mm-2139 === Subject: Modules over Noetherian rings fvFXAA0AAAB4-tSBtaI_1HFV2h5URmTb How to prove that every finitely generated module over Noetherian ring is finitely presented? Any references? === Subject: Re: Modules over Noetherian rings > How to prove that every finitely generated module over > Noetherian ring is finitely presented? Any references? > Let M be a finitely generated module over the the Noetherian ring A. Consider the kernel K of a surjection (!) A^n -> M. Then K is finitely generated over A. A classical reference is @BOOK{AtiyahMacDo, title = {Introduction to {C}ommutative {A}lgebra}, publisher = {Addison Wesley}, year = {1969}, author = {M.F. Atiyaha and I.G. Macdonald}, series = {Addison-Wesley Series in Mathematics}, address = {Reading, Massachusetts}, } J. === Subject: Cosine Product Limit 8B-Oyg0AAADYIzLnmXDIT1oPg_ifDqUn Hello! I want to evaluate the following limit: lim[x->0](1-cos(x)*cos(2x)*...*cos(nx))/x^2 How do I evaluat the cosine product: cos(x)*cos(2x)*...*cos(nx) using the Taylor series? Albert === Subject: Re: Cosine Product Limit > I want to evaluate the following limit: lim[x->0](1-cos(x)*cos(2x)*...*cos(nx))/x^2 > = lim(x->0) (1 - cos nx)/x^2 + lim(x->0) (cos nx)(1 - cos x *..* cos (n-1)x)/x^2 = n^2 / 2 + lim(x->0) (1 - cos x *..* cos (n-1)x)/x^2 = induction mumble mumble = sum(j=1,n) j^2 / 2 = n(n + 1)(2n + 1)/6 * (1/2) = same answer as Stephen with same comment > How do I evaluate the cosine product: cos(x)*cos(2x)*...*cos(nx) using > the Taylor series? > Masochistically, unless you're expecting us to do it you vile sadist. ;-) === Subject: Re: Cosine Product Limit > I want to evaluate the following limit: > lim[x->0](1-cos(x)*cos(2x)*...*cos(nx))/x^2 > = lim(x->0) (1 - cos nx)/x^2 > + lim(x->0) (cos nx)(1 - cos x *..* cos (n-1)x)/x^2 > = n^2 / 2 + lim(x->0) (1 - cos x *..* cos (n-1)x)/x^2 > = induction mumble mumble > = sum(j=1,n) j^2 / 2 = n(n + 1)(2n + 1)/6 * (1/2) > = same answer as Stephen with same comment > How do I evaluate the cosine product: cos(x)*cos(2x)*...*cos(nx) using > the Taylor series? > Masochistically, unless you're expecting us to do it you vile sadist. ;-) You only need it up to x^2 plus a O (x^4). cos x = 1 - x^2 / 2 + O (x^4) cos nx = 1 - n^2 x^2 / 2 + O (x^4) for any fixed n. Product from cos x to cos nx: 1 - (sum k^2 for 1 <= k <= n) * x^2 / 2 + O (x^4) Then the answer is the same. === Subject: Re: Cosine Product Limit , > How do I evaluate the cosine product: cos(x)*cos(2x)*...*cos(nx) using > the Taylor series? > Masochistically, unless you're expecting us to do it you vile sadist. ;-) You only need it up to x^2 plus a O (x^4). cos x = 1 - x^2 / 2 + O (x^4) > cos nx = 1 - n^2 x^2 / 2 + O (x^4) for any fixed n. Product from cos x to cos nx: 1 - (sum k^2 for 1 <= k <= n) * x^2 / 2 + O (x^4) Then the answer is the same. Right, and a student learns a lot more using that idea I think. === Subject: Re: Cosine Product Limit >I want to evaluate the following limit: lim[x->0](1-cos(x)*cos(2x)*...*cos(nx))/x^2 > Hint: L'Hospital and the definition of derivative. The answer I get is n (n+1) (2n+1) / 12. If this is homework, acknowledge sources of assistance in submitted work. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Cosine Product Limit <42AB3B53.7060201@netscape.net> 8B-Oyg0AAADYIzLnmXDIT1oPg_ifDqUn Hello Stephen! I applied 2 times L'Hospital and got a different result: [n(n+1)(2n+1)+6]/12 Also, I am not quite sure how to apply the definition of the derivative here. Do we mean the same thing: lim(x->x_0)[(f(x)-f(x_0))/(x-x_0)]=f'(x_0) ? Albert === Subject: Re: Cosine Product Limit > I want to evaluate the following limit: >> lim[x->0](1-cos(x)*cos(2x)*...*cos(nx))/x^2 Hint: L'Hospital and the definition of derivative. The answer I get is > n (n+1) (2n+1) / 12. I applied 2 times L'Hospital and got a different result: [n(n+1)(2n+1)+6]/12 Also, I am not quite sure how to apply the definition of the derivative >here. Do we mean the same thing: lim(x->x_0)[(f(x)-f(x_0))/(x-x_0)]=f'(x_0) ? > Apply L'Hospital once: In the denominator, you get 2x. In the numerator, you get a sum of products. Each term in the numerator is the product of an integer coefficient, one sine factor, and n-1 cosiine factors. On account of the sine factors, the numerator is 0 when x = 0. Hence, the new limit is one-half of the deriviative of the numerator. Take the derivative of the numerator: It is a sum of n^2 terms. There is a sine factor in all but n of these terms, hence these terms take the value 0 at x = 0. Setting x = 0 in the remaining n terms yields the sum of the squares of the first n positive integers; this sum is well-known to be n (n+1) (2n+1) / 6. (See, for example, http://mathworld.wolfram.com/PowerSum.html .) -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhattan === Subject: Re: Finding transfer function for this circuit <42ab3033$1@clarion.carno.net.au> Kb0T_QwAAACc9B9LpxfLjH0hHHYjPxft do it the same way as before, don't be lazy! the transfer function will change. === Subject: Re: Quantum Gravity? <42a9a957$72$fuzhry+tra$mr2ice@news.patriot.net> VrpLFAwAAACa6GkV_jrhgeb1Ocd9vzzh >The one question that usually arises is what about the entire >universe? > If it has a quantum (pure) state that evolves deterministically, >then absent an extra axiom like Projection, literally nothing will >happen in the universe so-modelled. > That's [i.e. The Problem of Time, which is what this is in reply to, as you would know if you hadn't deleted the rest of the quote above which said it was referring to that!] only true if it has an exact energy, Oh, for God's sake, people! Will you PLEASE look