functions as illustrated above are sufficient. To some extent I wonder why you seem set against the use of transfinite induction, vis-a-vis recursion schema. I used transfinite induction first, then for your satisfaction I would also like to use transfinite recursion schema, for induction's sake. I'm content using transfinite induction, and, transfinite induction is a generally accepted method of proof for each case of an index set that has a well-ordering by that ordering. http://eom.springer.de/t/t093700.htm In the provision of class functions G_1, G_2, G_3, for zero, successor, and limit ordinals, that seems an adequate outlay of machinery for transfinite induction and transfinite recursion. So, where the transfinite induction schema is to define, here each p_i, it is done as p_i = F(i) = G( {F(j), j < i} ). Aatu defines a transfinite recursion scheme: If G is a definable class function, there is a unique definable class function F such that, provably in ZFC, F(alpha) = G(F restricted to alpha) for all . Then, a transfinite recursion scheme is illustrated above. If you have something else in mind, please indicate an appropriate template. MoeBlee, I think definition of the recursive class functions suffices. Ross -- Finlayson Consulting === Subject: Re: Rational numbers, irrational numbers: each dense in real numbers >Let there be a family of choice functions F_(x,y) for each x,y E (0,1) >such that each C_(x,y) E F_(x,y) returns an irrational number p such >that y < p < x and p is irrational. Now you're taking yet another formulation. I started trying to figure out what you might mean, but later down in > your post I found that I need to ask you whether by '(0,1)' you mean > the real inveral (0 1) or whether you mean the set {0 1}. > In the above (0,1) indicates the open unit interval of the (standard) > real numbers. Okay. So when you write, for each x,y E (0,1) such that each C_(x,y) [...], I take it that you mean 'for each x and y such 0 =< x < y =<1 such that C((x y)) [...]'. In other words, C is a function whose domain is the set of open intervals that are subsets of (0 1). Is that correct? > More informally, the transfinite recursion theorem tells us that to > define a function on a well-ordered set < W,< >, it suffices to give a > definition of each function value F(t) in terms of all the previous > function values F(x) where x< t. --http://www.mscs.dal.ca/~selinger/courses/582W99/handouts/ordinals.pdf Indeed that is quite informal. The exact manner in which F(t) is defined in terms of function values F(x) where x Those notes indicate that Herb Enderton's text would indicate > requirements of a proof by transfinite recursion schema, besides > indicating that class functions as illustrated above are sufficient. Do you have Enderton's set theory book? If you do, then look in it and you will see exactly what I'm talking about. > To some extent I wonder why you seem set against the use of > transfinite induction, vis-a-vis recursion schema. Transfinite induction and transfinite recursion are very closely related, but they are different things. And I haven't said anything about being against any correct application of either of them. > I used > transfinite induction first, then for your satisfaction I would also > like to use transfinite recursion schema, for induction's sake. I'm > content using transfinite induction, and, transfinite induction is a > generally accepted method of proof for each case of an index set that > has a well-ordering by that ordering. You're very confused about the entire business. Though, below you do make some progress: > http://eom.springer.de/t/t093700.htm In the provision of class functions G_1, G_2, G_3, for zero, > successor, and limit ordinals, that seems an adequate outlay of > machinery for transfinite induction and transfinite recursion. So, > where the transfinite induction schema is to define, here each p_i, it > is done as p_i = F(i) = G( {F(j), j < i} ). Ah, now we're getting closer. What you probably mean is: For each ieX, F(i) = G({F(j) | j In the provision of class functions G_1, G_2, G_3, for zero, > successor, and limit ordinals, that seems an adequate outlay of > machinery for transfinite induction and transfinite recursion. So, > where the transfinite induction schema is to define, here each p_i, it > is done as p_i = F(i) = G( {F(j), j < i} ). Ah, now we're getting closer. What you probably mean is: For each ieX, F(i) = G({F(j) | j transfinite recursion (as long as we also have a correct definition of > G). Now tell me precisely what G is. MoeBlee I did already, G_1, G_2, and G_3 as above, for the zero ordinal, successor ordinals, and limit ordinals. So, then there is a well- ordered and more than countable subset of the irrationals P = {p_i, i E X, |X| > |N|}. Then, for each p_i it is shown, via another transfinite induction, that there is a distinct rational q_i, which readily leads to a contradiction if the rationals are countable, yet, if there's not a distinct q_i for each p_i, that as well readily leads to a contradiction if the rationals are dense in the reals. Now, it is well-known that the existence of uncountably many non-empty and non-degenerate (single point) partitions of the unit interval into subintervals would be contradictory to ZFC, inconsistent with ZFC and standard definitions of the real numbers. While that may be so, via the transfinite recursion schema as described above, the class functions are defined up to ordinals equivalent to the irrationals. Were they not defined, up to uncountably many ordinals, then there wouldn't be uncountably many irrationals in some interval of the reals. Ross -- Finlayson Consulting === Subject: Mathematics o Cartan's tetrads in Einstein's gravity 1 #1 OK -- then does A^a -> A^a - X^a under GCTs? Or under stretch-squeeze deformations of the manifold? Critical point. A real physical deformation always comes from either some dynamical change in the Tuv tensor. I am not talking about that at all. We are only talking about changes in POV or perspective of locally coincident observers in arbitrary relative motion, some feel g-forces, some do not - for a fixed geometrodynamical field configuration. Their momentary separation is small compared to the local radii of curvature. That's what locally coincident means operationally empirically in sense of Einstein's gedankenexperiment. The 2 always go together to keep e^a invariant - same as ihd/dx^u - (e/c)Au gauge invariant derivative on charged source fields What do this mean? Keep the components fixed? Or that the e^a transform as vectors under GCTs? Remember Einstein's basic physical local frame invariant is on some gauge orbit (local coincidence) physical event P (defined by the web of relations of geometrodynamic field to EM field etc as in Rovelli Ch 2) ds^2 = e^aea = (Minkowski LIF)abe^a^eb = (Curvilinear LNIF)uvdx^udx^v = guv(LNIF)dx^udx^v GCT's act on the u-indices. e^a = e^audx^u e^au is a 1st rank GCT covariant tensor therefore e^a is a GCT invariant scalar zero rank GCT tensor but it is a single Lorentz group vector (contravariant first rank Lorentz tensor). e^a = I^a(Minkowski) + A^a(non-inertial) such that A^a = 0 in both GIFs and LIFs, but not in GNIFs nor LNIFs (curvature does not matter here) under a GCT on the u-indices A^a -> A'^a = A^a + X^a I^a -> I'^a = I^a - X^a such that e^a -> e'^a = e^a X^a = 0 for 1905 SR GIF -> GIF' but X^a =/= 0 for GIF -> GNIF' - this is the beginning of 1916 GR, but still with zero curvature Cartan 2-form field R^a^b = 0 I^a is a 1905 SR affine invariant but it is not a non-affine invariant i.e. including GNIF's as in the Galilean relativity non-affine GIF to GNIF' transformation t -> t' z -> z' = z - (1/2)gt^2 gt/c << 1 I^a is the geometrodynamic field analog of ih(d/dx^a) in D^a = ihd/dx^a - (e/c)A^a in U(1) EM this U(1) gauge covariant derivative on a the charged complex number Psi field gives under a gauge transformation with phase chi D^aPsi -> D^aPsi' = e^ichiD^aPsi That is minimal coupling (equivalence principle) gives Pa -> Pu = e^auPa where the Pa generate the globally flat translation group T4, so Pu is the 4-momentum in the off-geodesic u-space using curvilinear guv metric, technically the tangent/co-tangent spaces depending on position of u lower or upper. LIFs are approximately geodesic a-spaces using constant Minkowski metric - I mean their origins in a region small compared to the radii of local curvature, i.e. 1905 SR works locally to a good approximation. Note that the local curvature tensor Rabcd =/= 0 in a LIF if Ruvwl =/= 0 in a coincident LNIF. This is basically from John A. Wheeler. So A^a (and also X^a - you were right) are basically the off-geodesic g-force inertial fields in non-inertial frames which for translational motions always need a non-gravity force to maintain them. For example in the text book SSS vacuum solution g00 = - 1/grr = 1 - rs/r rs/r < 1 area of closed 2D spherical surface surrounding point source = 4pir^2 is only for special static or shell (Wheeler) LNIF off-geodesic observers at fixed r without any circulating orbital angular momentum. I understand why basis vectors are coordinate-invariant under LLTs, but why should they be coordinate-invariant under GCTs? ? e^a transform as e^a' = [SO(1,3]^a'ae^a and each e^a is a GCT invariant (scalar). in a sense ds^a = e^a ds^2 = ds^adsa ds connects P with P' where their separation is small compared to local radii of curvature. In 1905 SR the radii of curvature are infinite. In our universe - the constant large-scale de Sitter radius of curvature is (Lambda)^-1/2 where the positive dark zero point energy density is (c^4/8piG)(Lambda) ~ hc/NLp^4 ~ 10^-29 gm/cc where Lp^2 = hG/c^3 ~ 10^-66 cm^2 & N ~ 10^122 BITS of Shannon classical (c-bit) information, i.e. Wheeler's IT FROM BIT from the future infinity (horizon) acting retro-causally in Fred Hoyle's teleology (Intelligent Universe, 1983) Lambda = Einstein's cosmological constant http://qedcorp.com/APS/ureye.gif The future de Sitter horizon is the world hologram and our 3D space volume without volume is its retrocausal hologram image with exactly N volume quanta in 1-1 correspondence with the N future area quanta. That is &L ~ (Lp^2L)^1/3 = size of quantum foam virtual bubble from Wigner L ~ N^1/2Lp world hologram from t'Hooft &L ~ N^1/6Lp ~ 10^-13 cm for cosmological L Note that the world hologram Hubble scale is N^1/2Lp ~ 10^61 10^-33 ~ 10^28 cm, the geometric mean between smallest and largest lengths in the universe is simply N^1/4Lp ~ (10^28 10^-33)^1/2 ~ 10^-2-10^-3 cm and the dark energy density is ~ hc/(geometric mean)^4. Note also that c/(geometric mean) ~ 10^12 Hertz. If A^a -> A^a - X^a under *deformations* (holding the local coordinates invariant) then the -X^a term would directly correspond to the first-order effect of the actual gravitational field in the 1915 theory. appearing under the action of the non-linear GCTs. No not for physical deformations, only for the non-physical GCT changes in local coordinates - on same gauge orbit. Analogy is to U(1) EM gauge transformations APsi -> (A + grad Chi)Psi igrad(e^iChi Psi) = [-(gradChi)Psi + igradPsi]e^iChi OK. *I suppose you can say that all the different points p on the same GCT gauge orbit P represent all possible subjective POVs of the infinity of possible locally coincident observers looking at the same physical P. The p's are the Shadows and P is the Platonic Idea (Allegory of the Cave, The Republic) Observables are GCT invariant. If, on the other hand, you get A^a -> A^a - X^a under GCTs alone, without deformation of the manifold, then there is nothing I can see in this -X^a term that corresponds to the actual non-tidal field of the 1915 theory. That's correct. X^a is simply a non-physical gauge artifact that is deformed by changes in the energy density etc of the source fields. OK. But at the same time, the components e^a do nonetheless change under GCTs, according to I^a -> I^a + A^a? Even in a flat spacetime? What happens is that e^a is GCT invariant, however, in general if you jump from a geodesic to a non-geodesic intersecting it at same P I^a -> I'^a = I^a + X^a 0 -> A'^a = - X^a e^a -> e'a = e^a === Subject: Who were the 25 Greatest Mathematicians? Who were the 25 Greatest Mathematicians? I'm sure there'd be much controversy over even the definition of great but I've tried to construct a Greatest list: http://james.fabpedigree.com/mathmen.htm To qualify, the person's work must have depth, breadth and historical importance. I'm not qualified to make such a list, but have tried to read opinions of those who are better qualified. I hope to hear more helpful criticism. I hope you'll report errors and omissions in my brief bios, and consider my explanations, but here is a summary of the list: 1. Johann Carl F. Gauss (1777-1855) 2. Archimedes of Syracuse (287-212 BC) 3. Isaac Newton (1642-1727) 4. Leonhard Euler (1707-1783) 5. Euclid of Alexandria (ca 322-275 BC) 6. Georg F. Bernhard Riemann (1826-1866) 7. Gottfried Wilhelm Leibniz (1646-1716) 8. Joseph-Louis Lagrange (1736-1813) 9. Jules Henri Poincare (1854-1912) 10. Pierre de Fermat (1601-1665) 11. Srinivasa Ramanujan (1887-1920) 12. Niels Henrik Abel (1802-1829) 13. David Hilbert (1862-1943) 14. Brahmagupta `Bhillamalacarya' (589-668) 15. Georg Cantor (1845-1918) 16. Leonardo `Fibonacci' Pisano (ca 1170-1245) 17. Carl G. J. Jacobi (1804-1851) 18. Evariste Galois (1811-1832) 19. Rene Descartes (1596-1650) 20. John von Neumann (1903-1957) 21. Augustin-Louis Cauchy (1789-1857) 22. Karl Wilhelm Theodor Weierstrass (1815-1897) 23. Blaise Pascal (1623-1662) 24. Arthur Cayley (1821-1895) 25. Aryabhatta (476-550) James Dow Allen === Subject: Re: Who were the 25 Greatest Mathematicians? A list of among the most famous (with a bias toward logic and set theory): 580-500 BC Pythagoras [ca] 405-355 BC Eudoxos 300-? BC Euclid [ca] 287-212 BC Archimedes 276-194 BC Eratosthenes [ca] 250-? Diophantos 1170-1250 Fibonacci [ca] 1501-1576 Cardano 1550-1617 Napier 1588-1648 Mersenne 1591-1661 Desargues 1596-1650 Descartes 1601-1655 Fermat 1623-1662 Pascal 1643-1727 Newton 1646-1716 Leibniz 1654-1705 J Bernoulli 1661-1704 L'Hopital 1667-1733 Saccheri 1667-1748 J Bernoulli 1667-1754 D'Moivre 1687-1759 N Bernoulli 1690-1764 Goldbach 1695-1726 N Bernoulli 1700-1782 D Bernoulli 1702-1761 Bayes 1707-1783 Euler 1710-1790 J Bernoulli 1717-1783 D'Alembert 1728-1777 Lambert 1736-1813 Lagrange 1749-1827 Laplace 1752-1833 Legendre 1759-1859 J Bernoulli 1768-1830 Fourier 1777-1855 Gauss 1781-1848 Bolzano 1789-1857 Cauchy 1790-1868 Mobius 1792-1856 Lobachevski 1802-1829 Abel 1802-1860 Bolyai 1804-1851 Jacobi 1804-1865 Hamilton 1805-1859 Dirichlet 1806-1871 DeMorgan 1809-1882 Liouville 1810-1893 Kummer 1811-1832 Galois 1815-1864 Boole 1815-1897 Weierstrass 1819-1903 Stokes 1821-1894 Chebyshev 1821-1895 Cayley 1822-1901 Hermite 1823-1891 Kronecker 1826-1866 Riemann 1831-1916 Dedekind 1834-1923 Venn 1838-1922 Jordan 1841-1902 Schroder 1842-1899 Lie 1845-1918 Cantor 1846-1927 Mittag-Leffler 1848-1925 Frege 1849-1913 Konig 1849-1917 F roebnius 1849-1924 Klein 1854-1912 Poincare 1856-1922 Markov 1858-1932 Peano 1859-1919 Hurwitz 1861-1931 Burali-Forti 1861-1947 Whitehead 1862-1943 Hilbert 1864-1909 Minkowski 1865-1963 Hadamard 1868-1937 Padoa 1868-1942 Hausdorff 1869-1951 Cartan 1871-1951 Zermelo 1871-1956 Borel 1872-1970 Russell 1874-1932 Baire 1874-1943 Hartogs 1874-1952 Huntington 1875-1932 Vitali 1875-1941 Lebesgue 1878-1956 Bernstein 1878-1956 Lukasiewicz 1878-1957 Lowenheim 1878-1973 Frechet 1879-1921 Jourdain 1880-1960 Veblen 1881-1966 Brouwer 1882-1935 Noether 1882-1964 Sheffer 1882-1969 Sierpinski 1883-1950 Luzin 1884-1944 G.D. Birkhoff 1885-1955 Weyl 1886-1939 Lesniewski 1887-1920 Ramanujan 1887-1963 Skolem 1888-1972 Courant 1888-1977 Bernays 1891-1965 Fraenkel 1891-1983 Vinogradov 1892-1945 Banach 1893-1924 Nicod 1894-1919 Suslin 1894-1964 Wiener 1895-1965 Rado 1896-1962 Ackermann 1896-1980 Kuratowski 1897-1940 Glivenko 1897-1954 Post 1898-1962 Artin 1898-1979 Hasse 1898-1980 Heyting 1900-1982 Curry 1902-1983 Tarski 1902-1984 Dirac 1903-1930 Ramsey 1903-1957 Von Neumann 1903-1979 Markov 1903-1987 Kolmogorov 1903-1989 Stone 1903-1995 Church 1904-1941 Lindenbaum 1904-1943 Presburger 1905-1976 Kalmar 1906-1976 Godel 1906-1992 Dieudonne 1906-1993 Tychonoff 1906-1993 Zorn 1907-1989 Rosser 1908-1931 Herbrand 1908-2000 Quine 1909-1945 Gentzen 1909-1967 Malcev 1909-1984 Ulam 1909-1994 Kleene 1909-2005 Mac Lane 1911-1995 R Robinson 1911-1996 G. Birkhoff 1912-1954 Turing 1912-1985 Goodstein 1913-1943 Teichmuller 1913-1975 Mostowski 1913-1996 Erdos 1913-1998 Eilenberg 1915-1975 Bar-Hillel 1916-2001 Shannon 1916 Halmos 1917-1988 Lyndon 1918-1974 A Robinson 1918-1978 Wang 1918-2001 Horn 1919 Smullyan 1919-1985 J Robinson 1920 Fraisse 1920-1998 Los 1923 Kreisel 1924 Mandelbrot 1926 Davis 1926 Ryll-Nardzewski 1927-1958 Taniyama 1928-1983 Bishop 1928 Haken 1928 Nash 1928 Grothendieck 1929 Hintikka 1930 Shimura 1930-1971 Montague 1931 Penrose 1932 Appel 1932 Scott 1933 Cohen 1937 Conway 1940 Kripke 1945 Shelah 1947 Chaitan 1947 Matiyasevich 1948 Friedman 1953 Wiles 1955 Woodin 1961 Baez If anyone has dates on the following, I'd appreciate knowing them: Aczel Addison Avigad Baldwin Barcan Bargaria Barwise Beeson Bell Beth Bishop Blass Bridges Bull Chang Chong Coquand Coxeter Craig Dekker Devlin Downey Easton Ehrenfeucht Enderton Ershov Feferman Fitch Gabbay Gitik Grassman Gurevich Hailperin Hardy Harrington Hasenjaeger [ ] - 2006 Henkin Hermite Hjorth Hodges Jech Jevons Kalicki Kanovei Kechris Keisler Klein Kochen Kunen Lambek Langer Lavine Lawvere LeBlanc Levy Lewis Lifsches Lorenzen Mahlo Maltsev Martin Martin-Lof McKinsey McNaughton Mendelson Miramanoff Monk J Moschovakis Y Moschovakis [ ]-1967 Myhill Nelson Pasch Poizat Parsons Pierce Pieri Pillay Platek Poncelet Prawitz Prikry Rasiowa Rescher Rice Richman N Shapiro S Shaprio Sheperdson Sikorski Silver Simpson Smorynski Soare Solovay Specker Spector Steel Tait Takeuti Troelstra Tukey Tung Turquette Van Den Dries Vaught Veldman Velleman Vesley Von Staudt Weil MoeBlee === Subject: Re: Who were the 25 Greatest Mathematicians? > If anyone has dates on the following, I'd appreciate knowing them: u for email) === Subject: Re: Who were the 25 Greatest Mathematicians? > A list of among the most famous (with a bias toward logic and set > theory): > MoeBlee Have you checked here for dates to some of these? There is an alphabetical listing which should make it easy. http://www-groups.dcs.st-and.ac.uk/~history/BiogIndex.html HTH ~A === Subject: Re: Who were the 25 Greatest Mathematicians? > Have you checked here for dates to some of these? There is an alphabetical listing which should make it easy. http://www-groups.dcs.st-and.ac.uk/~history/BiogIndex.html (though perhaps I goofed up on some of the dates and overlooked some?). MoeBlee === Subject: Re: Who were the 25 Greatest Mathematicians? The spacing didn't work right on that previous list. There's no signficance to be drawn from the erratic indenting of some of them names. MoeBlee === Subject: Re: Who were the 25 Greatest Mathematicians? > Who were the 25 Greatest Mathematicians? > I'm sure there'd be much controversy > over even the definition of great but > I've tried to construct a Greatest list: > http://james.fabpedigree.com/mathmen.htm > To qualify, the person's work must have > depth, breadth and historical importance. I'm not qualified to make such a list, but > have tried to read opinions of those > who are better qualified. I hope to hear > more helpful criticism. I hope you'll report errors and omissions > in my brief bios, and consider my explanations, > but here is a summary of the list: 1. Johann Carl F. Gauss (1777-1855) > 2. Archimedes of Syracuse (287-212 BC) > 3. Isaac Newton (1642-1727) > 4. Leonhard Euler (1707-1783) > 5. Euclid of Alexandria (ca 322-275 BC) > 6. Georg F. Bernhard Riemann (1826-1866) > 7. Gottfried Wilhelm Leibniz (1646-1716) > 8. Joseph-Louis Lagrange (1736-1813) > 9. Jules Henri Poincare (1854-1912) > 10. Pierre de Fermat (1601-1665) > 11. Srinivasa Ramanujan (1887-1920) > 12. Niels Henrik Abel (1802-1829) > 13. David Hilbert (1862-1943) > 14. Brahmagupta `Bhillamalacarya' (589-668) > 15. Georg Cantor (1845-1918) > 16. Leonardo `Fibonacci' Pisano (ca 1170-1245) > 17. Carl G. J. Jacobi (1804-1851) > 18. Evariste Galois (1811-1832) > 19. Rene Descartes (1596-1650) > 20. John von Neumann (1903-1957) > 21. Augustin-Louis Cauchy (1789-1857) > 22. Karl Wilhelm Theodor Weierstrass (1815-1897) > 23. Blaise Pascal (1623-1662) > 24. Arthur Cayley (1821-1895) > 25. Aryabhatta (476-550) James Dow Allen It's a good list, though it looks pretty much like a consensus, i.e., a lot of these people have been on many previous lists. And in how many cases do we have direct familiarity with the work of these people? E.g., I cannot say I have ever read a paper by Jacobi. It's striking also how many of these people straddled the fence between math and physics - I would include Gauss, Archimedes, Newton, Euler, Lagrange, Poincare, Hilbert, von Neumann, and others. Recent Fields Medalists are mostly not straddling both math and physics, with the exception of Edward Witten. Math has its roots, some so deep that they are taken for granted, in the physical world; it would be a shame in my view for the two fields to keep diverging, even though some of mathematical physics now is, frankly, ugly and numerical and approximative (because the really hard messy problems don't have analytic solutions). Marcus. === Subject: Re: Who were the 25 Greatest Mathematicians? Who were the 25 Greatest Mathematicians? > I'm sure there'd be much controversy > over even the definition of great but > I've tried to construct a Greatest list: > http://james.fabpedigree.com/mathmen.htm > To qualify, the person's work must have > depth, breadth and historical importance. I'm not qualified to make such a list, but > have tried to read opinions of those > who are better qualified. I hope to hear > more helpful criticism. I hope you'll report errors and omissions > in my brief bios, and consider my explanations, > but here is a summary of the list: 1. Johann Carl F. Gauss (1777-1855) > 2. Archimedes of Syracuse (287-212 BC) > 3. Isaac Newton (1642-1727) > 4. Leonhard Euler (1707-1783) > 5. Euclid of Alexandria (ca 322-275 BC) > 6. Georg F. Bernhard Riemann (1826-1866) > 7. Gottfried Wilhelm Leibniz (1646-1716) > 8. Joseph-Louis Lagrange (1736-1813) > 9. Jules Henri Poincare (1854-1912) > 10. Pierre de Fermat (1601-1665) > 11. Srinivasa Ramanujan (1887-1920) > 12. Niels Henrik Abel (1802-1829) > 13. David Hilbert (1862-1943) > 14. Brahmagupta `Bhillamalacarya' (589-668) > 15. Georg Cantor (1845-1918) > 16. Leonardo `Fibonacci' Pisano (ca 1170-1245) > 17. Carl G. J. Jacobi (1804-1851) > 18. Evariste Galois (1811-1832) > 19. Rene Descartes (1596-1650) > 20. John von Neumann (1903-1957) > 21. Augustin-Louis Cauchy (1789-1857) > 22. Karl Wilhelm Theodor Weierstrass (1815-1897) > 23. Blaise Pascal (1623-1662) > 24. Arthur Cayley (1821-1895) > 25. Aryabhatta (476-550) James Dow Allen It's a good list, though it looks pretty much like a consensus, > i.e., > a lot of these people have been on many previous lists. And in how > many cases do we have direct familiarity with the work of these > people? E.g., I cannot say I have ever read a paper by Jacobi. get his fundamenta nova seriously not only did he offer many important theorems but he developed many cool new _ways_ to prove things elliptic and theta functions product identities ways to tie number theory to analysis linear algebra ... jacobi most certainly is one to study source > It's striking also how many of these people straddled the fence > between math and physics - I would include Gauss, Archimedes, Newton, > Euler, Lagrange, Poincare, Hilbert, von Neumann, and others. Recent > Fields Medalists are mostly not straddling both math and physics, > with the exception of Edward Witten. Math has its roots, some so > deep that they are taken for granted, in the physical world; it > would be a shame in my view for the two fields to keep diverging, > even though some of mathematical physics now is, frankly, ugly and > numerical and approximative (because the really hard messy problems > don't have analytic solutions). kontsevich and drinfel'd have been riding the same edges building algebraic frameworks with strong physical properties alain connes' work has been much more explicit there is a lot going on these days on the boundary -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Who were the 25 Greatest Mathematicians? MY LIST > > Who were the 25 Greatest Mathematicians? > I'm sure there'd be much controversy > over even the definition of great but > I've tried to construct a Greatest list: > http://james.fabpedigree.com/mathmen.htm > To qualify, the person's work must have > depth, breadth and historical importance. > > I'm not qualified to make such a list, but > have tried to read opinions of those > who are better qualified. I hope to hear > more helpful criticism. > > I hope you'll report errors and omissions > in my brief bios, and consider my explanations, > but here is a summary of the list: some changes : 1. Georg F. Bernhard Riemann (1826-1866) 2. Yuri Matheyasevich 3. Andrew Wiles 4. Johann Carl F. Gauss (1777-1855) 5. Leonhard Euler (1707-1783) 6. Isaac Newton (1642-1727) 7. Pierre de Fermat (1601-1665) 8. Gottfried Wilhelm Leibniz (1646-1716) 9. Joseph-Louis Lagrange (1736-1813) 10. Evariste Galois (1811-1832) 10. Srinivasa Ramanujan (1887-1920) 10. G.H. Hardy & Littlewood 11. Niels Henrik Abel (1802-1829) 12. Carl G. J. Jacobi (1804-1851) 13. Perelmann 13. Fourier 14. Augustin-Louis Cauchy (1789-1857) 15. Karl Wilhelm Theodor Weierstrass (1815-1897) 15. Siegel 15. Hadamard 15. Rogers 16. John von Neumann (1903-1957) 16. Schrodinger 16. Smith 17. Rene Descartes (1596-1650) 18. Mandelbrot 19. Lorentz 20. Minkowski 21. Ackermann 21. Weber 22. Bessel 22. Meijer 23. Kepler 24. Leonardo `Fibonacci' Pisano (ca 1170-1245) 24. Archimedes of Syracuse (287-212 BC) 24. Lorenz 25. Robert Israel 26. tommy1729 27. F.A. Fajjal 27. Borwein & Borwein 28. quasi 29. Einstein 30. Gerry Myerson 31. Owl Hoot 32. Larry Hammick 33. John Napier 34. Timothy Golden 35. Diophantus 36. Stephen Wolfram 37. John Cleese .. 99999999999. pc with math software 999999999999. the milkman 99999999999999999997. cow 99999999999999999998. rock 99999999999999999999. Georg Cantor (1845-1918) 100000000000000000000. JSH 100000000000000000001. Homer Simpson (drunk) 100000000000000000002. people who talk a lot about 0.9999... (and similar bogus) 100000000000000000003. George Bush ? 100000000000000000004. Paris Hilton 100000000000000000005. Britney Spears (high) > 25. Aryabhatta (476-550) who ??? > James Dow Allen > tommy1729 === Subject: Re: Who were the 25 Greatest Mathematicians? MY LIST ... 28. quasi >29. Einstein ... > But I think I'll let Einstein cut in ahead of me. quasi === Subject: Re: Who were the 25 Greatest Mathematicians? > Who were the 25 Greatest Mathematicians? > I'm sure there'd be much controversy > over even the definition of great but > I've tried to construct a Greatest list: > http://james.fabpedigree.com/mathmen.htm > To qualify, the person's work must have > depth, breadth and historical importance. I'm not qualified to make such a list, but > have tried to read opinions of those > who are better qualified. I hope to hear > more helpful criticism. I hope you'll report errors and omissions > in my brief bios, and consider my explanations, > but here is a summary of the list: 1. Johann Carl F. Gauss (1777-1855) > 2. Archimedes of Syracuse (287-212 BC) > 3. Isaac Newton (1642-1727) > 4. Leonhard Euler (1707-1783) > 5. Euclid of Alexandria (ca 322-275 BC) > 6. Georg F. Bernhard Riemann (1826-1866) > 7. Gottfried Wilhelm Leibniz (1646-1716) > 8. Joseph-Louis Lagrange (1736-1813) > 9. Jules Henri Poincare (1854-1912) > 10. Pierre de Fermat (1601-1665) > 11. Srinivasa Ramanujan (1887-1920) > 12. Niels Henrik Abel (1802-1829) > 13. David Hilbert (1862-1943) > 14. Brahmagupta `Bhillamalacarya' (589-668) > 15. Georg Cantor (1845-1918) > 16. Leonardo `Fibonacci' Pisano (ca 1170-1245) > 17. Carl G. J. Jacobi (1804-1851) > 18. Evariste Galois (1811-1832) > 19. Rene Descartes (1596-1650) > 20. John von Neumann (1903-1957) > 21. Augustin-Louis Cauchy (1789-1857) > 22. Karl Wilhelm Theodor Weierstrass (1815-1897) > 23. Blaise Pascal (1623-1662) > 24. Arthur Cayley (1821-1895) > 25. Aryabhatta (476-550) James Dow Allen Yet another STUPID list from someone with obviously very limited knowledge of math and math history. I simply ask: Where is e.g. A. Grothendieck, P. Deligne and some of the other 20'th century Fields Medalists? Since 90%+ of all the mathematicians who have ever lived are alive today, one would expect a fair number of today's active mathematicians to be on the list just by statistical chance. Yet all of the choices on this list are dead. A very stupid list. I doubt that the O.P. can even name a dozen Fields Medalists along with their accomplishments. === Subject: Re: Who were the 25 Greatest Mathematicians? http://www.mathunion.org/Prizes/Fields/Prizewinners.html === Subject: Re: Who were the 25 Greatest Mathematicians? > Who were the 25 Greatest Mathematicians? > I'm sure there'd be much controversy Duh. Some people will think that that was your whole intention. > To qualify, the person's work must have > depth, breadth and historical importance. I don't see anyone who contributed to the computer proofs of the 4CT, which certainly qualifies as being historically important (although some people would disagree, calling the proof(s) infamous). The big names are: Kempe, Tutte, Heesch, Appel, Haken. Paul Erdos is also conspicuously missing, despite his thousands of collaborations and brilliancy. And don't even get me started about Archimedes Plutonium, discoverer of Archimedes Plutonium's Equation. --- Christopher Heckman > 1. Johann Carl F. Gauss (1777-1855) > 2. Archimedes of Syracuse (287-212 BC) > 3. Isaac Newton (1642-1727) > 4. Leonhard Euler (1707-1783) > 5. Euclid of Alexandria (ca 322-275 BC) > 6. Georg F. Bernhard Riemann (1826-1866) > 7. Gottfried Wilhelm Leibniz (1646-1716) > 8. Joseph-Louis Lagrange (1736-1813) > 9. Jules Henri Poincare (1854-1912) > 10. Pierre de Fermat (1601-1665) > 11. Srinivasa Ramanujan (1887-1920) > 12. Niels Henrik Abel (1802-1829) > 13. David Hilbert (1862-1943) > 14. Brahmagupta `Bhillamalacarya' (589-668) > 15. Georg Cantor (1845-1918) > 16. Leonardo `Fibonacci' Pisano (ca 1170-1245) > 17. Carl G. J. Jacobi (1804-1851) > 18. Evariste Galois (1811-1832) > 19. Rene Descartes (1596-1650) > 20. John von Neumann (1903-1957) > 21. Augustin-Louis Cauchy (1789-1857) > 22. Karl Wilhelm Theodor Weierstrass (1815-1897) > 23. Blaise Pascal (1623-1662) > 24. Arthur Cayley (1821-1895) > 25. Aryabhatta (476-550) === Subject: Re: Who were the 25 Greatest Mathematicians? Who were the 25 Greatest Mathematicians? > I'm sure there'd be much controversy Duh. Some people will think that that was your whole intention. To qualify, the person's work must have > depth, breadth and historical importance. I don't see anyone who contributed to the computer proofs of the 4CT, > which certainly qualifies as being historically important (although > some people would disagree, calling the proof(s) infamous). The big > names are: Kempe, Tutte, Heesch, Appel, Haken. Paul Erdos is also conspicuously missing, despite his thousands of > collaborations and brilliancy. And don't even get me started about Archimedes Plutonium, discoverer > of Archimedes Plutonium's Equation. --- Christopher Heckman 1. Johann Carl F. Gauss (1777-1855) > 2. Archimedes of Syracuse (287-212 BC) > 3. Isaac Newton (1642-1727) > 4. Leonhard Euler (1707-1783) > 5. Euclid of Alexandria (ca 322-275 BC) > 6. Georg F. Bernhard Riemann (1826-1866) > 7. Gottfried Wilhelm Leibniz (1646-1716) > 8. Joseph-Louis Lagrange (1736-1813) > 9. Jules Henri Poincare (1854-1912) > 10. Pierre de Fermat (1601-1665) > 11. Srinivasa Ramanujan (1887-1920) > 12. Niels Henrik Abel (1802-1829) > 13. David Hilbert (1862-1943) > 14. Brahmagupta `Bhillamalacarya' (589-668) > 15. Georg Cantor (1845-1918) > 16. Leonardo `Fibonacci' Pisano (ca 1170-1245) > 17. Carl G. J. Jacobi (1804-1851) > 18. Evariste Galois (1811-1832) > 19. Rene Descartes (1596-1650) > 20. John von Neumann (1903-1957) > 21. Augustin-Louis Cauchy (1789-1857) > 22. Karl Wilhelm Theodor Weierstrass (1815-1897) > 23. Blaise Pascal (1623-1662) > 24. Arthur Cayley (1821-1895) > 25. Aryabhatta (476-550) James Harris has: a short proof of FLT *defined* proof itself shown Galois theory ( 'as usually taught' ) to be false discovered a polynomial-time factoring algorithm discovered a new prime counting 'function' [sic] demonstrated that RH is probably false but that the first counterexample is so large we'll never find it. ...and much more... And he doesn't make the list ?!?! === Subject: Re: Who were the 25 Greatest Mathematicians? > Who were the 25 Greatest Mathematicians? Your mathematicians all seem to be, err..., dead... Here are two longer lists: http://nostalgia.wikipedia.org/wiki/Mathematician of both the living and the dead. === Subject: Re: Who were the 25 Greatest Mathematicians? > Who were the 25 Greatest Mathematicians? > I'm sure there'd be much controversy > over even the definition of great but > I've tried to construct a Greatest list: > http://james.fabpedigree.com/mathmen.htm > To qualify, the person's work must have > depth, breadth and historical importance. I'm not qualified to make such a list, but > have tried to read opinions of those > who are better qualified. I hope to hear > more helpful criticism. I hope you'll report errors and omissions > in my brief bios, and consider my explanations, > but here is a summary of the list: 1. Johann Carl F. Gauss (1777-1855) > 2. Archimedes of Syracuse (287-212 BC) > 3. Isaac Newton (1642-1727) > 4. Leonhard Euler (1707-1783) > 5. Euclid of Alexandria (ca 322-275 BC) > 6. Georg F. Bernhard Riemann (1826-1866) > 7. Gottfried Wilhelm Leibniz (1646-1716) > 8. Joseph-Louis Lagrange (1736-1813) > 9. Jules Henri Poincare (1854-1912) > 10. Pierre de Fermat (1601-1665) > 11. Srinivasa Ramanujan (1887-1920) > 12. Niels Henrik Abel (1802-1829) > 13. David Hilbert (1862-1943) > 14. Brahmagupta `Bhillamalacarya' (589-668) > 15. Georg Cantor (1845-1918) > 16. Leonardo `Fibonacci' Pisano (ca 1170-1245) > 17. Carl G. J. Jacobi (1804-1851) > 18. Evariste Galois (1811-1832) > 19. Rene Descartes (1596-1650) > 20. John von Neumann (1903-1957) > 21. Augustin-Louis Cauchy (1789-1857) > 22. Karl Wilhelm Theodor Weierstrass (1815-1897) > 23. Blaise Pascal (1623-1662) > 24. Arthur Cayley (1821-1895) > 25. Aryabhatta (476-550) James Dow Allen Any of these characters belong on your list? Famous mathematicians of the 15th/16th century Johannes Regiomontanus (1436-1476) Simon Stevin (1548-1620) Albrecht Durer (1471-1528) Nicolo Tartaglia (1500-1557) Geronimo Cardano (1501-1576) Jost Burgi (1552-1632) Luca Pacioli (1445-1514) Famous mathematicians of the 17th century Francois Vieta (1540-1603) John Napier (1550-1617) Galileo Galilei (1564-1642) Johannes Kepler (1571-1630) Bonaventura Cavalieri (1598-1647) James Gregory (1638-1675) Famous mathematicians of the 17th/18th century Jakob Bernoulli (1654-1705) Johan Bernoulli (1667-1748) Daniel Bernoulli (1700-1782) Famous mathematicians of the 18th century Brook Taylor (1685-1731) Moreau Maupertuis (1698-1759) Johann Heinrich Lambert (1728-1777) Gaspard Monge (1746-1818) Adrien Marie Legendre (1752-1833) Jean Baptiste Joseph de Fourier (1768-1830) Famous mathematicians of the 19th century Nikolai Ivanovich Lobachevski (1792-1856) Janos Bolyai (1802-1860) Frederich Wilhelm Bessel (1784-1846) Jakob Steiner (1796-1863) Peter Gustav Lejeune Dirichlet (1805-1859) Pafnuti Lvovich Chebyshev (1821-1894) Leopold Kronecker (1823-1891) Sophus Lie (1842-1899) Sonya Kovalevski (1850-1891) Famous mathematicians of the 19th century George Stokes (1819-1903) Richard Dedekind (1831-1916) Georg Frobenius (1849-1917) Felix Klein (1849-1925) Emmy Noether (1882-1935) Elie Cartan (1869-1951) Henri Leon Lebesge (1875-1941) Hermann Weyl (1885-1955) Jaques Salomon Hadamard (1865-1963) Stefan Banach (1892-1945) === Subject: Re: Who were the 25 Greatest Mathematicians? > Who were the 25 Greatest Mathematicians? > I'm sure there'd be much controversy > over even the definition of great but > I've tried to construct a Greatest list: > http://james.fabpedigree.com/mathmen.htm > To qualify, the person's work must have > depth, breadth and historical importance. I'm not qualified to make such a list, but > have tried to read opinions of those > who are better qualified. I hope to hear > more helpful criticism. I hope you'll report errors and omissions > in my brief bios, and consider my explanations, > but here is a summary of the list: 1. Johann Carl F. Gauss (1777-1855) > 2. Archimedes of Syracuse (287-212 BC) > 3. Isaac Newton (1642-1727) > 4. Leonhard Euler (1707-1783) > 5. Euclid of Alexandria (ca 322-275 BC) > 6. Georg F. Bernhard Riemann (1826-1866) > 7. Gottfried Wilhelm Leibniz (1646-1716) > 8. Joseph-Louis Lagrange (1736-1813) > 9. Jules Henri Poincare (1854-1912) > 10. Pierre de Fermat (1601-1665) > 11. Srinivasa Ramanujan (1887-1920) > 12. Niels Henrik Abel (1802-1829) > 13. David Hilbert (1862-1943) > 14. Brahmagupta `Bhillamalacarya' (589-668) > 15. Georg Cantor (1845-1918) > 16. Leonardo `Fibonacci' Pisano (ca 1170-1245) > 17. Carl G. J. Jacobi (1804-1851) > 18. Evariste Galois (1811-1832) > 19. Rene Descartes (1596-1650) > 20. John von Neumann (1903-1957) > 21. Augustin-Louis Cauchy (1789-1857) > 22. Karl Wilhelm Theodor Weierstrass (1815-1897) > 23. Blaise Pascal (1623-1662) > 24. Arthur Cayley (1821-1895) > 25. Aryabhatta (476-550) James Dow Allen 1. http://afrodita.rcub.bg.ac.yu/~flora/100.html 2. http://web.centre.edu/mat/century.html 3. http://www-groups.dcs.st-and.ac.uk/~history/BiogIndex.html See a pattern? ~A === Subject: Re: Who were the 25 Greatest Mathematicians? > 2. http://web.centre.edu/mat/century.html Focus on Sir Ronald Fischer. Quote from: http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Fisher.html [] Fisher was elected a Fellow of the Royal Society in 1929, was awarded the Royal Medal of the Society in 1938, and was awarded the Darwin Medal of the Society in 1948:- ... in recognition of his distinguished contributions to the theory of natural selection, the concept of its gene complex and the evolution of dominance. Han de Bruijn === Subject: Re: Who were the 25 Greatest Mathematicians? > Who were the 25 Greatest Mathematicians? > I'm sure there'd be much controversy > over even the definition of great but > I've tried to construct a Greatest list: > http://james.fabpedigree.com/mathmen.htm > To qualify, the person's work must have > depth, breadth and historical importance. I'm not qualified to make such a list, but > have tried to read opinions of those > who are better qualified. I hope to hear > more helpful criticism. I hope you'll report errors and omissions > in my brief bios, and consider my explanations, > but here is a summary of the list: 1. Johann Carl F. Gauss (1777-1855) > 2. Archimedes of Syracuse (287-212 BC) > 3. Isaac Newton (1642-1727) > 4. Leonhard Euler (1707-1783) > 5. Euclid of Alexandria (ca 322-275 BC) > 6. Georg F. Bernhard Riemann (1826-1866) > 7. Gottfried Wilhelm Leibniz (1646-1716) > 8. Joseph-Louis Lagrange (1736-1813) > 9. Jules Henri Poincare (1854-1912) > 10. Pierre de Fermat (1601-1665) > 11. Srinivasa Ramanujan (1887-1920) > 12. Niels Henrik Abel (1802-1829) > 13. David Hilbert (1862-1943) > 14. Brahmagupta `Bhillamalacarya' (589-668) > 15. Georg Cantor (1845-1918) > 16. Leonardo `Fibonacci' Pisano (ca 1170-1245) > 17. Carl G. J. Jacobi (1804-1851) > 18. Evariste Galois (1811-1832) > 19. Rene Descartes (1596-1650) > 20. John von Neumann (1903-1957) > 21. Augustin-Louis Cauchy (1789-1857) > 22. Karl Wilhelm Theodor Weierstrass (1815-1897) > 23. Blaise Pascal (1623-1662) > 24. Arthur Cayley (1821-1895) > 25. Aryabhatta (476-550) James Dow Allen > > 1. http://afrodita.rcub.bg.ac.yu/~flora/100.html > > 2. http://web.centre.edu/mat/century.html > > 3. http://www-groups.dcs.st-and.ac.uk/~history/BiogIndex.html > > See a pattern? > List #2 is rather strange. Darmarkar is supposed to be Narendra Karmarkar, I suppose. What on earth is R.L. Moore doing there? Several of the others are very iffy too, I think, nowhere near the same calibre as Weil, or Langlands, or Milnor, or Grothendieck, or Harish-Chandra, or Thom, or many others who did important pure math in the second half of the 20th century. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Who were the 25 Greatest Mathematicians? > >>2. http://web.centre.edu/mat/century.html > > List #2 is rather strange. Darmarkar is supposed to be Narendra Karmarkar, > I suppose. What on earth is R.L. Moore doing there? Several of the others > are very iffy too, I think, nowhere near the same calibre as Weil, or > Langlands, or Milnor, or Grothendieck, or Harish-Chandra, or Thom, or many > others who did important pure math in the second half of the 20th century. Han de Bruijn === Subject: Electromagnetics 7E Solutions hey , can i get the solutions manual for the 7e ? === Subject: Re: Complete Electronic (.pdf/doc) Solution Manuals. Get witihn 30 Minutes! can i get the 7E electromagetics solutions? === Subject: Re: Complete Electronic (.pdf/doc) Solution Manuals. Get witihn 30 Minutes! i have the following solution manuals.... and i have thousands of textbooks as well email me at diosbendit...@gmail.com if you want any of them email at diosbenditome (at) gmail (dot) com paypal paymets accepted only please email me rather than leaving a message here.. 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This includes the issue as to whether what we know of the life process itself can find rapprochement with modern quantum physics, or whether and how it needs to be extended. Given my own earlier decade-plus background as Director of the Cognitive Sciences program at SRI Intern'l in the '70's and early '80's, investigating remote viewing and other so-called paranormal phenomena, the life-science data I have to integrate all by themselves push the envelope (Proc. IEEE 64, 329 (1976); Jour. Sci. Exploration 10, 63 (1996). OK Unfortunately, as it now stands, mainstream physics reductionism is leading to an evermore complex picture of nature involving a forces, the implications of incorporating additional dimensions as in superstring theory, and so forth. Thus, in spite of efforts to develop a grand unified theory to simplify our picture of nature, the actual day-to-day work on this effort is complexifying faster than the hoped-for simplification. Therefore, not only are we missing holism on the grand scale, but a gratifying holism just for the physical sciences alone appears to be a rapidly accelerating goal post. Contemplation of such provocative issues in both the physical and life sciences led me into investigating an area of physics concerned with what is known as quantum vacuum fluctuations or zero-point energy, a universal background energy pervading all of space and associated with fluctuations of underlying space itself. Specifically, I began to consider the underlying quantum fluctuations as a fundamental stuff out of which a greater synthesis could be built. I hasten to add that I do not mean for such an approach to be simply reductionism on a grander scale, with no room for nonphysical factors to play a role. Rather, to the degree that energy is involved not only in physical but in nominally non- or para-physical phenomena (including, perhaps, such mundane phenomena as thought, charisma, etc., let alone psychokinesis), then such energy patterns might in principle emerge as a result of cohering or patterning the otherwise random, ambient zero-point energy. For me this hypothesis emerged when I considered how uneconomical Nature would have to be to posit, on the one hand, an all-pervading energetic field of ki or chi, as in the metaphysics of the martial arts and acupuncture, and, on the other hand, also posit an all-pervasive energetic field of quantum zero-point energy. It appeared to me to be more likely that we were dealing with a single underlying substructure which goes by various names in various cosmologies, depending on whether it is in its pre-manifest random form, or patterned at various hierarchical levels, including the purely material. In my professional area, I began with the pure physics side. In my first published study on the significance of zero-point energy for broad issues I showed that the basic stable states of matter are not merely inert, static structures, but rather depend on the presence of the underlying, sustaining zero-point energy which is continually being absorbed and re-emitted on a dynamic-balance basis. Pull the plug on the zero-point energy and all atomic structure would collapse (Physical Review D 35, p. 3266, 1987). In my second study I developed the idea, originally put forward as a hypothesis by the famous Russian physicist and human-rights advocate, Sakharov, that the gravity of Einstein's equations is not a separate, fundamental force, but rather is a pressure force derived from partial shielding of the ambient zero-point energy (Phys. Rev. A 39, p. 2333, 1989; 47, 3454, 1993). (A corollary to this view is that if one could cohere the otherwise random fluctuations of space, a host of interesting phenomena would follow.) thus quoth: energy due to minor microwave and radiant energy set at the same of a high energy dependent repell you from said dependent. a square within a square squared is 0 when the apllied force is in magnetics and quatum mechanics, but your technology is 100 years behind, good luck. thus: e-voting is for suckers!... on the other hand, you could learn a lot of electronics with an oscilloscope, coupled with the not-aforementioned textbook on ******* ********* ... and, of course, the memoir o'Michelson; you would certainly begin to feel-out the corner of that box, however convex or not. thus: a-hem; speaking for the other 6.5-billion hominids around here, if I may: where's that Monolith, when y'need it; at least, I could beat my head against it, to stop the sigint from Jack-is-the-box @MC-infra! Counting Down from the '60s with Jack in the Dreamstalk! did anyone ever review one of your touchy-feely-new-age textbooks on the theory of nothing -- could we have a link to the journal o'spooks & more spooks, Joe Newage & the Roswell Priesthoodies? did Lt.Col Corso actually file that first story with the agent-not-really-in-charge at the paper, or What?... may he rest in peace, in sprite of every thing! surely, you do not go so far as to vouchsafe with the paradigm, that all of our techniques were retrieved & reverse-engineered by nascent military-post-industrial-infotainment companies, that which Eisenhower warned against in the full paragraph, from that little pile of crap at the B-52 base, or did you already fall into that? don't struggle in the quicksand ... what is it, the backstroke?! > Suppressing indices for simplicity. This is analogous to a Yang-Mills > theory where the curvature two form field is R = DS i.e. curvature field 2-form = exterior covariant derivative of the spin > connection Yang-Mills potential with itself, i.e. in 1916 GR thus: the buckyfullerene challenge: uhm, carbondating?... so, not a textbook: A.A.Michelson's book on interferometry. thus: so, did anyone find a workable integer value about that, somewhat inconclusive (if not fully multivalued). >htpp://users.aol.com/atrupp/loschmid.htm thus: the hard part is folding them back up, thus: p-adics are like a box of chocolates ... inside or outside, shape, time-stamp, partitioning, pigeonholing et c. et c. et cetera. --14 Italian Senators Call for Cheney Impeachment Aug. 1, 2007 (EIRNS)- The Lyndon LaRouche Political Action Committee (LPAC) issued the following release today. Fourteen members of the Italian Senate have signed a call to the Members of Congress to support Rep. Kucinich's House Resolution 333 for the Impeachment of Dick Cheney. http://larouchepub.com/pr/2007/070801italian_senators_call.html === Subject: Re: Grade School Epistemology >I really find life is more pleasant since kill-filing >Lester's crude trolling for attention. I strongly >recommend it. > > Pleasant perhaps however somewhat less true. > > Tell us, is there a codex of proscribed works and authors which > offendeth thee? Yes, there is, and you're the sole occupant. Well, I guess I'll have to kill-file your new pseudonym, too. >It would be nice if people stopped quoting him. Or >better, if all of you would just remove sci.logic from >the subscription list of your replies. No one here >gives a about what he says anyway. > > So sci.logic doesn't care about truth? Or just no longer cares about > you? > > Art thou indeed Inquisitor General, Public Censor, and Keeper of the > Public Morals? Here's a thought: why not try reading sci.logic for a month or so, to get an idea of what the ng is about before posting your Sorry about removing sci.logic from your reply-to list. As you may recall, my usual news server doesn't allow cross-posting to more than three groups. Bye. -- hz === Subject: Re: Grade School Epistemology >I really find life is more pleasant since kill-filing >>Lester's crude trolling for attention. I strongly >>recommend it. >> >> Pleasant perhaps however somewhat less true. >> >> Tell us, is there a codex of proscribed works and authors which >> offendeth thee? Yes, there is, and you're the sole occupant. Well, I guess >I'll have to kill-file your new pseudonym, too. Gee. I had no idea I was so important. Wish I could say the same for you. >>It would be nice if people stopped quoting him. Or >>better, if all of you would just remove sci.logic from >>the subscription list of your replies. No one here >>gives a about what he says anyway. >> >> So sci.logic doesn't care about truth? Or just no longer cares about >> you? >> >> Art thou indeed Inquisitor General, Public Censor, and Keeper of the >> Public Morals? Here's a thought: why not try reading sci.logic for a month or >so, to get an idea of what the ng is about before posting your With the kind of crap you write I'm not surprized. >Sorry about removing sci.logic from your reply-to list. As you >may recall, my usual news server doesn't allow cross-posting >to more than three groups. Oh well I can always put it back. ~v~~ === Subject: Re: Grade School Epistemology <4701B376.60C02644@gmail.com> <47031A72.6F8EE5CB@gmail.com> may recall, my usual news server doesn't allow cross-posting >to more than three groups. Oh well I can always put it back. **************************************************************** Lester = stupid, shallow, whimsical, stubborn, haughty...and also a NG bully, now. Oh well...some leisure time he provides now and then. Tonio === Subject: Re: Grade School Epistemology >>Sorry about removing sci.logic from your reply-to list. As you >>may recall, my usual news server doesn't allow cross-posting >>to more than three groups. >> Oh well I can always put it back. >**************************************************************** >Lester = stupid, shallow, whimsical, stubborn, haughty...and also a NG >bully, now. So what's the complaint exactly? >Oh well...some leisure time he provides now and then. I've often found students who're too lazy or stupid to deal with truth prefer modern math where they're only required to count instead of think. ~v~~ === Subject: Re: Grade School Epistemology > Grade School Epistemology > ~v~~ > Religion is great, religion is good; and we thank it for our food. > Empirics is great, empirics is good; and we thank it for our food. > Modern math is great, modern math is good; and we thank it for our > food. > Amen. > ~v~~ > What's the beef? > Han de Bruijn >>******************************************************************** >>Lester can't understand any maths beyond gimme 2 (=1+1) cans of cat >>food, please, and thus has decided that it is as worth as religion. >>Go figure ;) >> Well the rationalization for religion, empirics, and modern math is >> purely utilitatrian. Apparently they aren't true but they work. So I >> expect cat food is about the most we can expect. **************************************************** I agree: that's the most you could possibly expect. And obviously the most you can offer. Cans of cat food are perfectly respectable if that's all modern math's got. Certainly good enough for grade schoolers and cats. Science however needs to be more inclusive. ~v~~ === Subject: Re: Grade School Epistemology >> Grade School Epistemology >> ~v~~ >> Religion is great, religion is good; and we thank it for our food. >> Empirics is great, empirics is good; and we thank it for our food. >> Modern math is great, modern math is good; and we thank it for our >> food. >> Amen. >> ~v~~ >******************************************************************* >Somebody learned a new fancy world: epistemology. >Too bad you haven't yet learned any maths...oh well. > One musn't burden oneself with the tedium of perfunctory knowledge. >>********************************************************** >>Wow! Perfunctory....cool! >>Indeed, you don't burden yourself with any knowledge at all, either >>perfunctory or deep, and good that you don't: it is a lost battle >>for you. >>Congratulations! :) >> In any event I find my world has gotten a lot newer and fancier since >> I learned to spell. >**************************************************** >Great for you. I, on the other hand, am more worried about maths than >spelling in a language that is neither my first nor my second one. >*wink* Just as mathematics isn't. [wink, wink] ~v~~ === Subject: Re: Number Theory based Series : deny > An ugly number is defined as one which is >> divisible by 2,3 or 5. >> 1 is by default and ugly number. >> 1,2,3,4,5,6,8,9,10,12,14,15 >> Find the nth ugly number. Is there a closed > form? >> Can anyone please help me derive the closed > form. >> I do understand that n = 2^x*3^y*5^z where > x,y,z >> belong to [0,inf) >> Any ideas? >>I use [x] for the floor function. >>To find the nth ugly number, let A=n-8[n/8]. >>Let f(x) = > -31/13440*x^8+145/2016*x^7-291/320*x^6+4297/720*x^5 > -13779/640*x^4+11875/288*x^3-121157/3360*x^2+10271/84 >> 0*x >>Then the nth ugly number is >>30[n/8]+f(A). >>a nice closed form, eh? (This assumes the > correction >> that an >>ugly number is NOT divisible by 2, 3 or 5.) >> Actually, it's wrong in 3 aspects ... >> Firstly, the OP corrected the problem, defining an >> ugly number as one >> which has no prime factors other than possibly > 2,3,5. >> Thus, for >> example, 14 is not an ugly number since it has a >> prime factor of 7. >> Secondly, even for the original problem, you are >> confusing ugly with >> not ugly. By my reading of the original >> (uncorrected problem) an >> ugly number is one which is either 1 or else has a >> prime factor of >> 2,3, or 5. If we change the class of 1 from ugly > to >> non-ugly, then you >> are counting the non-uglies, not the uglies > (although >> perhaps you >> could argue that ugliness, like beauty, is in the > eye >> of the >> beholder). >> Finally, even if we change 1 from ugly to > non-ugly, >> and then change >> all uglies to non-uglies and vice-versa, your > (rather >> ugly) formula, >> while amazing, is still not quite correct. Your >> formula is off by 1 >> whenever n is a multiple of 8. For example, for n > = >> 8, your formula >> gives 30 when it should give 29. However it's > easily >> corrected, as >> indicated below ... >> For clarity, and so as not to insult numbers which > I >> might need favors >> from in the future, I'll avoid using the term > ugly. >> Thus, consider the infinite sequence >> 1, 7, 11, 13, 17, 19, 23, 29, 31, > 37, >> 41, 43, 47, 49, 53, 59, 61, 67, 71, > 73, >> 77, 79, 83, 89, 91, 97, 101, 103, 107, > 109, >> ... >> obtained from the sequence of positive integers by >> deleting all >> multiples of 2, 3, or 5. >> Then the n'th term of the above sequence is given > by >> the closed form >> g(n) = 30*q(n) + f(r(n)) + q(r(n)-1) >> where f,q,r are defined as follows ... >> q(x) = floor(x/8) >> r(x) = x - 8*q(x) >> f(x) = > -(31/13440)*x^8+(145/2016)*x^7-(291/320)*x^6+(4297/720 >> )*x^5 > -(13779/640)*x^4+(11875/288)*x^3-(121157/3360)*x^2+(1 >> 0271/840)*x >> The above closed form is exactly the same as yours >> except for the term >> q(r(n)-1), which corrects the error when n is a >> multiple of 8. >> However, as discussed, it doesn't solve the OP's >> actual problem. >> Nevertheless, it is interesting (if only for the >> ugliness of f). >> quasi or in tommynotation deny (2n) U deny (3n) U deny (5n) >= deny(2^x 3^y 5^z) and it appears to have a closed form , even for 1 > variable. >in terms of a polynomial. >although we also need the floor function. No, Goddard's formula does not generate the > complement (in N) of {2^x*3^y*5^z} Rather it generates (with my minor correction) the > set of positive > integers relatively prime to 30. quasi >>i was rather referring to bart goddard's formula. >>yet your formula is intresting too. The formulas are the same (mine is his with a minor correction). Neither one is deny (2^x*3^y*5^z) Both try to generate the n'th term of the sequence of positive > integers which are relatively prime to 30. Goddard's formula is correct except when n is a multiple of 8. When n > is a multiple of 8, his formula is off by 1. All I did was make a minor modification, keeping the same essential > form but correcting the error for the case where n is a multiple of 8. My point is that neither of the functions has, as its positive integer > range, the complement of {2^x*3^y*5^z}. quasi This is very interesting. How did you arrive at this closed form? g(n) = 30*q(n) + f(r(n)) + q(r(n)-1) for nth number relatively prime with 30.