up what you're replying to before you reply (and try not to delete it!) The Problem of Time http://search.yahoo.com/search?p=%22The+Problem+of+Time%22 Energy and eigenvalues have nothing to do with nothing. The problem of time is related solely to diffeomorphism gauge invariance. === Subject: Re: pi in the sky BhnVAA0AAADpcidP-5AvYEnIArz5kqjP Keith, If the world is so logical, why is the book of tax regulations so think? And why, even when it's so thick, there are still people clever enough to carve legal tax shelters out of the existing rules? And wny can't they build a computer than can write and recognize hit songs like Michael Jackson or Mozart? === Subject: Re: Escultura a copycat of Archimedes Plutonium Re: I am a newly converted anti-Cantor kook! In sci.logic, a_plutonium@hotmail.com on 11 Jun 2005 09:11:24 -0700 >> ...the Adics solve all the old Number theory problems >> unsolvable to date. >> There's a difference between proving a conjecture like FLT false, and >> it being undecidable. There are no undecidable conjectures in mathematics. Goedel arrived > at undecideability in a false manner by using NaturalNumbers equal to > Finite Integers. That is a false and fake presumption. When the Natural > Numbers are taken for what they really are-- Infinite Integers (or > Adics) then Goedel's method fails and that all conjectures of > mathematics become decideable. And the axioms for these Infinite Integers are ... ? For the record, Peano's axioms can be expressed: [1] Zero is a Number (in N). [2] For all a, if a is a Number then a' is a Number. [3] There is no a such that 0 = a' (it is not the case that there exists an a such that 0 = a'). [4] For all a and b, if a and b are Numbers and a' = b', then a = b. [5] For any set S of Numbers, if 0 is in S and for all Numbers a, a in S implies a' in S, then one can deduce that S is the set of all Numbers. A modified variant can be had by replacing [3] with: [3'] There exists at least one a such that 0 = a'. which leads one to finite rings. One could in principle do something like the following. Definition: D1={1,2,3,4,5,6,7,8,9}. D={0} union D1. D1u=D1 union {ab: a in D1, b in D1u} union {a0: a in D1u} Nu= {0} union D1u. Any element of Nu is a Number. Arithmetic on these Numbers is done in the more or less usual fashion, by lining up digits on the *right*. Note that D1u may contain infinite sequences, which means Nu is not N. However, my definitions may need modification to model this properly. When it is taken for granted that Natural Numbers are equal to Infinite > Integers (Adics) then Cantor's many types of infinity fail and there > comes to be only one infinity, just as there is only really one > zero-ness there is only one infinite-ness. And Goedel's > undecide-ability fails. Not sure about undecidability, since that does not require infinity. When modern mathematics accepts the fakery of Natural-Numbers equal to > Finite Integers then it can derive other fakeries such as Cantor > transfinites and Goedel undecideability. It is easy to read Cantor or Goedel and see where they assume or > presume Natural Numbers are Finite Integers. OK, so help me out here. Where does Cantor assume Finite Integers in his system? Ditto for Goedel. Be specific. [.sigsnip] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Engineers vs physicists and other scientists > Both are beautiful, but laking in charm. At least they aren't strange :-) They are if they are in a quiche. === Subject: Re: Engineers vs physicists and other scientists > Of course it's a _correct_ formula; what is incorrect about it? Oh for the love of pete. === Subject: Re: Engineers vs physicists and other scientists > Don operates under weird algebra rules, one of which is > that parentheses change the meaning of an expression > in ways only he can fathom. So (m)a is not the same > as ma for instance. I think using his algebra one can say (f)a=(fm)a which is d+e[f]&(d+e[f]*k+l[m])*9+0&[a] which is [f]([f][m])*[a] which simplifies to f=(fm)*a which ends up simplified f-f=f-fm*a or f=ma I think === Subject: Re: Engineers vs physicists and other scientists > Yes Nathan; _IF_ the math is done right: Force (f) is equal to (f/a)a; not f=ma! (f)=f? Well, that's certainly... useless. > Weight (w) is equal to (w/g)g; not w=mg; because mass equals f/a, > and/or w/g; not m! (w)=w? Well, they must be breaking your doors down to get you to simplify math for everyone. === Subject: show that sum(1/(1+exp(n*x)), n=0..infinity) is uniform convergent how do I show that sum(1/(1+exp(n*x)),n=0..infinity) is uniform convergent? === Subject: Re: show that sum(1/(1+exp(n*x)), n=0..infinity) is uniform convergent Scoobidoo vient de nous annoncer : > how do I show that sum(1/(1+exp(n*x)),n=0..infinity) is uniform convergent? For x in [a, infty[, exp(n*x) >= exp(n*a), so : | 1/(1+exp(n*x) | =< 1/exp(a*n) =< ( 1/exp(a) )^n Since a>0, | 1/exp(a) | < 1, and so the gometric series sum( (1/exp(a) )^n ) converges, what proves that your series is normally convergent. === Subject: Re: show that sum(1/(1+exp(n*x)), n=0..infinity) is uniform convergent > how do I show that sum(1/(1+exp(n*x)),n=0..infinity) is uniform convergent? If you specify the common domain of the sequence of functions, you could use the inequality 1+z >= z for z>0 (z=exp(nx)) and some adapted geometric sequence to show the assertion. J. === Subject: Re: show that sum(1/(1+exp(n*x)), n=0..infinity) is uniform convergent on the interval [a,infinity] a>0 === Subject: Re: Mass is quantity; inertia is its measure > Unlike some other people around here claim: > space and time are measurable physical > quantities also. Okay, you're in a big empty room 40x40x10 feet in size... how do you measure how much space and time is in it? === Subject: Re: Mass is quantity; inertia is its measure > We're talking about mass being quantity, and inertia being its measure > aren't we? You are, nobody else wants to. > That's what I'm planning to continue on about until we all understand; > not according to the books, but as second nature. And after that are you going to learn everyone that the earth really IS flat? === Subject: Re: Mass is quantity; inertia is its measure >> The quantity of matter in a body is mass. >> Yes. Was there any need to continue? Yes we'll continue 'til we all get it straight: But not tonight. I'm > going to 'pack it in', nighty-night. None of us want to know what you and your boyfriend are doing in your personal time.. or maybe its not your boyfriend, maybe its the horse you keep hitched to your wagon backwards... === Subject: Re: A Hard Math Problem > How to prove the polynomial > 1 + x + x^2/2! + x^3/3! + ... + x^2n/(2n)! > has no real root? This is equivalent to finding a real number = log 0. Can you show log 0 is undefined? Bob Kolker === Subject: Re: A Hard Math Problem > How to prove the polynomial > 1 + x + x^2/2! + x^3/3! + ... + x^2n/(2n)! > has no real root? This is equivalent to finding a real number = log 0. Can you show log 0 > is undefined? What are you talking about? === Subject: Re: A Hard Math Problem !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > How to prove the polynomial >> 1 + x + x^2/2! + x^3/3! + ... + x^2n/(2n)! >> has no real root? This is equivalent to finding a real number = log 0. Uh no, it isn't. He is not talking about limits here. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: A Hard Math Problem >How to prove the polynomial > 1 + x + x^2/2! + x^3/3! + ... + x^2n/(2n)! > has no real root? >>This is equivalent to finding a real number = log 0. > Uh no, it isn't. He is not talking about limits here. Whoops. You are right. Bob Kolker > === Subject: Re: A Hard Math Problem SAlCkQwAAADOQfb8GuNFkWcQtC01OYCg > How to prove the polynomial > 1 + x + x^2/2! + x^3/3! + ... + x^2n/(2n)! > has no real root? > Let us denote f_n(x)= 1 + x + ... + x^n/n! Note that derivative of f_{n+1} is f_n. We prove by induction that f_2n(x)>0. For n=1 this is easy. Now assume that this holds for n-1. It means, that f'_(2n-1)>0 and f_(2n-1) is strictly increasing. Since f_(2n-1) is polynomial of odd degree, it has exactly one root x0. (Each polynomial of odd degree has at least one root, this one is increasing, therefore there is exactly one root.) It follows easily that f_(2n) has minimum at the point x0, and f_(2n)(x0) = f_(2n-1)(x0) + x0^2n/(2n)! = x0^2n/(2n)! > 0. (Clearly x0!=0.) Martin P.S. Another problem related to this polynomial was posted not very long time ago, by Dan Dima (Is this polynomial irreducible?) === Subject: Re: A Hard Math Problem BgL4OQ0AAAADekj6oAvKOMiQPVX22wm2 Hi ! Do you really think it's hard? I think we can solve it by a recurrence. Let Un(x) = 1 + x + x^2/2! + ... + x^(2n)/(2n)! We can see that for all n, x>=0 => Un(x)>0, easy right ? We have to prove for all x<0, Un(x)>0, right ? So we have : U0(x) = 1, U0(x)>0. Ok Then we suppose Un(x)>0. Un+1(x) = Un(x) + x^(2(n+1))/(2(n+1))!, right ? Now as 2(n+1) is even, x^(2(n+1))/(2(n+1))! is positive. We have Un(x)>0, it mean Un+1(x)>0. This is a recurrence reasoning, and it works fine. Conclusion : for all x, n>=0 Un(x)>0. And necessarily have no roots. === Subject: Re: A Hard Math Problem Sny74g0AAADy66iGh6ZMdSlIFta_KAXh > How to prove the polynomial > 1 + x + x^2/2! + x^3/3! + ... + x^2n/(2n)! > has no real root? g(x,n) = x^(2n)/(2n)! let u = 2n h(x,n) = x^(n)/n! x^u/u! => u/2 = n sum.. f(x)= g(x,0) + g(x,1/2) + g(x,3/2) + ... + g(x,n/2) = h(x,0) + h(x,1) + h(x,2) + ... + h(x,n) = 1+x+x^2/2!+x^3/3!+...+x^n/(n)! Erp.. why didn't you just write this in the first place.. 1 + x + x^2/2! + x^3/3! + ... + x^n/n! ? What's the sum of the first n terms from the taylor expansion of e^(k*x) w./ k of your choice, and does it have any real roots? -Mysid === Subject: Re: A Hard Math Problem >>How to prove the polynomial >> 1 + x + x^2/2! + x^3/3! + ... + x^2n/(2n)! >> has no real root? > g(x,n) = x^(2n)/(2n)! > let u = 2n > h(x,n) = x^(n)/n! x^u/u! => u/2 = n sum.. f(x)= g(x,0) + g(x,1/2) + g(x,3/2) + ... + g(x,n/2) > = h(x,0) + h(x,1) + h(x,2) + ... + h(x,n) > = 1+x+x^2/2!+x^3/3!+...+x^n/(n)! Erp.. why didn't you just write this in the first place.. 1 + x + x^2/2! + x^3/3! + ... + x^n/n! ? What's the sum of the first n terms from the taylor expansion > of e^(k*x) w./ k of your choice, and does it have any real > roots? -Mysid > 1+x has root -1, so n being even matters === Subject: Re: A Hard Math Problem GYng8QwAAAASJCdK-kCiIRtmF4RV4yrR > >>How to prove the polynomial >> 1 + x + x^2/2! + x^3/3! + ... + x^2n/(2n)! >> has no real root? >How to prove the polynomial >> 1 + x + x^2/2! + x^3/3! + ... + x^2n/(2n)! >> has no real root? > Erp.. why didn't you just write this in the first place.. > 1 + x + x^2/2! + x^3/3! + ... + x^n/n! > ? > Because polynomials of odd degree all have a real root, you have to write 1 + x + x^2/2! + x^3/3! + ... + x^n/n! for even n, has no real roots. > 1+x has root -1, so n being even matters > f(x) = 1 + x + x^2 / 2 + x^3 / 6 g(x) = 6f(x) = 6 + 6x + 3x^2 + x^3 g(-1) = 2; g(-2) = -2 Thus there's a real root between -1 and -2. === Subject: Re: A Hard Math Problem Sny74g0AAADy66iGh6ZMdSlIFta_KAXh > 1+x has root -1, so n being even matters Ok, that makes sense.. for finite n you want to start with '1' and work with pairs of terms