=> LCM(2,3,5) I could just see that q is the quotient and r is the reminder when n is divided by 8. I can just see that q(r(n) - 1) can only be zero or -1 since r(n) belongs [0,7]. I am still puzzled about the 8 and f(x). Naren. === Subject: Re: Number Theory based Series : deny >> > An ugly number is defined as one which is > divisible by 2,3 or 5. >> 1 is by default and ugly number. > 1,2,3,4,5,6,8,9,10,12,14,15 >> Find the nth ugly number. Is there a closed >> form? >> Can anyone please help me derive the closed >> form. >> I do understand that n = 2^x*3^y*5^z where >> x,y,z > belong to [0,inf) >> Any ideas? >>I use [x] for the floor function. >>To find the nth ugly number, let A=n-8[n/8]. >>Let f(x) = > -31/13440*x^8+145/2016*x^7-291/320*x^6+4297/720*x^5 > -13779/640*x^4+11875/288*x^3-121157/3360*x^2+10271/84 > 0*x >Then the nth ugly number is >>30[n/8]+f(A). >>a nice closed form, eh? (This assumes the >> correction > that an >ugly number is NOT divisible by 2, 3 or 5.) >> Actually, it's wrong in 3 aspects ... >> Firstly, the OP corrected the problem, defining an > ugly number as one > which has no prime factors other than possibly >> 2,3,5. > Thus, for > example, 14 is not an ugly number since it has a > prime factor of 7. >> Secondly, even for the original problem, you are > confusing ugly with > not ugly. By my reading of the original > (uncorrected problem) an > ugly number is one which is either 1 or else has a > prime factor of > 2,3, or 5. If we change the class of 1 from ugly >> to > non-ugly, then you > are counting the non-uglies, not the uglies >> (although > perhaps you > could argue that ugliness, like beauty, is in the >> eye > of the > beholder). >> Finally, even if we change 1 from ugly to >> non-ugly, > and then change > all uglies to non-uglies and vice-versa, your >> (rather > ugly) formula, > while amazing, is still not quite correct. Your > formula is off by 1 > whenever n is a multiple of 8. For example, for n >> = > 8, your formula > gives 30 when it should give 29. However it's >> easily > corrected, as > indicated below ... >> For clarity, and so as not to insult numbers which >> I > might need favors > from in the future, I'll avoid using the term >> ugly. >> Thus, consider the infinite sequence >> 1, 7, 11, 13, 17, 19, 23, 29, 31, >> 37, > 41, 43, 47, 49, 53, 59, 61, 67, 71, >> 73, > 77, 79, 83, 89, 91, 97, 101, 103, 107, >> 109, > ... >> obtained from the sequence of positive integers by > deleting all > multiples of 2, 3, or 5. >> Then the n'th term of the above sequence is given >> by > the closed form >> g(n) = 30*q(n) + f(r(n)) + q(r(n)-1) >> where f,q,r are defined as follows ... >> q(x) = floor(x/8) >> r(x) = x - 8*q(x) >> f(x) = > -(31/13440)*x^8+(145/2016)*x^7-(291/320)*x^6+(4297/720 > )*x^5 > -(13779/640)*x^4+(11875/288)*x^3-(121157/3360)*x^2+(1 > 0271/840)*x >> The above closed form is exactly the same as yours > except for the term > q(r(n)-1), which corrects the error when n is a > multiple of 8. >> However, as discussed, it doesn't solve the OP's > actual problem. >> Nevertheless, it is interesting (if only for the > ugliness of f). >> quasi >>or in tommynotation >>deny (2n) U deny (3n) U deny (5n) >>= deny(2^x 3^y 5^z) >>and it appears to have a closed form , even for 1 >> variable. >>in terms of a polynomial. >>although we also need the floor function. >> No, Goddard's formula does not generate the >> complement (in N) of >> {2^x*3^y*5^z} >> Rather it generates (with my minor correction) the >> set of positive >> integers relatively prime to 30. >> quasi i was rather referring to bart goddard's formula. yet your formula is intresting too. >> The formulas are the same (mine is his with a minor correction). >> Neither one is deny (2^x*3^y*5^z) >> Both try to generate the n'th term of the sequence of positive >> integers which are relatively prime to 30. >> Goddard's formula is correct except when n is a multiple of 8. When n >> is a multiple of 8, his formula is off by 1. >> All I did was make a minor modification, keeping the same essential >> form but correcting the error for the case where n is a multiple of 8. >> My point is that neither of the functions has, as its positive integer >> range, the complement of {2^x*3^y*5^z}. >> quasi This is very interesting. How did you arrive at this closed form? I stole it. More precisely, I took Bart Goddard's form and simply corrected a minor error. >g(n) = 30*q(n) + f(r(n)) + q(r(n)-1) for nth number relatively prime with >30.=> LCM(2,3,5) Note that it's not a closed form for your n'th term of your sequence, however it's almost a closed form for the n'th term of the complementary sequence from your original post, before you corrected it. >I could just see that q is the quotient and r is the reminder when n is >divided by 8. Right. Note that q(n) can also be written as [n/8] where the square brackets denote the floor function: [x] = the greatest integer <= x >I can just see that q(r(n) - 1) can only be zero or -1 since r(n) belongs >[0,7]. Right. It's zero unless n is a multiple of 8, in which case it's -1. >I am still puzzled about the 8 and f(x). > Good question. Since I just adapted Bart Goddard's formula, I'm sure Bart can probably give a better answer, but I'll give a very rough idea (with no details) of how I think it can be derived. However, if you want to take a shot at it yourself, I think it would be an interesting (although somewhat tedious) exercise to work through the details. If you have a computer algebra system such as Maple or Mathematica, you can be spared from some of the more tedious calculations. Here's the plan ... To get started, consider the following simple problem: How many positive integers between 1 and 1000 inclusive are multiples of 2 or 3 (or both)? Next, do the same for 2, 3, 5 -- that is, how many positive integers between 1 and 1000 inclusive are multiples of at least one of the numbers 2,3,5. The method is called the method of inclusion-exclusion. After that, change 1000 to an unknown, k say. Thus, try to answer the question: How many positive integers in the range 1 to k inclusive are multiple of at least one of the numbers 2,3,5. It's ok to use the floor function in building the answer, Let call the answer w(k). Then k - w(k) is the number of positive integers between 1 and k which are relatively prime to 2,3,5. Next, try to solve the equation k - w(k) = n for k in terms of n. In general, there may be more than one solution, in which case, you want the least solution, call it k1. Thus, assume you have found k1 as a function of n. Then k1 is the n'th term of the sequence of positive integers, relatively prime to 2,3,5. quasi === Subject: Re: Number Theory based Series : deny > > An ugly number is defined as one which is >> divisible by 2,3 or 5. >> 1 is by default and ugly number. >> 1,2,3,4,5,6,8,9,10,12,14,15 >> Find the nth ugly number. Is there a closed > form? >> Can anyone please help me derive the closed > form. >> I do understand that n = 2^x*3^y*5^z where > x,y,z >> belong to [0,inf) >> Any ideas? >>I use [x] for the floor function. >>To find the nth ugly number, let A=n-8[n/8]. >>Let f(x) = > -31/13440*x^8+145/2016*x^7-291/320*x^6+4297/720*x^5 > -13779/640*x^4+11875/288*x^3-121157/3360*x^2+10271/84 >> 0*x >>Then the nth ugly number is >>30[n/8]+f(A). >>a nice closed form, eh? (This assumes the > correction >> that an >>ugly number is NOT divisible by 2, 3 or 5.) >> Actually, it's wrong in 3 aspects ... >> Firstly, the OP corrected the problem, defining an >> ugly number as one >> which has no prime factors other than possibly > 2,3,5. >> Thus, for >> example, 14 is not an ugly number since it has a >> prime factor of 7. >> Secondly, even for the original problem, you are >> confusing ugly with >> not ugly. By my reading of the original >> (uncorrected problem) an >> ugly number is one which is either 1 or else has a >> prime factor of >> 2,3, or 5. If we change the class of 1 from ugly > to >> non-ugly, then you >> are counting the non-uglies, not the uglies > (although >> perhaps you >> could argue that ugliness, like beauty, is in the > eye >> of the >> beholder). >> Finally, even if we change 1 from ugly to > non-ugly, >> and then change >> all uglies to non-uglies and vice-versa, your > (rather >> ugly) formula, >> while amazing, is still not quite correct. Your >> formula is off by 1 >> whenever n is a multiple of 8. For example, for n > = >> 8, your formula >> gives 30 when it should give 29. However it's > easily >> corrected, as >> indicated below ... >> For clarity, and so as not to insult numbers which > I >> might need favors >> from in the future, I'll avoid using the term > ugly. >> Thus, consider the infinite sequence >> 1, 7, 11, 13, 17, 19, 23, 29, 31, > 37, >> 41, 43, 47, 49, 53, 59, 61, 67, 71, > 73, >> 77, 79, 83, 89, 91, 97, 101, 103, 107, > 109, >> ... >> obtained from the sequence of positive integers by >> deleting all >> multiples of 2, 3, or 5. >> Then the n'th term of the above sequence is given > by >> the closed form >> g(n) = 30*q(n) + f(r(n)) + q(r(n)-1) >> where f,q,r are defined as follows ... >> q(x) = floor(x/8) >> r(x) = x - 8*q(x) >> f(x) = > -(31/13440)*x^8+(145/2016)*x^7-(291/320)*x^6+(4297/720 >> )*x^5 > -(13779/640)*x^4+(11875/288)*x^3-(121157/3360)*x^2+(1 >> 0271/840)*x >> The above closed form is exactly the same as yours >> except for the term >> q(r(n)-1), which corrects the error when n is a >> multiple of 8. >> However, as discussed, it doesn't solve the OP's >> actual problem. >> Nevertheless, it is interesting (if only for the >> ugliness of f). >> quasi or in tommynotation deny (2n) U deny (3n) U deny (5n) >= deny(2^x 3^y 5^z) and it appears to have a closed form , even for 1 > variable. >in terms of a polynomial. >although we also need the floor function. No, Goddard's formula does not generate the > complement (in N) of {2^x*3^y*5^z} Rather it generates (with my minor correction) the > set of positive > integers relatively prime to 30. quasi >>i was rather referring to bart goddard's formula. >>yet your formula is intresting too. The formulas are the same (mine is his with a minor correction). Neither one is deny (2^x*3^y*5^z) Both try to generate the n'th term of the sequence of positive > integers which are relatively prime to 30. Goddard's formula is correct except when n is a multiple of 8. When n > is a multiple of 8, his formula is off by 1. All I did was make a minor modification, keeping the same essential > form but correcting the error for the case where n is a multiple of 8. My point is that neither of the functions has, as its positive integer > range, the complement of {2^x*3^y*5^z}. quasi >>This is very interesting. How did you arrive at this closed form? I stole it. More precisely, I took Bart Goddard's form and simply >corrected a minor error. >g(n) = 30*q(n) + f(r(n)) + q(r(n)-1) for nth number relatively prime with >>30.=> LCM(2,3,5) Note that it's not a closed form for your n'th term of your sequence, >however it's almost a closed form for the n'th term of the >complementary sequence from your original post, before you corrected >it. >I could just see that q is the quotient and r is the reminder when n is >>divided by 8. Right. Note that q(n) can also be written as [n/8] where the square >brackets denote the floor function: [x] = the greatest integer <= x >I can just see that q(r(n) - 1) can only be zero or -1 since r(n) belongs >>[0,7]. Right. It's zero unless n is a multiple of 8, in which case it's -1. >I am still puzzled about the 8 and f(x). Good question. Since I just adapted Bart Goddard's formula, I'm sure Bart can >probably give a better answer, but I'll give a very rough idea (with >no details) of how I think it can be derived. However, if you want to take a shot at it yourself, I think it would >be an interesting (although somewhat tedious) exercise to work through >the details. If you have a computer algebra system such as Maple or >Mathematica, you can be spared from some of the more tedious >calculations. Here's the plan ... To get started, consider the following simple problem: How many positive integers between 1 and 1000 inclusive are multiples >of 2 or 3 (or both)? Next, do the same for 2, 3, 5 -- that is, how many positive integers >between 1 and 1000 inclusive are multiples of at least one of the >numbers 2,3,5. The method is called the method of inclusion-exclusion. After that, change 1000 to an unknown, k say. Thus, try to answer the >question: How many positive integers in the range 1 to k inclusive are multiple >of at least one of the numbers 2,3,5. It's ok to use the floor function in building the answer, Let call the answer w(k). Then k - w(k) is the number of positive integers between 1 and k which >are relatively prime to 2,3,5. Next, try to solve the equation k - w(k) = n for k in terms of n. In general, there may be more than one solution, in which case, you >want the least solution, call it k1. Thus, assume you have found k1 as a function of n. Then k1 is the n'th term of the sequence of positive integers, >relatively prime to 2,3,5. Forget my proposed derivation above. It might work, but Bart Goddard's reply shows a far simpler, more natural approach. quasi === Subject: Re: Number Theory based Series : deny > Since I just adapted Bart Goddard's formula, I'm sure Bart can > probably give a better answer, I'm not sure what the question was, exactly, but I can explain the formula. A number is not divisible by 2, 3 or 5 iff it is prime to 30. There are 8 congruence classes prime to 30 mod 30. So every group of 30 consecutive integers contains exactly 8 ugly (in this sense) numbers. If I want the nth ugly number, I get 8 ugly numbers for each group of 30 integers, so I start with [n/8]*30, which gets me in the ballpark. The number A=n-8[n/8] is how many more ugly numbers I need (i.e., n= A mod 8). I used Maple to pass an 8th degree polynomial through the points (0,0), (1,1), (2,7), (3, 11), (4,13), (5,17), (6,19), (7,23), (8,29), and called the polynomial f(x). So if A=3, f(A) = 11. So [n/8]*30 + f(A) is the answer. quasi says there's an error in f(A). I didn't check and I didn't save the Maple worksheet. At first I thought I probably just typed something wrong, but looking at my last paragraph, I see that I'm not counting 29 mod 30. B. -- Cheerfully resisting change since 1959. === Subject: #44 Addition and Multiplication in Infinite Integers/P-adics ; new textbook; Mathematical-Physics (p-adic primer) for students of age 6 onwards I am learning also in the process of writing this textbook. I did not sit down and say last month that I have all ideas on this textbook ironed out and ready to go to write. So much of this textbook is a discovery process. One of the best outcomes of organizing your thoughts into a textbook is that it raises alot of questions for which forces you to discover an answer. Previous posts I said the Angle is important for either addition or multiplication. But now I am having doubts about bringing in angle for either multiplication or addition. The reason being is that my definition of addition and multiplication as Reals operation where the final answer is what digits remain the same after successive multiplying or adding yields an unexpected result that the answer never goes beyond .....999999. So I multiply any p-adics and the product never exceeds ....99999. Or, if I add any two P-adics, the summation never exceeds .....9999999. So, from the fact that the product nor summation ever exceeds the largest integer (with perhaps a finite string from radix point) of ....999999 then I need not even introduce Angles as part of the definition of add, multiply. And whether I need to introduce angles in the add or multiply of different Adic bases such as 5-adics with 17-adics. Well, I think that is straightened out by the idea that a different base is just a different sphere. A sphere at any instant of time is one specific P-adic such as all 10-adics or all 3-adics but not a mix of different types of bases. So we picture the different bases of P-adics as disjoint spheres, each having their own P-adic base. It is the symmetry that drives me to think that since add and multiply yield all answers equal to or less than the largest integer of that base-p-adic. So the ANGLE is more of a help but not intrinsic in the definition of add or multiply. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: #45 differences between the old math looking at P-adics and this textbook looking at P-adics; new textbook; Mathematical-Physics (p-adic primer) for students of age 6 onwards Somewhere in this textbook should be a chapter on the differences of the old math of how they looked upon P-adics and the way this textbook looks at P-adics. I should make a summary here and figure out in later editions where to place this chapter. Whether at the beginning or middle of the book. In summary form the old math community from Hensel to 1993 saw the P- adics as a mere *extension* of the Reals. As if the P-adics were just a different label for rational and irrational Reals. Instead of writing the Real number of (-1) they thought the 10-adic of ....99999 was just another way of writing (-1). They thought the 10-adic of ....0000001 was just another way of writing the Real Number 1.00000..... The trouble with much of 20th century mathematics was its terribly over attention and hype to Algebras. Where they hyped fields and rings of algebra, losing sight of other features that are more important. The major difference between the old way of looking at P-adics and the way this textbook teaches is that P-adics are different and independent of the Reals. Whereas from Hensel to 1993, all mathematicians had the opposite view that P-adics were dependent on the Reals since the Counting Numbers gave the prime base, and thus the P-adics were an extension of the Reals. Just as the Complex Numbers, were thought to be an extension of the Reals, where this book teaches that the Complex are part of the P-adics. So the major difference is that this book says the P-adics are independent and different from the Reals and that all Numbers in mathematics come from either P-adics or come from Reals, because all numbers have two and only two choices: Either they are infinite strings rightward or infinite strings leftward. There are no Doubly-infinites. Since all numbers are one or the other, then, all numbers are either a P-adic or a Real. Just as in from Wave and the two comprise all things. The same symmetry and dualism for P-adics and Reals as or Wave. And a great example and byproduct of this new found wisdom is that Geometry also has this symmetry of two things. Either it is Euclidean or it is the combination of Elliptic/Hyperbolic. So the Reals have to be the intrinsic coordinate system for one of those geometries and it is quite clearly Euclidean. And that leaves the P-adics to be the instrinsic and native numbers that compose all the points of both Elliptic combined with Hyperbolic geometry. A model for this combination of both Elliptic and Hyperbolic is the sphere semispheres and where +P- adics are Elliptic while -P-adics are Hyperbolic. I take note of a fanciful picture of the alleged old math way of picturing the p-adics as the frontis page of Koblitz's book p-adic Numbers, p-adic Analysis, and Zeta-Functions Springer 1984. It does not surprize me that the artist's conception of 3-adics is a circular disc shaped object. This book drives home the point that the P-adics form the surface of a sphere and are the points of that sphere. Whereas Reals form Euclidean geometry. I should make those statements stronger for alot of people think that Euclidean geometry just found the Reals as a convenient way of representing the points in Euclidean geometry. What duality in Quantum Mechanics implies is that mathematical Numbers and Geometry are integral parts of one another. That given P- adics, they create a Elliptic/Hyperbolic geometry or given a Elliptic/Hyperbolic geometry then the only Numbers that fit that geometry are P-adics. And so, an example of how wrong it is for the old fogey mathematicians to say the P-adics are an extension of the Reals is like saying the Elliptic or Riemannian geometry is an extension of Euclidean geometry. There is almost nothing similar between Elliptic geometry and Euclidean geometry So the big mistakes of mathematicians prior to 1993 concerning P-adics is that they thought P-adics were an extension of Reals when in fact they are independent and totally different. And secondly, they payed too much attention to the Algebraic field and ring features. And probably the extension idea fed into the Algebra hype and vice versa, so that no deep understanding of what the P-adics really were. So this textbook is the first book in the world that outlines the differences between Reals and P-adics. There are many new innovations in this textbook and the primary innovation is the ability to see all the P-adics in decimal notation. So that we can add, multiply, subtract and divide just as we do those operations in Reals. This book is not going to do operations in all sorts of bases such as 2-adics and 3-adics and 10-adics etc etc. This book allows a student to get it all from one single P-adic which is the decimal place just as Reals are written in decimal place value. This textbook once completed and after a few editions will replace the old textbooks, not only on P-adics but the silly notion of how the Reals are created from Counting Numbers to Rationals to Irrationals and finally a Dedekind Cut to form Reals. How much more simple it is to teach that the Reals are simply all possible digit arrangements of infinite rightward strings. The old way of developing the Reals gives no clue as to the other great Number system that is symmetrical to the Reals-- the P-adics. How much easier it is to say Reals are infinite rightwards and so there exists a equally large rival number system called P-adics which are infinite leftwards and you have the symmetry. Why did the old mathematicians stick to their oh, its a mere extension and never seem to have turned on the light that these are different and independent? If we look at the history of physics in the 20th century we can easily compare why and how a community falls off the correct path. In the 20th century there was a split between those physicists who were adoring of General Relativity that they pushed Quantum Mechanics aside, and so in the 20th century physics went wildly off course and off track as to true physics. Similarly, because so much focus by the old mathematicians on whether something is an Algebraic Field or Algebraic Ring that it set mathematics backward and unable to see correct mathematics. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: #45 differences between the old math looking at P-adics and this textbook looking at P-adics; new textbook; Mathematical-Physics (p-adic primer) for students of age 6 onwards you big dope; Hensel did not extend the reals with p-adics; they are entirely nonarchimedean; his only mistake was listing them rightwards, as if they were decimals! in deed, the real integers can be thought-of as infinite-adics, as is shown in most of the introductory p-adic thingies; just because they happen to be the archimedean, valuationable corner of adics, doesn't mean that adics extended them. no amount of revising your outline is ever going to result in doable (experimental) physics. well, you're probably here just to waste our time; success! thus: p-adics are like a box of chocolates ... inside or outside, shape, time-stamp, partitioning, pigeonholing et c. et c. et cetera. --14 Italian Senators Call for Cheney Impeachment Aug. 1, 2007 (EIRNS)- The Lyndon LaRouche Political Action Committee (LPAC) issued the following release today. Fourteen members of the Italian Senate have signed a call to the Members of Congress to support Rep. Kucinich's House Resolution 333 for the Impeachment of Dick Cheney. http://larouchepub.com/pr/2007/070801italian_senators_call.html === Subject: distribution of Mersenne prime exponents For the 44 known primes p such that 2^p - 1 is prime, I plotted the n'th known prime exponent against the integer n. So: 1 -> 2 2-> 3 ... 13 -> 521 ... 23 -> 11213 and so on up to n=44. I call the n'th (known) exponent E_n. The x-axis, showing 'n', is linear. The y-axis, showing E_n, is logarithmic. I got the data from Chris Caldwell's Prime Pages. I tried to get a good linear fit to log(E_n) in terms of n. This fit, g(n), can be expressed as: g(n) = 2.74* Pi^(n/3). (1) This is almost surely numerology, but it's easy to remember. I'm pretty sure 0.4 < g(n)/E_n < 2.5 holds for all or almost all n from 1 to 44. I wanted to know if the log(E_n) values would be likely to occur, as a statistician would look at it, from a Poisson process. Poisson processes are discussed, for example, in Sheldon Ross' ``Introduction to Probability Models. An example is counting Geiger counter clicks. Let T_n be the time at which the n'th click is heard, starting from an independently chosen point in time (say, the next time the phone rings). {T_n}_{n=1, 2, 3, ...} is well modeled by a Poisson process if conditions don't change and the half-life is very long. log( g(n)/2.74 ). [ H(n) itself is not the Poisson process. H(n) is the n'th arrival time for the Poisson process ... ] Viewing H(n) as a random variable, let G(n) = exp( 2.74*H(n)). I did a simulation of G(n), n=1,... 50. Actually, I did 2 or 3 until G(50) came pretty close to g(50). This gave a sample, say s_1, .... s_50. I also plotted s_1, ... s_45 on the same graph as E_n and g(n). I think the values log(E_n) stay too close, too often, to log(g(n)) for the log(E_n) to be likely in a Poisson process. This is said after seeing how s_1, ... s_45 wander away from g(n) quite a lot more than the E_n. There must be a goodness of fit test to a time-shifted Poisson process with unknown parameter lambda, but I haven't thought about that. The figure I uploaded to the Web has: - the true Mersenne exponents in blue, marked as stars. - the exponential fit, (1), in green marked by circles. - the Poisson process simulation in red marked by squares. By clicking on the image, one can zoom in or out. The image can be found here: < http://www.geocities.com/ezcos/MersenneDistrib.png > David Bernier P.S.: Maybe somebody did something similar previously. If so, I'd be interested in hearing about what was found, or conjectured. === Subject: Re: distribution of Mersenne prime exponents [...] > I wanted to know if the log(E_n) values would be likely to occur, as a > statistician would look at > it, from a Poisson process. Poisson processes are discussed, for > example, in Sheldon > Ross' ``Introduction to Probability Models. An example is counting > Geiger counter clicks. > Let T_n be the time at which the n'th click is heard, starting from an > independently chosen > point in time (say, the next time the phone rings). {T_n}_{n=1, 2, 3, > ...} is well > modeled by a Poisson process if conditions don't change and the > half-life is very > long. > > approximating > log( g(n)/2.74 ). > > [ H(n) itself is not the Poisson process. H(n) is the n'th arrival time > for the > Poisson process ... ] > > > Viewing H(n) as a random variable, > let G(n) = exp( 2.74*H(n)). > > I did a simulation of G(n), n=1,... 50. Actually, I did 2 or 3 until > G(50) came pretty close to > g(50). This gave a sample, say s_1, .... s_50. > > I also plotted s_1, ... s_45 on the same graph as E_n and g(n). > I think the values log(E_n) stay too close, too often, to log(g(n)) for > the > log(E_n) to be likely in a Poisson process. This is said after seeing > how s_1, ... s_45 wander away from g(n) quite a lot more than the > E_n. There must be a goodness of fit test to a time-shifted Poisson > process > with unknown parameter lambda, but I haven't thought about that. I've formulated a statistical inference problem where two parameters are unknown. P(t)_{t in R} is a Poisson process with parameter lambda, so that on average, lambda events occur per unit of time. There is a second parameter t_0. The m items of data are the first m arrival times after time t_0: T_1 < T_2< ... < T_m. So the Poisson process started in the distance past and one could use negative indices as well and obtain: T_{-2} < T_{-1} < T_{0} <= t_0 < T_1 < .... The statistical problem is to find estimators for lambda and t_0 based only on what is known: T_1, .... T_m . If one simplifies the problem with the assumption that t_0 = 0 always, an obvious estimator for lambda would be lambda^ = (T_m - T_1)/(m-1). Another estimator might be based on the sum of squares: sum_{j=2, ... m} (T_j - T_{j-1})^2 . David Bernier > The figure I uploaded to the Web has: > > - the true Mersenne exponents in blue, marked as stars. > - the exponential fit, (1), in green marked by circles. > - the Poisson process simulation in red marked by squares. > > By clicking on the image, one can zoom in or out. > > The image can be found here: > > < http://www.geocities.com/ezcos/MersenneDistrib.png > David Bernier > > P.S.: Maybe somebody did something similar previously. If so, I'd be > interested > in hearing about what was found, or conjectured. > === Subject: Point on a surface I'm trying to caluate the 3rd dimention of a point on a surface. So I've got N points (x,y,z) that are all on a surface, and I'm tying to find a formula to calculate 3rd dimention of another point (x,y,z) that is to say x and y are known and I'm looking for z (so that this point is then on the surface). Hope that is clear, Fraser === Subject: Re: Point on a surface P1=(x1,y1,z1) P2=(x2,y2,z2) P3=(x3,y3,z3) 3 points on surface (plane) N=(P1-P2)x(P3-P4) cross product P=(x,y,z) any point on surface (plane) (P-P2)*N=0 dot product equation of plane P4=(x4,y4,z4) point not on plane d=(P2-P4)*N/|N| distance from point to plane P5=P4+d N/|N| projection of point on plane Jon Giffen A.A.S. Electrical/Electronics Engineering Technology, Belmont Technical College. > I'm trying to caluate the 3rd dimention of a point on a surface. So I've got N points (x,y,z) that are all on a surface, and I'm tying > to find a formula to calculate 3rd dimention of another point (x,y,z) > that is to say x and y are known and I'm looking for z (so that this > point is then on the surface). Hope that is clear, Fraser > === Subject: Re: Point on a surface > I'm trying to caluate the 3rd dimention of a point on a surface. > > So I've got N points (x,y,z) that are all on a surface, and I'm tying > to find a formula to calculate 3rd dimention of another point (x,y,z) > that is to say x and y are known and I'm looking for z (so that this > point is then on the surface). > > Hope that is clear, Not very much, but I think that I understood it. If your N points are all on a surface, then there are real numbers _a_, _b_, _c_, and _d_ such that a*x + b*y + c*z = d whenever (x,y,z) is one of those points. Find them. Then, if _x_ and _y_ are known and you want to find _z_ such that (x,y,z) belongs to your surface, just solve the equation a*x + b*y + c*z = d. Jose Carlos Santos === Subject: Re: Point on a surface <5mh1tjF9ten4U1@mid.individual.net > I'm trying to caluate the 3rd dimention of a point on a surface. So I've got N points (x,y,z) that are all on a surface, and I'm tying > to find a formula to calculate 3rd dimention of another point (x,y,z) > that is to say x and y are known and I'm looking for z (so that this > point is then on the surface). Hope that is clear, Not very much, but I think that I understood it. If your N points are all on a surface, then there are real numbers a , > b , c , and d such that a*x + b*y + c*z = d whenever (x,y,z) is one > of those points. Find them. Then, if x and y are known and you want > to find z such that (x,y,z) belongs to your surface, just solve the > equation a*x + b*y + c*z = d. > Jose Carlos Santos Just as a note: JCS has assumed here that the surface is a plane. (and the solution he gives is correct for that surface) You also only need 3 non-collinear points on the plane to get the equation for it: If you call the points A, B, and C, then form the vectors AB and BC. Their cross product will be a normal to the plane . The equation of the plane is thus dx + ey + fz = g. Solve for g using any of the given points and the normal. If the surface is not a plane, then we need to know what sort of surface it is (e.g. a torus). I assume you don't have an equation describing the surface, since then your problem would be simply solving that equation for z (which may be easy or hard depending on the nature of the equation). If you don't know that, then the best you can do create some form of interpolating polynomial z = f(x,y) that agrees on the given points. If there are a lot of given points, finding such a polynomial will be possible, but will take a long time. === Subject: A complex transcendental equation Content-Length: 1307 Originator: rusin@vesuvius This is an expanded and corrected version of my previous post. This problem arises from applying the method of steepest descents to estimating quadrature error. Given a function g(z), (either g1(z)=log(e^z+1) or g2(z)=z/(1-e^-z)), I define p(z) = i ( z + g(z) ) and need to find the critical points of p(z), that is, solutions of p'(z)=0. For g1(z), it is easy to derive that the critical points are: -log(2)+(2k+1) pi i. The problem I have is for g2(z). In which case we can reexpress p'(z)=0 as the transcendental equation: 1+2 e^(2z)-e^z(3+z)=0. Using Mathematica's FindRoot routine and starting from -1 + 4 i I find that: -1.69329+i 4.93943, is an approximate solution of this equation. Now, this is approximately -1-log(2)+i pi^2/2 = -1.69315 + 4.93480 i (to 5 decimal places). Can anyone give a justification for these two numbers being so close? Ideas I have are: a) introduce a parameter then make an asymptotic expansion, perhaps using (e^z-1)/z ~ 1 b) use an expansion of g2(z) about i pi or 2 pi i c) solve using the product log (or Lambert W) function defined by W(z) e^W(z) = z, then use the expansion of W(z) W(z) ~ log z - log log z +... (see http://mathworld.wolfram.com/LambertW-Function.html) === Subject: radius of cylinder hi if a rectangular paper joining the 19cm edges(like a a4 sheet) to make a cylinder.what will be its radius of circular base?i tried but i could t get da correct answer. === Subject: Re: radius of cylinder >hi if a rectangular paper joining the 19cm edges(like a a4 sheet) to >make a cylinder.what will be its radius of circular base?i tried but i >could t get da correct answer. The circumference C and radius R of a right circular cylinder are related by the formula C = 2 pi R, so R = C / (2 pi) If the 19 cm edge you mentioned above forms the circumference of the cylinder, then R = 19 / (2 pi) = 3.02 cm (approximately) === Subject: Re: radius of cylinder- > >> hi if a rectangular paper joining the 19cm edges(like a a4 sheet) to >> make a cylinder.what will be its radius of circular base?i tried but i >> could t get da correct answer. > > The circumference C and radius R of a right circular cylinder are > related by the formula > > C = 2 pi R, so > > R = C / (2 pi) > > If the 19 cm edge you mentioned above forms the circumference of the > cylinder, then > > R = 19 / (2 pi) = 3.02 cm (approximately) > Not necessarily true: The pair of sides =not glued together= will make circular base and top of the cylinder. Only if these are 19 cm too then the diameter is 3.02 cm. sides are in the ratio of sqrt2 to 1. The sides are therefore 29.73 cm and 21.02 cm. Joining the long sides with an overlap of 0.02 cm yields a cylinder with a circumference of 21.00 cm, i.e. a radius of 3.342 cm. Joining the short sides with an overlap of 0.73 cm yields likewise a circumference of 29.00 cm, i.e. a radius of 4.616 cm. Now try and glue a Klein bottle from an A4 sheet and compute its surface area and its volume! Ciao: Johan E. Mebius === Subject: [] radius of cylinder > hi if a rectangular paper joining the 19cm edges(like a a4 sheet) to > make a cylinder.what will be its radius of circular base?i tried but i > could t get da correct answer. Here's your post with proper punctuation. to make a cylinder. What will be its radius of circular base? I tried but I could't get the correct answer. Please study it to know better how to punctuate your posts. Usually posts written like yours I dismiss as not capable of mathematics. === Subject: Re: [] radius of cylinder > hi if a rectangular paper joining the 19cm edges(like a a4 sheet) to >> make a cylinder.what will be its radius of circular base?i tried but i >> could t get da correct answer. Here's your post with proper punctuation. > a a4 sheet? >to make a cylinder. What will be its radius of circular base? I >tried but I could't get the correct answer. could't? >Please study it to know better how to punctuate your posts. Usually >posts written like yours I dismiss as not capable of mathematics. Since when are posts capable of mathematics? quasi === Subject: Re: [] radius of cylinder hi if a rectangular paper joining the 19cm edges(like >> a a4 sheet) to make a cylinder.what will be its radius >> of circular base?i tried but i could t get da correct answer. > Here's your post with proper punctuation. a a4 sheet) to make a cylinder. What will be its radius > of circular base? I tried but I could't get the correct > answer. I think his (her?) post needs a lot more editing work: I offer my greetings to sci.math and request that you give me some mathematical feedback on the following problem. Consider a rectangular sheet of paper, such as an a4 sheet of paper. The paper is rolled along one pair of parallel sides in order to join together the other pair of parallel sides to make a right circular cylinder. If the length of each of the sides that were not joined together is 19 cm, what is the radius of cylinder? I know the correct answer by using a non-mathematical method. However, I have not been able to obtain the correct answer by using a mathematical method. I would like someone to post, in this sci.math thread, a mathematical method for obtaining the radius of the base of the cylinder. Dave L. Renfro === Subject: Re: [] radius of cylinder Hi vaji.shan and welcome to the forum. I apologise for our rather rude friend below. With regard to your problem, in fact you have not given enough information to answer it. The 19cm edges only determine the height of the cylinder, not its base radius. What you would need to know is the length of the other edge, the one which you didn't join. For example, in the case of A4 paper, we join the edges of length 21cm, and the other edge has length 29.7cm, this is the one we need to work with. Now, the edge as mentioned about determines the circumference of the circle. The circumference and radius are related by the formula Circumference = 2 * Radius * Pi and so we can calculate that Radius = Circumference / (2 * Pi) In the case of A4 paper, this is about 4.73cm. Adam. > > hi if a rectangular paper joining the 19cm > edges(like a a4 sheet) to > make a cylinder.what will be its radius of circular > base?i tried but i > could t get da correct answer. > > Here's your post with proper punctuation. > > (like a a4 sheet) > to make a cylinder. What will be its radius of > circular base? I > tried but I could't get the correct answer. > > Please study it to know better how to punctuate your > posts. Usually > posts written like yours I dismiss as not capable of > mathematics. > > === Subject: Solution Manual Required Hello I really need solution manuals or student's solution manuals or instructor's solution manual for the following book: - Statistics for Engineers and Scientists by William Navidi (Second Edition) ISBN: 9780071102223 === Subject: Re: Solution Manual Required i have the following solution manuals.... and i have thousands of textbooks as well email me at diosbendit...@gmail.com if you want any of them email at diosbenditome (at) gmail (dot) com paypal paymets accepted only please email me rather than leaving a message here.. 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You're always mad. KM -- (-:alohacyberian:-) At my website view over 3,600 live cameras or visit NASA, the Vatican, the Smithsonian, the Louvre, CIA, FBI, and NBA, the White House, Academy Awards, 150 language translators! Visit Hawaii, Israel and more at: http://keith.martin.home.att.net/ === Subject: Re: Evidence that the USA is becoming a police state can't let go. Once again you blatantly demonstrate that you're very even-tempered. You're > always mad. KM Why are you responding on behalf of the other fellow? Like I said, once again you just gotta scratch that itch. Fact is, Keith, it's you who is always mad. Otherwise you'd see no need to insert yourself into someone else's dispute. It's been interesting for me to see this group- think. Can't wait to see your next little tantrum, === Subject: Re: Evidence that the USA is becoming a police state >Try not to be so pedantic when you reply again, assuming of course you >can't let go. Once again you blatantly demonstrate that you're very even-tempered. You're > always mad. KM > Why are you responding on behalf of the other fellow? Like I said, > once again you just gotta scratch that itch. Fact is, Keith, it's you > who is always mad. Otherwise you'd see no need to insert yourself into > someone else's dispute. It's been interesting for me to see this group- > think. Can't wait to see your next little tantrum, Hey guy, you're the one shooting blanks, virulently sure, but they're still blanks, with you saying the same things over and over and over and over again for days. Funny thing is, you've been criticising me in all kinds of posts to all kinds of people (in the third person 'he this, he that'), and now you blast *others* because they 'dare' speak of this? Get over yourself. You do have a redeeming quality, that of caring so very, very deeply. Could be a good thing. But offended for so long and so publicly? Mistake. Implies you've got hormones and peptides aboil that should be cavorting happily in that soup we'll call your current emotional state. Says you have issues not so much with me, but with how you think others need to see you. That even the small things get you into a state of near-hysteria and make you shriek, whine and bleat for days on end. It's your 'Waaa! Waaa!' moment. Do you see the humor in that, Flynn? No? Well, you've been begging for it, especially when you meet quips 'head on' (cough cough) with clever retorts like 'third grade' and 'I've got it all over you'. If you can't respond well, either just ignore it or laugh *with* the other. Best thing for you now is to go lie in that hole you've dug and rest-up. You deserve it. 