added to that, since each increment is 2 terms of higher powers then take cases -1 < x < 1; x <= -1 and x >= 1 -Mysid === Subject: Re: A Hard Math Problem aCFkEg0AAAD2ttg_o8dAFUCLmVeiVgvc Let f[n](x) = 1 + x + x^2/2! + x^3/3! + ... + x^2n/(2n)!. Since all the coefficients are positive with a leading term of 1, f[n](x)>0 when x>0 for all n. We proceed by mathematical induction. For the basis, note that f[0](x) = 1 > 0 for all x. Suppose f[n-1](x) > 0 for all x for some n>0. f[n](x) - f[n-1](x) = x^2n/(2n)! - x^(2n-1)/(2n-1)! = x^(2n-1)/(2n-1)! (x/(2n) - 1). So, f[n](x) - f[n-1](x) > 0 for x<0. Thus, f[n-1](x) > 0 for all x implies f[n](x) > 0 for all x (since it was shown above that f[n](x)>0 must be true for x>0). Therefore, by mathematical induction, f[n](x) > 0 for all n>=0. Thus f[n](x) has no real roots. === Subject: Re: A Hard Math Problem aCFkEg0AAAD2ttg_o8dAFUCLmVeiVgvc Argh, made a really stupid sign error. Please ignore my proof that I gave above. === Subject: Re: A Hard Math Problem Kb0T_QwAAACc9B9LpxfLjH0hHHYjPxft does not seem to be such hard problem. what have you tried? or are you interested in getting someone else to do your work? === Subject: Re: I know the sum, but not the numbers. >One way is to try to solve it with your pure brain >>What if my brain isn't pure? Intoxicated with beer and such? Nope. I don't drink. Alcohol. >Then it is significantly harder, but >I doubt pencil and paper will help. >Or intoxicated with dirty ideas? Not intoxicated. Just loaded :) >Try solving it in the nude. You know, it's about 85F in here. You may be on to something :) --julie === Subject: polynomial == algebraix polynomial ? All over the net, and in my texts, I see the term 'polynomial' defined as a sum of terms of the form 'a(sub)n x(sup)n', n=0,1,2... Yet I seem to remember learning that this was called an 'algebraic polynomial'. What do they call functions like y = sin(x) + sin(squared)(x) + sin(cubed)(x) y = sin (x) + sin(x-squared) + sin(x-cubed) Are these polynomials too? === Subject: Re: polynomial == algebraix polynomial ? All over the net, and in my texts, I see the term 'polynomial' defined > as a sum of terms of the form 'a(sub)n x(sup)n', n=0,1,2... Yet I seem to remember learning that this was called an 'algebraic > polynomial'. What do they call functions like > y = sin(x) + sin(squared)(x) + sin(cubed)(x) Polynomial in _what_ indeterminate? > y = sin (x) + sin(x-squared) + sin(x-cubed) No. > Are these polynomials too? === Subject: Pi necklace http://www.pisquaredoversix.force9.co.uk/Necklace.png -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Pi necklace UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK > http://www.pisquaredoversix.force9.co.uk/Necklace.png Looks similar to those WWJD (What Would Jesus Do) bracelets that were popular a while ago -- before it was discovered that the wearers were getting lead poisoning when the cheap plating wore off. Can I interst you in a WWGT necklace? (What Was God Thinking) -- > Clive Tooth > http://www.clivetooth.dk === Subject: Re: Topology > But the proof is not complete. The proof of what? I'm not following the thread. Only reading and replying to each post individually. So make each post stand alone and when replying keep the essential context, otherwise you're talking to yourself. > Consider a space X with topology Tao. > The Tao which is the eternal way, cannot be expressed in words. The Tao is not a Greek letter. Maybe it's a geek letter. ;-) So there's a topological space (X,tau) > Need to show that all sets U where for every x element of U there exist > a neighborhood N(x) contained in U is member of Tao. > Definitions of neighborhood vary from text to text. What's the definition of neighborhood you're using? for all x in A, some open U with x in U subset A ==> A open (1) Any problem with proving (1)? > The set of all such U obviously follow the axioms. But how do I know > that it's the same topology. How do I know that such sets is the > original Tao? > Call that collection of sets tau. You want to show tau = Tao. (1) shows that tau subset Tao. Now show Tao subset tau > Guys, don't you want to review this? > No, I've gone thru all this in detail proving most of the basic stuff from scratch. While you're at it, show the arbitrary intersection of closed sets is closed and the union of two closed sets is closed. This is simple set theory exercise. How facile are you with DeMorgan's rules? If not, then you better read up on them, because topology, ie general or point set topology, is set algebra feast complete with appetizers and second courses like Zorn's Lemma. === Subject: Re: Topology > http://at.yorku.ca/i/a/a/b/23.dir/ch1.htm After 1.1.2 It follows that a set U is open iff for every x element U there exist a > neighborhood N(x) that is contained in U. Now let's see if U is open then we need to find a neighborhood of N(x) > that contain x for every x element U. Let that N(x) be U itself. Then > we're done. Now the other way around. The set U is a union of neighborhoods (ie U = Union(x in U) N(x)), and thus a union of open sets (by definition of neighborhood). === Subject: Re: Topology 0Sa5_Q0AAADQO9xHiLixK7eiYfArq4ky > http://at.yorku.ca/i/a/a/b/23.dir/ch1.htm > After 1.1.2 > It follows that a set U is open iff for every x element U there exist a > neighborhood N(x) that is contained in U. > Now let's see if U is open then we need to find a neighborhood of N(x) > that contain x for every x element U. Let that N(x) be U itself. Then > we're done. > Now the other way around. The set U is a union of neighborhoods (ie U = Union(x in U) N(x)), and thus > a union of open sets (by definition of neighborhood). But I can see the path from here. Let's see, for every x in U there exist N(x) contained in U. N(x) contain an open set O that contain x. So the existence of N(x) in U simply means the existence of open neighborhood in U. Let S be the union of all such N(x). Of course, 1 x can have a bunch of N(x). Want to show that U is equal to S. If y is an element of S then y is an element of N(x) for some x in U. So y is an element of U. If x is an element of U then x is an element of N(x) for some x in U. So x is also an element of union of such N(x), namely S. By the way, what can I do in return. If you guys are learning like me, post your theorem here. My experience when I took graduate level math classes without prerequisites is that it's faster to study together than alone. Jonathan === Subject: Re: Topology >> http://at.yorku.ca/i/a/a/b/23.dir/ch1.htm >> After 1.1.2 >> It follows that a set U is open iff for every x element U there exist a >> neighborhood N(x) that is contained in U. >> Now let's see if U is open then we need to find a neighborhood of N(x) >> that contain x for every x element U. Let that N(x) be U itself. Then >> we're done. >> Now the other way around. >> The set U is a union of neighborhoods (ie U = Union(x in U) N(x)), and >> thus >> a union of open sets (by definition of neighborhood). But I can see the path from here. Let's see, for every x in U there exist N(x) contained in U. N(x) > contain an open set O that contain x. So the existence of N(x) in U > simply means the existence of open neighborhood in U. open need not be in quotes. in some topology texts, a neighborhood of x is simply an open set that contains x, so open neighborhood would be redundant. in this case, it's not. if G is open and x is in G, then G is a neighborhood of x. Let S be the union of all such N(x). Of course, 1 x can have a bunch of > N(x). Want to show that U is equal to S. If y is an element of S then y is an > element of N(x) for some x in U. So y is an element of U. If x is an element of U then x is an element of N(x) for some x in U. > So x is also an element of union of such N(x), namely S. > you can delete for some x in U, it doesn't make sense the way you've used it. this is a bit wordy, but correct. i know i'm guilty of that far too often. === Subject: Rovelli's Quantum Gravity Paul Rovelli calls the tetrad field the gravity field in his new book on Quantum Gravity. Z: OK, that makes sense to me. I agree with you that the non-trivial part B of the tetrad field may be the key to separating observer-dependent and objective aspects of the Einstein field, and arriving at a satisfactory understanding of gravitational energy. J: Rovelli starts from there, but he has no idea that ALSO the dimensionless Bu^a ~ BuLp(argVacuumODLRO)^,a for the 4D Diff(4) macro-quantum supersolid world crystal analogous to v = (h/m)GradargGroundstateODLRO for the 3D macro-quantum superfluid g(curved metric)uv ~(flat)uv + B(uIv) + B(uBv) for spin 2 Note that elastic linear B term and the plastic nonlinear B^2 term. Also, B while not a spin2 Diff(4) first rank tensor field all by itself, is a Minkowski spacetime spin 1 Yang-Mills vector field. Z: Maybe Alex can explain to you why he and I propose to draw a fundamental distinction between raw mathematical spacetime CSs, on the one hand, and physical spacetime CSs (moving CSs) that in GR represent observer reference frames, on the other. J: Yes, he did, but I am not convinced that his idea is any better than mine, which is that a local frame (inertial or non-inertial) is an equivalence class of local coordinate charts at P that do not change the motion of the detector. Similarly, I mean a local event as a coincidence of at least two processes, like Rovelli's collision of two point Rovelli's explanation of Einstein's solution of the hole problem. For example, any transition overlap function between 2 local charts for same coincident P that leaves c^2(LC)00^i, i = 1,2,3 invariant is not physical. Note that points A & B in Rovelli's are same P = {~|A,B,C ...} That is A ~ B if B = active Diff(4) on A etc. That is, think of EM and weak and strong fields Au ... as animals living on Leviathan the Great White Whale Moby Dick which is the geometrodynamic tetrad field e = I + B. The Ashtekar idea is a much more fancy way of doing what I have done more simply, i.e. the basic field is the tetrad = flat + curved, Z: Yes, but you have to be careful about how the coordinate-basis tetrad is physically interpreted. There is more than one possible interpretation. J: Rovelli et-al show a completely coordinate-free Cartan form tetrad e. In fact the tetrad e is a 1-form independent of coordinates. Torsion and curvature are 2-forms. No problem! e = eu^a&adx^u e is a dimensionless Cartan 1-form &a is a 1-co-form vector field basis in the tangent fiber with dimension 1/length dx^u is a 1-form basis in base space with dimension length You can make them anholonomic (non-coordinate) sticking in Lie brackets - all well-known techniques. e.g. [dx^a,dx^b] = Wab^cdx^c Wab^c = Jim Corum's object of anholonomy with dimension length Z: As I'm sure you know, the tetrad starts out simply as a coordinate-independent basis for a any 4D vector space defined at each point on the spacetime manifold. J: Yes, and what it represents physically is the state of linear elastic warping B(uIv) and nonlinear plastic cracking B(uBv) of the perfect Minkowski Planck lattice that is the pre-inflationary massless conformal false vacuum without any gravity and inertia at all! e.g. the state of warping and cracking of the 4D active