'A' for effort, I say. Man of the week, our Flynn. Nex === Subject: Re: Evidence that the USA is becoming a police state Try not to be so pedantic when you reply again, assuming of course you >can't let go. > Once again you blatantly demonstrate that you're very even-tempered. You're >always mad. KM > Why are you responding on behalf of the other fellow? Like I said, > once again you just gotta scratch that itch. Fact is, Keith, it's you > who is always mad. Otherwise you'd see no need to insert yourself into > someone else's dispute. It's been interesting for me to see this group- > think. Can't wait to see your next little tantrum, Hey guy, you're the .... <> Hey, Alan, you lost. Deal with it. Try being nice to the next person you feel like abusing, eh? === Subject: Re: Evidence that the USA is becoming a police state > and be nice, then no problem. It's typical how you bully-types stick > Calling others bully-types is an archetypal example of what psychologists and psychiatrists call projection and others refer to as the pot calling the kettle black. KM -- (-:alohacyberian:-) At my website view over 3,600 live cameras or visit NASA, the Vatican, the Smithsonian, the Louvre, CIA, FBI, and NBA, the White House, Academy Awards, 150 language translators! Visit Hawaii, Israel and more at: http://keith.martin.home.att.net/ === Subject: Re: Evidence that the USA is becoming a police state and psychiatrists call projection and others refer to as the pot calling > the kettle black. So then, by your reasoning, you are an ugly, potty-mouthed ignorant hysteric who is always angry and gullible due to mental instability, and an uneducated and stupid lying lunatic? Those are just *some* of the nice things you've had to say in this thread alone about others. Welcome to the club, Keith! I guess that makes Bush a brutal and cruel dictator. He's just projecting! No, the fact is you are wrong. Otherwise, no one could ever call an actual bully a bully. And when someone pops into a conversation to try to dictate his rules for posting and calls the two people lazy and stupid idiots for not following his until-then unstated rules of usenet, I can call him a bully without any second thoughts. Sometimes, it just is what it is. === Subject: Re: Evidence that the USA is becoming a police state asshole. Wanna be nice? Try it. > Why don't you take your own advice? Don't know how? KM > It's like this: I responded in kind; should he or you recognize that >and be nice, then no problem. It's typical how you bully-types stick >together and all, but you really don't know how to handle it when >someone shoves it right back at you. > Get over it. What I'd really like is evidence of your claims in the taser incident. > For me to go look them up is not completing the debate. > It is your burden to support your claims. No it isn't. This is usenet. You want to look into if further, have at > it. I don't need to spoon feed you. In any case, I already gave you a > link for you to start with. Why haven't you looked at it? Best of all is that when you make them to substantiate them. > Simply include a link with thehttp://partin it and we will all be > able to see your evidence. Google, dude. Google. You say > The guy planned to make a disturbance. He's apparently known for > it. > That's why he had the camera there and asked the woman if she was sure > she was taping it. Where is this information coming from? Like Dave told you above. It's been around. The link I gave you > earlier should have given you more than enough to start with. Pork is known to lie. In fact they are trained in it. In that they are the first in the line of justice then the entire system is built on lies. I myself have experienced this three-fold in Lincoln County, Maine. The freedoms that porkers take are invalid and so invalidate treat. -Tim === Subject: Re: Evidence that the USA is becoming a police state On Oct 3, 12:19 am, Timothy Golden BandTechnology.com >asshole. Wanna be nice? Try it. >Why don't you take your own advice? Don't know how? KM > It's like this: I responded in kind; should he or you recognize that >and be nice, then no problem. It's typical how you bully-types stick >together and all, but you really don't know how to handle it when >someone shoves it right back at you. > Get over it. > What I'd really like is evidence of your claims in the taser incident. >For me to go look them up is not completing the debate. >It is your burden to support your claims. No it isn't. This is usenet. You want to look into if further, have at > it. I don't need to spoon feed you. In any case, I already gave you a > link for you to start with. Why haven't you looked at it? > Best of all is that when you make them to substantiate them. >Simply include a link with thehttp://partinit and we will all be >able to see your evidence. Google, dude. Google. > You say > The guy planned to make a disturbance. He's apparently known for >it. >That's why he had the camera there and asked the woman if she was sure >she was taping it. > Where is this information coming from? Like Dave told you above. It's been around. The link I gave you > earlier should have given you more than enough to start with. Pork is known to lie. In fact they are trained in it. In that they are > the first in the line of justice then the entire system is built on > lies. I myself have experienced this three-fold in Lincoln County, > Maine. The freedoms that porkers take are invalid and so invalidate > treat. So this redirection means you were never interested in what actually happened? === Subject: Re: Total differential of y = f ( x, y ) ? > Is the total differential of y = f ( x, y ) equal to dy = dx @f /@x > + dy @f /@y which would mean > > dy = dx @f /@x / ( 1 - @f /@y ) ? > So you want to find y as a function of x that solves the equation y = f(x) and compute its derivative w.r.t. x. Just use the chain rule: dy/dx = @f/dx + @f/dy * dy/dx. So dy/dx = @f/dx / (1 - @f/dy). So, you're right. Here's an example: f(x,y) = x + y/2, y = f(x,y). dy/dx = 1/(1-1/2) = 2. Compare this with the direct calculation of dy/dx. Since y = x+y/2, y = 2x. Therefore dy/dx = 2. Careful though, y = f(x,y) need not have a solution. Ex. y = x + y. That leads to dy/dx = 1 / (1 - 1). -- Thomas Nordhaus === Summary: UDP === This server is temporarily under UDP because of recent massive hipcrime attacks, until the problem is solved by their administrators. We are sorry for the inconvenience. See news.admin.net-abuse.policy and news.admin.net-abuse.usenet for more details. === Subject: Re: Automorphisms and Inner Automorphisms > , > > Prove the Aut(G) and the Inn(G) is a group. > > Homework-wise, how far have you gotten? > > For example, in the case of Aut(G), if f is an > automorphism, and g is an > automorphism, what can you say about a o b (a > composed with b)? Is a o b > an automorphism? If yes, then Aug(G) is closed under > composition. If no, > then it's not, and Aug(G) can't be a group. So first, > you need to > convince yourself that Aut(G) is closed under > composition. > > Is composition associative? Is there an identity > element? Are there > inverses? > > That's plenty, hintwise. > > Then you do the same with Inn(G), except that the > elements of Inn(G) > have a particular form, so you can use that to prove > the group > properties. In addition, clearly Inn(G) is a subset of Aut(G), so you can use any results you know about quickly proving a subset of a group is a subgroup. For example, you need only show that it is non-empty and that a, b in Inn(G) implies a b^{-1} in Inn(G). === Subject: ,,...,.,.,,,.,.,NEWS ALERT,.,.,.,.,,,,....... ,,...,.,.,,,.,.,NEWS ALERT,.,.,.,.,,,,....... For those who like to see the action in motion, a free site dedicated to movies of legal age teens doing what comes natural. Best free movie site on the net. FOR WATCHING THIS CLIP, CLICK THIS LINK AND WHEN SITE OPENS, CLICK ON THE SECOND AD AT THE TOP. THE MOVIE WILL START AUTOMATICALLY. http://www.urlco.net/hotclips Tons Of Series Hardcore - pics and movies - all of babes in uniforms - doing the nasty to some lucky guy. http://www.urlco.net/sexygirls See big boobed bimbos doing everything you can imagine, from bouncing up and down to squeezing some lucky guy's cock! http://www.urlco.net/bigtits ---------------------------------------------------- === Subject: Sequence & arithmetic mean limit proof? I need to prove: if {s_n}_n>=0 is a real sequence such that lim s_n = +(infinity) then lim sigma_n = +(infinity) where the arithmetic means (sigma_n) are defined as: sigma_n = (s_0 +s_1 + ... + s_n)/(n+1) *note lim means limit as n->+(infinity) -------------------------------------- It seems like you get [s_0 + s_1 +...+(infinity)]/(infinity) = +infinity I think I should get the top approaches infinity quite a bit faster than the bottom does, since it diverges. maybe more like: [s_0 + s_1 +...+(infinity)]/n = +infinity But I'm not really sure how to approach this. === Subject: Re: Sequence & arithmetic mean limit proof? > I need to prove: if {s_n}_n>=0 is a real sequence such that lim s_n = +(infinity) then lim sigma_n = +(infinity) where the arithmetic means (sigma_n) are defined as: > sigma_n = (s_0 +s_1 + ... + s_n)/(n+1) *note lim means limit as n->+(infinity) > -------------------------------------- It seems like you get > [s_0 + s_1 +...+(infinity)]/(infinity) = +infinity I think I should get the top approaches infinity quite a bit faster than the bottom does, since it diverges. maybe more like: > [s_0 + s_1 +...+(infinity)]/n = +infinity But I'm not really sure how to approach this. Choose an arbitrary positive real value. Since lim s_n = +(infinity), there is a point at which all remaining terms of the sequence exceed the chosen value. You can then show that the partial sums of the series and the partial averages must have a lower bound, that they must eventually exceed zero, and that the averages must eventually exceed a given fraction (e.g. a half) of the chosen value. But since the chosen value was arbitrary, this means that the limit of the averages must exceed all real values, i.e. be infinite. === Subject: Re: Sequence & arithmetic mean limit proof? >I need to prove: if {s_n}_n>=0 is a real sequence such that lim s_n = +(infinity) then lim sigma_n = +(infinity) where the arithmetic means (sigma_n) are defined as: >sigma_n = (s_0 +s_1 + ... + s_n)/(n+1) *note lim means limit as n->+(infinity) >-------------------------------------- It seems like you get >[s_0 + s_1 +...+(infinity)]/(infinity) = +infinity No, you don't get that. That's meaningless. You need to start with the _definition_ of a_n tends to infinity. That clarifies exactly what you need to prove and also exactly what the hypothesis means. What _is_ that definition? >I think I should get the top approaches infinity quite a bit faster than the bottom does, since it diverges. maybe more like: >[s_0 + s_1 +...+(infinity)]/n = +infinity But I'm not really sure how to approach this. ************************ David C. Ullrich === Subject: College Physics, Volume 1: 7th Edition, by Serway, Faugh I need .pdf solutions for : top dollar College Physics, Volume 1: 7th Edition, by Serway, Faugh === Subject: i need the solution for Advanced Engineering Mathematics 9th edition by Erwin Kreyszig i need the solution for Advanced Engineering Mathematics 9th edition by Erwin Kreyszig please mail me if you have it thank you very much === Subject: finding the power of the number hi to all, consider (2^x=32) what will be the value of x dont say just 5 i want the steps to evaluate the expression === Subject: Re: finding the power of the number > hi to all, > > consider (2^x=32) > > what will be the value of x > dont say just 5 > i want the steps to evaluate the expression > > 2^x = 2*2*...*2 (x factors) 32 = 2*2*2*2*2 (5 factors) --------------------------- x = 5 This works easy but it would fail for 33 instead of 32. Here is the step-by-step procedure: 2^x = something // take logarithms on both sides x * log(2) = log(something) // divide by log(2) x = log(something) / log(2) Example (with log base 10): x = log(32)/log(2) = 1,50515/0,30103 = 5. Or with base 2: 5/1 = 5. Or with base e: ln(32)/ln(2) = 3,465735903/0,693147181 = 5. Rainer === Subject: Re: finding the power of the number > hi to all, consider (2^x=32) what will be the value of x > dont say just 5 > i want the steps to evaluate the expression > x = log(32)/log(2) === Subject: Re: finding the power of the number hi to all, consider (2^x=32) what will be the value of x > dont say just 5 > i want the steps to evaluate the expression > x = log(32)/log(2) thank u very much vimal developer engineer, Binary Karma Enterprise === Subject: Re: finding the power of the number : > x = log(32)/log(2) : thank u very much : vimal : developer engineer, : Binary Karma Enterprise You're an engineer (a development engineer, whatever that is) and you didn't know how to answer 2^x=32? I'm scared. Don't go building any bridges. Justin === Subject: solution manual Cc: algorithms does any one have the solution manual for Algorithm Design by Eva Tardos.... === Subject: unique factorization of elements of finite fields Cc: inge@science.uva.nl I am looking for an easy accessible proof of the following theorem (if this is true at all): in a finite field F of size p^n (p prime), every element e can be uniquely represented as e = c_1b_1 + ... c_nb_n where {b_1, ..., b_n} is a basis and c_i in Z/pZsubseteq F. === Subject: Re: unique factorization of elements of finite fields > I am looking for an easy accessible proof of the following theorem (if > this is true at all): in a finite field F of size p^n (p prime), > every element e can be uniquely represented as e = c_1b_1 + ... > c_nb_n where {b_1, ..., b_n} is a basis A basis of what? My guess is that you mean that it is a basis of F seen as a Z/pZ-vector space. > and c_i in Z/pZsubseteq F. This is trivially true, by the definition of basis. Perhaps that I misunderstood what you meant. Jose Carlos Santos === Subject: Re: unique factorization of elements of finite fields <5mhi0jFdk8jrU1@mid.individual.net > I am looking for an easy accessible proof of the following theorem (if > this is true at all): in a finite field F of size p^n (p prime), > every element e can be uniquely represented as e = c 1b 1 + ... > c nb n where {b 1, ..., b n} is a basis A basis of what? My guess is that you mean that it is a basis of F seen > as a Z/pZ-vector space. and c i in Z/pZsubseteq F. This is trivially true, by the definition of basis. Perhaps that I > misunderstood what you meant. Perhaps he is looking for a proof that GF(p^n) in fact forms a vector space over its unital sub-field??? 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The essential points. Using essential regions and for a even number ( a >= 16 ) we find the formula for the second derivative of (A_T)^ in each sub interval [ (k_0)^, (k_0 + 1)^ ] ( k_0 = 4, 5,..., ( a/2 ) - 1 ) of the interval [ 4^, ( a/2 )^ ] , a derivative which is continuous. In the expression of ( (A_T)^ )'' intervene x_(k_0) and y_(k_0) where these, are numeric values in homogeneous polynomials of degree two obtained from substituting in their variables the a_i coefficients of the f R^+ prime coding function. We call P_(k_0) := ( x_(k_0) , y_(k_0) ) an essential point. Fernando. P.S. Previous: 1.- Transporting the arithmetic. 2.- Codifying natural numbers. 3.- Maintaining the nomenclature. 4.- The x^ y^ plane. 5.- Hyperbolas in the x^ y^ plane. 6.- Choosing adequate R^+ coding functions. 7.- Vortex points. 8.- Semi vortex points. 9.- Characterizing primes. 10.- Brief and naive summary. 11.- R^+ prime coding function. 12.- Essential regions. 13.- Classifying square essential regions. 14.- Classifying triangular essential regions. 15.- The Es( k_0 ) set. 16.- Recognizing prime numbers via essential regions. 17.- Relating sum with product: the area. http://mathforum.org/kb/thread.jspa?threadID=1632449&messageID=5935848#59358 48 === Subject: Re: Quintic Splines, Solving rapidly On Oct 2, 5:34 am, spellu...@fb04373.mathematik.tu-darmstadt.de (Peter > thinking a little bit more on this I believe that your problem simply comes from > a wrong enumeration of the unknowns: This is a good observation. Rearrange the columns of the original system [ A -B ] [ u ] = [ d ] [ C A ] [ v ] [ 0 ] to interleave the u and v vectors, and then rearrange the rows the same way, and the system is transformed into a band matrix, which you can solve with a direct (non-iterative) general band solver from, e.g., LAPACK's DGBTRF and DGBTRS. Maybe this is what Peter was saying, because it results in a matrix with a total bandwidth of 7, as did Peter's. If you have the nine diagonals representing A, B, and C, you can form the band matrix by sprinkling those 9n values around appropriately in a matrix with 2n rows and 7 columns. If you use DGBTRF and DGBTRS, you will have to store this matrix in an array with 3 additional columns, which are used for pivoting. Dave === Subject: Re: Quintic Splines, Solving rapidly On Oct 2, 3:34 am, spellu...@fb04373.mathematik.tu-darmstadt.de (Peter thinking a little bit more on this I believe that your problem simply comes from > a wrong enumeration of the unknowns: The book covers how to use the the P(i) P''(i) and P''''(i) arrays where P''=u and p''''=v The P(i), P(i+1), P''(i), P''(i+1) and p''''(i) and p''''(i+1) is used to calculate coefficients A,B,C,D,E,F for the interpolation formula y(x-x(i))=(((((A*(x-x(i))+B)*(x-x(i)) +C)*(x-x(i))+D)*(x-x(i))+E)*(x-x(i))+F. A = (P''''(i+1)-p''''(i))/(120*(x(i+1)-x(i)) B = P''''(i)/24 . D = 0.5*P''(i) . F = P(i) > then you get a seven banded matrix for which standard Gaussian elimination > should work. Actually the values in the major diagonal of the -B matrix have a bigger magnitude than the values of A matrix's major diagonal. I am hoping it will converge at all. My test case is for i = 0 to N+1 where N is 90 x=4*i; y(x)=sin(x*PI/180) The difference between the points is only 4 degrees but B matrix cubes these values so they are relatively large. > you have of course 4 degrees of freedom in this system > and make the construction unique by appropriate additional requirements. > if you use separable boundary conditions, then the total system matrix remains > banded and only in the periodic case there is a coupling between the > first three and the last three components of the solution vector. > this requires a little modification in the Gaussian elimination like the > periodic cubic spline does. but nevertheless also here you can make full use of > sparsitiy ; only the last three columns of the U-part of the LU-decompositions > get filled and the last three rows of the L-part too. Yes, I know about eliminating the zeros by having arrays for the diagonals. BTW, this is what one customer does with the cubic splines ftp://ftp.deltamotion.com/public/movies/JAN-04%20VSS_0001.wmv It is a big 74 MB file to be prepared for it to take a second. You can see there are 6 axes geared to a feed chain or rollers that are moving the wood through the chipper heads. You can see I need to calculate the splines quickly. I have the arrays setup and can calculate the A,B,C,D,E,and F coefficients. Now I must solve for u and v or P'' and P''''. Give me a day or two. I'll be back. Peter Nachtwey === Subject: Re: Confusion on differentiable functions on R^2 > >> On Oct 1, 10:53 pm, The World Wide Wade > On Sep 28, 8:05 am, The World Wide Wade On Sep 27, 10:46 am, The World Wide Wade > The book I am reading says that a function F:R^2 -> C is >> differentiable at (a,b) if: >> F(x,y) = F(a,b) + A(x-a) + B(y-b) + ep(x,y) sqrt[(x-a)^2 + >> (y-b)^2] >> where ep(x,y) is continuous at (a,b). > No, you want ep(x,y) -> 0 at (a,b). >> That would be the case if there were no sqrt factor. As it is, if >> ep(x,y) is continuous, then the limit of the term ep(x,y) sqrt[...] >> as (x,y) -> (a,b) is 0. > No, the condition you need is ep(x,y) -> 0 at (a,b). Check the > one-variable definition of a derivative for comparison. >> The definition I've always used is that >> F(x,y) - [F(a,b) + A(x-a) + B(y-b)] >> converges to 0, [...] > For heaven's sake. The definition of differentiability > is not hard to find - you really could have just > looked it up instead of relying on what you recall > and what you seem to recall some professor emphasizing. > You didn't have a textbook this weekend but you clearly > had internet access - it took me about a minute to find > the definition at > > http://en.wikipedia.org/wiki/Derivative#The_total_derivative.2C_the_Jacobian . 2C_and_the_differential One thing I was confused with at one time was functions f:R^2 ->R such as: f(x,y) = 0 is x=y=0 and f(x,y) = r^{2*pi - theta} otherwise, where 0<=theta<2*pi and r>0 are the polar coordinates of (x,y). If I'm not mistaken, f is not continuous at (0,0) since f(x,y)>1/2 for (x,y) arbitrarily close to (0,0) in the euclidean norm. And yet, for a fixed theta, th_0, I believe f(.,.) along the half-ray theta = th_0 goes to zero as r goes to zero. I would wonder, is it fair that f is not continuous at (0,0)? Eventually, I got used to neighborhoods, general topology and how it relates to metric space topology and normed space topology. I looked at: http://en.wikipedia.org/wiki/Derivative#The_total_derivative.2C_the_Jacobian .2C_and_the_differential I noticed that they don't define lim_{ ||h|| -> 0, h in R^n} ... I think they should at least refer to open balls in the ||.|| norm around the origin in R^n... So F: R^2 -> R is differentiable at (a,b) if there exist real numbers f_x and f_y such that for every epsilon>0, there exists delta>0 such that ||(x,y)|| < delta implies: || F(a+x, b+y) - ( F(a,b) + f_x * x + f_y * y ) || / || (x,y) || < epsilon . Anyway, I think changing what's in Darren's book to show: ep(x,y) -> 0 as (x,y) -> (a,b) would be good enough. David Bernier === Subject: Re: Confusion on differentiable functions on R^2 On Oct 1, 10:53 pm, The World Wide Wade On Sep 28, 8:05 am, The World Wide Wade > On Sep 27, 10:46 am, The World Wide Wade >Darren >The book I am reading says that a function F:R^2 -> C is >differentiable at (a,b) if: >>F(x,y) = F(a,b) + A(x-a) + B(y-b) + ep(x,y) sqrt[(x-a)^2 + >(y-b)^2] >>where ep(x,y) is continuous at (a,b). >>No, you want ep(x,y) -> 0 at (a,b). >> That would be the case if there were no sqrt factor. As it is, if > ep(x,y) is continuous, then the limit of the term ep(x,y) sqrt[...] > as (x,y) -> (a,b) is 0. >> No, the condition you need is ep(x,y) -> 0 at (a,b). Check the > one-variable definition of a derivative for comparison. >The definition I've always used is that >F(x,y) - [F(a,b) + A(x-a) + B(y-b)] >converges to 0, >No, that occurs iff F is continuous at (a,b). > Check again. F being continuous at (a,b) means > F(x,y) - F(a,b) > converges to 0. > The obvious continues to escape you. And basic limit laws escape *you*. F being continous at (a,b) means lim F(x,y) = F(a,b), where the limit >is as (x,y) approaches (a,b). If F is continuous at (a,b), then lim (F(x,y) - F(a,b)) = lim F(x,y) - lim F(a,b) > = F(a,b) - F(a,b) = 0. So F being continous at (a,b) implies F(x,y)-F(a,b) converges to 0 as >(x,y) approaches (a,b), and the reverse is also seen to be true. > Let me repeat: For any choice of >> constants A and B, F is continuous at (a,b) iff F(x,y) - [F(a,b) + >> A(x-a) + B(y-b)] -> 0 as (x,y) -> (a,b). Do you really need to be >> shown the proof? Okay, let's try that one as well: lim (F(x,y) - [F(a,b) + A(x-a) + B(y-b)]) > = F(a,b) - F(a,b) - A(a-a) - B(b-b) = 0. More complicated, but that limit is also zero. However, this is not >the definition of continuity (which was what I was interpretting your >posts to be saying). >We are talking about the definition of differentiability. > Yes; that's why the A*(x-a) and B*(y-b) are there. > You really have to work to remain this ignorant; by now you could have >> consulted a text. I didn't have a textbook with me over the weekend; I had 90 tests to >grade, which were heavy enough to carry around. Now I have one with >me, which agrees with me on the definition of continuity. > Well then, let's use your definition of >> differentiability to prove some interesting things. For example, the >> graph of z = 0 has any *every* plane z = Ax + By as its tangent plane! >> After all, if F is identically 0, then > F(x,y) - [F(a,b) + Ax + By] = - [Ax + By] -> 0 > as (x,y) -> (0,0). [...] Finally you've added something intelligent to this thread, instead of >You're wrong, Bozo! (well, not literally, but that's what it's >amounted to). I hadn't been paying attention until just now. Skimming > through the great debate, You're wrong, Bozo! seems > perfectly appropriate. I do recall seeing a variation of the equation F(x,y) = F(a,b) + A * (x-a) + B * (y-b) + e1 * (x-a) > + e2 * (y-b) in my Calculus III class, but I don't seem to recall the professor >emphasizing the presence of the e1 and e2 functions.* For heaven's sake. The definition of differentiability > is not hard to find - you really could have just > looked it up instead of relying on what you recall > and what you seem to recall some professor emphasizing. The fact is that I haven't used that definition in a long while, if ever. So that definition didn't come to mind, but the concept of the tangent plane best approximates the function near (a,b) did. (And this is correct, and it's what I've been teaching when I've taught Calc III.) > You didn't have a textbook this weekend but you clearly > had internet access - it took me about a minute to find > the definition at http://en.wikipedia.org/wiki/Derivative#The_total_derivative.2C_the_J... . I got this >equation from the section on tangent planes and linearization and, >yes, it is a stronger condition than continuity (as differentiability >should be), and follows if the partial derivatives are continuous. But >all it appears to be is the partial derivatives are continuous >rewritten in a slightly different form. Whether it appears that way to you or not, that's not what it is. That's why I didn't say All it is is that the partial derivatives are continuous. It was based on reading through the proof that IF the partial derivatives are continuous, then the tangent plane approximates the surface according to the formula with the epsilons, and the fact that the epsilons come from the assumption that the partial derivatives are continuous. --- Christopher Heckman > For example, if f(x,y) satisfies (*) |f(x,y)| <= x^2 + y^2 then it's obvious that f is differentiable at the origin, > and it's equally obvious that (*) does not imply that f > itself is even continuous at any other point, to say nothing > of the partial derivatives. * Of course, back in those days, I didn't particularly like Calc III, >especially when I got to three-dimensional solids (which I realized >could get VERY messy) and Green's Theorem, etc. It wasn't until I took >a Complex Analysis course that I realized that Green's Theorem was >just the FTOC in a higher dimension. But I still had reservations >about Calc III; I didn't start teaching it until a couple of years >ago. I still haven't completely warmed to it, btw. ************************ David C. Ullrich === Subject: Re: Confusion on differentiable functions on R^2 > On Oct 1, 10:53 pm, The World Wide Wade On Sep 29, 10:08 am, The World Wide Wade >On Sep 28, 8:05 am, The World Wide Wade > On Sep 27, 10:46 am, The World Wide Wade >>> Darren > The book I am reading says that a function F:R^2 -> C is > differentiable at (a,b) if: >>F(x,y) = F(a,b) + A(x-a) + B(y-b) + ep(x,y) sqrt[(x-a)^2 + > (y-b)^2] >>where ep(x,y) is continuous at (a,b). >> No, you want ep(x,y) -> 0 at (a,b). >> That would be the case if there were no sqrt factor. As it is, if >>ep(x,y) is continuous, then the limit of the term ep(x,y) >>sqrt[...] >>as (x,y) -> (a,b) is 0. >No, the condition you need is ep(x,y) -> 0 at (a,b). Check the >> one-variable definition of a derivative for comparison. >The definition I've always used is that >F(x,y) - [F(a,b) + A(x-a) + B(y-b)] >converges to 0, > No, that occurs iff F is continuous at (a,b). > Check again. F being continuous at (a,b) means > F(x,y) - F(a,b) > converges to 0. The obvious continues to escape you. > > And basic limit laws escape *you*. > > F being continous at (a,b) means lim F(x,y) = F(a,b), where the limit > is as (x,y) approaches (a,b). If F is continuous at (a,b), then > > lim (F(x,y) - F(a,b)) = lim F(x,y) - lim F(a,b) > = F(a,b) - F(a,b) = 0. > > So F being continous at (a,b) implies F(x,y)-F(a,b) converges to 0 as > (x,y) approaches (a,b), and the reverse is also seen to be true. > > Let me repeat: For any choice of > constants A and B, F is continuous at (a,b) iff F(x,y) - [F(a,b) + > A(x-a) + B(y-b)] -> 0 as (x,y) -> (a,b). Do you really need to be > shown the proof? > > Okay, let's try that one as well: > > lim (F(x,y) - [F(a,b) + A(x-a) + B(y-b)]) > = F(a,b) - F(a,b) - A(a-a) - B(b-b) = 0. > > More complicated, but that limit is also zero. However, this is not > the definition of continuity (which was what I was interpretting your > posts to be saying). Of course it's not the definition of continuity. But it's equivalent to it. And it's a trivial exercise. And if you had had the wherewithal to see the 3 second solution, you would have noticed your definition of differentiability is equivalent to the definition of continuity. That was the point. >We are talking about the definition of differentiability. > Yes; that's why the A*(x-a) and B*(y-b) are there. You really have to work to remain this ignorant; by now you could have > consulted a text. > > I didn't have a textbook with me over the weekend; I had 90 tests to > grade, which were heavy enough to carry around. Now I have one with > me, which agrees with me on the definition of continuity. > > Well then, let's use your definition of > differentiability to prove some interesting things. For example, the > graph of z = 0 has any *every* plane z = Ax + By as its tangent plane! > After all, if F is identically 0, then F(x,y) - [F(a,b) + Ax + By] = - [Ax + By] -> 0 as (x,y) -> (0,0). [...] > > Finally you've added something intelligent to this thread, instead of > You're wrong, Bozo! (well, not literally, but that's what it's > amounted to). I see: the above is intelligent because I connected the dots for you, but No, the condition you need is ep(x,y) -> 0 at (a,b). Check the one-variable definition of a derivative for comparison. and Go back one variable: the above simplifies to [F(x) - [F(a) + A(x-a)]]/|x - a| -> 0; multiply and divide by (x-a) to see this occurs iff F'(a) = A. are unintelligent because you don't understand simple undergraduate mathematics. Characterizing my posts as nothing more than You're wrong, Bozo! is a lie. All in all, this has been quite a performance from someone who is entrusted with teaching calculus at the university level. > I do recall seeing a variation of the equation > > F(x,y) = F(a,b) + A * (x-a) + B * (y-b) + e1 * (x-a) > + e2 * (y-b) > > in my Calculus III class, but I don't seem to recall the professor > emphasizing the presence of the e1 and e2 functions.* I got this > equation from the section on tangent planes and linearization and, > yes, it is a stronger condition than continuity (as differentiability > should be), and follows if the partial derivatives are continuous. But > all it appears to be is the partial derivatives are continuous > rewritten in a slightly different form. > > * Of course, back in those days, I didn't particularly like Calc III, > especially when I got to three-dimensional solids (which I realized > could get VERY messy) and Green's Theorem, etc. It wasn't until I took > a Complex Analysis course that I realized that Green's Theorem was > just the FTOC in a higher dimension. But I still had reservations > about Calc III; I didn't start teaching it until a couple of years > ago. I still haven't completely warmed to it, btw. > > --- Christopher Heckman === Subject: Re: Confusion on differentiable functions on R^2 On Oct 2, 10:06 pm, The World Wide Wade On Oct 1, 10:53 pm, The World Wide Wade On Sep 29, 10:08 am, The World Wide Wade > On Sep 28, 8:05 am, The World Wide Wade > On Sep 27, 10:46 am, The World Wide Wade >>>Darren > The book I am reading says that a function F:R^2 -> C is > differentiable at (a,b) if: >>F(x,y) = F(a,b) + A(x-a) + B(y-b) + ep(x,y) sqrt[(x-a)^2 + > (y-b)^2] >>where ep(x,y) is continuous at (a,b). >>No, you want ep(x,y) -> 0 at (a,b). >> That would be the case if there were no sqrt factor. As it is, if >>ep(x,y) is continuous, then the limit of the term ep(x,y) >>sqrt[...] >>as (x,y) -> (a,b) is 0. >> No, the condition you need is ep(x,y) -> 0 at (a,b). Check the >>one-variable definition of a derivative for comparison. >The definition I've always used is that >F(x,y) - [F(a,b) + A(x-a) + B(y-b)] >converges to 0, >No, that occurs iff F is continuous at (a,b). > Check again. F being continuous at (a,b) means > F(x,y) - F(a,b) > converges to 0. > The obvious continues to escape you. And basic limit laws escape *you*. F being continous at (a,b) means lim F(x,y) = F(a,b), where the limit > is as (x,y) approaches (a,b). If F is continuous at (a,b), then lim (F(x,y) - F(a,b)) = lim F(x,y) - lim F(a,b) > = F(a,b) - F(a,b) = 0. So F being continous at (a,b) implies F(x,y)-F(a,b) converges to 0 as > (x,y) approaches (a,b), and the reverse is also seen to be true. > Let me repeat: For any choice of >constants A and B, F is continuous at (a,b) iff F(x,y) - [F(a,b) + >A(x-a) + B(y-b)] -> 0 as (x,y) -> (a,b). Do you really need to be >shown the proof? Okay, let's try that one as well: lim (F(x,y) - [F(a,b) + A(x-a) + B(y-b)]) > = F(a,b) - F(a,b) - A(a-a) - B(b-b) = 0. More complicated, but that limit is also zero. However, this is not > the definition of continuity (which was what I was interpretting your > posts to be saying). Of course it's not the definition of continuity. Your post came across as if that's what you intended, though. You >> No, that occurs iff F is continuous at (a,b). If you had said something like You only need continuity for that to happen, which is what you really meant, then I wouldn't have thought you were using a bizarre definition. Don't blame me for you not being able to communicate your ideas. > But it's equivalent > to it. And it's a trivial exercise. And if you had had the > wherewithal to see the 3 second solution, you would have noticed your > definition of differentiability is equivalent to the definition of > continuity. That was the point. >We are talking about the definition of differentiability. > Yes; that's why the A*(x-a) and B*(y-b) are there. > You really have to work to remain this ignorant; by now you could have >consulted a text. I didn't have a textbook with me over the weekend; I had 90 tests to > grade, which were heavy enough to carry around. Now I have one with > me, which agrees with me on the definition of continuity. > Well then, let's use your definition of >differentiability to prove some interesting things. For example, the >graph of z = 0 has any *every* plane z = Ax + By as its tangent plane! >After all, if F is identically 0, then > F(x,y) - [F(a,b) + Ax + By] = - [Ax + By] -> 0 > as (x,y) -> (0,0). [...] Finally you've added something intelligent to this thread, instead of > You're wrong, Bozo! (well, not literally, but that's what it's > amounted to). I see: the above is intelligent because I connected the dots for > you, but No, the condition you need is ep(x,y) -> 0 at (a,b). Check the > one-variable definition of a derivative for comparison. and Go back one variable: the above simplifies to [F(x) - [F(a) + > A(x-a)]]/|x - a| -> 0; multiply and divide by (x-a) to see this occurs > iff F'(a) = A. are unintelligent because you don't understand simple undergraduate > mathematics. which I learned only vaguely 17 years ago and haven't used since. I used to know a lot of history but haven't used that either; perhaps you would care to insult me about that? > Characterizing my posts as nothing more than You're > wrong, Bozo! is a lie. All in all, this has been quite a performance > from someone who is entrusted with teaching calculus at the university > level. And that remark would only come from an armchair quarterback. And it isn't actually a problem: when teaching Calc (that's what we teachers call Calculus for short) III at ASU, we don't bother to use that definition, and the definition of differentiability that I use --- The tangent plane best approximates the function near the point of tangency --- is correct, albeit a bit imprecise. But in terms of what I teach, it doesn't matter that I don't get all the details right on this one particular definition. And if you say I'm a lazy teacher for this, well, I'll let you step in and teach a semester for me, to show you how much work it is even to work at the level I do, while I do whatever it is that you do. And make more money during the interval. --- Christopher Heckman > I do recall seeing a variation of the equation F(x,y) = F(a,b) + A * (x-a) + B * (y-b) + e1 * (x-a) > + e2 * (y-b) in my Calculus III class, but I don't seem to recall the professor > emphasizing the presence of the e1 and e2 functions.* I got this > equation from the section on tangent planes and linearization and, > yes, it is a stronger condition than continuity (as differentiability > should be), and follows if the partial derivatives are continuous. But > all it appears to be is the partial derivatives are continuous > rewritten in a slightly different form. * Of course, back in those days, I didn't particularly like Calc III, > especially when I got to three-dimensional solids (which I realized > could get VERY messy) and Green's Theorem, etc. It wasn't until I took > a Complex Analysis course that I realized that Green's Theorem was > just the FTOC in a higher dimension. But I still had reservations > about Calc III; I didn't start teaching it until a couple of years > ago. I still haven't completely warmed to it, btw. === Subject: Re: Confusion on differentiable functions on R^2 >On Oct 2, 10:06 pm, The World Wide Wade [...] >> Of course it's not the definition of continuity. Your post came across as if that's what you intended, though. You >> No, that occurs iff F is continuous at (a,b). If you had said something like You only need continuity for that to >happen, which is what you really meant, Huh? Seems likely to me that he meant exactly what he said. What you say he should have said is exactly half of what he did say - why would that have been more clear? > then I wouldn't have thought >you were using a bizarre definition. Don't blame me for you not being able to communicate your ideas. Balderdash. >> But it's equivalent >> to it. And it's a trivial exercise. And if you had had the >> wherewithal to see the 3 second solution, you would have noticed your >> definition of differentiability is equivalent to the definition of >> continuity. That was the point. > We are talking about the definition of differentiability. >>Yes; that's why the A*(x-a) and B*(y-b) are there. >>You really have to work to remain this ignorant; by now you could have >>consulted a text. >> I didn't have a textbook with me over the weekend; I had 90 tests to >> grade, which were heavy enough to carry around. Now I have one with >> me, which agrees with me on the definition of continuity. >>Well then, let's use your definition of >>differentiability to prove some interesting things. For example, the >>graph of z = 0 has any *every* plane z = Ax + By as its tangent plane! >>After all, if F is identically 0, then >> F(x,y) - [F(a,b) + Ax + By] = - [Ax + By] -> 0 >>as (x,y) -> (0,0). [...] >> Finally you've added something intelligent to this thread, instead of >> You're wrong, Bozo! (well, not literally, but that's what it's >> amounted to). >> I see: the above is intelligent because I connected the dots for >> you, but >> No, the condition you need is ep(x,y) -> 0 at (a,b). Check the >> one-variable definition of a derivative for comparison. >> and >> Go back one variable: the above simplifies to [F(x) - [F(a) + >> A(x-a)]]/|x - a| -> 0; multiply and divide by (x-a) to see this occurs >> iff F'(a) = A. >> are unintelligent because you don't understand simple undergraduate >> mathematics. which I learned only vaguely 17 years ago and haven't used since. I >used to know a lot of history but haven't used that either; perhaps >you would care to insult me about that? If you insisted that the first president of the US was Lincoln, and claimed that someone who simply said that no, it was Washington was being unclear, you might get insulted for _that_. >> Characterizing my posts as nothing more than You're >> wrong, Bozo! is a lie. All in all, this has been quite a performance >> from someone who is entrusted with teaching calculus at the university >> level. And that remark would only come from an armchair quarterback. And it isn't actually a problem: when teaching Calc (that's what we >teachers call Calculus for short) III at ASU, we don't bother to use >that definition, and the definition of differentiability that I use >--- The tangent plane best approximates the function near the point of >tangency --- is correct, albeit a bit imprecise. But in terms of what >I teach, it doesn't matter that I don't get all the details right on >this one particular definition. And if you say I'm a lazy teacher for this, Supposing a history teacher said that the fact that he thought Lincoln was the first president didn't matter, because it didn't come up in the classes he was teaching. Lazy is probably not the word people would use for that. >well, I'll let you step in >and teach a semester for me, to show you how much work it is even to >work at the level I do, while I do whatever it is that you do. And >make more money during the interval. --- Christopher Heckman > I do recall seeing a variation of the equation >> F(x,y) = F(a,b) + A * (x-a) + B * (y-b) + e1 * (x-a) >> + e2 * (y-b) >> in my Calculus III class, but I don't seem to recall the professor >> emphasizing the presence of the e1 and e2 functions.* I got this >> equation from the section on tangent planes and linearization and, >> yes, it is a stronger condition than continuity (as differentiability >> should be), and follows if the partial derivatives are continuous. But >> all it appears to be is the partial derivatives are continuous >> rewritten in a slightly different form. >> * Of course, back in those days, I didn't particularly like Calc III, >> especially when I got to three-dimensional solids (which I realized >> could get VERY messy) and Green's Theorem, etc. It wasn't until I took >> a Complex Analysis course that I realized that Green's Theorem was >> just the FTOC in a higher dimension. But I still had reservations >> about Calc III; I didn't start teaching it until a couple of years >> ago. I still haven't completely warmed to it, btw. ************************ David C. Ullrich === Subject: Re: Confusion on differentiable functions on R^2 <39r6g3dssqcvotvq7d61vcvgdl4p0f87l7@4ax.com >On Oct 2, 10:06 pm, The World Wide Wade > [...] >> Of course it's not the definition of continuity. Your post came across as if that's what you intended, though. You >> No, that occurs iff F is continuous at (a,b). If you had said something like You only need continuity for that to >happen, which is what you really meant, Huh? Seems likely to me that he meant exactly what he said. But it wasn't to me, and that's the point; it doesn't matter whether you understand/understood it, because you weren't participating in the thread at the time. I mistakenly thought he was giving a definition, not pointing out a consequence. > What you say he should have said is exactly half of what he > did say - why would that have been more clear? then I wouldn't have thought >you were using a bizarre definition. Don't blame me for you not being able to communicate your ideas. Balderdash. > But it's equivalent >> to it. And it's a trivial exercise. And if you had had the >> wherewithal to see the 3 second solution, you would have noticed your >> definition of differentiability is equivalent to the definition of >> continuity. That was the point. >> We are talking about the definition of differentiability. >Yes; that's why the A*(x-a) and B*(y-b) are there. >You really have to work to remain this ignorant; by now you could have >>consulted a text. > I didn't have a textbook with me over the weekend; I had 90 tests to >> grade, which were heavy enough to carry around. Now I have one with >> me, which agrees with me on the definition of continuity. >Well then, let's use your definition of >>differentiability to prove some interesting things. For example, the >>graph of z = 0 has any *every* plane z = Ax + By as its tangent plane! >>After all, if F is identically 0, then > F(x,y) - [F(a,b) + Ax + By] = - [Ax + By] -> 0 >as (x,y) -> (0,0). [...] > Finally you've added something intelligent to this thread, instead of >> You're wrong, Bozo! (well, not literally, but that's what it's >> amounted to). > I see: the above is intelligent because I connected the dots for >> you, but > No, the condition you need is ep(x,y) -> 0 at (a,b). Check the >> one-variable definition of a derivative for comparison. > and > Go back one variable: the above simplifies to [F(x) - [F(a) + >> A(x-a)]]/|x - a| -> 0; multiply and divide by (x-a) to see this occurs >> iff F'(a) = A. > are unintelligent because you don't understand simple undergraduate >> mathematics. which I learned only vaguely 17 years ago and haven't used since. I >used to know a lot of history but haven't used that either; perhaps >you would care to insult me about that? If you insisted that the first president of the US was Lincoln, > and claimed that someone who simply said that no, it was > Washington was being unclear, you might get insulted for _that_. > Characterizing my posts as nothing more than You're >> wrong, Bozo! is a lie. All in all, this has been quite a performance >> from someone who is entrusted with teaching calculus at the university >> level. And that remark would only come from an armchair quarterback. And it isn't actually a problem: when teaching Calc (that's what we >teachers call Calculus for short) III at ASU, we don't bother to use >that definition, and the definition of differentiability that I use >--- The tangent plane best approximates the function near the point of >tangency --- is correct, albeit a bit imprecise. But in terms of what >I teach, it doesn't matter that I don't get all the details right on >this one particular definition. And if you say I'm a lazy teacher for this, Supposing a history teacher said that the fact that he thought > Lincoln was the first president didn't matter, because it > didn't come up in the classes he was teaching. I don't remember all of my Partial Differential Equations class. Should I not be allowed to teach Calculus because I don't remember how to do PDE's? The point of the remark about teaching (which you have missed) was that I haven't taught anything that was false; I went into as much detail as necessary for the purpose of that class, and that detail was correct. > Lazy is probably not the word people would use for that. Looks like you're calling me a crank here. --- Christopher Heckman >well, I'll let you step in >and teach a semester for me, to show you how much work it is even to >work at the level I do, while I do whatever it is that you do. And >make more money during the interval. > I do recall seeing a variation of the equation > F(x,y) = F(a,b) + A * (x-a) + B * (y-b) + e1 * (x-a) >> + e2 * (y-b) > in my Calculus III class, but I don't seem to recall the professor >> emphasizing the presence of the e1 and e2 functions.* I got this >> equation from the section on tangent planes and linearization and, >> yes, it is a stronger condition than continuity (as differentiability >> should be), and follows if the partial derivatives are continuous. But >> all it appears to be is the partial derivatives are continuous >> rewritten in a slightly different form. > * Of course, back in those days, I didn't particularly like Calc III, >> especially when I got to three-dimensional solids (which I realized >> could get VERY messy) and Green's Theorem, etc. It wasn't until I took >> a Complex Analysis course that I realized that Green's Theorem was >> just the FTOC in a higher dimension. But I still had reservations >> about Calc III; I didn't start teaching it until a couple of years >> ago. I still haven't completely warmed to it, btw. ************************ David C. Ullrich === Subject: Magic number...Don't scroll down in advance. Hello sir~ I can read your thinking. Pick a two-digit number between 1 and 50, using two different odd numbers. Just a moment, please... Don't scroll down. Did you decide your number ? and then Scroll down NOW. ............................................................................ ........... ............................................................................ ........... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Your number is 37. For reference, I saw magician David Blaine's show. === Subject: Re: Magic number...Don't scroll down in advance. > Hello sir~ > > I can read your thinking. > > Pick a two-digit number between 1 and 50, using two different odd numbers. > > Just a moment, please... > Don't scroll down. > > Did you decide your number ? > > and then Scroll down NOW. > > ............................................................................ . . > ......... > ............................................................................ . . > ......... > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > Your number is 37. > > For reference, I saw magician David Blaine's show. But my number was actually 13. === Subject: Re: Magic number...Don't scroll down in advance. Hello sir~ I can read your thinking. Pick a two-digit number between 1 and 50, using two different odd numbers. Just a moment, please... > Don't scroll down. Did you decide your number ? and then Scroll down NOW. ............................................................................ . . > ......... > ............................................................................ . . > ......... > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > Your number is 37. For reference, I saw magician David Blaine's show. But my number was actually 13. Mina didn't address You. Hero === Subject: Re: Magic number...Don't scroll down in advance. > I can read your thinking. Pick a two-digit number between 1 and 50, using two different odd numbers. OK. 10 possibilities, computing pseudorandom sum modulo 10, got 9, converting to largest possibility. I pick 39. > Your number is 37. Wow, almost right. Maybe sci.math is not the best place for this trick. -- Jens Kruse Andersen === Subject: Re: Magic number...Don't scroll down in advance. > Hello sir~ I can read your thinking. Pick a two-digit number between 1 and 50, using two different odd numbers. Just a moment, please... > Don't scroll down. Did you decide your number ? 13 and then Scroll down NOW. Your number is 37. Oopsie. For reference, I saw magician David Blaine's show. We must be a contrary bunch here in sci.math. This guy says that 9 out 10 times a random person will end up choosing 37, for no particular reason. http://www.expertvillage.com/videos/future-odd-numbers-explained.htm Successful psychics work by brushing past the misses and moving quickly on to the next thing. People later then only remember the hits. I've heard of cases where a stage psychic went through 20 letters of the alphabet (captured on videotape) ... this relative is connected with the letter B... I mean P... T... S... W... and yet the audience remembered that he amazingly zeroed right in on the correct letter. - Randy === Subject: Re: Magic number...Don't scroll down in advance. >Hello sir~ I can read your thinking. Pick a two-digit number between 1 and 50, using two different odd numbers. Just a moment, please... >Don't scroll down. Did you decide your number ? and then Scroll down NOW. ........................................................................... ............ >........................................................................... ............ >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >. >Your number is 37. I knew you were going to say that. >For reference, I saw magician David Blaine's show. Now that, I didn't know. -- Angus Rodgers Contains mild peril === Subject: Re: Magic number...Don't scroll down in advance. > Hello sir~ I can read your thinking. Pick a two-digit number between 1 and 50, using two different odd numbers. Don't you mean between 10 and 50 since you only want 2 digit numbers? Or are you designating the digits 1-9 as 01, 02, 03, etc.? > For reference, I saw magician David Blaine's show. David Blaine is . === Subject: Re: Magic number...Don't scroll down in advance. > On Oct 3, 8:56 am, mina_world > Hello sir~ I can read your thinking. Pick a two-digit number between 1 and 50, using two > different odd numbers. > > Don't you mean between 10 and 50 since you only want > 2 digit numbers? yeah. but asking it this way , makes it seem less magical. > > Or are you designating the digits 1-9 as 01, 02, 03, > etc.? > > For reference, I saw magician David Blaine's show. > > David Blaine is . yep. > so your suppose to pick a number between 1 and 50. wait actually 10 and 50 wait only the odd digit ones... no big deal ; there are not many possibilities 13 15 17 19 31 35 37 39 now most people wont pick a number ending on 5. why ? because that does not seem random , since it is clearly devisable by 5. therefore most people will pick between 13 17 19 31 37 39 ( even less possibilities ) most people dont pick a composite either since that also does not feel random , therefore they dont choose 39. 13 17 19 31 37 most people dont pick the digit 1 because that is the smallest they can pick , and therefore does not appear random --> 37 tommy1729 === Subject: Re: Magic number...Don't scroll down in advance. > Hello sir~ > > I can read your thinking. > > Pick a two-digit number between 1 and 50, using two different odd numbers. > > Just a moment, please... > Don't scroll down. > > Did you decide your number ? > Mine was 15. === Subject: Re: Magic number...Don't scroll down in advance. > Hello sir~ > > I can read your thinking. > > Pick a two-digit number between 1 and 50, using two > different odd numbers. > > Just a moment, please... > Don't scroll down. > > Did you decide your number ? > > Mine was 15. lol the real tommy1729 === Subject: Solution Manual for Physical Chemistry (7th) -- by P.W.Atkins I need the Solution Manual for Physical Chemistry (7th) by P.W.Atkins, can you please send it to me??? === Subject: Re: Many Solutions Manuals and Ebooks in Electronic (PDF)Format! I need the Solution Manual for Physical Chemistry (7th) by P.W.Atkins, can you please send it to me??? My e-mail address : noelying@gmail.com === Subject: Re: English translation of Weber's 1893 paper > Where can I find an English translation of Heinrich Weber's 1893 > paper? I think it is supposed to be the first publication of the > definition of the abstract field. No. Dedekind did use the german equivalent of field, which is 'Koerper', http://members.aol.com/jeff570/f.html He also says, that Eliakim Hastings Moore was doing Galois in 1893 too. Die allgemeinen Grundlagen der Galois'schen Gleichungstheorie. > Mathematische Annalen, volume 43 December 1893, Number 4, > pages 521-549http://resolver.sub.uni-goettingen.de/purl?GDZPPN002254670 Would The general foundation of Galois' equation theory be an > acceptable translation of the title? > A.N. Niel gave the translation already. With friendly greetings Hero === er ! hey asshole stop using my old profile. the REAL TOMMY1729 === > hey asshole stop using my old profile. the REAL TOMMY1729 I suggest reporting this unprecedented outrage to your mother. -- === > hey asshole stop using my old profile. > > the REAL > > TOMMY1729 hahaha i now control both profiles :-) take that you moron. i am in control of all tommy1729's. i can steal your identity too. hahaha this is the real tommy speaking , the one with the 1282 posts !!! game over cheater see you in hell the real tommy1729 Riemann , Matheyasevich and Wiles forever === Subject: Re: Power System Analysis and Design Fourth Edition J Duncan Glover Mail-To-News-Contact: abuse@dizum.com >Does any one have the FOURTH EDITION to Power System Analysis and >Design by J. Duncan Glover, Mulukutla S. Sarma, and Thomas J. Overbye. I feel so good to realize that (part of) the next generation of power engineers intends to attempt to get their degree by cheating rather than by learning the material. It's not bad enough that half the industry will be retiring in the next decade, but we'll apparently be replaced by incompetents. University of Pittsburgh EE professors should take note of this. If any University of Pittsburgh math professors are reading, maybe they should pass this information along to the EE department. Power system classes aren't that popular; I'd bet that there are fewer than 20 students who are taking a class using this text. -- Michael F. Stemper #include The FAQ for rec.arts.sf.written is at: http://www.geocities.com/evelynleeper/sf-written Please read it before posting. === Subject: Estimating an expected value by simulation Problem 25 in Introduction to Probability by Bertsekas describes the following problem, which can quite easily be solved. But how can you really use it? Problem: Let f(x) be a PDF such that for some nonnegative scalars a, b, c, we have f(x)=0 for all x outside the interval [a,b], and x*f(x)<=c for all x. Let Yi, i=1,...,n be independent R.V. with values generated as follows: a point (Vi,Wi) is chosen at random (according to a uniform PDF) within the rectangle whose corners are (a,0),(b,0),(a,c),(b,c) and if Wi<=Vi*f(Vi), the value of Yi is set to 1, and otherwise it is set to 0. Consider the random variable Z=(Y1+Y2+...+Yn)/n. Show that E[Z]=E[X]/c(b-a) and that var(Z)<=1/4n. In particular, var(Z)->0 as n->oo. As I said, solving it is straight forward. But what is really the application? I am guessing that you are seeking E[X], but instead do a simulation where you pick n random points in the given square, and then set the n indicators Yi according to the rule Wi Problem 25 in Introduction to Probability by Bertsekas describes the > following problem, which can quite easily be solved. But how can you > really use it? Problem: > Let f(x) be a PDF such that for some nonnegative scalars a, b, c, we > have f(x)=0 for all x outside the interval [a,b], and x*f(x)<=c for all > x. Let Yi, i=1,...,n be independent R.V. with values generated as > follows: a point (Vi,Wi) is chosen at random (according to a uniform > PDF) within the rectangle whose corners are (a,0),(b,0),(a,c),(b,c) and > if Wi<=Vi*f(Vi), the value of Yi is set to 1, and otherwise it is set to > 0. Consider the random variable Z=(Y1+Y2+...+Yn)/n. Show that E[Z]=E[X]/c(b-a) and that var(Z)<=1/4n. > In particular, var(Z)->0 as n->oo. As I said, solving it is straight forward. But what is really the > application? I am guessing that you are seeking E[X], but instead do a simulation > where you pick n random points in the given square, and then set the n > indicators Yi according to the rule Wi calculate E[X] by just taking E[Z]*c(b-a). But I guess we don't have > f(x), and hence cannot set the indicators. So I am lost in how this > thing can be used when estimating expected values by simulation. Is this > a general technique btw? I don't know that there necessarily has to be an intended application. Textbook problems can be artificial -- just for the reader to practise working with the techniques explained in the text. === Subject: Crossword Sudoku problem Since the Kakuro is much less prominent than its colleague Sudoko, I shortly repeat: It's like a crossword, only digits instead of letters, and the digit sum given as lead. Like Sudoku, only 1-9 and no repeat in a word. Of course the solution must be unique too. Problem 1: Give 6 leads in a 3*3 Kakuro with as many different digits in the solution as possible. Problem 2: Give the smallest Kakuro with every digit turning up. (1-digit leads are cheating :-) Please note that 12 15 18 v v v 6> 1 2 3 15> 4 5 6 24> 7 8 9 is no solution at all to either!  Because: 12 15 18 v v v 6> 1 2 3 15> 2 5 8 24> 9 8 7 -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de order stormed the surface where chaos set norm had there always been balance? ...surely not therein lies the beauty === Subject: Re: Crossword Sudoku problem > Since the Kakuro is much less prominent than its > colleague Sudoko, I shortly repeat: It's like a crossword, only digits instead of letters, > and the digit sum given as lead. Like Sudoku, only > 1-9 and no repeat in a word. Of course the solution > must be unique too. Problem 1: Give 6 leads in a 3*3 Kakuro with as many > different digits in the solution as possible. One of a number of possibilities: 10 15 20 v v v 7> ? ? ? 15> ? ? ? 23> ? ? ? > Problem 2: Give the smallest Kakuro with every digit > turning up. (1-digit leads are cheating :-) Please note that > 12 15 18 > v v v > 6> 1 2 3 > 15> 4 5 6 > 24> 7 8 9 is no solution at all to either! Because: 12 15 18 > v v v > 6> 1 2 3 > 15> 2 5 8 > 24> 9 8 7 -- > Hauke Reddmann <:-EX8 fc3a...@uni-hamburg.de > order stormed the surface where chaos set norm > had there always been balance? ...surely not > therein lies the beauty === Subject: WCCM8 & ECCOMAS 2008 - Mini-Symposium on Computational Bioimaging and Visualization - Announce & Call for Contributions Mini-Symposium on Computational Bioimaging and Visualization Within the WCCM8 & ECCOMAS 2008 International Conference (http://www.iacm-eccomascongress2008.org ) Venice, Italy, 30 June - 5 July 2008 The use of computational procedures based on medical images to model and visualize represented objects requires a high level of research investment, due mainly to the strong demands of clinical partners. These procedures can have different goals, such as shape reconstruction and visualization, segmentation, motion and deformation analyses, registration, simulation, etc. The main goal of this Mini- symposium is to promote the debate between researchers involved in the related fields (Bio-medical Image Acquisition, Analysis and Processing; Multi-scale Shape and Motion Reconstruction; Multi-scale Modeling and Analysis; Scientific Visualization, Bio-medical Software Libraries, Environments; etc.), in order to set the major lines of developments for the near future. Mini-symposium Organizers: Jo.8bo Manuel R.S. Tavares (tavares@fe.up.pt, FEUP, Porto Portugal) Renato Natal Jorge (rnatal@fe.up.pt, FEUP, Porto, Portugal) Thomas J.R. Hughes (tjr hughes@hotmail.com, ICES, University of Texas at Austin, USA) Chandrajit Bajaj (bajaj@cs.utexas.edu, ICES, University of Texas at Austin, USA) IMPORTANT DATES Deadline for presenting a one page abstract: December 15th., 2007 Acceptance of the contributions: January 31th., 2008 Deadline for submitting the final abstract: February 28th., 2008 === Subject: Stable finding of orthogonal vector to a normal in 3D? Fellows, I'm struggling with seems to be a trivial problem but currently I'm not sure that it has a stable solution. The problem is this; assume that I have a point in 3D and corresponding normal of an unknown surface at that point. Equipped with this input I need to construct a local orthogonal basis of a plane that is tangential to the surface, i.e. finding the binormal and tangent vectors. The thing that makes it somewhat challenging is that I don't want to have an iterative solution or performing some kind of test if it's a valid solution. This results in that, as far as I've found out so far, there is always a special case where the method fails or become numerically instable (the method will be implemented on a computer). E.g. to use Gram-Schmidt, I need an additional vector, not parallel to the normal which just brings me back to the original problem of finding a non-parallel vector, thus Gram-Schmidt is not applicable. Another way may be to randomly select a vector, but this may of course result in a vector that is close to parallel to the normal, thus failing. To summarize, the problem boils down to finding a second vector that is non-parallel to the normal, preferably orthogonal, without any iteration or testing. Having such a vector there is a plethora of methods that may be used to create an orthogonal basis. Any input on this problem would be greatly appreciated. === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? > Fellows, > I'm struggling with seems to be a trivial problem but currently I'm > not sure that it has a stable solution. The problem is this; assume > that I have a point in 3D and corresponding normal of an unknown > surface at that point. Equipped with this input I need to construct a > local orthogonal basis of a plane that is tangential to the surface, > i.e. finding the binormal and tangent vectors. > To summarize, the problem boils down to finding a second vector that > is non-parallel to the normal, preferably orthogonal, without any > iteration or testing. Having such a vector there is a plethora of > methods that may be used to create an orthogonal basis. > Since nobody has mentioned this: Suppose N=(a,b,c) then u=(-b,a,0) is orthogonal to N. Then the cross product v=uXN will be orthogonal to both u and N. (v works out to be (ac, bc, -a^2-b^2)). These vectors can then be normalized. === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? Ola, >... assume > that I have a point in 3D and corresponding normal of an unknown > surface at that point. Equipped with this input I need to construct a > local orthogonal basis of a plane that is tangential to the surface, I probably don't understand your question correctly but suppose that I first translated your normal vector to start at the origin (by subtracting the 3D point component wise) and end up with a vector x=[0.3 0.4 0.866]^t then if I perforn a QR decomposition on x using modified Gram-schmidt then I get Q=[-0.300 -0.400 -0.866 -0.400 0.877 -0.266 -0.866 -0.266 0.423] where the original x appears (scaled) in column 1 and columns 2 and 3 contain 2 vectors orthogonal to x and to each other. These last 2 vectors form a orthogonal basis for all vectors laying in the plane orthogonal to the translated x and, hence, form a basis for the translated plane tangential to the surface. Isn't that what you are (mostly) looking for? > to finding a second vector that > is non-parallel to the normal, preferably orthogonal, without any > iteration or testing. I used Matlab's QR routine for the above example. Hope this (somehow) helps === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? > Ola, ... assume > that I have a point in 3D and corresponding normal of an unknown > surface at that point. Equipped with this input I need to construct a > local orthogonal basis of a plane that is tangential to the surface, I probably don't understand your question correctly but > suppose that I first translated your normal vector to start at the > origin (by subtracting the 3D point component wise) > and end up with a vector > x=[0.3 0.4 0.866]^t then if I perforn a QR decomposition on x using modified Gram-schmidt > then I get > Q=[-0.300 -0.400 -0.866 > -0.400 0.877 -0.266 > -0.866 -0.266 0.423] where the original x appears (scaled) in column 1 and columns 2 and 3 > contain 2 vectors orthogonal to x and to each other. > These last 2 vectors form a orthogonal basis for all vectors laying in > the plane orthogonal to the translated x > and, hence, form a basis for the translated plane tangential to the > surface. Isn't that what you are (mostly) looking for? to finding a second vector that > is non-parallel to the normal, preferably orthogonal, without any > iteration or testing. I used Matlab's QR routine for the above example. Hope this (somehow) helps Hi Basically your on the right track, the only thing I'm not sure about at the moment is how complex a QR-decomposition is in terms of computing (it will be implemented on a computer), if it's an iterative process etc. Considering that I will perform this operation in the order of 25 - 100 million times a second it cannot be to complex. That's why I would like to avoid iterations and conditional checks/ branches if at all possible. === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? > Basically your on the right track, the only thing I'm not sure about > at the moment is how complex a QR-decomposition is in terms of > computing (it will be implemented on a computer), if it's an iterative > process etc. Considering that I will perform this operation in the > order of 25 - 100 million times a second it cannot be to complex. > That's why I would like to avoid iterations and conditional checks/ > branches if at all possible. > > Householder QR is 2n^2(m-n/3) flops but Q is not given directly. mgs is 2mn^2 , Q's are found directly but is less stable in the sense that if Q_ is the computed Q and Q_^TQ_ = I + E then with householder ||E|| ~ u with mgs ||E|| ~ u*condition(A) u is machine epsilon in size. === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? > Fellows, > I'm struggling with seems to be a trivial problem but currently I'm > not sure that it has a stable solution. The problem is this; assume > that I have a point in 3D and corresponding normal of an unknown > surface at that point. Equipped with this input I need to construct a > local orthogonal basis of a plane that is tangential to the surface, > i.e. finding the binormal and tangent vectors. > > The thing that makes it somewhat challenging is that I don't want to > have an iterative solution or performing some kind of test if it's a > valid solution. This results in that, as far as I've found out so far, > there is always a special case where the method fails or become > numerically instable (the method will be implemented on a computer). > E.g. to use Gram-Schmidt, I need an additional vector, not parallel to > the normal which just brings me back to the original problem of > finding a non-parallel vector, thus Gram-Schmidt is not applicable. > > Another way may be to randomly select a vector, but this may of course > result in a vector that is close to parallel to the normal, thus > failing. > > To summarize, the problem boils down to finding a second vector that > is non-parallel to the normal, preferably orthogonal, without any > iteration or testing. Having such a vector there is a plethora of > methods that may be used to create an orthogonal basis. > > Any input on this problem would be greatly appreciated. > Given a nonzero vector V in R^n, you want to find another nonzero vector W that is orthogonal to V. Let j be the index that minimizes |V_j|, and if e_j is the j'th standard unit vector take W = e_j - c V where c = V_j/||V||^2. Actually any j such that |V_j|^2 <= ||V||^2 / 2 would do. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? On Oct 3, 9:34 pm, Robert Israel > Fellows, > I'm struggling with seems to be a trivial problem but currently I'm > not sure that it has a stable solution. The problem is this; assume > that I have a point in 3D and corresponding normal of an unknown > surface at that point. Equipped with this input I need to construct a > local orthogonal basis of a plane that is tangential to the surface, > i.e. finding the binormal and tangent vectors. The thing that makes it somewhat challenging is that I don't want to > have an iterative solution or performing some kind of test if it's a > valid solution. This results in that, as far as I've found out so far, > there is always a special case where the method fails or become > numerically instable (the method will be implemented on a computer). > E.g. to use Gram-Schmidt, I need an additional vector, not parallel to > the normal which just brings me back to the original problem of > finding a non-parallel vector, thus Gram-Schmidt is not applicable. Another way may be to randomly select a vector, but this may of course > result in a vector that is close to parallel to the normal, thus > failing. To summarize, the problem boils down to finding a second vector that > is non-parallel to the normal, preferably orthogonal, without any > iteration or testing. Having such a vector there is a plethora of > methods that may be used to create an orthogonal basis. Any input on this problem would be greatly appreciated. > Given a nonzero vector V in R^n, you want to find another nonzero vector W > that is orthogonal to V. Let j be the index that minimizes |V_j|, and > if e_j is the j'th standard unit vector take W = e_j - c V where > c = V_j/||V||^2. Actually any j such that |V_j|^2 <= ||V||^2 / 2 > would do. > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Hi Robert of j:th component of V? Finding the minimizing j:th component, considering that it will be implemented on a computer, would need an exhaustive search to find the j:th component, thus basically the same as iteration and testing. It might be the case that it turns out that this can't be avoided though. === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? > Fellows, > I'm struggling with seems to be a trivial problem but currently I'm > not sure that it has a stable solution. The problem is this; assume > that I have a point in 3D and corresponding normal of an unknown > surface at that point. Equipped with this input I need to construct a > local orthogonal basis of a plane that is tangential to the surface, > i.e. finding the binormal and tangent vectors. The thing that makes it somewhat challenging is that I don't want to > have an iterative solution or performing some kind of test if it's a > valid solution. This results in that, as far as I've found out so far, > there is always a special case where the method fails or become > numerically instable (the method will be implemented on a computer). > E.g. to use Gram-Schmidt, I need an additional vector, not parallel to > the normal which just brings me back to the original problem of > finding a non-parallel vector, thus Gram-Schmidt is not applicable. Another way may be to randomly select a vector, but this may of course > result in a vector that is close to parallel to the normal, thus > failing. To summarize, the problem boils down to finding a second vector that > is non-parallel to the normal, preferably orthogonal, without any > iteration or testing. Having such a vector there is a plethora of > methods that may be used to create an orthogonal basis. Any input on this problem would be greatly appreciated. > Given a vector N , just take any vector M, which is not in the same direction or directly opposite, thus there is no number r with r times M = N, than M x N is orthogonal to N. In order to get such a vector, such a 'any vector', just take N and add a number to one of its components. So M = N + ( 0, 3, 0, ...,0). This should work without testing, if i'm not mistaken. With friendly greetings Hero === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? Fellows, > I'm struggling with seems to be a trivial problem but currently I'm > not sure that it has a stable solution. The problem is this; assume > that I have a point in 3D and corresponding normal of an unknown > surface at that point. Equipped with this input I need to construct a > local orthogonal basis of a plane that is tangential to the surface, > i.e. finding the binormal and tangent vectors. The thing that makes it somewhat challenging is that I don't want to > have an iterative solution or performing some kind of test if it's a > valid solution. This results in that, as far as I've found out so far, > there is always a special case where the method fails or become > numerically instable (the method will be implemented on a computer). > E.g. to use Gram-Schmidt, I need an additional vector, not parallel to > the normal which just brings me back to the original problem of > finding a non-parallel vector, thus Gram-Schmidt is not applicable. Another way may be to randomly select a vector, but this may of course > result in a vector that is close to parallel to the normal, thus > failing. To summarize, the problem boils down to finding a second vector that > is non-parallel to the normal, preferably orthogonal, without any > iteration or testing. Having such a vector there is a plethora of > methods that may be used to create an orthogonal basis. Any input on this problem would be greatly appreciated. Given a vector N , just take any vector M, which is not in the same > direction or directly opposite, thus > there is no number r with r times M = N, > than > M x N is orthogonal to N. In order to get such a vector, such a 'any vector', just take N and > add a number to one of its components. > So M = N + ( 0, 3, 0, ...,0). > This should work without testing, if i'm not mistaken. With friendly greetings > Hero Hi Hero found out myself (I hope). Adding an arbitrary offset vector to the normal won't do, as I'll explain below. Given your example: M = N + ( 0, 3, 0, ...,0) assume that N happens to be equal to e.g. N = ( 0, 6, 0, ...,0) => M = ( 0, 9, 0, ...,0) which is parallel to N i.e. M = 3 * N. In general, this method will fail as soon as the random offset is the same vector, down to scale, as the normal or become numerically instable if it's nearly the same. === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? >Fellows, >I'm struggling with seems to be a trivial problem but currently I'm >not sure that it has a stable solution. The problem is this; assume >that I have a point in 3D and corresponding normal of an unknown >surface at that point. Equipped with this input I need to construct a >local orthogonal basis of a plane that is tangential to the surface, >i.e. finding the binormal and tangent vectors. > The thing that makes it somewhat challenging is that I don't want to >have an iterative solution or performing some kind of test if it's a >valid solution. This results in that, as far as I've found out so far, >there is always a special case where the method fails or become >numerically instable (the method will be implemented on a computer). >E.g. to use Gram-Schmidt, I need an additional vector, not parallel to >the normal which just brings me back to the original problem of >finding a non-parallel vector, thus Gram-Schmidt is not applicable. > Another way may be to randomly select a vector, but this may of course >result in a vector that is close to parallel to the normal, thus >failing. > To summarize, the problem boils down to finding a second vector that >is non-parallel to the normal, preferably orthogonal, without any >iteration or testing. Having such a vector there is a plethora of >methods that may be used to create an orthogonal basis. > Any input on this problem would be greatly appreciated. Given a vector N , just take any vector M, which is not in the same > direction or directly opposite, thus > there is no number r with r times M = N, > than > M x N is orthogonal to N. In order to get such a vector, such a 'any vector', just take N and > add a number to one of its components. > So M = N + ( 0, 3, 0, ...,0). > This should work without testing, if i'm not mistaken. With friendly greetings > Hero Hi Hero found out myself (I hope). Adding an arbitrary offset vector to the > normal won't do, as I'll explain below. Given your example: M = N + ( 0, 3, 0, ...,0) assume that N happens to be equal to e.g. N = ( 0, 6, 0, ...,0) => M = ( 0, 9, 0, ...,0) which is parallel to N i.e. M = 3 * N. In general, this method will > fail as soon as the random offset is the same vector, down to scale, > as the normal or become numerically instable if it's nearly the same. > First i want to repeat: x-product only in 3D. First component of N equals zero, M = N + (1, 0 , 0 ) First component of N not equal to zero and N = ( a, b, c) , so M = ( 2 * a , b , c ) And then M x N is perpendicular to both. May be this will work without a test: N = ( a, b , c ) so M = ( c, 2 * a , 3 * b ) With friendly greetings Hero === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? > Fellows, >I'm struggling with seems to be a trivial problem but currently I'm >not sure that it has a stable solution. The problem is this; assume >that I have a point in 3D and corresponding normal of an unknown >surface at that point. Equipped with this input I need to construct a >local orthogonal basis of a plane that is tangential to the surface, >i.e. finding the binormal and tangent vectors. > The thing that makes it somewhat challenging is that I don't want to >have an iterative solution or performing some kind of test if it's a >valid solution. This results in that, as far as I've found out so far, >there is always a special case where the method fails or become >numerically instable (the method will be implemented on a computer). >E.g. to use Gram-Schmidt, I need an additional vector, not parallel to >the normal which just brings me back to the original problem of >finding a non-parallel vector, thus Gram-Schmidt is not applicable. > Another way may be to randomly select a vector, but this may of course >result in a vector that is close to parallel to the normal, thus >failing. > To summarize, the problem boils down to finding a second vector that >is non-parallel to the normal, preferably orthogonal, without any >iteration or testing. Having such a vector there is a plethora of >methods that may be used to create an orthogonal basis. > Any input on this problem would be greatly appreciated. > Given a vector N , just take any vector M, which is not in the same >direction or directly opposite, thus >there is no number r with r times M = N, >than > M x N is orthogonal to N. > In order to get such a vector, such a 'any vector', just take N and >add a number to one of its components. >So M = N + ( 0, 3, 0, ...,0). >This should work without testing, if i'm not mistaken. > With friendly greetings >Hero Hi Hero found out myself (I hope). Adding an arbitrary offset vector to the > normal won't do, as I'll explain below. Given your example: M = N + ( 0, 3, 0, ...,0) assume that N happens to be equal to e.g. N = ( 0, 6, 0, ...,0) => M = ( 0, 9, 0, ...,0) which is parallel to N i.e. M = 3 * N. In general, this method will > fail as soon as the random offset is the same vector, down to scale, > as the normal or become numerically instable if it's nearly the same. > First i want to repeat: x-product only in 3D. First component of N equals zero, M = N + (1, 0 , 0 ) > First component of N not equal to zero and N = ( a, b, c) , so M = ( 2 > * a , b , c ) And then M x N is perpendicular to both. May be this will work without a test: > N = ( a, b , c ) so M = ( c, 2 * a , 3 * b ) With friendly greetings > Hero Hi Hero If I understand you correctly, given N = ( a, b , c ) create M = ( c, 2 * a , 3 * b ), i.e. basically rearrange and scale a, b, c. If that's what you mean, assume we calculate the dot-product between N, M, i.e. M*N: M*N = a*c + 2*a*b + 3*b*c For these to be parallel or near parallel, the dot-product will be zero or close to zero, thus if the method should work it must be impossible to select a, b, c (a, b, c not all equal to zero) such that a*c + 2*a*b + 3*b*c = 0 It seems obvious to me that there must exist such a real numbers a, b, c so that the dot-product is equal to zero, i.e. M and N are parallel. One example of such numbers are to select a=b=0 and an arbitrary value to c (not equal to zero). This means that the method fails if the normal is parallel to the z-axis. === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? >> Fellows, >> I'm struggling with seems to be a trivial problem but currently I'm >> not sure that it has a stable solution. The problem is this; assume >> that I have a point in 3D and corresponding normal of an unknown >> surface at that point. Equipped with this input I need to construct a >> local orthogonal basis of a plane that is tangential to the surface, >> i.e. finding the binormal and tangent vectors. >The thing that makes it somewhat challenging is that I don't want to >> have an iterative solution or performing some kind of test if it's a >> valid solution. This results in that, as far as I've found out so far, >> there is always a special case where the method fails or become >> numerically instable (the method will be implemented on a computer). >> E.g. to use Gram-Schmidt, I need an additional vector, not parallel to >> the normal which just brings me back to the original problem of >> finding a non-parallel vector, thus Gram-Schmidt is not applicable. >Another way may be to randomly select a vector, but this may of course >> result in a vector that is close to parallel to the normal, thus >> failing. >To summarize, the problem boils down to finding a second vector that >> is non-parallel to the normal, preferably orthogonal, without any >> iteration or testing. Having such a vector there is a plethora of >> methods that may be used to create an orthogonal basis. >Any input on this problem would be greatly appreciated. > Given a vector N , just take any vector M, which is not in the same >direction or directly opposite, thus >there is no number r with r times M = N, >than > M x N is orthogonal to N. > In order to get such a vector, such a 'any vector', just take N and >add a number to one of its components. >So M = N + ( 0, 3, 0, ...,0). >This should work without testing, if i'm not mistaken. > With friendly greetings >Hero > Hi Hero > found out myself (I hope). Adding an arbitrary offset vector to the >normal won't do, as I'll explain below. Given your example: > M = N + ( 0, 3, 0, ...,0) > assume that N happens to be equal to e.g. > N = ( 0, 6, 0, ...,0) => M = ( 0, 9, 0, ...,0) > which is parallel to N i.e. M = 3 * N. In general, this method will >fail as soon as the random offset is the same vector, down to scale, >as the normal or become numerically instable if it's nearly the same. > First i want to repeat: x-product only in 3D. First component of N equals zero, M = N + (1, 0 , 0 ) > First component of N not equal to zero and N = ( a, b, c) , so M = ( 2 > * a , b , c ) And then M x N is perpendicular to both. May be this will work without a test: > N = ( a, b , c ) so M = ( c, 2 * a , 3 * b ) With friendly greetings > Hero Hi Hero > If I understand you correctly, given N = ( a, b , c ) create M = ( c, > 2 * a , 3 * b ), i.e. basically rearrange and scale a, b, c. If that's > what you mean, assume we calculate the dot-product between N, M, i.e. > M*N: M*N = a*c + 2*a*b + 3*b*c For these to be parallel or near parallel, the dot-product will be > zero or close to zero, thus if the method should work it must be > impossible to select a, b, c (a, b, c not all equal to zero) such that a*c + 2*a*b + 3*b*c = 0 It seems obvious to me that there must exist such a real numbers a, b, > c so that the dot-product is equal to zero, i.e. M and N are parallel. > One example of such numbers are to select a=b=0 and an arbitrary value > to c (not equal to zero). This means that the method fails if the > normal is parallel to the z-axis. > Oaky, one more try: As N is not equal to zero, calculate the length | N | and divide: N1 = N / | N | Now add 10 to the first component of N1. It's a nice problem. Hero === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? > Oaky, one more try: > As N is not equal to zero, calculate the length | N | and divide: > N1 = N / | N | > Now add 10 to the first component of N1. It's a nice problem. Hero This will not work, when N = ( 1, 0 , 0 ). Now, i'll take more care, and so more time. See You perhaps Hero === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? > It's a nice problem. > What about this one: N is given and not zero M = ( 1, 0 , 0 ) x N + ( 0, 1 , 0 ) x N Now i hope M has a different direction from N, so M x N is perpendicular to N. Hero === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? > It's a nice problem. What about this one: > N is given and not zero > M = ( 1, 0 , 0 ) x N + ( 0, 1 , 0 ) x N > Now i hope M has a different direction from N, so > M x N is perpendicular to N. > Even more beautiful: M is already perp to N. So for any given vector N in R x R x R, N not zero is M = ( 1, 0 , 0 ) x N + ( 0, 1 , 0 ) x N perpendicular to N. Good night Hero === Subject: Re: Stable finding of orthogonal vector to a normal in 3D? Fellows, > I'm struggling with seems to be a trivial problem but currently I'm > not sure that it has a stable solution. The problem is this; assume > that I have a point in 3D and corresponding normal of an unknown > surface at that point. Equipped with this input I need to construct a > local orthogonal basis of a plane that is tangential to the surface, > i.e. finding the binormal and tangent vectors. The thing that makes it somewhat challenging is that I don't want to > have an iterative solution or performing some kind of test if it's a > valid solution. This results in that, as far as I've found out so far, > there is always a special case where the method fails or become > numerically instable (the method will be implemented on a computer). > E.g. to use Gram-Schmidt, I need an additional vector, not parallel to > the normal which just brings me back to the original problem of > finding a non-parallel vector, thus Gram-Schmidt is not applicable. Another way may be to randomly select a vector, but this may of course > result in a vector that is close to parallel to the normal, thus > failing. To summarize, the problem boils down to finding a second vector that > is non-parallel to the normal, preferably orthogonal, without any > iteration or testing. Having such a vector there is a plethora of > methods that may be used to create an orthogonal basis. Any input on this problem would be greatly appreciated. Given a vector N , just take any vector M, which is not in the same > direction or directly opposite, thus > there is no number r with r times M = N, > than > M x N is orthogonal to N. In order to get such a vector, such a 'any vector', just take N and > add a number to one of its components. > So M = N + ( 0, 3, 0, ...,0). > This should work without testing, if i'm not mistaken. > Sorry MxN only works in 3D, so M = N + ( 0, 3, 0). Hero === Subject: Vector spaces over real numbers here are two questions which turned up solving some homework problems for a basic linear algebra course: 1) Are the rational numbers a vector space over the real numbers? I suppose, they are not: For instance, sqrt(2) is a real number and 1 is a rational number, but the product sqrt(2)*1 (multiplication with a scalar) is not rational any longer, which violates closure of multiplication with scalars. Am I correct? We didn't really require closure for addition and multiplication explicitelly for vector spaces, however, I think this was either forgotten or implied--but AFAIK it is part of the definition, isn't it? 2) Do some simplification on a expression containing some real scalars and two variables denoting vectors. Not really difficult, however, I had to do a step analogous to (variables denote vectors and (-a) is the additive invers vector of a): (-1)*a=(-a), (given that, of course, 1 is the multiplicative neutral scalar) This looks of course obvious, but I can't remember we did specify/postulate anything like this for vector spaces over the reals... I could image (although I have not yet any concrete examples in mind) one can construct vector spaces over real numbers which do not satisfy this requirement--or am I wrong here, can this be proven out of the axioms? Daniel Kraft -- by MSN, associate ICQ with stress--so please use good, old E-MAIL! === Subject: Re: Vector spaces over real numbers > > here are two questions which turned up solving some homework problems > for a basic linear algebra course: > > 1) Are the rational numbers a vector space over the real numbers? More interesting in a number of ways is the issue of the reals as a vector space over the rationals. === Subject: Re: Vector spaces over real numbers > >> here are two questions which turned up solving some homework problems >> for a basic linear algebra course: >> 1) Are the rational numbers a vector space over the real numbers? > > More interesting in a number of ways is the issue of the reals as a > vector space over the rationals. Well, this was the problem statement; and as the rationals are no vector space over the reals, it's of course not quite interesting ;) What do you think are the interesting properties of reals over rationals? Daniel -- by MSN, associate ICQ with stress--so please use good, old E-MAIL! === Subject: Re: Vector spaces over real numbers here are two questions which turned up solving some homework problems >> for a basic linear algebra course: > 1) Are the rational numbers a vector space over the real numbers? More interesting in a number of ways is the issue of the reals as a > vector space over the rationals. Well, this was the problem statement; and as the rationals are no vector > space over the reals, it's of course not quite interesting ;) What do you think are the interesting properties of reals over rationals? They are always good for some counter-examples; e.g. I just looked up my scribbled remarks in my Algebra by Serge Lang (1969 edition); I was startled when I couldn't solve the following exercise (p. 383): 20. (Tate) Let E,F be complete normed vector spaces over the real numbers. Let f:E->F be a map having the following property. There exists anumber C>0 such that for all x,y in E, we have | f(x+y) - f(x) - f(y) | <= C. Show that there exists a unique linear map g:E->F such that |g-f| is bounded (i.e. |g(x)-f(x)| is bounded a s a function of x). Generalize to the bilinear case. [Hint: Let g(x) = lim_{ntoinfty} f(2^n x)/2^n . ] I was even more startled when I found a simple counterexample based on viewing R as a Q-vector space :) hagman === Subject: Re: Vector spaces over real numbers > 2) Do some simplification on a expression containing some real scalars > and two variables denoting vectors. > > Not really difficult, however, I had to do a step analogous to > (variables denote vectors and (-a) is the additive invers vector of a): > > (-1)*a=(-a), > (given that, of course, 1 is the multiplicative neutral scalar) > > This looks of course obvious, but I can't remember we did > specify/postulate anything like this for vector spaces over the reals... > I could image (although I have not yet any concrete examples in mind) > one can construct vector spaces over real numbers which do not satisfy > this requirement--or am I wrong here, can this be proven out of the axioms? Ok, found the prove myself! So this part is solved ;) -- by MSN, associate ICQ with stress--so please use good, old E-MAIL! === Subject: Re: Vector spaces over real numbers days. My association with the Department is that of an alumnus. >here are two questions which turned up solving some homework problems >for a basic linear algebra course: 1) Are the rational numbers a vector space over the real numbers? I suppose, they are not: For instance, sqrt(2) is a real number and 1 >is a rational number, but the product sqrt(2)*1 (multiplication with a >scalar) is not rational any longer, which violates closure of >multiplication with scalars. Am I correct? If we are assuming the usual product of real numbers as being the scalar multiplication, yes, you are correct. (In fact, there can be no way to define scalar multiplication in order to make the rationals into a vector space over the reals: there aren't enough rationals, but you are probably not expected to be able to argue that way). >We didn't really require closure for addition and multiplication >explicitelly for vector spaces, however, I think this was either >forgotten or implied--but AFAIK it is part of the definition, isn't it? It was in fact said explicitly, you probably just didn't notice. They probably talked about operations, and that name already implies closure (just like saying that f is a function from A to B implies that the values of f are taken in B, and not somewhere else). >2) Do some simplification on a expression containing some real scalars >and two variables denoting vectors. Not really difficult, however, I had to do a step analogous to >(variables denote vectors and (-a) is the additive invers vector of a): (-1)*a=(-a), >(given that, of course, 1 is the multiplicative neutral scalar) This looks of course obvious, but I can't remember we did >specify/postulate anything like this for vector spaces over the reals... No, this is not part of the usual axioms of a vector space, but it can be derived from them. > I could image (although I have not yet any concrete examples in mind) >one can construct vector spaces over real numbers which do not satisfy >this requirement--or am I wrong here, can this be proven out of the axioms? This can be proven from the axioms. HINT: -a is the unique vector that added to a will give a result of 0 (the zero vector). What do you get if you add (-1)*a to a? Subhint: write the second a as 1*a. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Vector spaces over real numbers > >> here are two questions which turned up solving some homework problems >> for a basic linear algebra course: >> 1) Are the rational numbers a vector space over the real numbers? >> I suppose, they are not: For instance, sqrt(2) is a real number and 1 >> is a rational number, but the product sqrt(2)*1 (multiplication with a >> scalar) is not rational any longer, which violates closure of >> multiplication with scalars. Am I correct? > > If we are assuming the usual product of real numbers as being the > scalar multiplication, yes, you are correct. (In fact, there can be no > way to define scalar multiplication in order to make the rationals > into a vector space over the reals: there aren't enough rationals, but > you are probably not expected to be able to argue that way). > >> We didn't really require closure for addition and multiplication >> explicitelly for vector spaces, however, I think this was either >> forgotten or implied--but AFAIK it is part of the definition, isn't it? > > It was in fact said explicitly, you probably just didn't > notice. They probably talked about operations, and that name already > implies closure (just like saying that f is a function from A to B > implies that the values of f are taken in B, and not somewhere else). Yeah, this might be. >> 2) Do some simplification on a expression containing some real scalars >> and two variables denoting vectors. >> Not really difficult, however, I had to do a step analogous to >> (variables denote vectors and (-a) is the additive invers vector of a): >> (-1)*a=(-a), >> (given that, of course, 1 is the multiplicative neutral scalar) >> This looks of course obvious, but I can't remember we did >> specify/postulate anything like this for vector spaces over the reals... > > No, this is not part of the usual axioms of a vector space, but it can > be derived from them. > >> I could image (although I have not yet any concrete examples in mind) >> one can construct vector spaces over real numbers which do not satisfy >> this requirement--or am I wrong here, can this be proven out of the axioms? > > This can be proven from the axioms. > > HINT: -a is the unique vector that added to a will give a result of 0 > (the zero vector). What do you get if you add (-1)*a to a? Subhint: > write the second a as 1*a. again at distributivity laws... However, *this* is something we for sure didn't yet talk about (no proves, just definitions and examples yet). But maybe they thought nobody would question (-1)*a=-a :D I could ask my colleagues tomorrow, but probably this would make them look at me strangely from now on Daniel -- by MSN, associate ICQ with stress--so please use good, old E-MAIL! === Subject: Complex calculus 1. If u(z) is a harmonic & bounded function with 0 < lz-al < R, show that limz->a f(z) exists. 2. u(z), v(z) are harmonic functions in simple connected domain D, u(z)v(z) = 0 in D show that u(z)=0 in D or v(z)=0 in D (cf) u(z) = u(x+yi) = f(x,y) is harmonic if f xx and f yy are continuous and f xx ^2 + f yy ^2 = 0 Help~! === Subject: Re: Complex calculus > 1. If u(z) is a harmonic & bounded function with 0 < lz-al < R, show > that limz->a f(z) exists. What's f? > 2. u(z), v(z) are harmonic functions in simple connected domain D, > u(z)v(z) = 0 in D > show that u(z)=0 in D or v(z)=0 in D hint: The set where a continuous function is 0 is closed. So you have two closed sets in D whose union is D. One of them must have non-empty interior, ... > (cf) u(z) = u(x+yi) = f(x,y) is harmonic if f xx and f yy are > continuous and f xx ^2 + f yy ^2 = 0 > > > Help~! > === Subject: Re: Complex calculus On 10 4 , 2 43 , The World Wide Wade 1. If u(z) is a harmonic & bounded function with 0 < lz-al < R, show > that limz->a f(z) exists. What's f? 2. u(z), v(z) are harmonic functions in simple connected domain D, > u(z)v(z) = 0 in D > show that u(z)=0 in D or v(z)=0 in D hint: The set where a continuous function is 0 is closed. So you have > two closed sets in D whose union is D. One of them must have non-empty > interior, ... (cf) u(z) = u(x+yi) = f(x,y) is harmonic if f xx and f yy are > continuous and f xx ^2 + f yy ^2 = 0 Help~! > - - I let f be the function with f(x,y) = u(x+yi) === Subject: Linear Algebra http://img2.dcinside.com/viewimage.php?id=mathematics&no=29bcc427b48377a16fb 3dab004c86b309d6c5a9806fc4546edd11ff3b570daa86b8b67ff315297880cb0270277a34c4 5 673842aceb1b8073c5b295501d19dd3420096cf14e15d4a5910f0c0c8d4732ed728e2c76b576 8 4182040145d49197ae9ee37fd&f_no=a15910ab1d27b34a8e333e4d9a16ccb6df563fdeedeca c Hm... I've sloved all the homework without these two. === Subject: Re: Linear Algebra > http://img2.dcinside.com/viewimage.php?id=mathematics&no=29bcc427b483... Hm... I've sloved all the homework without these two. > ********************************************************* The site you've posted looks highly suspicious. Probably nobody will attempt to open such a file. Try to send over your question in another format. Tonio === Subject: Re: Linear Algebra Hm... I've sloved all the homework without these two. > ********************************************************* > The site you've posted looks highly suspicious. Probably nobody will > attempt to open such a file. > Try to send over your question in another format. > Tonio I swear. It's not about virus or something like . I just linked the file from the homework site. === Subject: Method to solve cylindrically symmetric laplacian for a cone? I'd like to find out the electric potential of a cone at a given potential. It might even be a bit more general where I have a cone-like cylindrically symmetric shape. So the laplacian is zero outside and the boundary conditions are on a finite cone. How would I solve this PDE? I know undergrad mathematics for physicists and I've seen the most basic examples of laplacian solution in a cylinder, Green functions and Bessel functions. Anton === Subject: Re: I need Classical Mechanics Solutions Manual (Goldstein) I am Master student and hope to support me with Goldstein solutions 2nd or 3rd Ed, thank you alot if you send the solutions on shahin.said@gmail.com === Subject: Bourbaki's street address? University of Nancago, I presume he gave a specific street address. Does anyone know what it was? TIA === Subject: Re: Bourbaki's street address? > University of Nancago, I presume he gave a specific street address. Does > anyone know what it was? > > TIA > > Does the book Nicolas Bourbaki: Faites et legende contain important documents from Bourbaki's life? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: fitting a curve to data I want to start with any curve and deduce the equation that describes it. My data points are spaced at some resolution. The curve is a best fit to the points. What area of math deals with this problem? I want the length of the curve minimized, and the distance from the curve to the data points minimized, both. I want to derive the equation of a boat hull. === Subject: Re: fitting a curve to data > I want to start with any curve and deduce the equation that describes it. > My data points are spaced at some resolution. The curve is a best fit to > the points. What area of math deals with this problem? I would just call it curve fitting; see e.g. http://en.wikipedia.org/wiki/Curve_fitting I want the length of the curve minimized, and the distance from the curve to > the data points minimized, both. I want to derive the equation of a boat hull. To find a best fit curve you must already have in mind some functional form for the equation of the curve (e.g. quadratic curve, exponential curve, ellipse, or whatever). Otherwise the possibilities are limitless: infinitely many curves will fit the data points exactly. Minimizing the curve length and the distance from the curve to the data points does not seem a great idea: the result would be a series of straight line segments joining the points. Do you have any particular reason to prefer any type of curve over any other? If not then the default choice might be a polynomial; e.g. y = a*x^2 + b*x + c, or y = a*x^3 + b*x^2 + c*x + d, etc. === Subject: Re: fitting a curve to data >> I want to start with any curve and deduce the equation that describes it. >> My data points are spaced at some resolution. The curve is a best fit to >> the points. >> What area of math deals with this problem? I would just call it curve fitting; see e.g. > http://en.wikipedia.org/wiki/Curve_fitting > I want the length of the curve minimized, and the distance from the curve >> to >> the data points minimized, both. >> I want to derive the equation of a boat hull. To find a best fit curve you must already have in mind some > functional form for the equation of the curve (e.g. quadratic curve, > exponential curve, ellipse, or whatever). Otherwise the possibilities > are limitless: infinitely many curves will fit the data points > exactly. Minimizing the curve length and the distance from the curve to the > data points does not seem a great idea: the result would be a series > of straight line segments joining the points. Do you have any particular reason to prefer any type of curve over any > other? If not then the default choice might be a polynomial; e.g. y > = a*x^2 + b*x + c, or y = a*x^3 + b*x^2 + c*x + d, etc. > Another way: to get an approximating polynomial surface download www.estlab.com/PolSurf.exe Arto Huttunen === Subject: Re: Minimal Ellipse by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.13 primary) with ESMTP id l93GAPK12858 perimeter problems as different object functions. So we are now in a situation to address both these problems, it posted to smr also: Narasimham === Subject: Re: Series pattern Consider the series with j running over the positive integers: >Sum_j 1 / ( (k*j - 1) * (k*j) * (k*j + 1) ) > Also worth noting that this is Sum_j zeta(2j + 1) / k^(2j + 1) A further thought is that this does not work for k=1: the first term in the initial series would be 1/(0*1*2) while the zeta function series becomes a sum of numbers above 1. But Sum_j 1 / ( (j - 1) * j * (j + 1) ) does converge when j runs over the integers from 2 upwards. The partial sum from 2 to n is 1/4 - 1/(2 * n * (n+1)) so it converges to 1/4. So Sum_{j>1} [zeta(2j + 1) - 1] = 1/4 === Subject: #46 Listing of the major mistakes of the old math and their Reals extensions; new textbook; Mathematical-Physics (p-adic primer) for students of age 6 onwards These are the major mistakes of the old math where they hyped on Algebraic Fields and Rings when they should have focused on Geometry as to what Numbers are native to what geometry. (1) By old math I mean all math before 1993 when I started posting that the Counting Numbers were really the P-adics. (2) The mistake that old math had the notion they were tops over physics, when in fact, mathematics is a tiny subset compartment of physics. (3) The mistake that mathematics had only one type of Number and all others were side-extensions of Reals (4) The Peano axioms and the development of the Reals from Naturals then Rationals then Irrationals and finally a Dedekind Cut and then the Complex Number extension, that entire program is deeply flawed because it is never linked by any solid anchor of geometry itself. There are two separate, independent and different geometries as Euclidean and then the bundled Elliptic plus Hyperbolic. in the Universe. Thus, Numbers have duality and come in two types. This is something the old math never had the tiniest of idea, that Numbers must come in two types. Just as geometry comes in two types. (6) Symmetry plays a huge role in physics and so it should in mathematics. Since Reals are infinite rightward strings, then any intelligent person would easily come to realize the other major type of number is infinite leftward strings. Symmetry would also say that Doubly Infinites is nonsense. (7) So we have two great systems of Numbers which are independent of one another just as there are two great systems of Geometry and are independent of one another. These are Euclidean compared to Elliptic/Hyperbolic. And the Numbers are Reals compared to P-adics. (8) Knowing these facts then we go back and look at Natural Numbers and Complex Numbers and ask ourselves, are these part of Reals or are they a part of P-adics. As for the Counting Numbers or Natural numbers the answer is quite simple. The Natural Numbers or Counting Numbers are a subset of the P-adics and are not members of the Reals. One of the very first proof theorems of mathematics is that of the Infinitude of Primes. How can you have infinitude of primes if the Reals are only finite portion string leftwards? You can not. The P-adics are infinite leftward string and so you can have an infinity of primes in P-adics. (9) Although the Complex number of (i) which is also called imaginary number since in Reals it is the square root of (-1) had evolved in the history of mathematics to make the Reals complete to the operations of roots. However, the P-adics have a native numbers of square root of (-1) such as the P-adic of (-)....000001 itself. When we take the root of the P-adic (-)....000001 we end up with the same number since all operations on negative P-adics ends up with a negative signed answer. (10) Instead of wasting much of the 20th century on Algebras, they should have spent that time on finding what MODEL best models Riemannian and Lobachevskian geometry. I ended up finding this best model for it is the sphere itself where one hemisphere is Elliptic geometry (Riemannian) and the antipodal hemisphere becomes Hyperbolic geometry (Lobachevskian). Having thus found the world's best Model for NonEuclidean geometry, I can use that as the anchor for revealing much of the structure of the P-adics. Such things as ...99999 although it behaves like the Real (-1) it is altogether different and it is the world's largest integer. I called it the Infinity Integer. (11) And probably the most annoying mistake of the 20th century which annoyed and frustrated me to no end in the 1990s and 2000s decade was the base dependency of the P-adics of the 20th century. This base dependency of 3-adics versus 5-adics without students ever seeing all the P-adics in their mind's-eye all at once is largely responsible for why math was retarded in the 20th century. It would be like all students before they reached University were taught how to turn one number in base 2 into some other base such as 10, when in fact, math is more than simply changing bases. So this textbook shows any student who reads and learns from it, what the P-adics are in full sight. Just as we learn the structure of Reals as the Decimal Reals, this textbook teaches the structure of P-adics as Decimal P-adics. And let the ivory towered hair twirling professor of mathematics worry about his/her nitpicking and nattering Algebra on P-adics, for they are so lost in fog that they cannot explain or teach the truth. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: #46 Listing of the major mistakes of the old math and their Reals extensions; new textbook; Mathematical-Physics (p-adic primer) for students of age 6 onwards On Oct 3, 9:54 am, Archimedes Plutonium These are the major mistakes of the old math where they hyped on === Subject: #47 By opening up the Decimal P-adics reveals the mistakes of the old math; Listing of the major mistakes of the old math and their Reals extensions; new textbook; Mathematical-Physics (p-adic primer) for students of age 6 onwards (11) And probably the most annoying mistake of the 20th century which > annoyed and frustrated me to no end in the 1990s and 2000s decade was > the base dependency of the P-adics of the 20th century. This base > dependency of 3-adics versus 5-adics without students ever seeing all > the P-adics in their mind's-eye all at once is largely responsible for > why math was retarded in the 20th century. It would be like all students > before they reached University were taught how to turn one number in > base 2 into some other base such as 10, when in fact, math is more than > simply changing bases. So this textbook shows any student who reads and > learns from it, what the P-adics are in full sight. Just as we learn > the structure of Reals as the Decimal Reals, this textbook teaches the > structure of P-adics as Decimal P-adics. And let the ivory towered hair > twirling professor of mathematics worry about his/her nitpicking and > nattering Algebra on P-adics, for they are so lost in fog that they > cannot explain or teach the truth. > Let me talk alot more about the last point I made above in (11). Once I or anyone else opens up the P-adics to easy operations of add, multiply, subtract and divide, where it is as easy as in Reals, then the vast amount of progress in understanding the P-adics comes about. When school children ask their teacher what is 1/2 of .... 999999 and what angle does it represent and when they ask is this number of 5....000000 larger than 49....9999999, then we have opened up the P-adics. Taken the P-adics away from the ivory towered nattering nutter who is lost in some silly algebra. Once we get schoolchildren talking about prime numbers that come in strings of three instead of the mere twin-primes such as 29 and 31. Where we have Triplet primes because the P-adics can construct such triplet primes, do we make vast progress in mathematics. Once we have schoolchildren able to Add, to Multiply to Subtract and Divide P-adics almost as easily as they do Reals, then the world of mathematics is better off. But that will not happen unless we give them P-adics without base. Where we give them the Decimal P-adics just as mathematics education is now entirely devoted to Decimal Reals. People and students alike have to be able to see in their mind's eye what something is. They can never do that if they are never taught the Decimal Reals but taught silly base 2 and base whatever without ever a anchor. The theorems of mathematics are base independent, meaning that all you need is Decimal Reals. Likewise for P-adics in that the 20th century had little to no progress with P-adics and were seen as some remote and arcane rather useless tool. They were seen as useless because no-one could picture them. No-one could visualize them. All could visualize the Reals because they are the points in a line-segment of Euclidean Geometry. So this textbook clears up P-adics and thrusts them forward as the great Number system that rivals the Reals. This textbook allows anyone who reads it, to have a full picture in their minds what the P-adic numbers are. The best the old math did was a frontis page in a book by Koblitz showing a crude artist sketch of 3-adics. That is the pitiful best that the old math could do for P-adics. How do I arrest control of the P-adics from those nattering nutter algebraists who spend most of their time twirling their hair? I do it by giving young students the way to ADD, to MULTIPLY to Subtract and Divide and to Exponentiate and to take Roots and to perform almost all the operations that exist in Reals. So if a student can do a operation with Reals, they can do it with Decimal P-adics. This sort of reminds me of political powers such as dictators who steal away the freedom of speech and expression of its people and when some technology comes along that the dictators cannot control and then there is a outburst of freedom of speech and expression. So long as the P-adics were in control of ivory towered math professors who could only see them as based on primes and algebras and would not allow students to develop the P-adics to where they could see them all and operate on them, well, P-adics were going nowhere under the control of math professors. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Discount on New Motherboards Big discount on brand new motherboards http://cheaphardware.blogspot.com/ === Subject: Conditioned diffusions Content-Length: 713 Originator: rusin@vesuvius Hi: Given a diffusion w and a time in [0,T]. Let w1 be the conditioned diffusion obtained from w, based on some event. Define w2 = set of paths in w and not in w1. Questions. 1. Confirm that w2 is a conditioned diffusion. 2. What is the dependency between w1 and w2? Seems w1 and w2 are mutually exclusive (correct term?) in [0,T] --- is this true? Any reference to mutually exclusive stochastic processes and two mutually exclusive conditioned diffusions? This is not for a specific task---so I cannot specify a concrete example. Just thought about it and could not recall ever coming across mutually exclusive stochastic processes. I am stumped by the concept of mutually exclusive over time. Best Nag === Subject: Peter Pace has Evidence that the USA is becoming a police state > http://home.comcast.net/~plutarch/PoliceState.html Here is a thread which has the only substantial quotes of Paces substantial statement on October 1 2007 as he left the chairman position of the joint chiefs of staff unvoluntarily. He made some strong comments on the media as well and I wish to find a full transcript of his statement. He is the strongest argument that the USFG is sovietizing. Police state or secret state matters not. The department of misinformation is alive and well even though it is exposing that one. And finally I have a record of it to reference: http://www.youtube.com/watch?v=UNt3U8JAm5g# Please listen to Pace. This is not a read between the lines sort of speech. It is diplomatic and strong. It is beyond suggestive while attempting to be respectful, yet that respect is clearly a facade of professionalism. The police state? Call it what you like; it's much much more than just a police state. If we thought the Soviets had it bad... Just think what our manipulative systems are capable of with their payouts and shutups and conversation steering phonies performing puppetry in the free press; the actors just doing their jobs.... Fortunateley we just need a few good men, and I believe that Pace is one of them. Plus he's got the dirt. The dirty dirty dirt. Pace's got the dirty dirt. dirty dirt dirt... Still I'm hunting for a transcript... DOD has not yet posted one and I doubt they ever will. -Tim === Subject: Length of the solution vector (of linear systems) I would like to now if the question about the length (euclidean norm) of a solution vector of a linear system is simpler to answer than the question about the solution itself. Hence, is it possible to determine the length of the solution vector without computing the solution? Does anybody nows something about it? Thomas === Subject: Re: Length of the solution vector (of linear systems) I would like to now if the question about the length (euclidean norm) > of a solution vector of a linear system is simpler to answer than the > question about the solution itself. Hence, is it possible to determine > the length of the solution vector without computing the solution? Does > anybody nows something about it? It is certainly easy to get an upper bound using Hadamard's estimate for determinant, but if you are looking for an EXACT answer, then I suspect the only one is : solve the system. === Subject: Work Systems: The Methods, Measurement & Management of Work i would like to get the Work Systems: The Methods, Measurement & Management of Work solution manual. what year is this solutions for? === Subject: Re: Work Systems: The Methods, Measurement & Management of Work >i would like to get the Work Systems: The Methods, Measurement & > Management of Work solution manual. what year is this solutions for? > since you do no work, you cannot have it. === Subject: Re: Another example of Schuh as moron (was Another Synthol moron) the map, but the view you describe is not the majority one. I admit to not having taken a poll of mathematicians the world over, but the books I've read on the subject by mathematicians -- everything from Morris Kline's OOP college text Mathematics for the Non- Mathematician and Overcoming Math Anxiety to Innumeracy and Goedel, Escher, Bach -- suggest or say outright that modern mathematicians recognize that math is an intellectual creation of human beings..the Bible (math) was written by man, not handed down from God (nature).... === Subject: Re: Another example of Schuh as moron (was Another Synthol moron) > I admit to not having taken a poll of mathematicians the world over, > but the books I've read on the subject by mathematicians -- everything > from Morris Kline's OOP college text Mathematics for the Non- > Mathematician and Overcoming Math Anxiety to Innumeracy and > Goedel, Escher, Bach -- suggest or say outright that modern > mathematicians recognize that math is an intellectual creation of > human beings..the Bible (math) was written by man, not handed down > from God (nature).... > Morris Kline's views have been roundly rejected by many in the mathmatics community, and Mathematics: The Loss of Certainty in paticular has gotten scathing reviews. His positions are not the mainstream, and his books seem to have misled you about what mathematicians in general tend to think on these issues. As for Goedel, Escher, Bach, Hofstadter is a cog-sci professor and while he is a clever fellow I don't think he's the best place to go to find out what mathematicians think so-called mathematicals are. === Subject: Re: Another example of Schuh as moron (was Another Synthol moron) community, and Mathematics: The Loss of Certainty in paticular has > gotten scathing reviews. His positions are not the mainstream, and his > books seem to have misled you about what mathematicians in general tend > to think on these issues. Really! Wow, I can't believe it...makes so much sense.... So mathematicians are really latter-day Platonists??? Do they also think numbers have a literal physical existence somewhere, or that mathematical/logical laws are universal, timeless, and out there awaiting discovery?? > As for Goedel, Escher, Bach, Hofstadter is a cog-sci professor and while > he is a clever fellow I don't think he's the best place to go to find out > what mathematicians think so-called mathematicals are. Hmm. Could you recommend some, then? I'm going to have to visit a math professor soon, then...this is a real paradigm shift for me.... === Subject: Re: Another example of Schuh as moron (was Another Synthol moron) community, and Mathematics: The Loss of Certainty in paticular has > gotten scathing reviews. His positions are not the mainstream, and his > books seem to have misled you about what mathematicians in general tend > to think on these issues. Really! Wow, I can't believe it...makes so much sense.... So mathematicians are really latter-day Platonists??? Do they also think numbers have a literal physical existence somewhere, or that mathematical/logical laws are universal, timeless, and out there awaiting discovery?? > As for Goedel, Escher, Bach, Hofstadter is a cog-sci professor and while > he is a clever fellow I don't think he's the best place to go to find out > what mathematicians think so-called mathematicals are. Hmm. Could you recommend some, then? I'm going to have to visit a math professor soon, then...this is a real paradigm shift for me.... === Subject: Platonism > So mathematicians are really latter-day Platonists??? Do they also > think numbers have a literal physical existence somewhere, or that > mathematical/logical laws are universal, timeless, and out there > awaiting discovery?? Many mathematicians are Platonists, in the sense of a particular kind of mathematical realism, but no one thinks that means numbers are physical objects. http://en.wikipedia.org/wiki/Philosophy_of_mathematics#Mathematical_realism http://en.wikipedia.org/wiki/Platonic_idealism === Subject: Re: Another example of Schuh as moron (was Another Synthol moron) fine. Hehe, so I've had it wrong all along! > I'm not a weightlifter. I just work out regularly, but I don't compete. You don't have to compete officially to be a weightlifter. I follow all the official rules -- no lifting of butt or foot, etc. -- so my gyms lifts are what I'd be lifting if I competed officially. You're a runner even if you don't race, by the same estimation...heck, I'm a smoker even if I don't smoke, since all that from cars and smokers comes to me anyway.... > I was thinking about Goedel, but I think that the first actual proof > was Turing's. But I'm not sure (and I do not care enough either way, to > try and understand what Google has to say about the matter). It was Goedel and not Turing. Turing's most famous for his thoughts on computer science. > You don't know if pure mathematics points to something in real world > or not. Many things which started as pure math (non Euclidean > geometries, for example) were thought to be perfectly abstract, but > nowadays they are used to describe reality. That is true. But insofar as a tree falling in a forest with no observers to note the fact has not, from our perspective, fallen, what is pure math right now remains pure until applications are found. > And it's not an exception we are talking about here. Plenty of math > started as an abstract idea, which later could be used to describe > reality. Sierpinski's triangle, for example. You are using algorithms > based on fractal theory while watching porn on the net, so now > Sierpinski points toward jiggly boobs. He should be glad, I think. ;-) Indeed, and like much else, words like pure are only a momentary convenience. Much remains pure right now, however, for lack of an application. > That's the only requirement for math to be valid. Pointing at anything > real isn't required. Yes -- and so math is not about what's real, but it is entirely made up. Q.E.D. > It's not as precise as that. You can build strength in muscles without > increasing it's cross-section and even as the cross-section slightly > decreases. Well, then, how's that, now...if the equation describes muscle strength, then how can the muscle get stronger outside the parameters of the equation?? > You could write it, but it doesn't mean that it would necessarily be > solvable. Really?? You think such an equation would be too complex?? > Ramanujan, I think. He learned it too. I don't know what you mean by learn, then...needless to say, some people obviously have an innate facility with numbers -- a special friendship with them, as was said of Ramanujan by his British benefactor -- which is what I was talking about.... > He's got talent, but he had to put some gross amount of hours into > developing it. Yes, but it's not learning, unless by learning you wish to encompass all formative experiences, including self-reflection.... > I could agree with that. I never doubted we disagree, except over words to use. > I think that they discovered zero, not invented it. Well you're still going in circles...you have basically conceded how mathematics is invented in some of your comments, but now you wish to resort back to that infantile Platonism.... > No, I don't think so. Their knowledge about Universe should be more or > less the same as ours (except things where any of us is wrong, because > as long as you can be wrong in many unique ways, you can be right in > only one way). Do you think God thinks in terms of numbers and equations? No. Absolutely not. We think in such terms because our minds are built in such as way as to perceive the world thus. An alien intelligence truly alien would be nowhere like ours. Rendezvous with Rama by Arthur C. Clarke goes into this, though not very artfully, as I've complained previously. > I don't think so. Their poetry would be haiku, but their math and > physics shouldn't be. Europeans can understand Japanese math with no > problem. European poets understand Japanese haiku with no problems, either (insofar as any artist understands another, which isn't to say that some artists don't have their preferences or outright dislikes). No, a truly alien lifeform would almost of necessity perceive things differently such as to make no sense to us. Heck, I wonder if there's been a science fiction story with the premise that aliens *are* indeed contacting us, only we don't recognize their signals...Douglas Adams had a bit of this in one of his Hitchhiker's books.... > Then he's not superintelligent. ;-) Maybe we're not! I mean, it took several tens of thousands of years between cave art and zero, after all.... > But it's not like children's math is distinctly different from adult's > math. It's fundamentally the same thing, even if most children wouldn't > understand more sophisticated examples of adult's mathematics. There's no such thing as children's math and adult's mathematics...there would be a difference between alien math (perception of physical patterns) and human math (perception of physical patterns)...consider that if a dog were sentient like us, there would still be no explaining red or blue or any other color to him...ditto a truly alien intelligence.... > It was this guyhttp://en.wikipedia.org/wiki/Jacques_Lacan, but I can't ROTFLOL! Mais bien sur, notre ami Jacques Lacan!! I didn't know he had a paper about i being a phallic something or other...interesting!! > You mean an accident, when his penis becomes an imaginary number? Quite a > sad story, I'd say. ;-( Well, it'd be great schtuff if only Douglas Adams was still around.... > So they can't see his penis, even on a clear day? What would that mean to a possibly sexless species, anyway.... > Ouch. The guy is doomed, if you asked me. ;-) You know, there's more to sex than imaginary numbers.... > -- > Andrzej Rosa 1127R === Subject: Alien math. (was Re: Another example of Schuh as moron (was Another Synthol moron)) [...] >> I'm not a weightlifter. I just work out regularly, but I don't compete. > > You don't have to compete officially to be a weightlifter. I follow > all the official rules -- no lifting of butt or foot, etc. -- so my > gyms lifts are what I'd be lifting if I competed officially. Go then, and compete. You can do token lifts in squat and deadlift if all you care about is your bench. _Then_ I'll call you a weightlifter, but until you have an official result to post, you are another talker, who likes to play with weights on his free time, just like me. > You're a runner even if you don't race, by the same estimation...heck, Sure. And everybody is a winner too. We all deserve medals. > I'm a smoker even if I don't smoke, since all that from cars and > smokers comes to me anyway.... No, you are not. The amount of smoke you get this way is several orders of magnitude lower than what real smoker inhales. Buy a pack and smoke one cigarette, pulling the smoke deep in your lungs, just to have an idea what it takes to earn the title! BTW - have a bucket handy, in case you bite more than you can chew. ;-) >> I was thinking about Goedel, but I think that the first actual proof >> was Turing's. But I'm not sure (and I do not care enough either way, to >> try and understand what Google has to say about the matter). > > It was Goedel and not Turing. Turing's most famous for his thoughts > on computer science. Rightly so, because what I was referring to, was his work on computability. http://en.wikipedia.org/wiki/Alan_Turing#University_and_his_work_on_computab ility He went on to prove that there was no solution to the Entscheidungsproblem by first showing that the halting problem for Turing machines is undecidable: it is not possible to decide, in general, algorithmically whether a given Turing machine will ever halt. [...] >> That's the only requirement for math to be valid. Pointing at anything >> real isn't required. > > Yes -- and so math is not about what's real, but it is entirely made > up. > > Q.E.D. It all depends what definition of real you want to use. Triangle isn't real, because there exist exactly 0 real world examples of mathematical triangle. But we can still use our knowledge of various facts mathematicians discovered about triangles. >> It's not as precise as that. You can build strength in muscles without >> increasing it's cross-section and even as the cross-section slightly >> decreases. > > Well, then, how's that, now...if the equation describes muscle > strength, then how can the muscle get stronger outside the parameters > of the equation?? You answered your own question. There must exist parameters which aren't accounted for by the equation. >> You could write it, but it doesn't mean that it would necessarily be >> solvable. > > Really?? You think such an equation would be too complex?? It could be. It depends on how close to reality you'd want to get. [...] >> He's got talent, but he had to put some gross amount of hours into >> developing it. > > Yes, but it's not learning, unless by learning you wish to > encompass all formative experiences, including self-reflection.... If you spend hours solving mathematical problems, you learn math. No matter if you are guided by a tutor or do it all by yourself, you still learn how to solve mathematical problems. I simply tried to point, that even the most gifted people in the human history weren't miraculously born with their mathematical prowess. They developed it. It's a skill, among other things. So, when you say I don't know much math, because I was never born with math in me, you are making a wrong assumption, that gifted people were born with math in them. They weren't. They spent ages getting math into them. Almost nobody is born genius, but even if you are born with a potential to become one, you have to put in the hours needed. [...] >> I think that they discovered zero, not invented it. > > Well you're still going in circles...you have basically conceded how > mathematics is invented in some of your comments, but now you wish to > resort back to that infantile Platonism.... The problem is, that when it comest to math, Platonism doesn't look all that infantile too me. ;-( >> No, I don't think so. Their knowledge about Universe should be more or >> less the same as ours (except things where any of us is wrong, because >> as long as you can be wrong in many unique ways, you can be right in >> only one way). > > Do you think God thinks in terms of numbers and equations? > > No. Absolutely not. I don't believe in God. > We think in such terms because our minds are built in such as way as > to perceive the world thus. An alien intelligence truly alien would > be nowhere like ours. Rendezvous with Rama by Arthur C. Clarke goes > into this, though not very artfully, as I've complained previously. Read Solaris, if you are interested in what somebody with brains had to say about the topic. >> I don't think so. Their poetry would be haiku, but their math and >> physics shouldn't be. Europeans can understand Japanese math with no >> problem. > > European poets understand Japanese haiku with no problems, either > (insofar as any artist understands another, which isn't to say that > some artists don't have their preferences or outright dislikes). > > No, a truly alien lifeform would almost of necessity perceive things > differently such as to make no sense to us. It shouldn't matter how they perceive things. If they can build space ships, they need to be able to _predict_ the behavior of things, and here math comes handy. For example they know that their spaceship burns five gazylions of antigravitons per parsec, so they can cruise through half the Galaxy on a full tank. But if they have six fingers instead of five, chances are they commonly use dodecimal numbering system instead of decimal one, like we do. > Heck, I wonder if there's > been a science fiction story with the premise that aliens *are* indeed > contacting us, only we don't recognize their signals... Sure. Really have a look at Solaris. [...] >> But it's not like children's math is distinctly different from adult's >> math. It's fundamentally the same thing, even if most children wouldn't >> understand more sophisticated examples of adult's mathematics. > > There's no such thing as children's math and adult's > mathematics...there would be a difference between alien math > (perception of physical patterns) and human math (perception of > physical patterns)...consider that if a dog were sentient like us, > there would still be no explaining red or blue or any other > color to him...ditto a truly alien intelligence.... I think that you've chosen your example poorly. After all we know, can describe, and perceive full spectrum of electromagnetic radiation, not just the visible part of it. So perception matters relatively little, if you are able to think in a logical way. >> It was this guyhttp://en.wikipedia.org/wiki/Jacques_Lacan, but I can't > > ROTFLOL! Mais bien sur, notre ami Jacques Lacan!! > > I didn't know he had a paper about i being a phallic something or > other...interesting!! Lacan's ideas centered on Freudian concepts such as the unconscious, the castration complex, ~~~~~~~~~~~~~~~~~~ [...] -- Andrzej Rosa 1127R === Subject: Re: Alien math. (was Re: Another example of Schuh as moron (was Another Synthol moron)) all you care about is your bench. _Then_ I'll call you a weightlifter, > but until you have an official result to post, you are another talker, > who likes to play with weights on his free time, just like me. I'm a weightlifter because I lift weights at respectable, upper beginner-lower intermediate level poundages. It's all very simple. I'm not a hard-core lifter, not competitive lifter, but a real lifter all the same given my level of athletic performance. > Sure. And everybody is a winner too. We all deserve medals. Whoa, that's quite an overbite there -- talk about mood swings! No, not everybody is a winner, because winners have to win. But everybody who runs and lifts -- for real, not some little skip between the parking lot and Walmart carrying groceries -- is a runner and lifter. > No, you are not. The amount of smoke you get this way is several orders > of magnitude lower than what real smoker inhales. Buy a pack and smoke > one cigarette, pulling the smoke deep in your lungs, just to have an > idea what it takes to earn the title! I think it's the cumulative effects that are so debilitating...wasn't second-hand smoke even worse than getting it first-hand oneself?? Though I ride my bike alot, commuting and whatnot, I never call it a healthy thing to do because I'm in traffic almost all the time! > BTW - have a bucket handy, in case you bite more than you can chew. ;-) > Rightly so, because what I was referring to, was his work on > computability. http://en.wikipedia.org/wiki/Alan_Turing#University_and_his_work_on_c... > He went on to prove that there was no solution to the > Entscheidungsproblem by first showing that the halting problem for > Turing machines is undecidable: it is not possible to decide, in > general, algorithmically whether a given Turing machine will ever halt. Ah, not quite proving that some things in math will never be known. > It all depends what definition of real you want to use. Triangle > isn't real, because there exist exactly 0 real world examples of > mathematical triangle. Well, I don't know what you mean by a mathematical triangle, then...do you mean the perfect Platonic form of a triangle? > But we can still use our knowledge of various > facts mathematicians discovered about triangles. Strictly speaking, they didn't discover them...mathematics is a language...it was made up by human beings...mathematical facts aren't discovered any more than grammatical facts of English usage were discovered by philologists.... > You answered your own question. There must exist parameters which > aren't accounted for by the equation. Well, I said if -- if the equation describes muscle strength -- if it did, then it does so completely; otherwise, it's not an equation of muscle strength! Have you ever heard of an equation imperfectly describing what it's supposed to describe? > It could be. It depends on how close to reality you'd want to get. I can't wait for computer games to get super realistic so that programmers will be forced to investigate the issue! Either that, or something arising out of steroids and medical science.... > If you spend hours solving mathematical problems, you learn math. Ah, semantics indeed! In a sense, you're correct with that sentence. But in the context of our discussions about talent, I think you're simply having another conversation altogether -- related, but not the same. Yes, one learns by doing...for some things, however, some people know intuitively...no amount of aimless musing could make most four year- olds come up with multiplication on their own, whereas a guy in statistics is said to have done that (Pearson, I think his name is).... > No > matter if you are guided by a tutor or do it all by yourself, you still > learn how to solve mathematical problems. I simply tried to point, that > even the most gifted people in the human history weren't miraculously > born with their mathematical prowess. They developed it. It's a skill, > among other things. You're confusing aptitude with skill. I'm talking about aptitude, talent, innate affinities...you're talking skill, and knowledge, and problem sets and solutions. > So, when you say I don't know much math, because I was never born with > math in me, you are making a wrong assumption, that gifted people were > born with math in them. They weren't. They spent ages getting math > into them. Believe you me, I'm a proponent of the everybody can math! school (yes, math as a verb)...but I also recognize that some have more of an aptitude for it than others -- just like with weightlifting, say. > Almost nobody is born genius, but even if you are born with a potential > to become one, you have to put in the hours needed. Again, aptitude and skill. I'm talking about aptitude. > The problem is, that when it comest to math, Platonism doesn't look all > that infantile too me. ;-( Platonic forms are appealing to those aspects of our personalities which still yearn for absolutes and certainty. Such are the desires of children and many a flag-waving patriot. To be sure, many of my frustrations with weightlifting and why my body reacts as mysteriously as it does arises from just such infantilism. > I don't believe in God. Neither do I -- but such a being, a super-intelligence, would not think in terms of numbers and mathematics. It would know precisely how long or how wide without hair-splitting numbers...it is our weakness in perception which makes mathematics a necessary invention. > Read Solaris, if you are interested in what somebody with brains had to > say about the topic. Hey, wasn't that a movie?? Two movies, actually...a French one and an American remake recently? Do you mean to suggest A.C. Clarke didn't have brains on the topic? The book put me to sleep, but the ideas were interesting...until then, I'd not considered that math could be different for aliens.... > It shouldn't matter how they perceive things. If they can build space > ships, they need to be able to _predict_ the behavior of things, and > here math comes handy. But that's precisely my point -- they would perceive things in such a way as to not need math. I mean, that could be the case, particularly if they're not carbon-based life-forms (an even further stretch of the imagination, I agree)...every monkey thinks everyone else is a monkey, every thief thinks everyone else is a thief...I think we shouldn't be so sure that math is universal throughout the universe for all intelligent life, or even necessary or existent for all intelligent life...would telepathic beings have need for words and language? Almost certainly not. So why math?? Indeed, what kind of a civilization would telepathic beings have?? Would such beings even perceive themselves to be individuals, if their telepathy was totally global?? I wish I had enough math and science to adequate address these issues in fiction. As it is, I'm only an English major. > For example they know that their spaceship burns > five gazylions of antigravitons per parsec, so they can cruise through > half the Galaxy on a full tank. Again, you're assuming they need to think in terms of decimals and fractions...and again I can only offer you the example of color-blind dogs or insects that perceive colors we do not: it is entirely possible extraterrestrial life-forms do not need math for their technology.... > But if they have six fingers instead of five, chances are they commonly > use dodecimal numbering system instead of decimal one, like we do. Yes -- and you should also agree, then, that if they do *not* have brains like we do, they may use an entirely different mode of perceiving the physical world, and utilizing the physical world, other than mathematics...perhaps we see in patterns (mathematics) because of our brains (which I say is, in its higher-order functions, a pattern- generating machine -- I do believe I'm the only one to say this; at least I don't remember having read that anywhere else), whereas an alien brain...or brains, even, for the one individual organism...would...like, perceive things on a totally different basis.... Heck, just look at the natural languages, and how some allegedly don't have a future tense...certainly Chinese doesn't have conjugations and inflections.... > Sure. Really have a look at Solaris. Hmm, okay...though I'm really not expecting much.... > I think that you've chosen your example poorly. After all we know, can > describe, and perceive full spectrum of electromagnetic radiation, not > just the visible part of it. So perception matters relatively little, if you are able to think in a > logical way. I think you're missing the forest for the trees...a dog is naturally color-blind...no matter how many equations you show him, he will never know colors...likewise, our brains may just be such that some things are beyond our perception, even if we have the mathematics for it (spatial dimensions beyond three, for instance).... > Lacan's ideas centered on Freudian concepts such as the unconscious, the > castration complex, > ~~~~~~~~~~~~~~~~~~ Yeah, but to connect that with the imaginary number i???? That would be quite the coup de main! > -- > Andrzej Rosa 1127R === Subject: Re: Alien math. (was Re: Another example of Schuh as moron (was Another Synthol moron)) [...] > I'm a weightlifter because I lift weights at respectable, upper > beginner-lower intermediate level poundages. It's all very simple. > I'm not a hard-core lifter, not competitive lifter, but a real lifter > all the same given my level of athletic performance. So I'm strongman, because I load and carry stuff from time to time. I'm a strongman of a first floor level, turn to left at the end of a corridor. >> Sure. And everybody is a winner too. We all deserve medals. > > Whoa, that's quite an overbite there -- talk about mood swings! > > No, not everybody is a winner, because winners have to win. And to be called just like some sportsmen, you gotta compete at the sport. > But > everybody who runs and lifts -- for real, not some little skip between > the parking lot and Walmart carrying groceries -- is a runner and > lifter. So, you have a cutoff too? So do have plenty of people who bother competing. You know, that regular training isn't the same as doing an activity on occasion. They know, that actually going out there and facing the bar isn't the same as training. >> No, you are not. The amount of smoke you get this way is several orders >> of magnitude lower than what real smoker inhales. Buy a pack and smoke >> one cigarette, pulling the smoke deep in your lungs, just to have an >> idea what it takes to earn the title! > > I think it's the cumulative effects that are so debilitating... Poor Einstein, Bohr, Oppenheimer (all *very* heavy smokers). They could be so much brighter without all those nasty cumulative effects. > wasn't > second-hand smoke even worse than getting it first-hand oneself?? No, it wasn't. They found that if a smoker inhales a smoke, then some part of toxins actually goes into his body and he doesn't breathe them out, but that's it. The rest was a lot of bollocks. > Though I ride my bike alot, commuting and whatnot, I never call it a > healthy thing to do because I'm in traffic almost all the time! There was a study on this too, and they found that if you cycle the same distance as you drive in a car, on the highway (with bikes being much slower), you still had lower concentration of toxins in your blood at the end of the road. So cycling even in traffic is still healthy. [...] >> It all depends what definition of real you want to use. Triangle >> isn't real, because there exist exactly 0 real world examples of >> mathematical triangle. > > Well, I don't know what you mean by a mathematical triangle, > then...do you mean the perfect Platonic form of a triangle? No, I mean real triangle, as derived from the axioms of euclidean geometry (for example). We know a lot of useful stuff about such triangles, and we know them for sure. On the other hand, we can say relatively little about real life examples of a triangle. >> But we can still use our knowledge of various >> facts mathematicians discovered about triangles. > > Strictly speaking, they didn't discover them... You mean that mathematicians invented that triangles happen to have the sum of their angles equal 180 degrees? And before they invented this fact, triangles could have random sum of their angles? > mathematics is a > language...it was made up by human beings...mathematical facts They are _facts_, not facts. They can afford it, because these facts happen in a very well defined world. > aren't discovered any more than grammatical facts of English usage > were discovered by philologists.... Stupid Pythagoras. He should invent his equation without squares. It would make all the calculations so much simpler. ;-) >> You answered your own question. There must exist parameters which >> aren't accounted for by the equation. > > Well, I said if -- if the equation describes muscle strength -- if > it did, then it does so completely; otherwise, it's not an equation of > muscle strength! Actually that's typical. All science works this way, and they aren't all that fussy about strictly naming their equations. Especially because most of the time you don't know your unaccounted parameters, so you can't say in which conditions your equation will stop working precisely. > Have you ever heard of an equation imperfectly > describing what it's supposed to describe? All the time. That's why mathematicians do not regard reality to be a worthy topic of study. [I agree with your aptitude vs skill here] >> Read Solaris, if you are interested in what somebody with brains had to >> say about the topic. > > Hey, wasn't that a movie?? Two movies, actually...a French one and an > American remake recently? The first was Russian. > Do you mean to suggest A.C. Clarke didn't have brains on the topic? > The book put me to sleep, but the ideas were interesting...until then, > I'd not considered that math could be different for aliens.... Well, it's not like I'm picking on Clarke. It's just that no other author think up anything comparable to Golem or Summa Technologiae. >> It shouldn't matter how they perceive things. If they can build space >> ships, they need to be able to _predict_ the behavior of things, and >> here math comes handy. > > But that's precisely my point -- they would perceive things in such a > way as to not need math. It could be, as long as they were some sort of alien elves, who do not build or produce complicated things. Once you start building complex stuff, you need your tools, math among them. > I mean, that could be the case, particularly > if they're not carbon-based life-forms (an even further stretch of the > imagination, I agree)...every monkey thinks everyone else is a monkey, > every thief thinks everyone else is a thief...I think we shouldn't be > so sure that math is universal throughout the universe for all > intelligent life, or even necessary or existent for all intelligent > life...would telepathic beings have need for words and language? > Almost certainly not. So why math?? I think that even telepathic beings would need a language, just like all technologically advanced civilizations would need a math. > Indeed, what kind of a civilization would telepathic beings have?? > Would such beings even perceive themselves to be individuals, if their > telepathy was totally global?? I wish I had enough math and science to adequate address these issues > in fiction. As it is, I'm only an English major. Everybody is only something. Especially those, who strive to improve themselves. >> For example they know that their spaceship burns >> five gazylions of antigravitons per parsec, so they can cruise through >> half the Galaxy on a full tank. > > Again, you're assuming they need to think in terms of decimals and > fractions...and again I can only offer you the example of color-blind > dogs or insects that perceive colors we do not: it is entirely > possible extraterrestrial life-forms do not need math for their > technology.... So how do they know the effective range of their saucers? >> But if they have six fingers instead of five, chances are they commonly >> use dodecimal numbering system instead of decimal one, like we do. > > Yes -- and you should also agree, then, that if they do *not* have > brains like we do, they may use an entirely different mode of > perceiving the physical world, and utilizing the physical world, other > than mathematics...perhaps we see in patterns (mathematics) because of > our brains (which I say is, in its higher-order functions, a pattern- > generating machine -- I do believe I'm the only one to say this; at > least I don't remember having read that anywhere else), whereas an > alien brain...or brains, even, for the one individual > organism...would...like, perceive things on a totally different > basis.... You mean, without patterns? Unlikely. Patterns let you predict outcomes. You see sun rising in the East and setting in the West, and you see a pattern there, so you expect the sun to do the same the next day. It is very useful thing to have. [...] >> Sure. Really have a look at Solaris. > > Hmm, okay...though I'm really not expecting much.... It's just probably the best sci-fi book ever written, so you shouldn't expect much. ;-) >> I think that you've chosen your example poorly. After all we know, can >> describe, and perceive full spectrum of electromagnetic radiation, not >> just the visible part of it. >> So perception matters relatively little, if you are able to think in a >> logical way. > > I think you're missing the forest for the trees...a dog is naturally > color-blind...no matter how many equations you show him, he will never > know colors... Because dogs are stupid. But you could explain colors to a (color) blind intelligent person. Using examples, analogies, experiments and so on, but he would be able to understand that contrary to him, you see colors. > likewise, our brains may just be such that some things > are beyond our perception, even if we have the mathematics for it > (spatial dimensions beyond three, for instance).... There are things in our _knowledge_ that stretch our ability to intuitively comprehend them. Take Aussies, for example. Do you want me to really grasp the fact, that they walk upside down? No way! Yet, we know it and comprehend it, and are able to predict outcomes of such crazy things like traveling around the world. [...] -- Andrzej Rosa 1127R === Subject: Alien color vision > I think you're missing the forest for the trees...a dog is naturally > color-blind...no matter how many equations you show him, he will never > know colors... Dogs seem to lack a green color cone. They are similar to humans who suffer from a from red-green color-blindness (like my father--such people are actually common), but they can see color. Note that from the point of view of animals (the tetrachromats) with four types of cones, we would count as color-blind also. === Subject: manual solutions of Elementary Principles of Chemical Processes Cc: modernbook@hotmail.com Please send me the manual solutions of Elementary Principles of Chemical Processes 3rd eddition by felder, as soon as possible. === Subject: Re: manual solutions of Elementary Principles of Chemical Processes > Please send me the manual solutions of Elementary Principles of > Chemical Processes 3rd eddition by felder, as soon as possible. > which one? there are three, all different. === Subject: Re: Siegel's critique of Lang On Sep 20, 7:49 pm, Timothy Murphy > Where did you see this, as a matter of interest. I've heard (from someone who saw it) that Mordell loved to show people this letter. === Subject: Re: Siegel's critique of Lang > > =20 > lang was never on great terms with weil > Although they did collaborate on that great song,=20 > It's a Lang Weil to Tipperary. >> I still can't help pronouncing his name as in Auld Lang=20 >> Syne.