Diff(4) invariant supersolid Higgs vacuum at P is given by the base space reciprocal lattice vector field eu(P) = eu^a&a = (Iu^a + Bu^a)&a where &a = nab&^b and [eu(P)] is a length^-1 i.e. eu - &u = Bu = crystal distortion reciprocal lattice phonon wave vector away from equilibrium along the u axis (u = 0,1,2,3), e^u - dx^u = B^u is the lattice distortion with dimension of length. nab = flat Minkowski metric of pre-inflation unstable false conformal massless twistor vacuum without gravity and inertia because vacuum ODLRO = 0 but Ashtekar & Co do not use local gauge invariance & vacuum ODLRO directly the way I do. Z: OK. J: They do not seem to even be conscious of vacuum ODLRO - but maybe I have underestimated them there. Z: Maybe not. J: We will see as I read Rovelli's book. Z: OK. Z: What they are doing in loop gravity is overly-complicated (complexification of the tetrad, taking self-dual instanton part etc. Where all the Pundits go wrong is not realizing that there even is a macro-quantum theory whose rules are different from micro- quantum theory. I get diff(4) invariance + background independence non-perturbatively trivially! Z: I'm inclined to agree. J: Micro-QM is linear, nonlocal, unitary with signal locality. Macro-QM is nonlinear, local, non-unitary with signal nonlocality. Gravity + dark energy is a More is different macro-quantum phenomenon. Vacuum ODLRO makes quantum theory (QT) and general relativity (GR) completely consistent with each other. Z: I think it's an interesting and very promising approach. J: BTW John Baez in Ch 5 last part also waffles on gravity energy problem, p. 452 Notes to Part III Z: YES. So does Penrose in Road to Reality. J: I think I gave the real answer as to why this is really a pseudo-problem: BTW from my Rovelli Notes The non-trivial warp part B of the Einstein-Cartan tetrad components e = I + B comes into being and becoming from the simultaneous local gauging and spontaneous vacuum symmetry breaking of the spacetime translation group T4, therefore there is no reason at all, from NoetherÍs theorem, to imagine that either total energy or total linear momentum of the pure gravity field as the spacelike integral of a local density should be conserved as the lapse function time pushes forward from one spacelike slice to another in the ADM 3+1 foliation. Indeed, the Hamiltonian of the pure gravity field is strictly zero because of the constraint structure of EinsteinÍs field equation. For example, the Wheeler-DeWitt equation in the superspace whose points are 3D geometries is H(pure gravity)(BIT Wave Function of Multiverse) = 0 leading to the ñproblem of timeî in idealistic non-Bohmian attempts at quantum gravity that are only thoughtlike BIT with no rocklike hidden variable IT. In the Bohmian interpretation, the super-geodesic equation of motion for the IT hidden variable is separate from the above Wheeler-DeWitt equation for the pilot BIT wave functional that need not depend on any time parameter. All you can say is Sum of all Tuv = 0 locally, i.e. matter + gravity + exotic zpf vacua Z: Or, more generally, you can introduce *some* kind of source field for vacuum G_uv. This provides an objective localizable physical container for the energy lost through gravitational radiation -- which is not the situation in orthodox GR. J: No, you are confusing apples (zero point dark energy) and oranges (classical gravity waves). Residual zero point energy exotic vacua of both dark energy (Pioneer 10 & 11 anomaly a_g = -cH) and dark matter (Galatic Halos flat stellar rotation curves) & stability of extended electron Bohm hidden variable and Ken Shoulders mesoscopic charge clusters does give a local tuv(zpf) but that has nothing at all to do with classical gravity waves in weak field approximation on a flat background. With TOTAL Tuv^;v = 0 Therefore, any total energy-momentum = 0 Universe is a free lunch. In my view the search for quantum gravity is a serious error. Gravity is emergent in the inflationary vacuum phase from micro- quantum theory to macro-quantum theory. God plays dice in the unstable pre-inflationary micro-quantum vacuum that has no gravity and no inertia in it. The rules change completely in the Big Bang. God loads the dice significantly in order for gravity and dark energy to emerge into Being and Becoming. Our post-inflationary expanding accelerating universe in the multiverse of parallel worlds next door is a vibrating ñsuper-solidî or ñworld crystalî. It's certainly a promising alternative to the usual perturbative approach. The macro-quantum ghost of the departed Maxwellian aether. It's not departed, we are fish swimming in it, or rather the fish are themselves phase ripples in it - solitons i.e. vacuum geons like Chapline & Laughlin's dark energy stars on micro-scale of mass without mass (Wheeler) in which the zero point energy false vacuum cores give effective strong short range gravity G* ~ 10^40G on the fermi scale of 10^-13 cm ~ e^2/mc^2. The only flag I would raise here is about the stability of the vacuum LRO. According to your BEC model, why is the gravitational vacuum observed to be so stable? What exactly would it take to disrupt and destabilize this LRO? This means you still do not understand the key idea. Ask why is the superconductor ground state stable? It's the same problem! The Goldstone phase of Vacuum ODLRO is rigid that's part of More is different. That's why space-time is stiff. If you do not do your homework and read the key superfluid papers in PW Anderson's A Career in Theoretical Physics you will never really understand the idea here. Soft condensed matter physicists are very familiar with this idea. And what about light propagation? How does light propagate through the vacuum BEC? Could the characteristics of the vacuum BEC be responsible for the permeability and dielectric constant of the vacuum? That's Puthoff's dead end. I will not go there. I get guv that's enough. Light is ds^2 = 0. I follow Einstein there. Hal has wasted 20 years or more on that and has gotten nowhere important. I think Ibison is finally persuading him to look in my direction? === Subject: is C2xD5 isomorphe with D10? is C2xD5 isomorphe with D10? C2 is the cyclic group orden 2 and D5 is the diedergroup orden 2*5 === Subject: Re: is C2xD5 isomorphe with D10? 