=20 > =20 > Why wouldn't it be pronounced as in Auld Lang Syne? I've never > heard it pronounced any other way in English (and though he was=20 > born in France, he spent most of his life in the US). Just don't > pronounce his first name as Surge. >> Well, Wikipedia tells me his parents have been American, but get no hint,= >> where is family was from ... > > I don't know how Wikipedia tells you his parents were American. It doesn't > tell me that. You are right, I mixed something up, Wiki does not even indicate that. >> But for the topic: the German http://de.wikipedia.org/wiki/Serge_Lang tel= >> ls >> a bit more ('making' him a kind man, but he was a roughneck AFAIK) and th= >> is >> Mordell hatte Serge Langs Diophantine Geometry verrissen und Siegel ausserte >> sich zustimmend, er vergleicht diese Richtung, mit dem Wildern von Schweinen >> in einem schoenen Garten oder dem Aufmarsch der SA-Sturmtruppen. >> If one ever knows that Siegel left Europe 1940 because of the Nazis and on >> the other hand Lang was a political engaged for the 'Leftists' only could= >> be >> astoned ... > > Lang also left Europe in 1940 because of the Nazis. I doubt that this had > much to do with his relations with Siegel and Mordell. Hm, that is not my point: the indirect citation translates to English as Mordell had torn Serge Lang's Diophantine Geometry to pieces and Siegel agreed, he compares this direction with poaching of pigs in a beautiful garden or the parade of the SA storm troops. The SA was the armed and uniformed branch of the NSDAP (in brown shirts). === Subject: Re: Siegel's critique of Lang >> But for the topic: the German http://de.wikipedia.org/wiki/Serge_Lang >> tel= >> ls >> a bit more ('making' him a kind man, but he was a roughneck AFAIK) and >> th= >> is >> Mordell hatte Serge Langs Diophantine Geometry verrissen und Siegel >> ausserte >> sich zustimmend, er vergleicht diese Richtung, mit dem Wildern von >> Schweinen >> in einem schoenen Garten oder dem Aufmarsch der SA-Sturmtruppen. >> If one ever knows that Siegel left Europe 1940 because of the Nazis and >> on >> the other hand Lang was a political engaged for the 'Leftists' only >> could= >> be >> astoned ... > > Lang also left Europe in 1940 because of the Nazis. I doubt that this > had > much to do with his relations with Siegel and Mordell. > > Hm, that is not my point: the indirect citation translates to English as > > Mordell had torn Serge Lang's Diophantine Geometry to pieces and Siegel > agreed, he compares this direction with poaching of pigs in a beautiful > garden or the parade of the SA storm troops. > > The SA was the armed and uniformed branch of the NSDAP (in brown shirts). Yes, I know. One might have hoped that Siegel would have the good sense to avoid such inflammatory metaphors, but apparently not... -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Siegel's critique of Lang <3qPIi.22207$j7.399622@news.indigo.ie> <5mfmnmFd85irU1@mid.individual.net> <5mhlqqFdftasU1@mid.individual.net > =20 >> lang was never on great terms with weil > Although they did collaborate on that great song,=20 > It's a Lang Weil to Tipperary. >> I still can't help pronouncing his name as in Auld Lang=20 >> Syne.=20 > =20 > Why wouldn't it be pronounced as in Auld Lang Syne? I've never > heard it pronounced any other way in English (and though he was=20 > born in France, he spent most of his life in the US). Just don't > pronounce his first name as Surge. >> Well, Wikipedia tells me his parents have been American, but get no hint,= > where is family was from ... I don't know how Wikipedia tells you his parents were American. It doesn't > tell me that. You are right, I mixed something up, Wiki does not even indicate that. > But for the topic: the Germanhttp://de.wikipedia.org/wiki/Serge_Langtel= >> ls >> a bit more ('making' him a kind man, but he was a roughneck AFAIK) and th= >> is > Mordell hatte Serge Langs Diophantine Geometry verrissen und Siegel ausserte >> sich zustimmend, er vergleicht diese Richtung, mit dem Wildern von Schweinen >> in einem schoenen Garten oder dem Aufmarsch der SA-Sturmtruppen. > If one ever knows that Siegel left Europe 1940 because of the Nazis and on >> the other hand Lang was a political engaged for the 'Leftists' only could= >> be >> astoned ... Lang also left Europe in 1940 because of the Nazis. I doubt that this had > much to do with his relations with Siegel and Mordell. Hm, that is not my point: the indirect citation translates to English as Mordell had torn Serge Lang's Diophantine Geometry to pieces and Siegel > agreed, he compares this direction with poaching of pigs in a beautiful > garden or the parade of the SA storm troops. The SA was the armed and uniformed branch of the NSDAP (in brown shirts). just to continue some of the biographical trivia here it is not as well known that lang actually helped uncover cia collaborations with columbia university ( the cia is the central intelligence agency of the us known for such wonderful accomplishments as usama bin laden the prevalence of crack cocaine in certain communities ... ) lang requested certain accounting documents when rumors became widespread on campus he was denied access and later investigations showed that there was indeed cia activities with the school and some of its programs -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: Siegel's critique of Lang <3qPIi.22207$j7.399622@news.indigo.ie> <5mfmnmFd85irU1@mid.individual.net> <5mhlqqFdftasU1@mid.individual.net > it is not as well known that lang > actually helped uncover cia collaborations with columbia university > http://www.dipmat.unipg.it/~mamone/sci-dem/nuocontri/mamone_05.htm With friendly greetings Hero === Subject: Re: Siegel's critique of Lang On Sep 20, 10:49 pm, Timothy Murphy was disgusted with the way in which my own contributions to the > subject had been disfigured and made unintelligible. My feeling is > very well expressed when you mention Rip van Winkle! Where did you see this, as a matter of interest. I know that Mordell had it in for Lang, > I thought because Lang had made disparaging remarks > about an author Mordell admired (Bachmann?). Mordell has written a scathing review of one of Lang's books, > and make a large number of copies of the review > which he gave to all and sundry. But I'm surprised Siegel would feel the same way. Even if Lang is just flat Siegel with such lively language is entitled. The statement says more about Siegel than it says about Lang. I know neither of these people in terms of readings. Take though for instance Boas- here is a man who took his math to a compassionate level. Any who lack this compassion can be criticized as pigs uprooting the garden; blind to their lack of attention and insensitive. I myself am quite capable of what might seem insensitive criticism. We need to study Lang... Got any quotes there? Just one good one is enough to expose a true emotional rivalry. Otherwise I would gander that Siegel's level of maturity is simply more developed. Then again as I read galathea's first post perhaps the whole thing is a compliment to Lang. He digs down to the roots. Such cryptic behavior is a good sign. The sincerity and honesty balanced with truth perhaps. -Tim === Subject: Re: Siegel's critique of Lang !! In looking up some results of Carl Siegel (1896-1981), I came across a > !! > !! Here's an excerpt from Siegel's letter ... > !! > !! When I first saw [Lang's Diophantine geometry], about a year ago, I > !! was disgusted with the way in which my own contributions to the > !! subject had been disfigured and made unintelligible. My feeling is > !! very well expressed when you mention Rip van Winkle! > !! > !! The whole style of the author contradicts the sense for simplicity and > !! honesty which we admire in the works of the masters in number theory - > !! Lagrange, Gauss, or on a smaller scale, Hardy, Landau. Just now Lang > !! has published another book on algebraic numbers which, in my opinion, > !! is still worse than the former one. I see a pig broken into a > !! beautiful garden and rooting up all flowers and trees. > !! > !! What do people think? > !! > !! Did Lang drain all the life from the subject? > In Serge's world view, the way you choose to look at a theorem is > often more important than the theorem itself. As the title > suggests, Serge's outlook on number theory was decidedly geometric. > While others at the time shared this viewpoint (e.g. Weil, Tate, > Serre), it is easy to forget that others did not, as Mordell's > review of the earlier Diophantine Geometry attests. tate said: Wonderful as it was, our graduate training with Artin was almost > totally one-dimensional and non-geometric: number fields and > function fields in one variable. In the following years, Serge > helped me become comfortable with higher dimensional things, > such as the Jacobian, Picard, and Albanese varieties over > arbitrary ground fields and also with reduction mod p, which > was not such a simple matter working with Weil's Foundations in > those days before schemes. it is commonly said that functional programming is hard > that few understand how to do it immediately > and that you have to be a skilled functional coder to even read the stuff but functional programming is the pinnacle of abstraction and reuse object oriented procedural programming > cannot reach the levels of reuse for a simple reason: > imperative procedures associates semantic preconditions to state > forcing preconditions and postconditions on any imperative process Finally I find a detail to dispute galathaea on!!! Suppose that the layer of procedure is such that the output is in a new domain. Then all such functions who output to the same domain are not conditional. It is merely the quality of the data in the new domain which is of interest. For instance functions who print their structures to stdout so long as they identify their output appropriately within the new domain have approached a new domain cleanly and unconditionally. Beyond that transducers are fine domain boundaries. In physics we see that the classical forces have some relation to an arithmetic product yet their output domain is in acceleration, not position. Hence an arithmetic domain change of great interest is present. The torques which I study are nearby. -Tim it is interesting that functional programming > corresponds most naturally to category theoretic foundations algebraic geometry is essentially a functional approach to number theory > it looks to extract invariants describing a geometric figure > through analysis of the functions that transform it or are defined using it lang gives motivation > but it is the motivation of abstractions that guide much of his work's organisations i think it is quite telling to see > wiles > silverman > tate > rohrlich > and other luminaries in the fields of geometry > regularly reference lang's works in their papers when i look at at more modern books on the topic > hindry and silverman's Diophantine Geometry for instance > i cannot help but see lang's structure reiterated lang was dense > but his advanced treatises were never really written as introductions > those who shared the same visions of structure > have found his work immensely useful and important just to fill in the earlier abbreviated quote by siegel > i think it becomes clear that the issue was ultimately this move to abstraction When I first saw [Lang's Diophantine geometry], about a year ago, I was > disgusted with the way in which my own contributions to the subject had > been disfigured and made unintelligible. My feeling is very well expressed > when you mention Rip van Winkle! The whole style of the author contradicts the sense for simplicity and honesty > which we admire in the works of the masters in number theory - Lagrange, Gauss, > or on a smaller scale, Hardy, Landau. Just now Lang has published another book > on algebraic numbers which, in my opinion, is still worse than the former one. > I see a pig broken into a beautiful garden and rooting up all flowers and trees. Unfortunately there are many fellow-travellers who have already disgraced a > large part of algebra and function theory; however, until now, number theory > had not been touched. These people remind me of the impudent behaviour of the > national socialists who sang: Wir werden weiter marschieren, bis alles in > Scherben zerfallt!'' I am afraid that mathematics will perish before the end of this century if the > present trend for senseless abstraction - as I call it: theory of the empty > set - cannot be blocked up. ... i think senseless abstraction is the key sentiment here... -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar === Subject: Re: It is irrational to put rational and irrational numbers on the same line. >It is irrational to put rational and irrational numbers on the same line. They aren't put there...they ARE there. It's not a matter of choice, it is a mathematical fact. > Imaginary numbers have their own axis. If irrational numbers cannot be derived from rational numbers; why are > they said to be inbetween? Because the real numbers (the union of the rationals and the irrationals) form a field. For example, 1.4 < sqrt(2) < 1.5 is a provable fact. And what do you mean when you say irrational numbers cannot be derived from rational numbers? You can easily operate on two rationals to generate an irrational. For example, 2^0.5 is irrational, whilst 2 and 0.5 are each rational. And although the rationals are closed under addition, subtraction, multiplication and division, this is not true for irrationals. You can have two irrational numbers x & y, such that x+y is rational. The irrationals are not closed under ANY operation I am aware of, not even infinitely more numerous than rational) is in a sense less complete than rationals. === Subject: Re: It is irrational to put rational and irrational numbers on the same line. <200710021709.l92H9dsC192238@walkabout.empros.com> Ifirrationalnumbers cannot be derived fromrationalnumbers; ... > I assume you meanirrational*real* numbers, right? >> All real numbers, including theirrationalreal numbers, can be >> derived fromrationalnumbers, so your premise is false, and your >> posting is moot. >How ARE irrationals DERIVED 'on' the line...? Irrationals are a subset of the reals. The reals aren't derived on a > line; they're defined as equivalence classes of Cauchy sequences of > rationals. They are not fractions, are they? No, they're equivalence classes of Cauchy sequences of rationals. -- > Michael F. Stemper > #include Ghosts crowd the young child's fragile eggshell mind. For goodness sake... OK. 'Line' you have a problem with, so, think AXIS. Just think... The standard NUMBER LINE (mathematics term, if unfamiliar to you) is made up of integers, their rational fractions AND...??? Please rephrase your Cauchy'ism. Adam Lewis, Perth. === Subject: Re: It is irrational to put rational and irrational numbers on the same line. > The standard NUMBER LINE (mathematics term, if unfamiliar to you) > is made up of integers, their rational fractions AND...??? The stuff in between those fractions. Duh. -- Jesse F. Hughes I think the problem for some of you is that you think you are very comprehend and there is the problem. -- James S. Harris === Subject: Re: It is irrational to put rational and irrational numbers on the same line. <200710021709.l92H9dsC192238@walkabout.empros.com >> Ifirrationalnumbers cannot be derived fromrationalnumbers; ... >> I assume you meanirrational*real* numbers, right? >> All real numbers, including theirrationalreal numbers, can be >> derived fromrationalnumbers, so your premise is false, and your >> posting is moot. >How ARE irrationals DERIVED 'on' the line...? Irrationals are a subset of the reals. The reals aren't derived on a > line; they're defined as equivalence classes of Cauchy sequences of > rationals. >They are not fractions, are they? No, they're equivalence classes of Cauchy sequences of rationals. -- > Michael F. Stemper > #include Ghosts crowd the young child's fragile eggshell mind. For goodness sake... OK. 'Line' you have a problem with, so, think AXIS. Just think... The standard NUMBER LINE (mathematics term, if unfamiliar to you) > is made up of integers, their rational fractions AND...??? Please rephrase your Cauchy'ism. Adam Lewis, Perth.- ************************************************************ Read again, think, read again...THINK! For the n-th time you've been given a nice, correct and accurate answer to one of your questions, don't be then AGAIN so haughty, stubborn and stupid as to throw it away with contempt...again. We mathematicians, and mathematics in general, can live perfectly well without any real line, which is ONLY a geometrical device intended to help our intuition to get along with newly learnt stuff. Real numbers can be constructed as equivalence sets of Cauchy sequences of rational numbers. Period. We DO NOT need to place these elements in any line whatsoever, or axis or nothing. Now, my rather intense and clearly pretty dense boy: are you happier now than before? There, we don't need any freaking line at all to work, learn and enjoy of the beauty of mathematics in general, and of this part of analysis in particular! Tonio === Subject: Re: Principles of Induction in non well-founded set theories? > By a non well-founded set theory (NWFST) I mean a set theory without an Axiom Of Foundation. Such a theory may or may not have an axiom from which we can prove the negation of the Axiom Of Foundation. My question is: Are the Principles of Induction true in NWFSTs, and are they provable? To go into more detail, I will work in NBG without the Axiom Of Foundation. We define an Ordinal, as a class X with the following three properties: (1) For any two members of X, these two members are comparable by the membership relation, or equal. > (2) Any subset of X has a least element (i.e. an element Y such that for any member Z of X, we have Z = Y or Z is not a member of Y). nonempty subset! > (3) Any element of X is a subset of X. We define the class On as the class of all ordinal sets. We define the successor function s(X) = X U {X}. We define K as the class of all elements of On which are 0 or the successor of some element of On. We define w as the set of all elements whose successor is a subset of K. It is possible to show (using the Axiom Of Infinity) that w is a set, and in common language it is the set of natural numbers. It has the empty set as a member, and the empty set is denoted 0. By the Principles of Induction I mean the following: First Principle of Induction: Given any class S containing 0, such that whenever k is a member of S then so is s(k), then w is a subset of S. Second Principle of Induction: Given any class S containing 0, such that whenever k is an element of w and a subset of S, then k is an element of S, we have that w is a subset of S. Principle of Transfinite Induction: Given any class S containing 0, such that whenever k is a subset of S, then k is an element of S, we have that On is a subclass of S. Since you leave out Foundation, there might in principle exist a set X such that {X} = X. Such an X is an ordinal. A property holding for On{X} seems to be provable by T.I., hence the principle doesn't work. === Subject: Re: Principles of Induction in non well-founded set theories? > Since you leave out Foundation, there might in principle exist a set X > such > that {X} = X. > Such an X is an ordinal. No, it isn't. For one thing, it isn't well-ordered by the element relation. Indeed, the element relation isn't even a (strict) linear order on this set, since it is reflexive. -- Jesse F. Hughes Knowing about logic is not the same as being in touch with reality. -- David Kastrup === Subject: Re: Principles of Induction in non well-founded set theories? > (1) For any two members of X, these two members are comparable by the membership relation, or equal. > (2) Any subset of X has a least element (i.e. an element Y such that for any member Z of X, we have Z = Y or Z is not a member of Y). nonempty subset! (3) Any element of X is a subset of X. > Since you leave out Foundation, there might in principle exist a set X > such > that {X} = X. > Such an X is an ordinal. regularity, we prove that no ordinal is a member of itself. MoeBlee === Subject: Re: Principles of Induction in non well-founded set theories? > However we have implicitly assumed the Axiom Of > Foundation, in assuming that T has a smallest member > > T is a nonempty subset of w. So T has a least member. > I don't see > where the axiom of regularity is required for that. > > MoeBlee > > If we don't use the axiom of regularity, what result would you quote in order to show that T has a least member, and how would you prove that result? === Subject: Re: Principles of Induction in non well-founded set theories? <21696803.1191364646713.JavaMail.jakarta@nitrogen.mathforum.org > However we have implicitly assumed the Axiom Of > Foundation, in assuming that T has a smallest member T is a nonempty subset of w. So T has a least member. > I don't see > where the axiom of regularity is required for that. MoeBlee If we don't use the axiom of regularity, what result would you quote in order to show that T has a least member, and how would you prove that result? w is the set of natural numbers. Every nonempty subset of w has a least member (we will have proven that already). T is a nonempty subset of w. So T has a least member. MoeBlee === Subject: Re: Principles of Induction in non well-founded set theories? > w is the set of natural numbers. Every nonempty > subset of w has a > least member (we will have proven that already). No, we haven't proven that already! This is what I am trying to say, the proof of this result Every nonempty subset of w has a least member which you take as given, appears to require the axiom of regularity! === Subject: Re: Principles of Induction in non well-founded set theories? <20856135.1191398150340.JavaMail.jakarta@nitrogen.mathforum.org w is the set of natural numbers. Every nonempty > subset of w has a > least member (we will have proven that already). No, we haven't proven that already! But we can. > This is what I am trying to say, the proof of this result Every nonempty subset of w has a least member which you take as given, appears to require the axiom of regularity! To prove that every nonempty subset of w has a least member does not require the axiom of regularity. It looks like Jesse F. Hughes gives a proof a few posts down, so I won't bother giving another one at this juncture. MoeBlee === Subject: Re: Principles of Induction in non well-founded set theories? >> w is the set of natural numbers. Every nonempty >> subset of w has a >> least member (we will have proven that already). No, we haven't proven that already! This is what I am trying to say, > the proof of this result Every nonempty subset of w has a least > member which you take as given, appears to require the axiom of > regularity! You're simply mistaken. Define w as usual, so that w is the least set containing 0 and closed under successor. By this definition, w clearly satisfies the principle of induction: (Ax)( 0 in x & (An)( n in x -> s(n) in x ) ) -> w c x. It is not hard to prove the well-ordering principle from induction. Let us take for granted that we have a proof that w is linearly-ordered by the element relation. Let S c w and suppose S does not have a least element. Let T c w be the set: T = { n in w | (A m <= n) not m in S } Clearly, 0 is in T, since 0 cannot be in S (it would be the least element). Suppose that n is in T. If n + 1 is *not* in T, then n + 1 is the least element of S. Hence, if n is in T, so is n + 1 and consequently w = T. Thus, S = {}. No regularity used in that proof. To be fair, we just assumed that w is linearly ordered, but I promise that *that* proof does not depend on regularity. -- Jesse F. Hughes Casting [Demi] Moore as a woman who has come to the New World so that she can 'worship without fear or persecution' in _The_Scarlet_Letter_ is like casting Bruce Willis as Young Rene Descartes. -Joe Queenan === Subject: Re: Principles of Induction in non well-founded set theories? > >> w is the set of natural numbers. Every nonempty >> subset of w has a >> least member (we will have proven that already). No, we haven't proven that already! This is what I > am trying to say, > the proof of this result Every nonempty subset of > w has a least > member which you take as given, appears to require > the axiom of > regularity! > > You're simply mistaken. > > Define w as usual, so that w is the least set > containing 0 and closed > under successor. This was not the definition given in my original post. However, by that definition it is clearly true that the two properties you mention (w contains 0 and w is closed under successor) are true. > By this definition, w clearly > satisfies the > principle of induction: > > (Ax)( 0 in x & (An)( n in x -> s(n) in x ) ) -> w c > c x. Why does w clearly satisfy the principle of induction? The standard proof of the POI (which again I gave in my original post) relies on the Well Ordering Principle (the thing you are trying to prove). I do not know of any other proof of POI which does not rely on the fact that any nonempty subset of w has a least member. The rest of your post recounts the standard proof that the WOP follows from the POI. This is well-known to me and I do not disagree with you on this. However the POI must be demonstrated before this proof is undertaken. === Subject: Re: Principles of Induction in non well-founded set theories? <2719469.1191412806150.JavaMail.jakarta@nitrogen.mathforum.org >> w is the set of natural numbers. Every nonempty >> subset of w has a >> least member (we will have proven that already). > No, we haven't proven that already! This is what I > am trying to say, >the proof of this result Every nonempty subset of > w has a least >member which you take as given, appears to require > the axiom of >regularity! You're simply mistaken. Define w as usual, so that w is the least set > containing 0 and closed > under successor. This was not the definition given in my original post. However, by that definition it is clearly true that the two properties you mention (w contains 0 and w is closed under successor) are true. Right, and your definition is equivalent with the ordinary definition Jesse mentioned. > By this definition, w clearly > satisfies the > principle of induction: (Ax)( 0 in x & (An)( n in x -> s(n) in x ) ) -> w c > c x. Why does w clearly satisfy the principle of induction? For all x, if 0ex and x is closed under successor, then w subset of x. That's right from the ordinary definition of w. And your definition is equivalent though it's a little more roundabout. The principle of weak mathematical induction falls right out of the definition of w, like a peach into your lap. > The standard proof of the POI (which again I gave in my original post) relies on the Well Ordering Principle (the thing you are trying to prove). I do not know of any other proof of POI which does not rely on the fact that any nonempty subset of w has a least member. But now you do know of a proof of weak mathematical induction that doesn't mention the least member property. > The rest of your post recounts the standard proof that the WOP follows from the POI. This is well-known to me and I do not disagree with you on this. However the POI must be demonstrated before this proof is undertaken.- And now you see that weak induction is demonstrated. MoeBlee === Subject: Re: Principles of Induction in non well-founded set theories? > On Oct 3, 4:59 am, Adam Burley >> w is the set of natural numbers. Every > nonempty >> subset of w has a >> least member (we will have proven that > already). > No, we haven't proven that already! This is > what I >am trying to say, >the proof of this result Every nonempty subset > of >w has a least >member which you take as given, appears to > require >the axiom of >regularity! > You're simply mistaken. > Define w as usual, so that w is the least set >containing 0 and closed >under successor. This was not the definition given in my original > post. However, by that definition it is clearly true > that the two properties you mention (w contains 0 and > w is closed under successor) are true. > > Right, and your definition is equivalent with the > ordinary definition > Jesse mentioned. > Can you prove this though? By the way, I liked the analogy like a peach into your lap. That was funny. === Subject: Re: Principles of Induction in non well-founded set theories? <21832724.1191438470756.JavaMail.jakarta@nitrogen.mathforum.org Right, and your definition is equivalent with the > ordinary definition > Jesse mentioned. Can you prove this though? I think so. I quickly ran it through my my creaky old brain yesterday and it checked out. I'll look it over more carefully this evening and show you the proof. However, it might require some previous lemmas that I'd proven before; so, in case those lemmas are not also in your basket of the previously proven, then I'd have to start an outline from some juncture at which we both already agree. And, in any case, why would you want a definition of w that isn't equivalent with the ordinary definition? MoeBlee === Subject: Re: Principles of Induction in non well-founded set theories? > On Oct 3, 12:07 pm, Adam Burley > >Right, and your definition is equivalent with the >ordinary definition >Jesse mentioned. Can you prove this though? > > I think so. I quickly ran it through my my creaky old > brain yesterday > and it checked out. I'll look it over more carefully > this evening and > show you the proof. However, it might require some > previous lemmas > that I'd proven before; so, in case those lemmas are > not also in your > basket of the previously proven, then I'd have to > start an outline > from some juncture at which we both already agree. > > And, in any case, why would you want a definition of > w that isn't > equivalent with the ordinary definition? > > MoeBlee > > I mean, clearly it is equivalent with the ordinary definition. However it is a question of whether it is possible to prove the equivalence, without using induction. I think the reason why Godel uses this definition is that it doesn't require the Axiom Of Infinity to prove that w exists (as a class: obviously you still need the Axiom Of Infinity to show that w is a set). === Subject: Re: Principles of Induction in non well-founded set theories? <17749952.1191440842158.JavaMail.jakarta@nitrogen.mathforum.org I mean, clearly it is equivalent with the ordinary definition. However it is a question of whether it is possible to prove the equivalence, without using induction. Even in proving that ordinary_w = your_w, we can use whatever induction theorems we've proven. So for ordinary_w, we'll have proven induction theorems and we can use them to prove ordinary_w = your_w. But we can't use induction theorems for your_w until we prove them for your_w (though once we prove that ordinary_w = your_w then of course we can use the induction theorems for your_w too). > I think the reason why Godel uses this definition is that it doesn't require the Axiom Of Infinity to prove that w exists (as a class: obviously you still need the Axiom Of Infinity to show that w is a set Just for my orientation, what book are you using or what paper by Godel are you using? MoeBlee === Subject: Re: Principles of Induction in non well-founded set theories? >> > w is the set of natural numbers. Every nonempty > subset of w has a > least member (we will have proven that already). >> No, we haven't proven that already! This is what I >> am trying to say, >> the proof of this result Every nonempty subset of >> w has a least >> member which you take as given, appears to require >> the axiom of >> regularity! >> >> You're simply mistaken. >> >> Define w as usual, so that w is the least set >> containing 0 and closed >> under successor. This was not the definition given in my original post. Sorry, I came in late. > However, by that definition it is clearly true that the two > properties you mention (w contains 0 and w is closed under > successor) are true. > By this definition, w clearly >> satisfies the >> principle of induction: >> >> (Ax)( 0 in x & (An)( n in x -> s(n) in x ) ) -> w c >> c x. Why does w clearly satisfy the principle of induction? The > standard proof of the POI (which again I gave in my original post) > relies on the Well Ordering Principle (the thing you are trying to > prove). I do not know of any other proof of POI which does not rely > on the fact that any nonempty subset of w has a least member. By definition of w. Let's be more explicit. One familiar form of the axiom of infinity is: (E x)( 0 in x & (A n)(n in x -> s(n) in x) ) Call a set inductive if it contains 0 and is closed under successor. Let S be any inductive set and define w = intersection { T c S | T is inductive }. It is easy to show that w is inductive. We want to show that if x is *any* inductive set, then w c x. This is the principle of induction. Let x be an inductive set. Show that x n S is also inductive. Conclude that w c (x n S) c x. The rest of your post recounts the standard proof that the WOP > follows from the POI. This is well-known to me and I do not disagree > with you on this. However the POI must be demonstrated before this > proof is undertaken. There is, by the way, one little sticking point in that proof. We want to show that the element relation is a linear order on w. I skipped this in my previous post, but showing that the element relation is anti-reflexive and anti-symmetric does require more work when we don't have regularity. We have to show (by induction) that for every element n in w, it is not the case that n is in Tc(n), where Tc is transitive closure. -- It has been shown that no man can sit down to write without a very profound design. Thus to authors in general trouble is spared. A novelist, for example, need have no care of his moral. It is there -- that is to say, it is somewhere -- and the moral and the critics can take care of themselves. --E.A. Poe === Subject: Re: Principles of Induction in non well-founded set theories? > >> > w is the set of natural numbers. Every nonempty > subset of w has a > least member (we will have proven that > already). >> No, we haven't proven that already! This is what > I >> am trying to say, >> the proof of this result Every nonempty subset > of >> w has a least >> member which you take as given, appears to > require >> the axiom of >> regularity! >> >> You're simply mistaken. >> >> Define w as usual, so that w is the least set >> containing 0 and closed >> under successor. This was not the definition given in my original > post. > > Sorry, I came in late. > > However, by that definition it is clearly true that > the two > properties you mention (w contains 0 and w is > closed under > successor) are true. > By this definition, w clearly >> satisfies the >> principle of induction: >> >> (Ax)( 0 in x & (An)( n in x -> s(n) in x ) ) -> w > c >> c x. Why does w clearly satisfy the principle of > induction? The > standard proof of the POI (which again I gave in my > original post) > relies on the Well Ordering Principle (the thing > you are trying to > prove). I do not know of any other proof of POI > which does not rely > on the fact that any nonempty subset of w has a > least member. > > By definition of w. Let's be more explicit. One > familiar form of the > axiom of infinity is: > > (E x)( 0 in x & (A n)(n in x -> s(n) in x) ) > I am using NBG, which has a different form of the Axiom of Infinity, namely: (E x)(~(Em(x)) & (A y)(y in x -> (E z)(z in x & y c z & ~(y = z)))) I am not sure whether this Axiom can be used for the below argument. > Call a set inductive if it contains 0 and is closed > under successor. > Let S be any inductive set and define > > w = intersection { T c S | T is inductive }. > We need to show that w is unique for any choice of inductive set S. Such a proof would seem to require the principle of induction. > It is easy to show that w is inductive. We want to > show that if x is > *any* inductive set, then w c x. This is the > principle of induction. > > Let x be an inductive set. Show that x n S is also > inductive. > Conclude that w c (x n S) c x. > What makes you think that w c (x n S)? I think that there you have assumed that w c x. The rest of your post recounts the standard proof > that the WOP > follows from the POI. This is well-known to me and > I do not disagree > with you on this. However the POI must be > demonstrated before this > proof is undertaken. > > There is, by the way, one little sticking point in > that proof. We > want to show that the element relation is a linear > order on w. I > skipped this in my previous post, but showing that > the element > relation is anti-reflexive and anti-symmetric does > require more work > when we don't have regularity. We have to show (by > induction) that > for every element n in w, it is not the case that n > is in Tc(n), where > Tc is transitive closure. > > -- > It has been shown that no man can sit down to write > without a very profound > design. Thus to authors in general trouble is spared. > A novelist, for example, > need have no care of his moral. It is there -- that > is to say, it is somewhere > -- and the moral and the critics can take care of > themselves. --E.A. Poe I gave a proof in my original post that membership is anti-reflexive from the so-called Second POI. So don't worry about those matters. === Subject: Re: Principles of Induction in non well-founded set theories? >> By definition of w. Let's be more explicit. One >> familiar form of the >> axiom of infinity is: >> >> (E x)( 0 in x & (A n)(n in x -> s(n) in x) ) >> I am using NBG, which has a different form of the Axiom of Infinity, namely: (E x)(~(Em(x)) & (A y)(y in x -> (E z)(z in x & y c z & ~(y = z)))) I am not sure whether this Axiom can be used for the below argument. So, from this axiom, we get a set x that has an infinite chain y0 c y1 c y2 c ... where each inclusion is strict. Here's how I would try to proceed, but I don't know if it works. First, define a subset of x that is exactly the chain {y0, y1, ...}. I confess I don't see how to do that yet, but if it can be done, then I hope we can use the axiom of replacement to define an inductive set. Once we have an inductive set, we're on our way. Like I said, however, I am not at all convinced that this would work with the axiom you give. >> Call a set inductive if it contains 0 and is closed >> under successor. >> Let S be any inductive set and define >> >> w = intersection { T c S | T is inductive }. >> We need to show that w is unique for any choice of inductive set > S. Such a proof would seem to require the principle of induction. No, it follows from the definition. See below. >> It is easy to show that w is inductive. We want to >> show that if x is >> *any* inductive set, then w c x. This is the >> principle of induction. >> >> Let x be an inductive set. Show that x n S is also >> inductive. >> Conclude that w c (x n S) c x. >> What makes you think that w c (x n S)? I think that there you have > assumed that w c x. x n S is an inductive subset of S. w is the intersection of all inductive subsets of S. Hence w c (x n S). This is also how one shows that the definition of w does not depend on S. Suppose S' is another inductive set, with w' defined as the intersection of all inductive subsets of S'. We want to show that w = w'. Now, w' n w is an inductive subset of S, so w c (w' n w) and hence w c w'. A symmetric argument shows w' c w. -- You got more out of it than I put into it last night. Who were you thinking of when we were loving last night? -- Texas Tornadoes === Subject: Re: Principles of Induction in non well-founded set theories? <7559082.1191438317586.JavaMail.jakarta@nitrogen.mathforum.org I am using NBG, which has a different form of the Axiom of Infinity, namely: (E x)(~(Em(x)) & (A y)(y in x -> (E z)(z in x & y c z & ~(y = z)))) What does ~Em(x) mean? Do you mean ~Mx (I.e., x is not a set)? Moeblee === Subject: Re: Principles of Induction in non well-founded set theories? > On Oct 3, 12:04 pm, Adam Burley > > I am using NBG, which has a different form of the > Axiom of Infinity, namely: (E x)(~(Em(x)) & (A y)(y in x -> (E z)(z in x & y c > z & ~(y = z)))) > > What does ~Em(x) mean? > > Do you mean ~Mx (I.e., x is not a set)? > > Moeblee > > Sorry about that. By Em i mean Empty. So ~(Em(x)) means x is not empty. Probably I should have put ~(x = 0), that would have been clearer. === Subject: Re: Principles of Induction in non well-founded set theories? <3784309.1191440681389.JavaMail.jakarta@nitrogen.mathforum.org By Em i mean Empty. Okay. The axiom doesn't specify also that x is a set? MoeBlee === Subject: Re: Principles of Induction in non well-founded set theories? <7559082.1191438317586.JavaMail.jakarta@nitrogen.mathforum.org > I am using NBG, which has a different form of the Axiom of Infinity, namely: (E x)(~(Em(x)) & (A y)(y in x -> (E z)(z in x & y c z & ~(y = z)))) What does ~Em(x) mean? Do you mean ~Mx (I.e., x is not a set)? Or rather, should it be Mx (i.e., x is a set)? MoeBlee === Subject: How to check if a point is on a line? I have a line L with the endpoints: A = (3,0), B = (5,6). I also have the point: P = (4,3). I would now like to check if the point P is on L. Normally I would insert P into: y = ax+b and checking id the two sides are equal. But this assumes that I know 'b' and 'a'. Is there some tricky way to decide if P is on L knowing only the sign of 'a'? === Subject: Re: How to check if a point is on a line? > I have a line L with the endpoints: A = (3,0), B = (5,6). I also have > the point: P = (4,3). > > I would now like to check if the point P is on L. > > Normally I would insert P into: > > y = ax+b > > and checking id the two sides are equal. But this assumes that I know > 'b' and 'a'. > > Is there some tricky way to decide if P is on L knowing only the sign of > 'a'? If you wanted to be mathematically correct in the terminology, you probably should call L a line segment, not a line. As for the question at hand, three points are collinear if the slope is equal between each of the points. Dave === Subject: Re: How to check if a point is on a line? > I have a line L with the endpoints: A = (3,0), B = (5,6). I also have > the point: P = (4,3). > > I would now like to check if the point P is on L. > > Normally I would insert P into: > > y = ax+b > > and checking id the two sides are equal. But this assumes that I know > 'b' and 'a'. > > Is there some tricky way to decide if P is on L knowing only the sign of > 'a'? If the points are (x1,y1), (x2,y2) and (x3,y3) and all different then they are colinear if and only if the determinant of M is zero where M123 = [[ x1 - x2 y1 - y2 ] [ x1 - x3 y1 - y3 ]] This remains true under any permutation of the points 1,2, and 3 === Subject: Re: How to check if a point is on a line? > I have a line L with the endpoints: A = (3,0), B = (5,6). I also have > the point: P = (4,3). I would now like to check if the point P is on L. Normally I would insert P into: y = ax+b and checking id the two sides are equal. But this assumes that I know > 'b' and 'a'. Is there some tricky way to decide if P is on L knowing only the sign of > 'a'? Another way of representing the line L is the set of points A*r + B*(1-r) where r is any real number (for 0 <= r <= 1, you get the line segment between A and B). So one approach would be to write down the corresponding equation in x, solve for r, and see if it is satisfied for y. That is, solve A_x*r + B_x*(1-r) = P_x: 3*r + 5*(1-r) = 4 3r + 5 - 5r = 4 -2r = -1 r = 0.5 You'll find that P_y satisfies A_y*r + B_y*(1-r) = P_y for this value of r. So the answer is yes, P lies on the line segment between A and B, exactly half way between A and B. - Randy === Subject: Re: How to check if a point is on a line? > I have a line L with the endpoints: A = (3,0), B = > (5,6). I also have > the point: P = (4,3). > > I would now like to check if the point P is on L. > > Normally I would insert P into: > > y = ax+b > > and checking id the two sides are equal. But this > assumes that I know > 'b' and 'a'. > > Is there some tricky way to decide if P is on L > knowing only the sign of > 'a'? The best way would be to FIND a and b! You know that (3, 0) is on the line so x=3, y=0 must satisfy the equation: 3= a(0)+ b. You know that (5, 6) is on the line so x= 5, y= 6 must satisfy the equation: 6= a(5)+ b. Solve those two equations for a and b. (You could of course find a directly: it is the slope of the line (y1- y0)/(x1- x0).) === Subject: Re: How to check if a point is on a line? > I have a line L with the endpoints: A = (3,0), B = (5,6). I also have > the point: P = (4,3). I would now like to check if the point P is on L. Normally I would insert P into: y = ax+b and checking id the two sides are equal. But this assumes that I know > 'b' and 'a'. Is there some tricky way to decide if P is on L knowing only the sign of > 'a'? But you do know a and b: 3a+b = 0 5a+b = 6 So a is 3 and b is -9 y = 3x - 9 Your point (4,3) is on this line. You say A and B are 'endpoints', so I think you mean that the line isn't infinite. (4,3) is fine though and obviously between these points. === Subject: Re: Is the phonon as phony as Uncle Al is? > loschmidt has fascinated me for some time now > being not only the first to give the form for benzene > (prior to kekule and the ourobouros) > but also the count of atoms in a volume of gas > and numerous other firsts partially lost to rewritten history Why will you not spell his name properly? Johann Josef Loschmidt. -- Michael Press === Subject: Re: Is the phonon as phony as Uncle Al is? > On Oct 2, 9:10 am, Timothy Golden BandTechnology.com >On Sep 28, 3:45 am, Timothy Golden BandTechnology.com > Why acoustic waves do not >suffer the same propagation rate as heat is a fine restatement of the >problem. Particularly a small acoustic source can be regarded as an >attempt at approaching equivalence. Yet superposition is well >maintained under the mechanical system. Furthermore Kittel's text does >not once enter into this headspace that we are now in. If he does not >touch this and he is a base layer then all the higher up stuff can >avoid it too right? > nobody touches it because mechanics does not do thermodynamics > that's the real elephant in the room > it's the biggest unsolved problem in physics >and it has been for 150 years > avoid is the appropriate term to use >because it is an active process > boltzmann's h-theorem is regularly used > to bury the disagreements between the two theories > inside a broken derivation > it's errors are very applicable to this discussion on phonons > and so can serve as a good case study > the problem in both circumstances > is the move from local to global dynamical principles > the global principles mix local states > in effect assuming the behavior expected > and losing the state distinguishability information >that are crucial to local dynamical evolution > in the phonon case > they do what is common in quantum mechanics > they go to a spectral analysis >and consider only eigenfunctions > these eigenfunctions are not the only dynamical solutions >they merely form a basis of solutions > consistent with the global restrictions > it is these solutions which are the phonons >( at least - once you move to the fock space > and reinterpret excitations as creation/annihilations ) > other dynamical solutions can be discovered > as local perturbations of basis elements > or purely through initial conditions evolutions > outside the basis set > either way > these solutions do not obey the global constraints > and require more information for their specification and analysis > these are the acoustic solutions > the acoustic/thermic divide >is just another example of the local/global divide > behind the dynamics/thermodynamics inconsistency >and the reason you won't get this explanation from many >is just that it is not taught > physics departments don't teach the problems with thermodynamics > the only major investigator in the twentieth century > that made any deep attacks on the issue >was ilya prigogine >and he didn't succeed >( though he made some very telling associations to decoherence > that set it up as a much more information theoretic issue ) > local solutions can propagate faster > because they involve less mass in the movement >global solutions are excitations of the lattice > coordinated modes > dressed in their avoidances > -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- >galathaea: prankster, fablist, magician, liar In researching the H-theorem I came across Loschmidt's paradox > http://en.wikipedia.org/wiki/Loschmidt%27s_paradox > which of course I take interest in though I see no need to rely upon > time reversal for a true theory. > So in some ways I do not understand the topic at all. > It's as if there is some circular reasoning going on. > The generalization from a gas to a solid makes no sense to me. > It is even puzzling to me how hydrodynamics and gas dynamics can be so > closely linked as to be cross-compatible in the design of vanes and > hulls. > The adiabatic process is also puzzling- warm air rises right? > What is the natural temperature differential in a sealed adiabatic > column of air? > as they compose a lower density volume? This problem is nonexistent in > existing thermodynamics as far as I can tell. Is there a gravitational > link right here in this simplistic construction? If there is no > natural temperature difference then perhaps the compensation is self- > I am open to being corrected if I am wrong about this. i am not sure if you have seen my past postings on loschmidt > but you may want to look them up > (also look up autymn's - or she'll get pissed ;) loschmidt has fascinated me for some time now > being not only the first to give the form for benzene > (prior to kekule and the ourobouros) > but also the count of atoms in a volume of gas > and numerous other firsts partially lost to rewritten history a very intelligent man > who made the cardinal sin of questioning the absoluteness of the > second law > and thus became someone talked about only awkwardly > apologetically > like tesla for gas columns > there appears to be several structural phases i do suspect loschmidt was correct > that the natural tendency by gravity is to have a hotter lower > section > because of the acceleration of falling adding energy this is a metastable state > though > and when the temperature differential becomes too large > advection events start mixing the gas advection is driven by uneven planar sections > and requires macroscopic threshold differentials along these sections > (dependent upon the characteristic mass of the gas > and the interatomic forces involved) > so there will be state prior to advection > and since it is dependent upon fluctuation size > it is metastable i've posted to usenet in the past > a series of simulations on gas column temperature differentials > which demonstrated in an interaction-free gas (gravity and collisions > only) > the increase in column temperature the lower in position it was a simple simulation > that reproduced loschmidt's predictions > and matched well the calculations inhttp://users.aol.com/atrupp/loschmid.htm you might also be interested in actual measurements on such columns > by dr roderich graeff > also validating loschmidt's claims -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar Hi galathaea. You are so well read. I think the perpetual energy generation claim is a sign that there is a mistake. I'm having a difficult time thinking in terms of gas dynamics and would prefer to step back to the solid form. I have not gotten to your past posts or Autymn's but it is something I'll try to get to. Andreas Trupp's site seems substantial. His lack of fear is commendable. Still, I am more interested in getting to an alternate form and believe that the alternate form may tie gravity and thermo at a more basic level of interaction via torque. The extension of the colliding gas molecules does not translate directly to solids does it? Geeze look at how interconnected all of this is and then the lead in to time as well... Down here everything is open and overwhelming. Anyhow a linkage from solid to gas may hold better than the existing interpretation from gas to solid. I look back and reread your statement on decoherence above here and Ilya Prigogine. I don't understand the part about local versus global propagation. I can accept some latency and filtering and that sort of limitation, but to try to see a local perturbation as not rippling through the structure is beyond me unless rotation is the type of perturbation. This then begs that we throw atomic bonding in with the rest and so the overwhelming extent of openness in this problem. Still, what would be better than an atomic bonding model that was the same mechanism as gravity and thermodynamics? By trying to join two things at a time we likely can't manage to squeeze in a third and so the state of current physics. Going for it here in the truest sense is to take it all in one fell swoop. Of course you know that I work within the polysign spacetime paradigm. It is there in some old substrate theories that I have exposed differential torque. In effect if we were to treat complex products as natural we then see a correspondence to the second derivative just as the line products. However the action is rotational. P3 is a natural part of spacetime as it is part of electromagnetic behavior. To throw even more twist in I don't wish to rule out higher signs in this current situation. I think there are too many dynamics to do it in T3. What I am describing is far more than I am capable of, yet it is here that I want to form a progression. This suggests going back to the pure math and I probably will shortly. Still the empirical beliefs are guidance. -Tim === Subject: Re: Is the phonon as phony as Uncle Al is? On Oct 2, 7:13 am, Timothy Golden BandTechnology.com >it's errors are very applicable to this discussion on phonons > and so can serve as a good case study its, retard > these eigenfunctions are not the only dynamical solutions >they merely form a basis of solutions > consistent with the global restrictions Basis is such a dumb name. Why not stick with factor? > local solutions can propagate faster > because they involve less mass in the movement >global solutions are excitations of the lattice Fast is not a speed. Geeze Autymn, I thought you were a decent poster. You must have some > emotional issue with galathaea. > I know nothing of this but welcome you to give a more substantial > opinion on this topic. Anyone really. she doesn't like it when i make mistakes i don't either and these are aesthetic mistakes so they're serious bizness but i think i differ with her on proper possessive pronoun spelling i don't like the standard way but i've been avoiding converting over completely i think both should have apostrophe in good parallel with Lincoln's gun and Lincoln's gone we already have means of distinguishing if it is syntactically related to an object or predicate and i think the irritations invoked by mistakes is pathological so i will probably convert completely over more consciously however she is absolutely right fast is not a speed it is how tightly something is held aesthetically that is a much more powerful meaning and my use was only damaging to the language she was looking out for me so i wouldn't become ugly -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: Is the phonon as phony as Uncle Al is? I'll apologize for Tttteslaaaa, just as soon as I ZZZZAPPPP you all into a temporary metastable state (and you configure how to take it back from thereat .-) so, did anyone find a workable integer value about that, somewhat inconculusive (if not fully multivalued). > and matched well the calculations inhttp://users.aol.com/atrupp/loschmid.htm you might also be interested in actual measurements on such columns > by dr roderich graeff > also validating loschmidt's claims thus: cool; you could have all of the parts descend from the rafters on pulleys; c a r e f u l l y ! like, the first rainy day of the season, is Try Out the Air Bag Day; the hard part is folding them back up, just like a parachute (unless you keep every thing well below the speedlimit: no tachyons; it's the law !-) thus: although I didn't see the movie, I somehow had the realization, just a little while of a time ago, that p-adics are like a box of chocolates ... inside or outside, shape, time-stamp et c. et c. et cetera. > since then is based on the (false) assumption that pi*i is real. He --Fox promotes Cheeny Admin. war with Iran: http://larouchepub.com/other/2007/3437fox_poop_cheney.html --14 Italian Senators Call for Cheney Impeachment Aug. 1, 2007 (EIRNS)- The Lyndon LaRouche Political Action Committee (LPAC) issued the following release today. Fourteen members of the Italian Senate have signed a call to the Members of Congress to support Rep. Kucinich's House Resolution 333 for the Impeachment of Dick Cheney. http://larouchepub.com/pr/2007/070801italian_senators_call.html === Subject: Re: Is the phonon as phony as Uncle Al is? NB: could have been in *American Scientist*. > so, did anyone find a workable integer value > about that, somewhat inconculusive (if not fully multivalued). thus: what I'm saying: study the p-adics; tell us what happenned to you! thus: you big dope; Hensel did not extend the reals with p-adics; they are entirely nonarchimedean; his only mistake was listing them rightwards, as if they were decimals! in deed, the real integers can be thought-of as infinite-adics, as is shown in most of the introductory p-adic thingies; just because they happen to be the archimedean, valuationable corner of adics, doesn't mean that adics extended them. no amount of revising your outline is ever going to result in doable (experimental) physics. well, you're probably here just to waste our time; success! thus: p-adics are like a box of chocolates ... inside or outside, shape, time-stamp, partitioning, pigeonholing et c. et c. et cetera. --14 Italian Senators Call for Cheney Impeachment Aug. 1, 2007 (EIRNS)- The Lyndon LaRouche Political Action Committee (LPAC) issued the following release today. Fourteen members of the Italian Senate have signed a call to the Members of Congress to support Rep. Kucinich's House Resolution 333 for the Impeachment of Dick Cheney. http://larouchepub.com/pr/2007/070801italian_senators_call.html === Subject: question about automorphisms of the complex number field Hi there, I'd like to know if there is an easy way of proving the following theorem(I'm thinking in the lines of constructing the automorphism): If theta is a non-trivial involutive automorphism of C (C= complex number field), then its fixed field is isomorphic to R (R= real number field). Moreover, this isomorphism is a restriction on R of an automorphism of C that is conjugate to complex conjugation in the automorphism group of C. Any help would be much appreciated. === Subject: Re: Software for calculating/simulating air-flow dynamics. Approved-Poster: approved I'm seeking software (downloads that runs on Windows XP) that > simulates air-flow dynamics; the input field being an x,y scatter- > graph where each point on the graph is the simulation framework for > air-flow dynamics i.e., how the air flow would behave in flowing > through a series of obstacles. > Any recommendations for simple to use software? > I suggest looking into fluid dynamics in MatLab/Simulink at Mathworks: http://www.mathworks.com/ Mathematica will work equally well: http://mathworld.wolfram.com/ .They are both simple to use software products, once you learn them. The problem you are addressing, however, is not a simple one. === Subject: Re: New list and email of contat for solutions manual I need the solution manual for Macroeconomics by Andrew B. 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Luis Rodrigus Valencia.pdf 2 MB [Soluciones a los problemas] Suplemento Calculo Infinitesimal Calculus- Michael Spivak.pdf 8 MB [Solution Manual] CD Physics - Halliday, Resnick and Walker?s - Fundamentals of Physics 1, 2, 3 and 4 (4th ed.)(over 2000pages).rar 43 MB [Solutions Manual] Classical Electrodynamics - 2nd Ed. John David Jackson byKasper van Wijk.pdf 1 MB [Solutions Manual] Communication Systems Engineering Proakis J (2002).pdf 2 MB [Solutions Manual] [Instructors] Advanced Engineering Mathematics 8Ed - Erwin Kreyszig.pdf 19 MB [Solutions Manual] [Instructors] Calculus 5Th Ed James Stewart .pdf 75 MB [Solutions Manual] [Instructors] Introduction to Linear Algebra--3rd Edition - Gilbert Strang.pdf 500 KB [Solutions Manual] [Instructors] Physics by Resnick Halliday Krane, 5th Ed. Vol 2.pdf 1 MB [Solutions Manual] Anton Bivens Davis CALCULUS early transcendentals 7th edition.rar 11 MB [Solutions Manual] Applied Statistics and Probability for Engineers 3rd Ed. 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PLEASE DONT REPLY HERE AND SEND ME EMAIL AT diosbenditome (at) gmail (dot) com === Subject: Re: Solution Manual for Calculus. WANTED > check this out http://solutionmanuals.spaces.live.com/ maximum pay of 8 bucks i have the following solution manuals.... and i have thousands of textbooks as well email me at diosbendit...@gmail.com if you want any of them email at diosbenditome (at) gmail (dot) com paypal paymets accepted only please email me rather than leaving a message here.. Chemical and Engineering Thermodynamics- 3rd Edition- Solutions Manual.rar 11 MB Prentice Hall - Solutions Manual; Communication Systems Engineering (McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition (Stephen J Chapman).pdf 5 MB [eBook.med] Prentice.Hall- Digital image processing - Gonzalez 2Ed- Solutions Manual (2002).pdf 2 MB [Ejercicios propuestos y sus soluciones] Algebra Lineal - Juan de Burgos -.pdf 7 MB [Instructor's Solutions Manual] Introduction to Electrodynamics - 3rd ed. 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Devore.rar 4 MB Probability Random Variables and Stochastic Processes Solutions Manual.Papoulis.McGraw Hill.2002.pdf 16 MB Problemas resueltos de Estad?stica I.rar 143 KB Rubik - Solucao Do Cubo Magico.pdf 224 KB Schaums Mathematical Handbook of Formulas and Tables.pdf 26 MB Signal Processing and Linear Systems - B P Lathi - Solutions Manual.pdf 11 MB Solution Manual to engineering fluid mechanics 7e.pdf 4 MB Solution To Two-Dimensional Incompressible Navier-Stokes Equations- Maciej Matyka.pdf 373 KB Thomas' Calculus, Early Trascendentals 10th ed Instructors Solutions Manual.pdf 19 MB Wankat & Oreovicz - Teaching Engineering.rar 911 KB Wiley - Pozar - Microwave Engineering 3ed - Solutions Manual.rar 11 MB Wiley Chemical And Engineering Thermodynamics 3Ed Solutions Manual.pdf 11 MB Zwillinger D. et al - CRC Standard Probability and Statistic Tables and Formulae (1999).pdf 9 Mb analytical mechanics.rar askeland science and engineering of materials solutions.rar classical dynamics 5e thornton.rar crowe 7e engineering fluid mechanics.rar eng mech dyn bedford and fowler.rar eng mech statics bedford 4e.rar feedback control of dynamic systems.rar fox 6th fluid mech solutions.rar 13 Mb fund of ther open chs.rar 32 Mb fundamentals of heat and mass transfer solutions.rar 27 Mb fundamentals of machine component design 3e solutions.rar 13 Mb Fundamentals of Thermal-Fluid Sciences.rar 9 Mb heat transfer 2e solutions.rar 41 Mb hibbler 10th statics.rar 30 Mb hibbler dynamics 10E.rar 93 Mb Introduction Fluid Mechanics, 6th Edition Fox,McDonald, & Pritchard.rar 7 Mb materials science and engineering an intro 6E callister.rar 3 Mb mech eng design solutions.rar 77 Mb merian eng mech ... === Subject: Instructor and Student solutions manual for Discrete Math, Kenneth H Rosen Cc: neo__world@hotmail.com Instructor and Student solutions manual for Discrete Math, Kenneth H Rosen, 6th or 5th ed.?? In pdf? let me know at neo__world@hotmail.com (that's 2 underscores) let me know method of payment! === Subject: please please mr. stumper - before we ask our students to solve the country's power system problems - lets expose them to as many problems/solutions as possible - you would seem to suggest that we should limit them to the 6 or 7 problems/solutions that we dole out as assignments. they deserve more credit than. that's the problem with our ed system. many teachers on power trips are very good at holding students back, acting omnipotent on a high horse, when in reality we are furnished with solution manuals, power point presentations, etc. from which we transcribe several problems/solutions on the board and dole out solutions as if we know better than our students what they need to do in order to prepare for the exam (as if we invented the these problems/ solutions like we somehow own them). But we refuse to share this abundance of information with our students. so I ask you mr stemper --- who has the advantage??? we have the advantage!!! we are the ones with all the information!!!! So WHO ARE THE CHEATERS Mr. Stemper??? we are the cheaters!!!! the students are in the dark until they find help elsewhere from some other resource beyond their text book. Publishers love people like you who support their monopoly on information. Education is certainly not their goal - profit is their motive - hence they publish new editions each year, shuffling the problems and solutions in the back of each chapter, to keep you and me in the know and our students in the dark. Why do you think the problems/soultions in each new edition remain so riddled with as many mistakes as the old editions??? Because education and refinement are not their goal - profit is their motive - and power is your motive - whether you realize this or not. What would you teach this student from pittsburgh by punishing him for looking for information online??? would you be teaching him how to survive in the real world?? you suggest that by SIMPLY LOOKING online for solutions to HW (solutions which are not provided in the book) this student is an incompetent cheater. the real struggle SHOULD BE for students to understand and articulate the info not to find the info... I would have to say that you are part of the problem mr. stemper. the expensive world of academia is very heirarchial and elitist. one would assume that you like it that way. you like to monoplize and dole out the information as you see fit - you are complicit with the publisher - you and i struggled to find the information some 20 years ago, so we should make our students struggle just as hard!!! How innovative!!! Having light may we we pass it onto others (just as long as the students don't become smarter than us!!!) i for one hope that our power system engineers have unlimited access to boatloads of problems and solutions to study at home. oh yea mr. stemper, thank you and your generation for keeping our country's power grid in such wonderful condition over the past two decades. Wonderful job mr. stemper. I would suggest you find a job in security, think about the state of the U.S. ed system, and leave the engineering to the innovators and workers. those of us actively seeking solutions with an open mind. oh yea and before you jump to any other conclusions --- I do NOT condone cheating on exams. === Subject: Re: please > please mr. stumper - before we ask our students to solve the country's > power system problems - lets expose them to as many problems/solutions > as possible - you would seem to suggest that we should limit them to > the 6 or 7 problems/solutions that we dole out as assignments. they > deserve more credit than. that's the problem with our ed system. many > teachers on power trips are very good at holding students back, acting > omnipotent on a high horse, when in reality we are furnished with > solution manuals, power point presentations, etc. from which we > transcribe several problems/solutions on the board and dole out > solutions as if we know better than our students what they need to do > in order to prepare for the exam (as if we invented the these problems/ > solutions like we somehow own them). But we refuse to share this > abundance of information with our students. so I ask you mr stemper > --- who has the advantage??? we have the advantage!!! we are the ones > with all the information!!!! So WHO ARE THE CHEATERS Mr. Stemper??? > we are the cheaters!!!! the students are in the dark until they find > help elsewhere from some other resource beyond their text book. > Publishers love people like you who support their monopoly on > information. Education is certainly not their goal - profit is their > motive - hence they publish new editions each year, shuffling the > problems and solutions in the back of each chapter, to keep you and me > in the know and our students in the dark. Why do you think the > problems/soultions in each new edition remain so riddled with as many > mistakes as the old editions??? Because education and refinement are > not their goal - profit is their motive - and power is your motive - > whether you realize this or not. What would you teach this student > from pittsburgh by punishing him for looking for information online??? > would you be teaching him how to survive in the real world?? you > suggest that by SIMPLY LOOKING online for solutions to HW (solutions > which are not provided in the book) this student is an incompetent > cheater. the real struggle SHOULD BE for students to understand and > articulate the info not to find the info... I would have to say > that you are part of the problem mr. stemper. the expensive world of > academia is very heirarchial and elitist. one would assume that you > like it that way. you like to monoplize and dole out the information > as you see fit - you are complicit with the publisher - you and i > struggled to find the information some 20 years ago, so we should make > our students struggle just as hard!!! How innovative!!! Having light may we we pass it onto others (just as long as the > students don't become smarter than us!!!) i for one hope that our > power system engineers have unlimited access to boatloads of problems > and solutions to study at home. oh yea mr. stemper, thank you and your > generation for keeping our country's power grid in such wonderful > condition over the past two decades. Wonderful job mr. stemper. I > would suggest you find a job in security, think about the state of the > U.S. ed system, and leave the engineering to the innovators and > workers. those of us actively seeking solutions with an open mind. > oh yea and before you jump to any other conclusions --- I do NOT > condone cheating on exams. > I have the Manual Solutions in front of me, ask me a question and I will give you the answer for $5. === Subject: Analysis with Riemann integral.. Hello sir~ f : [0,1] -> R is continuous. For all n in N, Show that [int{0 to 1} f(x).(x^n) dx]^2 <= {1/(2n+1)}.[int{0 to 1} {f(x)}^2 dx] ----------------------------------------------------------- Let's go... [f(x) + (x^n).t]^2 >= 0 and this is continuous on [0,1]. For all t in R, 0 <= int{0 to 1} [f(x) + (x^n).t]^2 dx = [int{0 to 1} (f(x))^2 dx] + [2{int{0 to 1} f(x).x^n dx}.t + [{int{0 to 1} x^(2n) dx}.t^2] = [int{0 to 1} (f(x))^2 dx] + [2{int{0 to 1} f(x).x^n dx}.t + [(1/(2n+1)).t^2] By discriminant, [int{0 to 1} f(x).x^n dx]^2 - (1/(2n+1)).[int{0 to 1} (f(x))^2 dx] <= 0. Thus, [int{0 to 1} f(x).(x^n) dx]^2 <= {1/(2n+1)}.[int{0 to 1} {f(x)}^2 dx]. It seems ...nice. If you had other advice, I hope to know that. === Subject: Re: Analysis with Riemann integral.. > f : [0,1] -> R is continuous. For all n in N, Show that > [int{0 to 1} f(x).(x^n) dx]^2 <= {1/(2n+1)}.[int{0 to 1} {f(x)}^2 dx] [f(x) + (x^n).t]^2 >= 0 and this is continuous on [0,1]. For all t in R, 0 <= int{0 to 1} [f(x) + (x^n).t]^2 dx 0 <= int{0 to 1} [f(x) - (x^n).t]^2 dx > = [int{0 to 1} (f(x))^2 dx] + [2{int{0 to 1} f(x).x^n dx}.t > + [{int{0 to 1} x^(2n) dx}.t^2] = [int{0 to 1} (f(x))^2 dx] - [2{int{0 to 1} f(x).x^n dx}.t + [{int{0 to 1} x^(2n) dx}.t^2] > = [int{0 to 1} (f(x))^2 dx] + [2{int{0 to 1} f(x).x^n dx}.t + > [(1/(2n+1)).t^2] = int{0 to 1} (f(x))^2 dx - [2{int{0 to 1} f(x).x^n dx}.t + (1/(2n+1)).t^2 2t.integral(0,1) f(x).x^n dx <= integral(0,1) f(x)^2 dx + t^2 / (2n + 1) Let t = s.integral(0,1) f(x) x^n dx 2s.(integral(0,1) f(x).x^n dx)^2 <= integral(0,1) f(x)^2 dx + t^2 / (2n + 1) (2s - s^2 / (2n + 1)).(integral(0,1) f(x).x^n dx)^2 <= integral(0,1) f(x)^2 dx s(4n + 2 - s)/(2n + 1)).(integral(0,1) f(x).x^n dx)^2 <= integral(0,1) f(x)^2 dx (integral(0,1) f(x).x^n dx)^2 <= (2n + 1)/s(4n + 2 - s)).integral(0,1) f(x)^2 dx Set s = 2n + 1 (integral(0,1) f(x).x^n dx)^2 <= (1/(2n + 1)).integral(0,1) f(x)^2 dx > By discriminant, [int{0 to 1} f(x).x^n dx]^2 - (1/(2n+1)).[int{0 to 1} (f(x))^2 dx] <= 0. > Whoa! Justify that inequality. > Thus, [int{0 to 1} f(x).(x^n) dx]^2 <= {1/(2n+1)}.[int{0 to 1} {f(x)}^2 dx]. It seems ...nice. > If you had other advice, I hope to know that. > === Subject: Re: Analysis with Riemann integral.. > Hello sir~ > > f : [0,1] -> R is continuous. > > For all n in N, > Show that > [int{0 to 1} f(x).(x^n) dx]^2 <= {1/(2n+1)}.[int{0 to 1} {f(x)}^2 dx] How about for f(x) = x and n = 3? I get 1/25 = 1/21 in that case. === Subject: Re: Open to all advanced Mathematics Tests > I am interested in taking a relatively serious Math test. > I would definitely need a level above the generalGRE However, those are mostly geared to students currently enrolled in a program. I was thinking more about 'open' tests; like the gre subject exams: anyone can take them as long as they register. Any information's appreciated. > http://www.mathpropress.com/competitions.html http://www.kalva.demon.co.uk/putnam.html > Dave L. Renfro === Subject: Re: A combinatorial puzzle. > On Sep 29, 9:31 pm, Bill Taylor Synchronicity! Or, the long arm of co-incidence. I had recently been spending some spare hours on > a standard piece of theoretical math, when lo and behold, > our department got a general enquiry from a member > of the public who wanted help with this problem > that had been bothering him for years. > His enquiry was passed on to me. And OC, it turned out that it was intimately connected > with the stuff I had just been studying! SYNCHRONICITY. Anyway, here is his problem. He has 16 people in a golf tournament. > He wishes to split them up into 4 groups of 4, > on each day, 4 playing together at a time. > He wishes to timetable this procedure > over five days, (or maybe 6, if necessary), > subject to the priviso that... ...every player must be in the same 4-group > as every other player, exactly once. That is, every pair-of-players must occur > exactly once somewhere in the tournament. EITHER produce such a timetable > OR prove that none can exist. issues came up under that particular problem name. --- Christopher Heckman This particular problem could as well be a case of a complete set of mutually orthogonal latin squares (MOLS) of order 4: 1 2 3 4 1 1 1 1 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 2 2 2 2 3 4 1 2 2 1 4 3 4 3 2 1 1 2 3 4 3 3 3 3 2 1 4 3 4 3 2 1 3 4 1 2 1 2 3 4 4 4 4 4 4 3 2 1 3 4 1 2 2 1 4 3 Each entry indicates which foursome that player is assigned on respective days. === Subject: nike cheap sell wholesale shoes jordan shox air max force one gucci prada Hello.every one, our company have much cheap product . please visit our web, And chose the product of your like ,let go ,hope you shiping happy ,Go www.cheapest-sell.cn Hotmail:cheapestsell@hotmail.com Yahoo: cheapestsell@yahoo.com.cn www.cheapest-sell.cn Nike Air Jordan 1 ,Nike Air Jordan 2 Shoes,Nike Air Jordan 3,Nike Air Jordan 4 Shoes ,Nike Air Jordan 5 Chaussure Shoes,Nike Air Jordan 6 Catalog ,Nike Air Jordan 7 Shoes Catalog , Nike Air Jordan 8 Customized ,Nike Air Jordan 9 Shoes Customized ,Nike Air Jordan 10 Wholesalers,Nike Jordan 11 Shoes Wholesalers,Nike Air Jordan 12 retailer,Nike Air Jordan 13 Shoes Factory ,Nike Air Jordan 14 Shoes Sell,Nike Air Jordan 16 Exporter ,Nike Air Jordan 17 Shoes Exporter, Nike Air Jordan 18 Offer, Nike Air Jordan 19 Shoes Offer,Nike Air Jordan 20 Manufacture,Nike Jordan 21 Shoes Manufacture,Nike Jordan 22 CUSTOMIZED , We Import&Export&Trading&Retail&sell&buy&distribution& Wholesale Nike footwear and Nikes Sneakers Jordans Sneakers to this market: USA,America,US,United States,UK,England,United Kingdom,IT,Italy, NT,Netherlands,China,Chinese,Germany,DE,Greece,GR,France, FR,Spain,Portugal,Switzerland,Switzerland,Brazil,Chile,Peru,C Korea,Australia,Hongkong,Canada,Mexico,Etc Nike shoes | china nike shoes | air jordan sneakers | cheap gucci shoes | cheap prada sneakers | gucci sneakers | mix jordan sneakers | chanel sandals | gucci sandals | dior sandals | wholesale jordan sneakers | nike running shoes | nike stock shoes | air jordan at whlesale price | nike shoes air jordan supplier from in china | Lacoste Trainers | puma trainers | louis vuitton purse | prada purse | gucci handbags | chanel purse | coach purse | nike bas kerball shoes | Nike Sneakers| cheap nike sneakers | nike shoes from china | nike replica | copy nike sneakers | nike factory stores | nike stores Nike wholesale - Nike shoes wholesale nike jordan sneakers wholesale( www.cheapest-sell.cn) We (http: //www.cheapest-sell.cn) Wholesale Cheap Jordan Shoes,Michael Jordan Shoes,Nike Jordan Basketball shoes, ... 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I've got it, do you have a solution manual for Fluid Mechanics (1st > ED) by Cengel and Cimbala? You have the pdf 11th edition Dynamics R.C. Hibbeler Engineering Mechanics?? .. i need it.. could you send it to me please? === Subject: Re: Probability of exceeding a specific value >> < snip COIN-TOSSING in a > 1949 issue of the US Proceedings of the National Academy of Sciences. > It can be found in PDF format here: > < http://www.pnas.org/cgi/reprint/35/10/605.pdf> during which one partly has been leading [ partly should be party]. > With their Theorem 2, one can calculate probabilities such as: > Given a coin-tossing game with 10,000 tosses. What is the > probability that A is ahead 9000 times given that A is ahead > by +2 after 10,000 tosses? > They also write: > ``If a coin is tossed once a second for a total of 365 days, > the probability that one of the players will lead for more > than 364 days and 10 hours is about 0.05! >> One preselected player? No, I don't think so; it's the scenario you describe below (either player). >> Or at the end one or the other of the two players >> will have led for more than ... > > Unless my numbers are completely off-beam it must be the latter. As > the number of tosses goes to infinity the probability that a > preselected player is ahead for more than a fraction R of the time > played should be 1/2 * (1 - CDF(R) + CDF(1 - R)) (CDF as defined a > little way above), which here comes out to about 0.025. In my computation using the CDF you found, it had escaped me that the player who is behind at the end could still be ahead for more than 364 days and 10 hours. And as you mentioned, I also found that this second way of having one or the other player lead for at least 364 days and 10 hours has a probability that is a tiny fraction of 0.05. According to their definition of being ahead, player #1 could be leading at 0, 2, 4, 6, .... 2n of the 2n tosses, each with a probability of 1/(n+1). With an n of 10,000 or so, I thought of the (easily derived) test of pseudo-random number generators where one has an array a[i] , i=0...n with a[i] counting the games where player #1 is ahead 2*i times. I was wondering if this test would be easy to pass and I have no idea. David Bernier === Subject: Re: Comprehensive Solution Manual for Textbooks Updated list per August 20th, 2007 I have the current edition of these comprehensive solutionsmanualfor > the following textbooks in electronic format (PDF/Word). The solutionsmanualare comprehensive with answers to both even & odd problems in > the text. The price is US$35 for each. There are two methods of payments. One is through PAYPAL (if you have > a Paypal account) and another through PROPAY (if you don't have a > Paypal and you only have a credit/debit card). Email me at sbooks4sale[at]hotmail[dot]com if you are interested. If > you could not find the book you are looking for, please let me know, I ****************************** AccountingChapters 1-13 - Charles T. Horngren et al (7th edition) > (ISBN: 0132249952)AccountingChapters 12-25 - Charles T. Horngren et al (7th edition) > (ISBN: 0132249960)AccountingChapters 1-25 - Charles T. Horngren et al (7th edition) > (ISBN: 0132439603)AccountingChapters 1-26 - Charles T. Horngren et al (6th edition) > (ISBN: 0131088513)AccountingConcepts and Applications - Steve Albrecht (10th edition) > (ISBN: 0324376154)AccountingConcepts and Applications - Steve Albrecht (9th edition) > (ISBN: 0324187564)AccountingInformation Systems - James Hall (5th edition) (ISBN: > 0324312954)AccountingInformation Systems - Marshall Romney, Paul Steinbart (10th > edition) (ISBN: 0131475916)AccountingInformation Systems - Ulric J. Gelinas (7th edition) (ISBN: > 0324378823) > Additional Calculus Topics - Raymond Barnett (11th edition) (ISBN: > 0132318229) > AdvancedAccounting- Floyd Beams (9th edition) (ISBN: 0131851225) > Advanced Calculus - G. B. Folland (1st edition) (ISBN: 0130652652) > Dunlap (1st edition) (ISBN: 0534392946) > Analytical Mechanics - Grant Fowles, George Cassiday (7th edition) > (ISBN: 0534494927) > Applied Algebra - Darel Hardy (1st edition) (ISBN: 0130674648) > Applied Linear Algebra - Chehrzad Shakiban, Peter J. Olver (1st > edition) (ISBN: 0131473824) > Applied Multivariate Statistical Analysis - Richard A. Johnson (6th > edition) (ISBN: 0131877151) > Applied Partial Differential Equations - Richard Haberman (4th > edition) (ISBN: 0130652431) > Auditing and Assurance Services - Alvin A. Arens et al (11th edition) > (ISBN: 0131867121) > Auditing and Assurance Services - Alvin A. Arens et al (12th edition) > (ISBN: 0135132126) > Auditing Assurance and Risk - W. Robert Knechel, Steve Salterio, Brian > Ballou (3rd edition) (ISBN: 0324313187) > Auditing Cases - Mark Beasley (3rd edition) (ISBN: 0131494910) > Auditing: A Business Risk Approach - Larry E. Rittenberg (6th edition) > (ISBN: 0324375581) > Automation, Production Systems, and Computer-Integrated Manufacturing > - Mikell P. Groover (2nd edition) (ISBN: 0130889784) > Biochemistry - Mary Campbell (4th edition) (ISBN: 0534405215) > Biochemistry (with Lecture Notebook) - Mary Campbell (4th edition) > (ISBN: 0534391818) > Biology - Neil A. Campbell (Test Bank only w/ TestGen Software) (7th > edition) (ISBN: 080537146X) > Business Law and the Legal Environment - Jeffrey F. Beatty (4th > edition) (ISBN: 0324303971) > Business Law and the Regulation of Business - Richard A. Mann (9th > edition) (ISBN: 0324537131) > Business Law Principles for Today's Commercial Environment - David P. > Twomey (2nd edition) (ISBN: 0324303947) > Business Law: Text and Exercises - Roger LeRoy Miller (5th edition) > (ISBN: 032464096X) > Business Statistics: A Decision Making Approach - David F. Groebner > (7th edition) (ISBN: 0132416921) > Calculus - Dale Varberg (9th edition) (ISBN: 0131429248) > Calculus and Its Applications - Larry Goldstein (11th edition) (ISBN: > 0131919636) > Calculus Early Transcendentals - Henry Edwards (7th edition) (ISBN: > 0131569899) > Calculus for Business, Economics, Life Sciences & Social Sciences - > Raymond Barnett (11th edition) (ISBN: 0132328186) > Calculus for the Life Sciences - Marvin L. Bittinger (1st edition) > (ISBN: 0321279352) > Calculus With Applications - Margaret L. Lial et al (8th edition) > (ISBN: 0321228146) > Calculus with Applications for the Life Sciences - Raymond N. > Greenwell (1st edition) (ISBN: 0201745828) > Capital Budgeting and Long-Term Financing Decisions - Neil Seitz (4th > edition) (ISBN: 0324258089) > Cases in ManagementAccountingand Control Systems - Brandt Allen (4th > edition) (ISBN: 0135704251) > Chemistry : An Introduction to General, Organic, Biological Chemistry > - Karen Timberlake (9th edition) (ISBN: 0805330151) > edition) (ISBN: 0534408966) > CMOS Circuit Design, Layout, and Simulation - David E. Boyce et al > (1st edition) (ISBN: 0780334167) > CollegeAccounting1-12 - Jeffrey Slater (9th edition) (ISBN: > 0131071696) > CollegeAccounting1-25 - Jeffrey Slater (10th edition) (ISBN: > 0132286386) > CollegeAccountingChapters 1-15 - James Heintz (19th edition) (ISBN: > 0324382499) > CollegeAccountingChapters 1-27 - James Heintz (19th edition) (ISBN: > 0324376162) > CollegeAccountingChapters 1-9 - James Heintz (19th edition) (ISBN: > 0324382480) > College Geometry - David C. Kay (2nd edition) (ISBN: 0321046242) > College Physics - Jerry D Wilson (6th edition) (ISBN: 0131495798) > Comparative InternationalAccounting- Christopher Nobes (9th > edition) (ISBN: 0273703579) > Complex Variables With Applications - A. David Wunsch (3rd edition) > (ISBN: 0201756099) > Computer Algorithms - Allen Van Gelder, Sara Baase (3rd edition) > (ISBN: 0201612445) > Computer Networking with Internet Protocols - William Stallings (1st > edition) (ISBN: 0131410989) > Computer Networking: A Top-Down Approach - James F. Kurose (4th > edition) (ISBN: 0321497708) > Computer Networking: A Top-Down Approach Featuring the Internet - > James F. Kurose (3rd edition) (ISBN: 0321227352) > Computer Organization and Architecture - William Stallings (7th > edition) (ISBN: 0130351199) > Computer Organization and Architecture: Designing for Performance - > William Stallings (7th edition) (ISBN: 0131856448) > Computer Systems Organization & Architecture - John D. Carpinelli (1st > edition) (ISBN: 0201612534) > Concepts in Federal Taxation 2007 - Kevin Murphy (14th edition) (ISBN: > 0324313527) > Concepts in Federal Taxation 2008 - Kevin Murphy (15th edition) (ISBN: > 0324640153) > Concepts In Systems and Signals - John D. Sherrick (2nd edition) > (ISBN: 0131782711) > Concrete Structures - Mehdi Setareh (1st edition) (ISBN: 0131988271) > Contemporary Financial Management - Charles Moyer (10th edition) > (ISBN: 0324289081) > Contemporary Financial Management Fundamentals - Charles Moyer (1st > edition) (ISBN: 0324015771) > Cornerstones of ManagerialAccounting- Maryanne M. Mowen (2nd > edition) (ISBN: 0324379609) > Corporate Finance - Jonathan Berk (1st edition) (ISBN: 0321415116) > Corporate Financial Management - Douglas R. Emery (3rd edition) > (ISBN: 0132278723) > CostAccounting- Charles T. Horngren, George Foster, Srikant M. Datar > (12th edition) (ISBN: 0131495380) > CostAccounting: Traditions & Innovations - Jesse Barfield (5th > edition) (ISBN: 032418090X) > Course in Probability - Neil Weiss (1st edition) (ISBN: 0201774712) > Cryptography and Network Security - William Stallings (4th edition) > (ISBN: 0131873164) > Data and Computer Communications - William Stallings (8th edition) > (ISBN: 0132433109) > Derivatives Markets - Robert L. McDonald (2nd edition) (ISBN: > 032128030X) > Differential Equations - John Polking (2nd edition) (ISBN: 0131437380) > Differential Equations and Linear Algebra - Jerry Farlow (2nd edition) > (ISBN: 0131860615) > Differential Equations Computing and Modeling - Henry Edwards (4th > edition) (ISBN: 0136004385) > Differential Equations With Boundary Value Problems - John C. Polking > (2nd edition) (ISBN: 0130911062) > Digital & Analog Communication Systems - Leon Couch (7th edition) > (ISBN: 0131424920) > Digital Communications - John Proakis (4th edition) (ISBN: 0072321113) > Digital Design - Morris Mano (4th edition) (ISBN: 0131989243) > Digital Signal Processing - John Proakis (4th edition) (ISBN: > 0131873741) > Digital Systems: Principles and Applications - Ronald Tocci et al > (10th edition) (ISBN: 0131725793) > Discrete and Combinatorial Mathematics - Ralph P. Grimaldi (5th > edition) (ISBN: 0201726343) > Discrete Mathematics - Edgar G. Goodaire, Michael M Parmenter (3rd > edition) (ISBN: 0131679953) > Discrete Mathematics - Otto, Eynden, Dossey, Spence (4th edition) > (ISBN: 0321079124) > Discrete Mathematics - Otto, Eynden, Dossey, Spence (5th edition) > (ISBN: 0321305159) > Discrete Mathematics - Richard Johnsonbaugh (6th edition) (ISBN: > 0131176862) > Economic Development - Michael Todaro, Stephen Smith (9th edition) > (ISBN: 0321278887) > Economic Growth - David Weil (1st edition) (ISBN: 0201680262) > Economics of Money, Banking, and Financial Markets, Update - Frederic > Mishkin (7th edition) (ISBN: 0321331850) > Economics Today: The Macro View - Roger Miller (13th edition) (ISBN: > 0321278992) > Economics Today: The Micro View - Roger Miller (13th edition) (ISBN: > 0321278984) > Electrical Engineering: Principles and Applications - Allan R. > Hambley (4th edition) (ISBN: 0131989227) > Electrical Machines, Drives and Power Systems - Theodore Wildi (6th > edition) (ISBN: 0131776916) > Electronics and Computer Math - Bill Deem (8th edition) (ISBN: > 0131711377) > Electronics Fundamentals: Circuits, Devices and Applications - Thomas > Floyd (7th edition) (ISBN: 013219709X) > Elementary Differential Equations - Werner E. Kohler, Lee W.Johnson > (1st edition) (ISBN: 0201709260) > Elementary Differential Equations With Boundary Value Problems - Lee > Johnson et al (1st edition) (ISBN: 0321121643) > Elementary Linear Algebra with Applications - Bernard Kolman (9th > edition) (ISBN: 0132296543) > Elementary ... read more ? Hey, I am looking for the solution manual for: Discrete and Combinatorial Mathematics - Ralph P. Grimaldi (5th edition) (ISBN: 0201726343) Please reply ASAP. === Subject: Linear Dependence Let a set S be linearly dependent. Is every subset of S linearly dependent? I know this is not true, however, I cannot think of a counterexample. Does anyone have any good ideas? === Subject: Re: Linear Dependence <21427019.1191473073521.JavaMail.jakarta@nitrogen.mathforum.org>, > Let a set S be linearly dependent. > Is every subset of S linearly dependent? > > I know this is not true, however, I cannot think of a counterexample. Does > anyone have any good ideas? Why do you think it not true? If there is no linear combination of vectors in a set which add up to the zero vector why do you think leaving some of the vectors out will produce what all of them together cannot produce. Similarly, if a vector v is /not/ a linear combination of all the vectors in some set S, how can it be a linear combination of vectors in any proper subset of S? === Subject: Re: Two results of set geometry How does one determine the size of a finite set? > How does one determine the size of N? > before we try to generalize things. >> No, let's integrate those questions and ask them as one. How do we >> generalize the way we measure finite sets so as to give the most >> intuitively appealing results for infinite sets? > > > Before we can generalize the way we measure finite sets > we need to decide how we measure a finite set. No, before we can choose which way to generalize for the infinite case, we need to consider all the ways that works correctly,and therefore equivalently, in the finite case. Then, choose the best approach. > Noting that by finite sets you mean sets > that can be indexed by N, we have to answer the two questions > > How does one determine the size of a subset of N with > a largest element? > How does one determine the size of N? > Right, there are several choices. > > We can note that a set mapped through an order-preserving formula f from >> some initial finite segment n of N has a range defined by that mapping, >> f(n)-f(0). Whatever that formula is, if we are given a range of values >> [x,y] we can determine the number of elements within that range in this >> set. If g is the inverse formula to f, such that f(g(x))=g(f(x))=x, then >> we can say that floor(g(y))-floor(g(x))+1 is the count of the set. > > This procedure > assigns a count to the intitial segments of N that > have an end. This does not assign a count to N. > This does not answer the question > > How does one determine the size of N? > > So why not say The count of a set is the maximum of > the set of counts of the elements of the set ? >> Okay. > > N has no count. How does one determine the size of > N. > One declares it as some kind of unit, much like you already do. One can do more with it than cardinality. > > >> not just a natural, but a >> T-riffic or adic number, when the set is complete. > That word complete again. What do *you* mean by this > word. Does a set have to have a last element > to be complete? What about sets that are not complete? >> Please make your question mean something related to what's on your mind. > I am convinced that you only want to deal > with sets with a last element, but that you do not > want to say the only real sets are sets > that have a last element. So you say instead > the only real sets are complete sets. > You use the word complete to > mean a set that acts like it has a last element > but you have not found a set that does not have > a last element that acts like it has a last element. > Thus, the only definition you can find for complete > is a complete set is a set with a last element. > You do not wish to admit this, so evade any question > about what you mean by complete. > - William Hughes >> Even an open real set has a well-defined limit. > > An irrelevant remark. well-defined limit > does not mean the same thing as last element. It means, at the very least, the penultimate element, since only the endpoint is missing from a real open interval. No? > Rather than evading the question: > What do you mean by complete? > why don't you try answering it? > > - William Hughes > > What, exactly, is the question? In what context did I say complete, again? Peace, Tony === Subject: Re: Two results of set geometry > You have stated it, but that does not make it true. And since you have > stated a lot of stuff that is flat out not true, pardon us if we reject > it until it is proven. >> Prove the Axiom of Choice. > If we use the axiom of choice, then ordinarily we use it as an axiom. >> So, you don't prove it, you assume it. > > Why don't you read the REST of my post? > > You leave it so obvious that you read posts just to answer them line- > by-line without considering how sentence relates to the next or to > consider understanding the content of the post in context. It's > irritating. > You may find it irritating, and I sometimes do, when the beginning of the response leads to some whole elaborated misconception that changes the topic, but I also like to make the conversation seem more... conversational. A little interaction helps keep the poitn from being lost. > Or, we can prove conditionals of the form AC -> P. >> Conditional or countable choice is a theorem, not an axiom. > > This has NOTHING to do with conditional (maybe you mean 'dependent') > or countable choice. > Conditional, countable, dependent choice - basically the same thing. True. Uncountable choice, and extrapolation of universal well ordering. Not. >> Uncountable >> sets cannot be well-ordered. > > In Z set theory there is not a theorem that if S is uncountable then > there does not exist a well ordering of S. > No, in fact, there is an axiom defined exactly for the purpose of proving that uncountable sets ARE ALWAYS well-orderable, since it is not derivable from basic fundamental principles of the language of comparison. In other words, it's artificial, much like my little rubber duckie.... >> We went through that with the H-riffics, >> and it applies to the p-adics. > > Whatever you went throught doesn't refute that in Z ser theory there > is not a theorem that if S is uncountable then there does not exist a > well ordering of S. > No, there exists the contrary statement, without justification. You know that. > In any case, we are > up front as to what our axioms are and that we don't prove them > (except in the trivial sense of a one-line proof with the axiom as the > only line) but that we prove all other claimed theorems from axioms > and as little as modus ponens. > >> Yes, I understand that. Your job is deduction. But, inductive logic >> plays a part here, too. I don't mean inductive proof, since that is a >> misnomer and is really part of deduction, being a form of proof, but >> rather, something else, less formalizable. I refer to the less conscious >> part of logic, where we weigh what we sense and formulate facts, and >> compare facts over time and formulate rules regarding those facts. The >> rules you are working with were developed using logical induction, >> intuition coupled with checking the results. Developing math is a >> science, involving trial and error, and peer review. The rules are not >> handed down by god. So, some of us see fit to work on the rules >> themselves, rather than just accept the rules that have come up in the >> last century or so. I appreciate your interest in reviewing my system >> deductively, once I get it together, inductively. That would be very > > Why don't you find out first what a formal system IS? Why don't you > find out first what the criteria of evaluation or review for a > proposed deductive system ARE before you start asking people to apply > those criteria to evaluate or review your proposed system? > > You're full of misconceptions about nearly every aspect of > mathematical logic and set theory. You've even posted misconceptions > in your post here. Discount me at your own will. That's your choice. I don't need it but I'll take it. > > As to systems and rules, I have always said that I don't disqualify > any alternative formal system that might be proposed merely on grounds > of being alternative, and I don't disqualify any intuitions, informal > brainstorming or methods of imagination to arrive eventually at an > alternative formulation. But I don't conflate the intuitions and > informal brainstorming with a formal system itself. Yet, you disqualify infinite-case inductive proof as a means to solve infinitary questions. I guess you never read that in a book. Whatever. > > MoeBlee > Peace, O Anti-anti-Cantorian, Tony === Subject: Re: Two results of set geometry > WM in convinced that _assuming_ that > there are infinite sets, still /the union of finite sized elements > cannot result in an infinite sized element/. >> They cannot, if each is a proper subset of some other in the set. That >> is the case, is it not? > No, it's not. > >> Are you sure? > > Yes. > > w is an obvious counterexample. Only by axiomatic/definitionistic declaration without justification. > >> It's the case that each is a proper subset of some other. I'm sure you >> don't dispute that. The first statement was my own, and I know your >> opinion of it. However, I have mine. If no subset includes any more than >> a finite number of elements, then the set cannot either. > > Yes, obviously if every subset of S is finite, then S is finite. > Is N then finite? >> Virgil just >> necessarily larger, then every element in N. I thought that was a step >> in the right direction. > > Whatever it is, it doesn't dispute that there are sets such that every > member of the set is finite, every member is a proper subset of > another member of the set, and the union of the set is infinite. > Omega is not the union of the set? It's not, like, the limit, Man? Wow, like, what the hell IS it? I mean, c'mon, get real. >> If each is the last plus 1, then the nth is n, and there cannot be n of >> them without n being in the set. That's basic logic. No non-logical (ala >> Ross) axioms required. It's pre-axiomatic, and therefore has precedence. > > If n ranges over only natural numbers, then what you mentioned is a > theorem of Z set theory. That's very excellent. It's constructive. But if n ranges over cardinal numbers, then I > know of no system of logic that takes what you just mentioned as > basic. That's because of what the system of cardinal numbers misses, as far as distinctions between numbers goes. But you're always welcome to point one out if it exists or > create one of your own. > > MoeBlee > > (Yes, I know, very) Peace, Tony === Subject: Re: Two results of set geometry > >> It really has nothing to do with the reals or uncountability. It's about >> power set, and N=S^L in general for other number bases. Sure, the power >> set is always larger than the root set. I agree with that. It's obvious. >> But, the reals are not a power set of the naturals, as far as I see. > No, but the set of reals bijects with the power set of the set of > naturals. So if the power set of the set of naturals is uncountable > then the set of reals is uncountable. > >> IF uncountably infinite is defined as more than countably infintite, > > 'uncountably infinite' is NOT defined as 'more than countably > infinite'. Oh. Then what makes the power set of a countably infinite set uncountable, again? It's proven to be greater, no? Through some mechanism not unsimilar to infinite-case induction? I mean, pulllease... > >> like countably infinite is defined as more than finite, sure. > > 'countably infinite' is NOT defined as 'more than finite'. Greater than OR EQUAL? Perhaps? Tink 'bouddit. > > So, I don't know what your point is. > See? It's right here, between 0 and 1! :) >> Pulllllease! I've had about enough of that. Really, it's a waste of >> time.... > > However you think of your time does not affect that the set of real > numbers bijects with the power set of the set of natural numbers which > does not biject with the set of natural numbers. > > MoeBlee > > > Or even, flyject! Peace, Tony === Subject: Re: Two results of set geometry > > Please, WM, let's have the substantiation. Please tell us even one > mathematical text that declares that if every member of S is finite > then the union of the cardinalities of members of S is finite. > >> please be patient. i'll try too. > > You're joking or trolling? I searched for the context of that little snippage. Didn't find anything. Got a reference? > > So there be no doubt, WM claims the following is a general principle > of mathematics (I've offered this formulation previously, and > requested that he correct me if it does not represent his claim; and > he still may do so if he likes.): > > Ax(xeS -> x is finite) -> U{card(x) | xeS} is finite. > > MoeBlee > > That's been the discussion for the last 3+ years. Peace, Tony === Subject: Re: Two results of set geometry <46f275de@news2.lightlink.com> <46f28785@news2.lightlink.com> <46f3275c@news2.lightlink.com> <46f7df75@news2.lightlink.com> <46f93b76@news2.lightlink.com> <46fbc39c@news2.lightlink.com> <46fd0a1d@news2.lightlink.com> <46fdc383@news2.lightlink.com> > It can be proven that c has the same cardinality as the power set of omega. > As cardinality is defined, in conjunction with Cantor's diagonal > argument, which is flawed, yes. > A thought struck me while showering. Why are we discussing diagonal > arguments? To prove that c has the same cardinality as the power set > of omega all we need is a bijection between the power set of omega and > the set of real numbers. William Hughes kindly gave the outline in a > previous post. Let I(x,n) be an indicator function. x in P(N). n in N. > I(x,n) = 1 if n in x. > = 0 if n not in x. Now a bijection between the power set of omega and the set of real > numbers is given by.. f : P(omega) -> R > f(x) maps to the equivalence class of Cachy sequences containing ( I(x, > 0), I(x,0)+I(x,1)/3, I(x,0)+I(x,1)/3+I(x,2)/9, ... ) And I think that's it.. right? Yes, that is it. === Subject: Re: Two results of set geometry > Oh. So, in set theory, a solid square of finite dimensions is not >> infinitely tall as far as the number of points on either of the vertical >> sides? > A square of finite dimensions is not infinitely tall. Using number of points > as a measure of height is nonsensical. > Oh. So, the edge is not an uncountable set of points? > I am saying that a sqaure of finite dimensions is not infinitely tall. > >> You are being woefully unimaginative. > > Because I think that a square of finite dimensions is not > infinitely tall? So you think that a square that is 1 foot > in height in width is in fact infinitely tall? If measured in infinitesimal units. Do you think a square that is one foot tall is one mile tall? > > Which is all fine and dandy, assuming you actually define what you mean > by count and continuum. You apparently are not using any standard > definition for either. >> What is the standard definition of count? > You count something by starting at 1, and ending at some natural number n. > What happens if I start at ....77777? > I don't know. But that is not counting as it is typically understood. >> I'm not interested in being typical. I'll leave that job to you. Is it >> not true that every adic number has a successor, even ...999? > > What does that have to do with counting? When you count something, > do you start at 1 or not? You are claiming to count the points from > 0 to 1. Is 0 the first point, or is it point ....7777777? > All this blather about adics is just you avoiding answering > a simple question about what points you are counting. I am talking about the points which supposedly constitute the segment. Is it a set of points, or not? > > Can you define any of your terms? >> E 0 >> E 1 >> 0<1 >> x E y x This does not define anything specific. The rationals meet this definition. > >> Fine. The rationals are an actually infinite number, and as long as all >> bit positions are finite, that's ultimately all you have anyway. I'm not >> saying that's all there is, but I am saying that's all that lies within >> the infinite tree. > > So Big'Un can be the count of the rationals between 0 and 1? > Look, you simply are not addressing the questions I am asking. > You claim that Big'Un is the count of points betewen 0 and 1. > I ask which points. Given your answer above, apparently it does > not matter. I could count the rational points, or all the > rational points whose denominator is a power of 2, and in both > cases the answer is Big'Un. > All I am saying is that the above leads to the conclusion that there are more than any finite number of numbers in the interval, and calling that number Big'Un. If you can do it with omega, I can do it with Big'Un. > So are you just talking about rationals? The reals, which do not contain > any infinitesimals also meet this definition. Subsets of the rationals also meet > this definition. So what points are you talking about? >> I am talking about all points between 0 and 1. Irrational numbers exist, >> but as WM points out, only in a potential sort of way, never being fully >> expressed numerically. When we go to actually infinitesimal units of >> measure, actually infinite bit positions, then we have included all >> standard reals, and not before. > > But you still have not defined all points. There are no > infinitesimal points in any standard system of points. > > What is |[0,1)|? > >> Sorry, |{x: xeR ^ 0<=x<1}|, the count of reals in the unit interval. > >> You have not defined what that is yet, so you have > not defined Big'Un or Lil'Un yet. I know it is the count, but you > have never defined what that is. >> It's simply, geometrically, the count of points in any finite unit interval. > > You still have not defined count of points. Do you really > think that you can just keep saying count of points and not have > people notice that you have never defined what it means? > 0 is a point 1 is a point between any two points lies another point. It's a primitive, an atom of geometrical space. Go ask Hilbert what he means by a point. He never defines it. His axioms are schlock anyway. They need to be generalized and boiled down. :) >> There, now we have a count of reals per unit interval, and a size for >> each one. Of course, that violates the continuum, so we can treat each >> infinitesimal real spot as a unit interval of its own, and generate >> sub-infinitesimal differences, essentially the squares of infinitesimals. > No, we have nothing until you define what |[0,1)| means. > >> Look, N=omega? Guess what? |[0,1) in R| = Big'Un. > > That is not the definition of N. So |[0,1) in R| has > no definition apparently. It can be abbreviated as Big'Un, > but it in no way relates to anything else. It cannot > be computed or calculated. > No, but it can be used in calculations. As far as I'm concerned, the unit cube has 12* Big'Un edge points, 6*Big'Un^2 surface area, and Big'Un^3 volume. > You suppose? Don't you know what you have considered? In any case you are not > talking about points as understood by everyone else. So all your talk about > Big'Un being the count of points is meaningless until you define what you > mean by count and points. >> Meaning is in the eye of the beholder. > > More accurately, meaning is in the eye of the beholders. > Meaning is conveyed from one to another. Given that you > seem incapable of conveying whatever meaning you think > exists in your ideas, perhaps there is no meaning at all. > > Have you ever tried to convey to a cat the fine points of chess? > So what is (1+1-Lil'Un)/2? Is it not a real number? Is it not between > 1-Lil'Un and 1? >> Yes it is. It's a sub-real. I admit, it's an entirely different system, >> and not likely to appeal to those steeped in the standard teaching. Sorry. > Which just further stresses the point that you have totally failed to > unambiguously define the location of Big'Un. The reals you are counting > apparently have almost nothing to do with the standard reals, which are closed > under addition and division by 2. > Sure, by standard definitions. My reals are points on the line, each >> distinguishable point corresponding to a distinguishable real value. > No, your reals are Tony-Points on a Tony-Line. There is not first point > after the first point in a line, according to the standard definitions of points > and lines. Yet in your line there is a point that corresponds to Lil'Un, > the first number after 0. >> You're allowed to have Lil'Un/2, if you really have any use for it. > So in what sense is Lil'Un the first real after 0? Why isn't Lil'Un/2 the > first real after 0? When you started counting the reals, why did you > pick Lil'Un as the first to count after 0, instead of any of the infinite > number of other reals between 0 and Lil'Un? >> Because Lil'Un is the inverse of the number of reals. No difference less >> than Lil'Un can distinguish any two reals. In fact, Lil'Un can't. It's a >> bit like omega, in that sense, and almost as unsatisfying for that >> reason, but there is no getting around that with questions of infinity. >> So, if you can declare omega to be the least ordinal greater than all >> finite successors, then I can consider Big'Un to be the count of reals >> per unit interval, and Lil'Un to be the least difference between reals. > > There is no least difference between reals. For any two reals x and y, > x-y is a real, because the reals are closed under subtraction, > and (x-y)/2 is a real, because the reals are closed under > division by 2. So stop talking about the reals because you > clearly are talking about something else. What do you think I'm talking about? > >> Is it really any more strange than the standard transfinitology? At >> least it combines count and measure to some degree. > > Your ideas are inconsistent and incoherent. Who cares about strange? > > Who knows, as you refuse to define what anything means. But there are no > adjacent points, so your talk about Lil'Un being the first point after 0 > is nonsense according to any standard notion of points, so there obviously > is a huge difference between your notions and the standard notions. >> Don't stop the presses. > > Funny, just below you claim that I can cite no difference between > your points and standard points, yet here you gladly admit > that here is a huge difference between your point and standard > points. a huge difference between your notions and the standard notions > > > Fine. I can see you don't like infinitesimals unsmoothing things that >> way, but they can be further divided. > I don't like people using perfectly standard words for non standard concepts > and trying to pretend that they are in fact the same. >> You were unable to cite a difference between them, so how do you know >> they are not the same? > > I have cited many a difference between them? Do you bother > to read? Just above you claimed that Lil'Un is the smallest > difference between two reals, when in the standard reals, > there does not exist any smallest difference. > I was asking for a difference between an infinitesimal segment and a point, I believe. >> Two standard reals are equal if there is no >> finite difference between them. Countable segments of my real >> infinitesimals correspond to Robinson's halos, finite differences being >> only possible with an infinite number of intermediate reals. Surely, you >> don't contend that there are any two distinct reals with fewer than an >> infinite number of reals between them? > > You are the one claiming to be able to count reals one by one. > You claim that there is a first distinct real after 0, aka Lil'Un. > What about the infinite number of reals between 0 and Lil'Un? Why > don't you count them when you count the reals between 0 and 1. > Or if you are going to now claim that Lil'Un is not a real, why > are you counting it at all? It's the first point in the interval. It's less than any real, being infinitesimal, but if there are an infinite number of reals in the interval, each occupies an infinitesimal portion of that segment. > > Look, the whole question here is about your claim of counting > the reals in the interval [0,1). Until you define what that > means, everything else you say is meaningless. > > > Right. That's why I am using the word count, instead of number or > But you need to define what you mean by count. So far you have been > unable to do so. >> I have been doing it for years. Fine. I don't have a complete theory. >> I'm not a professional mathematician. Whaddya want? > > You have never defined count that I have seen that would apply > to the reals in [0,1). > > > Um, no. I don't really see it as a proof about the reals, anyway, but >> about power set. > It is a proof about power sets. But there are other proofs that relate > the reals to power sets. Remember, you cannot just pick and choose the > conclusions you like, but instead have to go where the logic leads you. So why did you mention unary if you do not even know how to represent > the real numbers in unary? >> Because the proof is about power set, but later interpreted to say >> something about the reals. The power set proof does not work in unary. >> What makes you think the reals are power set to the naturals? > > What do you mean by the power set proof does not work in unary? > Can you explain that, or is that just some random phrase that > popped out of your head. I could go deep into the relationship between power set, binary logic, and binary strings, and the generalization of all three to include larger alphabets and multivalued logic, but I don't care to right now. > > And it is very easy to show that the reals have the same cardinality > as the power set of the naturals, so it follows that the reals > have a greater cardinality than the naturals. > > > The list of all binary strings of width x has length 2^x. As x grows, >> 2^x-x grows. It's the difference between power set and root. > That is all fine and dandy for finite x, but as I tried to explain to Allan, > you cannot just assume that the standard mathematical operators are even > defined once you start talking about infinite numbers. >> Yes, you can, until you run into a contradiction. That's a lot better >> approach than simply making up new rules willy nilly because no one can >> ever check your results. > > No, you can't because those operations have actual meanings, that > just do not make sense when applied to infinite numbers. Sure they do. x>0 <-> x+y>y. aleph_0+1>aleph_0. No problem. > > Just consider the simple case of 2^x where x is an integer. > This has a very simple meaning, that is captured by the > following program: > > result=1; > for (int i=0; i result*=2; > return result; It would be a lot faster to say result<<=1 rather than result*=2. > > But this is meaningless if x is infinite, because our > program now contains an infinite loop. It does not end. > It never produces a result, because that is what infinite > means. Infinite loops happen all the time. So what? > > It is also worth noting, that this program does not tell > us what 2^3.5 is. When we start allowing new types of > numbers, we need to expand the definition of our operations. > Until we do that, the operation is meaningless. Sure. > > If you want any of this to make sense, the first thing you need to do is > define count. You then need to precisely define what you are counting. > You have done neither, and instead just declare Big'Un is the Tony-count of > the set of the Tony-reals between 0 and 1. Stephen >> It's a work in progress. Caution. Reduced speed ahead. Traffic merging. > > But you have already drawn your conclusions, even though you still > have not figured out what your premises are. Perhaps you should > consider a career in politics. > > Stephen > > > Would you vote for me? Peace, Tony === Subject: Re: Two results of set geometry >> > Because Lil'Un is the inverse of the number of reals. No difference less > than Lil'Un can distinguish any two reals. In fact, Lil'Un can't. It's a > bit like omega, in that sense, and almost as unsatisfying for that > reason, but there is no getting around that with questions of infinity. > So, if you can declare omega to be the least ordinal greater than all > finite successors, then I can consider Big'Un to be the count of reals > per unit interval, and Lil'Un to be the least difference between reals. >> >> There is no least difference between reals. For any two reals x and y, >> x-y is a real, because the reals are closed under subtraction, >> and (x-y)/2 is a real, because the reals are closed under >> division by 2. So stop talking about the reals because you >> clearly are talking about something else. > What do you think I'm talking about? You are apparently talking something that has a least difference. There is no least difference between reals, so you are not talking about the reals. I have no idea what you think you are talking about, and you have failed to define what you are talking about despite numerous requests. >> >> You are the one claiming to be able to count reals one by one. >> You claim that there is a first distinct real after 0, aka Lil'Un. >> What about the infinite number of reals between 0 and Lil'Un? Why >> don't you count them when you count the reals between 0 and 1. >> Or if you are going to now claim that Lil'Un is not a real, why >> are you counting it at all? > It's the first point in the interval. But there is no first point in the interval. You yourself admitted there is a Lil'Un/2, and a Lil'Un/4, and an infinite number of other Lil'Uns. > It's less than any real, being > infinitesimal, but if there are an infinite number of reals in the > interval, each occupies an infinitesimal portion of that segment. If Lil'Un is an infinitesimal, then it is not a real, and you should not be counting it when you count the reals. If Lil'Un is a real, then you yourself have admitted that there are an infinte number of reals between Lil'Un and 0, in which case your count of the reals arbirtarily skips an infinite number of reals. Can you explain this? Either your count of the reals includes elements that are not reals, or else it arbitrarily skips infinite numbers of reals. All your talk of infinite squares and adics is totally beside the point. Stephen === Subject: Re: Two results of set geometry > Oh. So, in set theory, a solid square of finite dimensions is not >> infinitely tall as far as the number of points on either of the vertical >> sides? > A square of finite dimensions is not infinitely tall. Using number of points > as a measure of height is nonsensical. > Oh. So, the edge is not an uncountable set of points? > I am saying that a sqaure of finite dimensions is not infinitely tall. > No foolin? What about the infinite square you use for your proof? > > What infinite square? I have never seen a proof that used an > infinite square. > The answer was probably in your snip. > Which is all fine and dandy, assuming you actually define what you mean > by count and continuum. You apparently are not using any standard > definition for either. >> What is the standard definition of count? > You count something by starting at 1, and ending at some natural number n. > What happens if I start at ....77777? > I don't know. But that is not counting as it is typically understood. > >> For every element in the set exists a successor. What is not counting? > > For every element in what set? And no, it is not counting if you > just start at some random point. If someone says they counted 100 > dollars, don't you assume that they in fact started from 1? > Not if I have 100 people each counting different piles of bills. Then I just add their 100 to whatever count I have so far. > Who knows if it is wrong, as it is not defined. You cannot say that > Big'Un is the count of the TonyReals from 0 to 1 until you define > what the TonyReals are. And your definition of the TonyReals cannot > include Big'Un. > I define them as points in the interval. Big'un is a marriage of count >> and measure, for the uncountable. > Points in the interval? What does that mean? What interval? What points? > Can you define any of your terms? >> E 0 >> E 1 >> 0<1 >> x E y x This does not define anything specific. The rationals meet this definition. > So are you just talking about rationals? The reals, which do not contain > any infinitesimals also meet this definition. Subsets of the rationals also meet > this definition. So what points are you talking about? >> Those that produced an infinite set of elements per unit interval of >> value, Duh. Why do I feel Like I landed on the Land of The Lost? >> I think I'll stop now. > > Again, you have not explained what points you are talking about. > Your thinking is so muddled you easily forget the subject at > hand. You claimed that Big'Un was the count of points from 0 > up to but not including 1. What I have been asking you is > what points are you counting? You claim to be counting something. > What are you counting? You clearly are not counting points > as understood by standard geometry. So what are you counting? > > Stephen > Points. Peace, Tony === Subject: Re: Two results of set geometry >> Oh. So, in set theory, a solid square of finite dimensions is not >> infinitely tall as far as the number of points on either of the vertical >> sides? > A square of finite dimensions is not infinitely tall. Using number of points > as a measure of height is nonsensical. >> Oh. So, the edge is not an uncountable set of points? > I am saying that a sqaure of finite dimensions is not infinitely tall. >> You are being woefully unimaginative. Picture a square of infinite >> dimensions, consisting of unit squares of uncountable number. Now, zoom >> infinitely out, so that you can actually see the square, such that it >> appears to be one unit in size. > > I actually LOL'd. That's good for you. > > Picture a rock so heavy that it cannot be lifted. Now imagine lifting > it. > Okay. Now what. Shall I picture accelerating it infinitely quickly for only a moment? Did that even take any power? >> At this point, the unit squares of which >> it consists have shrunk to infinitesimal elements, as the infinite >> square has shrunk to a finite size. > > Which is to say: Picture something which is infinite. Now imagine it > is not infinite. > Right. Did you try? >> When measured in such infinitesimal units, the finite square is >> infinitely large. > > Your analogy fails here. A finite square has edges; an infinite > square (tiling of the plane with unit squares) does not. No, it doesn't. Your conception of an infinite square fails. > > Peace, Tony === Subject: Re: Two results of set geometry > >> If we assert that f(x)>g(x) for all x > (n in N), and that for all >> infinite counts c, c > (n in N), then f(c)>g(c). > Whoa there, cowboy. How do you deduce that > f(c) > g(c) for c where c > n for all n in N? >> Yeeeehaaa! Man, that was one fast pig! Thought she was gonna win me the >> race, then she dumped me in this here mud puddle. Sorry, Pardner, 'bout >> yer pants. What was ya sayin'? All the squealin' kinda drowned you out. >> (spits out a mouthful of pig and reloads with a wad O' chaw) >> Oh, yeah. So, anyways... >> If EkeN AxeN (x>k -> f(x)>g(x)), and AneN c>n, then c>k, and ce{x: x>k}, >> so f(c)>g(c). See? > > Counterexample: > > Let B be your Big'un. Let f(x) = (B - 1 - x). Let g(x) = x. > > For all x in N, f(x) > g(x). For all x in N, B > x. But f(B) = -1, so > f(B) < g(B). > > Infinite induction is not consistent with your own vaguely described > notions. Your claims about it amount to it applies to all such > functions f and g, except when it doesn't. > > Perhaps I failed to mention it that time, but I've said it many times before. Infinite-case inductive proof of inequalities depends on the limit of the difference being greater than 0. lim(x->oo: (B-1-x)-x)= lim((x->oo: B-1-2x). Is B>2*oo? No. The limit is not greater than 0. Besides, I have never used an infinite value in my examples of infinite-case induction. They are always in the form of finite formulas, such as x^2>2x for x>2. So, you're kind of reading a whole new idea into it that I never suggested. That's like me applying infinite-case induction to set theory and claiming set theory is therefore inconsistent. Peace, Tony === Subject: Re: Two results of set geometry <46e97aef@news2.lightlink.com> <46ec8caf@news2.lightlink.com> <46eee267@news2.lightlink.com> <46f943d1@news2.lightlink.com> <46fe8a3d@news2.lightlink.com> <4703b317@news2.lightlink.com >> If we assert that f(x)>g(x) for all x > (n in N), and that for all >> infinite counts c, c > (n in N), then f(c)>g(c). > Whoa there, cowboy. How do you deduce that > f(c) > g(c) for c where c > n for all n in N? >> Yeeeehaaa! Man, that was one fast pig! Thought she was gonna win me the >> race, then she dumped me in this here mud puddle. Sorry, Pardner, 'bout >> yer pants. What was ya sayin'? All the squealin' kinda drowned you out. >> (spits out a mouthful of pig and reloads with a wad O' chaw) > Oh, yeah. So, anyways... > If EkeN AxeN (x>k -> f(x)>g(x)), and AneN c>n, then c>k, and ce{x: x>k}, >> so f(c)>g(c). See? Counterexample: Let B be your Big'un. Let f(x) = (B - 1 - x). Let g(x) = x. For all x in N, f(x) > g(x). For all x in N, B > x. But f(B) = -1, so > f(B) < g(B). Infinite induction is not consistent with your own vaguely described > notions. Your claims about it amount to it applies to all such > functions f and g, except when it doesn't. > Perhaps I failed to mention it that time, but I've said it many times > before. I.e., it applies to all such functions f and g, except when it doesn't. > Infinite-case inductive proof of inequalities depends on the > limit of the difference being greater than 0. lim(x->oo: (B-1-x)-x)= > lim((x->oo: B-1-2x). > Not every expression of the form (lim x->oo f(x)) actually exists. For example, lim x->oo sin(x) doesn't exist; nor does lim n->oo (-1)^n. So, it is reasonable to ask: How do you know /what/ lim x->oo: B - 1 - 2x is? How do you know it even exists? The usual way to answer this question would be to say: L is the limit of B - 1 - 2x if, and only if, for every real number e, there is a real number d such that |L - (B - 1 - 2x)| < e for all x > d. Suppose e is 1/2. If L = (B-1-2*oo), then what d satisfies the above equation? > Is B>2*oo? No. Only you can answer the question, is B > oo?. Typically, it is not the case that lim n->oo f(x) = f(oo), because oo is not typically taken to be a number. > The limit is not greater than 0. How do you know it exists at all? Is sin(oo) > 0? > Besides, I have never used an infinite value in my examples of > infinite-case induction. They are always in the form of finite formulas, > such as x^2>2x for x>2. I'm just pointing out that your system isn't very consistent: Infinite induction always works, except when it doesn't. > So, you're kind of reading a whole new idea into > it that I never suggested. That's like me applying infinite-case > induction to set theory and claiming set theory is therefore inconsistent. > Well, infinite case induction is provably not a theorem of set theory. There's no way to prove much about f and g in infinite induction: it works when it does, and it doesn't when it doesn't. === Subject: Re: Two results of set geometry > >> Fine. The rationals are an actually infinite number, and as long as all >> bit positions are finite, that's ultimately all you have anyway. I'm not >> saying that's all there is, but I am saying that's all that lies within >> the infinite tree. > > So from the fact that .1010010001... is in the infinite tree > we conclude that not only are all the initial segments of . > 1010010001... > rational but so is .1010010001... > > - William Hughes > > > That is not a fact, but an assumption. All of that string does not exist in a tree with only finite bit positions. Peace, Tony === Subject: Re: Two results of set geometry <46f275de@news2.lightlink.com> <46f28785@news2.lightlink.com> <46f3275c@news2.lightlink.com> <46f7df75@news2.lightlink.com> <46f93b76@news2.lightlink.com> <46fbc39c@news2.lightlink.com> <46fd128a@news2.lightlink.com> <46fe6887@news2.lightlink.com> <4703b0e6@news2.lightlink.com >> Fine. The rationals are an actually infinite number, and as long as all >> bit positions are finite, that's ultimately all you have anyway. I'm not >> saying that's all there is, but I am saying that's all that lies within >> the infinite tree. So from the fact that .1010010001... is in the infinite tree > we conclude that not only are all the initial segments of . > 1010010001... > rational but so is .1010010001... - William Hughes That is not a fact, but an assumption. All of that string does not exist > in a tree with only finite bit positions. The string .1010010001... does exist is the tree. This follows directly from the definition of when a string is in the tree. Spare me the There is no node of the tree at which .101001001... exists. This is true, but does not show that .1010010001... is not in the tree. - William Hughes === Subject: Re: Two results of set geometry > There are no infinitely tall squares in set theory, so I have no > idea what you are WM are talking about. A square has four sides. > >> Oh. So, in set theory, a solid square of finite dimensions is not >> infinitely tall as far as the number of points on either of the vertical >> sides? > > Are you saying size is a measure of height of the square > or is the number of points in the square? Again, you're being > vague and inconsistent with the term size. > No, I am not being inconsistent. I am saying there are Big'un points per unit interval. On the unit square, that means a side is Big'Un points long, and 1 unit long, and they are the same thing. > > The figures that WM has been talking about have two sides, the > left side, and the diagonal, as far as I can see. There is no bottom, > which of course is the simple fact that the two of you seem totally > incapable of understanding. > >> Incapable of accepting is what I think you mean. When we say two numbers >> are equal, and derive this truth inductively, I think WM and I both >> agree the equality holds in the infinite case, where such a case exists. > > And of course it's not enough to simply agree whether > something is true. You have to prove it true, and people > have to agree that the proof is correctly derived. Oh. Well, then, I disagree with the Axiom of Choice. Prove uncountable choice to me. Give it a shot. Demonstrate it, and convince me. Please! I am no more obligated to prove my basic assumptions than yourself. > > Neither of you have done that. Until you do, we're going > to keep pointing out that you have no justification for > believing it. > Justify uncountable choice. Give it a shot. Peace, Tony === Subject: Re: Two results of set geometry > A set is finite if and only if the elements of the > set have a maximum count. > >> Not in my mind. There are uncountable sets with a last element, and >> therefore a count of sorts at that point. > > And how does one count the elements in those finite > uncountable sets? > They're not finite sets, even if they have a finite range of values. We assign a number of reals per unit interval, and relate the size of the set to that measure. > >> Take the 10-adics, for instance. > > What makes you think the p-adics are uncountable? The fact that there exist elements within the set infinitely separated. What makes you think they are not? > >> If we say that there are omega digits to any given one, we >> have 10^omega elements in the set. The string ...999 is the omega^10th >> element in the count. > > Again, how are you counting the elements of the set of digits? You mean how do I index the bits? > >> Likewise, the H-riffics produce an uncountable >> linear sequence of reals in any given interval. > > You never proved that the H-riffics are uncountable, and in > fact several of us proved rather trivially that they are only > a countable set. But do keep trying. > You proved that your misconception of the set was countable. Peace, Tony === Subject: Re: Two results of set geometry > So WM's claim is simply: If there WERE infinitely many natural numbers > and an infinite set of all natural numbers, there would have to be an > infinite natural number (set) in the set of all natural numbers; > otherwise the set could not be infinite. >> The argument is based on more than that. In order for a set defined by >> the equality between count and value for every included element, as is >> the case with N (1st is 1, etc), to ever contain some infinite count of >> values it would have to also contain an infinite value. There is no >> omega'th without omega being an element. It's infinite-case induction. I >> think that what WM and I agree on. > So asked if omega has an infinite number of elements you > reply with a proof that omega does not have an element with > an infinite count. Your reply is true, but not on point. >> There is no point with omega. That's the point. > > The answer to the question > > Does set A have an infinite number of elements? > > does not depend on the answer to the question > > Does set A have an element with an infinite count? > > So the answer to the second question is irrelevant to answering > the first question. > Not in my opinion. Whether A is actually infinite, i.e., uncountable, depends on exactly that. > The number of elements in omega is equal to the supremum > of the set of counts of the elements. Knowing that no count is > infinite does not tell you that the supremum is not infinite. > - William Hughes >> That depends how you define supremum, I guess. > > > Let's stick with the usual definition, the supremum of a set A is the > smallest x, such that if a is an element of A, then > x is greater than or equal to a. It is easy to show that > any non-empty set of ordinals bounded above has a supremum. > What if there is no bound? What if omega is an artificial creation that doesn't actually correspond to a number? Then it's not a supremum. But, as I said below, if you insist on using it as a number, at least allow normal numerical results to apply. >> We can say there is this >> number omega which is at least as large as any finite count, but not >> call it a count, so it's a supremum of all finite counts. We can talk >> about the relationship between formulaically related sets over that >> range. That's fine. > > And does not involve any contradiction. Except that it goes beyond bijection-means-equalivalence. The formulaic relationship over whatever range you consider gives the relative size of sets. > >> But, we can't treat omega as strictly finite or >> infinite. > > Yes we can. We do not need a maximum count to define the number > elements in omega (this number is not a T-natural). > We can say that the number of elements > in omega is strictly greater than any finite number. No, as Virgil recently admitted, we can say it is at larger than each finite number, but not necessarily greater than every finite number. > Thus, omega has an infinite number of elements. It's Dedekind infinite. I'll give you that. > > > - Wiliam Hughes > Peace, Tony === Subject: Re: Two results of set geometry <46fd38cf@news2.lightlink.com> <46fe7599@news2.lightlink.com> <4703aee7@news2.lightlink.com Thus, omega has an infinite number of elements. It's Dedekind infinite. I'll give you that. fortiori, it's infinite. MoeBlee === Subject: Re: Two results of set geometry <46fd38cf@news2.lightlink.com> <46fe7599@news2.lightlink.com> <4703aee7@news2.lightlink.com So WM's claim is simply: If there WERE infinitely many natural numbers > and an infinite set of all natural numbers, there would have to be an > infinite natural number (set) in the set of all natural numbers; > otherwise the set could not be infinite. >> The argument is based on more than that. In order for a set defined by >> the equality between count and value for every included element, as is >> the case with N (1st is 1, etc), to ever contain some infinite count of >> values it would have to also contain an infinite value. There is no >> omega'th without omega being an element. It's infinite-case induction. I >> think that what WM and I agree on. > So asked if omega has an infinite number of elements you > reply with a proof that omega does not have an element with > an infinite count. Your reply is true, but not on point. >> There is no point with omega. That's the point. The answer to the question Does set A have an infinite number of elements? does not depend on the answer to the question Does set A have an element with an infinite count? So the answer to the second question is irrelevant to answering > the first question. Not in my opinion. We are discussing number of elements in N. Since you do not have a definition for the number of elements in N, your opinion is irrelevant. - William Hughes === Subject: Re: Two results of set geometry >> It can be proven that c has the same cardinality as the power set of omega. > As cardinality is defined, in conjunction with Cantor's diagonal > argument, which is flawed, yes. > A thought struck me while showering. Why are we discussing diagonal > arguments? To prove that c has the same cardinality as the power set > of omega all we need is a bijection between the power set of omega and > the set of real numbers. William Hughes kindly gave the outline in a > previous post. > > Let I(x,n) be an indicator function. x in P(N). n in N. > I(x,n) = 1 if n in x. > = 0 if n not in x. > > Now a bijection between the power set of omega and the set of real > numbers is given by.. > > f : P(omega) -> R > f(x) maps to the equivalence class of Cachy sequences containing ( I(x, > 0), I(x,0)+I(x,1)/3, I(x,0)+I(x,1)/3+I(x,2)/9, ... ) Undefined verbiage. For a number such as pi there will always be undefined bits, and the value will not be reached in any finite number thereof. > > And I think that's it.. right? Whither diagonalisation? > Where you say that the limit of a sequence, such as 0.0101010101010..., is a value, such as 1/3, that means that once the sequence is complete, that's the value reached. But, the sequence is never complete. For every finite n, 2^n is finite. The nodes and paths of the infinite binary tree are as countable as the levels, though both are larger sets, one the level of the power set of levels. Peace, Tony === Subject: Re: Two results of set geometry <46f275de@news2.lightlink.com> <46f28785@news2.lightlink.com> <46f3275c@news2.lightlink.com> <46f7df75@news2.lightlink.com> <46f93b76@news2.lightlink.com> <46fbc39c@news2.lightlink.com> <46fd0a1d@news2.lightlink.com> <46fdc383@news2.lightlink.com> <4703ad64$1@news2.lightlink.com> It can be proven that c has the same cardinality as the power set of omega. > As cardinality is defined, in conjunction with Cantor's diagonal > argument, which is flawed, yes. > A thought struck me while showering. Why are we discussing diagonal > arguments? To prove that c has the same cardinality as the power set > of omega all we need is a bijection between the power set of omega and > the set of real numbers. William Hughes kindly gave the outline in a > previous post. Let I(x,n) be an indicator function. x in P(N). n in N. > I(x,n) = 1 if n in x. > = 0 if n not in x. Now a bijection between the power set of omega and the set of real > numbers is given by.. f : P(omega) -> R > f(x) maps to the equivalence class of Cachy sequences containing ( I(x, > 0), I(x,0)+I(x,1)/3, I(x,0)+I(x,1)/3+I(x,2)/9, ... ) Undefined verbiage. What is undefined? > For a number such as pi there will always be > undefined bits, and the value will not be reached in any finite number > thereof. What do bits have to do with the mapping given? What value is supposed to be reached in a finite number of bits? To be honest Tony my post was not directed at you. I was posting for someone who actually understands mathematics to read and check. I really doubt that you have a firm enough grasp of Cauchy sequences to even understand my post, let alone judge the validity. You seem to think that a convergent sequence should reach its limit in a finite number of terms. If that's the case then wow. Just wow. === Subject: Re: Two results of set geometry <46f275de@news2.lightlink.com> <46f28785@news2.lightlink.com> <46f3275c@news2.lightlink.com> <46f7df75@news2.lightlink.com> <46f93b76@news2.lightlink.com> <46fbc39c@news2.lightlink.com> <46fd0a1d@news2.lightlink.com> <46fdc383@news2.lightlink.com> <4703ad64$1@news2.lightlink.com > Where you say that the limit of a sequence, such as 0.0101010101010..., > is a value, such as 1/3, that means that once the sequence is complete, > that's the value reached. Piffle. There is nothing in the definition of limit that uses *any* concept of complete. > But, the sequence is never complete. Irrelevant. (Doubly so a long as you refuse to disclose what you mean by complete). - William Hughes === Subject: Re: Two results of set geometry We have proved that the power set is larger than the set. Period. > > No, we have _not_ proved that. Period. > > We have proven that a particular set is uncountable. You are _utterly_ > mistaken in thinking we have proved anything about power set in > general. Sorry. > We have proven something about power set in particular, and languages in general. It's N=S^L. See below, where you say, Let f be an arbitary function from a set S into its power set, and then go on to derive a contradiction. > Where is the flaw? What does unary have to do with anything? What do > squares have to do with anything? >> Squares have to do with diagonals convering the full set of rows as well >> as columns. You know that. They're equal. To 1-1 covers both, or neither. > > This is incoherent. I can't even parse this paragraph gramatically, > let alone mathematically. > > Try again. This paragraph is the raison d'etre of this post. Write > _clearly_. What is the flaw in Cantor's diagonal argument? > >> It really has nothing to do with the reals or uncountability. > It's a proof that a particular set is uncountable and as a corollary > the reals are uncountable. How is it nothing to do with the real or > uncountability? >> It proves it larger than a known countably infinite set. It doesn't mena >> it's uncountable, though such sets exist. > > An uncountable set is a set with a cardinality larger that countable. > So this comment of yours makes no sense. In fact you even said when > > The proof that you posted is a proof that a particular set is > uncountable. As a corollary, the reals are uncountable. I do not > understand how this really has nothing to do with the reals or > uncountability. > >> It's about power set, and N=S^L in general for other number bases. > No, it isn't. The above proof is not a proof of Cantor's theorem. It > is a proof that a particular set is uncountable. >> Prove Cantor's Theorem, diagonalley. > > More distraction? You're refusing to acknowledge your error by making > a tedious demand of me. Fine. > > Let f be an arbitary function from a set S into its power set. > > Consider T = { x in S : x not in f(x) } > > For a contradiction suppose T is in the image of f. Then for some y in > S we have f(y) = T. > > If y is in T then y is in f(y) therefore y is not in T. Contradiction. > If y is not in T then y is not in f(y) therefore y is in T. > Contradiction. > Therefore T is not in the image of f. So, you proved a contradiction between |P(S)|=|S|. Like I said.... > > T is a subset of S not in the image of f. Therefore f is not > surjective. > Therefore no function from a set into its power set is surjective. > Therefore there is no bijection between a set and its power set. > > QED. QUOD erat demonstrandum? That the power set is larger than the set? That's exactly what I said the proof was about. You seem to understand that. Perhaps English is the problem. > > This is a diagonal proof of Cantor's Theorem. What you posted was > _not_ a proof of Cantor's Theorem. Apparently you do not understand > this. > I gave an informal description of exactly what you just spewed. Nice copying. >> Sure, the power set is always larger than the root set. I agree with that. It's obvious. >> But, the reals are not a power set of the naturals, as far as I see. > Who has said the reals are a power set of the naturals? >> CH rests upon that premise, that c=2^omega, asking whether there is >> exists some number between the two. Of course, there does. > > CH is the proposition that there is no cardinality between the > cardinality of the integers and the cardinality of the reals. You say > of course there is. What is that cardinality? I say it's a question about sizes of sets. I don't use raw cardinality for that measure. As far as cardinality is concerned, I doubt there is anything between those points, because of the type of argument accepted. If the regular field axioms are extended to infinite set sizes, clearly there are sizes not detected by bijection alone. > > And you still haven't answered - who has said the reals are a(sic) > power set of the naturals? What does this have to do with Cantor's > diagonal argument or Cantor's theorem? Are you daft? You insist it's a proof of the uncountability of the reals, but what you presented was a proof that |N|<|P(N)|. Do you not equate c with 2^aleph_0? Wake up. > > Do you know what a power set is? > Very well, thank you. Do you know what it's a special case of? > And, more importantly, what does this have to do with the flaw in > Cantor's diagonal argument? > The fact that 2^aleph_0 is conflated with c. Power set is power set. Continuum is another matter. Peace, Tony === Subject: Re: Two results of set geometry <46f1f9c1@news2.lightlink.com> <46f275de@news2.lightlink.com> <46f28785@news2.lightlink.com> <46f3275c@news2.lightlink.com> <46f7df75@news2.lightlink.com> <46f93b76@news2.lightlink.com> <46fbc39c@news2.lightlink.com> <46fd0a1d@news2.lightlink.com> <46fdc383@news2.lightlink.com> <4703ac1e@news2.lightlink.com > snipped context restored. > here TO gives his rendition of cantor's diagonal argument which he claims is flawed. > this is a proof that the set of sequences of two symbols is not countable. Originally, it postulated a countably infinite list of all countably > infinite strings of two symbols. It demonstrated a contradiction to the > countability of the list, by showing that for any such list, there > will always be strings not on the list, as can be generated by taking > the bitwise inverse of the diagonal of any such list, being bitwise > different from every string already on the list. So, there had to be > more than a countably infinite of strings, which of course had to be > uncountably infinite. > We have proved that the power set is larger than the set. Period. No, we have _not_ proved that. Period. We have proven that a particular set is uncountable. You are _utterly_ > mistaken in thinking we have proved anything about power set in > general. Sorry. We have proven something about power set in particular, and languages in > general. It's N=S^L. See below, where you say, Let f be an arbitary > function from a set S into its power set, and then go on to derive a > contradiction. The proof I gave is a proof of Cantor's theorem. The proof you gave is a proof that a particular set is uncountable. These are different proofs of different theorems. > Where is the flaw? What does unary have to do with anything? What do > squares have to do with anything? >> Squares have to do with diagonals convering the full set of rows as well >> as columns. You know that. They're equal. To 1-1 covers both, or neither. This is incoherent. I can't even parse this paragraph gramatically, > let alone mathematically. Try again. This paragraph is the raison d'etre of this post. Write > _clearly_. What is the flaw in Cantor's diagonal argument? No answer. > The above proof is not a proof of Cantor's theorem. It > is a proof that a particular set is uncountable. >> Prove Cantor's Theorem, diagonalley. More distraction? You're refusing to acknowledge your error by making > a tedious demand of me. Fine. Let f be an arbitary function from a set S into its power set. Consider T = { x in S : x not in f(x) } For a contradiction suppose T is in the image of f. Then for some y in > S we have f(y) = T. If y is in T then y is in f(y) therefore y is not in T. Contradiction. > If y is not in T then y is not in f(y) therefore y is in T. > Contradiction. > Therefore T is not in the image of f. So, you proved a contradiction between |P(S)|=|S|. Like I said.... I proved Cantor's Theorem (that there is no bijective function between a set and its power set), because you asked me to. This is a _different_ proof from the one you gave. That you don't understand this shows just how deep your lack of understanding is. > T is a subset of S not in the image of f. Therefore f is not > surjective. > Therefore no function from a set into its power set is surjective. > Therefore there is no bijection between a set and its power set. QED. QUOD erat demonstrandum? That the power set is larger than the set? > That's exactly what I said the proof was about. You seem to understand that. > Perhaps English is the problem. You said that the proof _you_ gave is a proof of Cantor's Theorem. You were wrong. > This is a diagonal proof of Cantor's Theorem. What you posted was > _not_ a proof of Cantor's Theorem. Apparently you do not understand > this. I gave an informal description of exactly what you just spewed. Nice > copying. No, you did not. I just spewed a proof of Cantor's Theorem. You gave an informal description of a proof that the set of sequences of two symbols is not countable. What I just spewed is not a formalisation of your informal description. It is a different proof of a different theorem. > And you still haven't answered - who has said the reals are a(sic) > power set of the naturals? What does this have to do with Cantor's > diagonal argument or Cantor's theorem? Are you daft? You insist it's a proof of the uncountability of the > reals, but what you presented was a proof that |N|<|P(N)|. You proved that the set of sequences of 2 symbols is not countable. As a corollary the reals are uncountable. I proved that a set is not bijectable with its power set. These are different proofs of different theorems. Cantor's first diagonal argument is not a proof of Cantor's Theorem. That you think it is shows that you are very, very confused.