5fo8Jw0AAACGZghYfDob_TWf13DbgmeN No, C2 xD5 has two elements of order 2, while D10 only has one - therefore they are not isomorphic. === Subject: Re: is C2xD5 isomorphe with D10? sAS5-AwAAABlKnmtMjBbYHvhxI6W0cAg Of course, presumably D5 is just C5, so C2xD5 is actually abelian, whereas D10 isn't. That's another way to distinguish them. === Subject: Re: is C2xD5 isomorphe with D10? sAS5-AwAAABlKnmtMjBbYHvhxI6W0cAg > No, C2 xD5 has two elements of order 2, while D10 only has one - > therefore they are not isomorphic. I don't think this is right. I think you'll find C2xD5 has one element of order 2, whereas D10 has five. === Subject: Re: optimal path to connect the dots > I am a programmer writing code to compress a 3D triangle mesh. I need > to be able to decompress it randomly, i.e. get the vertices of any > triangle in unit time. I don't think sci.math is the right newsgroup, comp.compression is likely much more suitable. That said, I don't think this discussion is likely to cause too much off-topic noise here :) > Suppose there are 1 million vertices and they are stored in a linear array. A triangle is a set of three vertices, so the most naive solution is 60 > bits per triangle, i.e. three 20-bit integers to get the vertices from > the array. Can I do better than that? Yes, you can definitely do better than that. Since you're doing a very fine mesh, it is very likely that you can store almost all of your triangles in strips, a flattened and equilateral view of which might be (in fixed-width font): *---*---*---*... /12/34/56/... *---*---*---*... Each successive triangle differs only in one vertex from the preceding triangle. Not only that, if you have many adjacent strips you can simple encode the vertices in adjacent lines. Most triangles would then not need any extra storage. The arrangement of the vertex array itself would encode most of your triangles. If you have a very complex 3D surface topology, you would gain less from this approach since many of the strips are likely to be short. However, it would still be far better than 60 bits per triangle. - Tim === Subject: Basic Topology Questions _3xkiw0AAACBZ92GvM1ol-YotzWA65o2 Hello everyone, I'm trying to brush up on my topology.. I found an excellent online book called Topology Without Tears. It can be found here: http://uob-community.ballarat.edu.au/~smorris/topbookchaps1-8.pdf Anyways, I've been reading it and doing exercises, but apparently the solutions aren't posted anywhere. This is a little frustrating since I know I'm getting some of the answers wrong! Two of the most basic excercises I know I'm getting wrong. I'd love to have someone check my answers, since these test the most basic definitions of topology. The questions I'm having problems with are in exercises 1.1 3 and 1.1 9. 1.1 9 states. Let R be te set of reals. Exactly 3 of the following ten collection of subsets are topologies. Which 3? i) R, null, and intervals (a,b) for which a and b are real with a < b. I say no since the union of 2 nonintersecting intervals is not in the set. ii) R, null, and intervals (-r, r) for r any positive real. I say yes .. unions and intersections still lead to intervals (-r, r). iii) R, null, and intervals (-r, r) for r any positive rational. Yes for the same reason as ii. iv) R, null, and intervals [-r, r] for r any positive rational. Yes for the same reason as ii. v) R, null, and intervals (-r, r) for r any positive irrational. Yes for the same reason as ii. vi) R, null, and intervals [-r, r] for r any positive irrational. Yes for the same reason as ii. vii) R, null, and intervals [-r, r) for r any positive real. Yes for the same reason as ii. viii) R, null, and intervals (-r, r] for r any positive real. Yes for the same reason as ii. ix) R, null, and all intervals [-r, r] and (-r, r) for r any positive real. Yes because of iv and ii, and intersections of [-r, r] and (-r, r) is (-r, r), while unions are [-r, r]. x) R, null, and every interval [-n, n] and (-r, r) for n any positive integer and r any positive real. Yes because of an argument similar to ix Now on the other hand, as I think about it more, it seems that ii-x all have the property that the infinite intersection of all their subsets is just 0 .. but that shouldn't matter since we only consider finite intersections. I must clearly be getting something very wrong since only 3 of them are topologies but I find that 9 of them are! Any help would be really appreciated.. I feel that I'm hampered by progressing further in this book when I'm stuck on such a simple problem. I'm also stuck on the easy looking 1.1 3 but I'll type that one out later :) Julien === Subject: Re: Basic Topology Questions >1.1 9 states. Let R be te set of reals. Exactly 3 of the following ten >collection of subsets are topologies. Which 3? >i) R, null, and intervals (a,b) for which a and b are real with a < b. >I say no since the union of 2 nonintersecting intervals is not in the >set. > > Correct. >ii) R, null, and intervals (-r, r) for r any positive real. >I say yes .. unions and intersections still lead to intervals (-r, r). > > U{(-1+1/n, 1-1/n): n in N, n>0} >iii) R, null, and intervals (-r, r) for r any positive rational. >Yes for the same reason as ii. > > Even worse: rationals approach reals. >iv) R, null, and intervals [-r, r] for r any positive rational. >Yes for the same reason as ii. > > U{[-1+1/n, 1-1/n]: n in N, n>0} >v) R, null, and intervals (-r, r) for r any positive irrational. >Yes for the same reason as ii. > > Aargh: irrationals approach rationals >vi) R, null, and intervals [-r, r] for r any positive irrational. >Yes for the same reason as ii. > > No >vii) R, null, and intervals [-r, r) for r any positive real. >Yes for the same reason as ii. > > [-1+1/n. 1-1/n) >viii) R, null, and intervals (-r, r] for r any positive real. >Yes for the same reason as ii. > > No >ix) R, null, and all intervals [-r, r] and (-r, r) for r any positive >real. >Yes because of iv and ii, and intersections of [-r, r] and (-r, r) is >(-r, r), while unions are [-r, r]. > > OK >x) R, null, and every interval [-n, n] and (-r, r) for n any positive >integer and r any positive real. >Yes because of an argument similar to ix > > OK -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and New York === Subject: Re: Basic Topology Questions <42AB9792.6050101@netscape.net> _3xkiw0AAACBZ92GvM1ol-YotzWA65o2 >ii) R, null, and intervals (-r, r) for r any positive real. >I say yes .. unions and intersections still lead to intervals (-r, r). U{(-1+1/n, 1-1/n): n in N, n>0} don't get, however is ii) Doesn't this counterexample just leave us with the set (-1, 1). which is still in the candidate topology? Julien === Subject: Re: Basic Topology Questions >>ii) R, null, and intervals (-r, r) for r any positive real. >>I say yes .. unions and intersections still lead to intervals (-r, r). >> U{(-1+1/n, 1-1/n): n in N, n>0} don't get, however is ii) Doesn't this counterexample just leave us with the set (-1, 1). which >is still in the candidate topology? > Oops. You are correct. I don't know what I was thinking. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor in Central New Jersey and Manhatttan === Subject: Re: Basic Topology Questions > Hello everyone, I'm trying to brush up on my topology.. I found an excellent online > book called Topology Without Tears. It can be found here: http://uob-community.ballarat.edu.au/~smorris/topbookchaps1-8.pdf Anyways, I've been reading it and doing exercises, but apparently the > solutions aren't posted anywhere. This is a little frustrating since I > know I'm getting some of the answers wrong! Two of the most basic excercises I know I'm getting wrong. I'd love to > have someone check my answers, since these test the most basic > definitions of topology. The questions I'm having problems with are in exercises 1.1 3 and 1.1 > 9. 1.1 9 states. Let R be te set of reals. Exactly 3 of the following ten > collection of subsets are topologies. Which 3? > i) R, null, and intervals (a,b) for which a and b are real with a < b. > I say no since the union of 2 nonintersecting intervals is not in the > set. ii) R, null, and intervals (-r, r) for r any positive real. > I say yes .. unions and intersections still lead to intervals (-r, r). iii) R, null, and intervals (-r, r) for r any positive rational. > Yes for the same reason as ii. show that an irriational interval can be obtained using union. iv) R, null, and intervals [-r, r] for r any positive rational. > Yes for the same reason as ii. consider the union of [-r, r] for r in (0, 1) and rational. v) R, null, and intervals (-r, r) for r any positive irrational. > Yes for the same reason as ii. let r_n be positive, irrational and r_n -> 1 from below. what is the union of (-r_n, r_n)? vi) R, null, and intervals [-r, r] for r any positive irrational. > Yes for the same reason as ii. construct an open interval using union.. vii) R, null, and intervals [-r, r) for r any positive real. > Yes for the same reason as ii. construct an open interval using the union viii) R, null, and intervals (-r, r] for r any positive real. > Yes for the same reason as ii. same as above ix) R, null, and all intervals [-r, r] and (-r, r) for r any positive > real. > Yes because of iv and ii, and intersections of [-r, r] and (-r, r) is > (-r, r), while unions are [-r, r]. i think this is a topology, but you need to show more here. x) R, null, and every interval [-n, n] and (-r, r) for n any positive > integer and r any positive real. > Yes because of an argument similar to ix a little more work here as well. Now on the other hand, as I think about it more, it seems that ii-x all > have the property that the infinite intersection of all their subsets > is just 0 .. but that shouldn't matter since we only consider finite > intersections. I must clearly be getting something very wrong since only 3 of them are > topologies but I find that 9 of them are! i've got 3. my advice is to be careful, try out as many different constructions as you can using unions and finite intersections. if you see a restrition like rational endpoints see if you can use the allowed set operations to get an interval with irrational endpoints. === Subject: Re: Basic Topology Questions JdyOWwwAAAC8AhuHITILJrgANkHxqjXJ Don't forget that arbitrary (including infinite) unions of open sets are open. Using this you can get (-r, r) where r is a positive irrational number to be the union of intervals of the type (-q, q) for q rational. This shows iii is not a topology. The same approach will work for many of the rest. === Subject: Re: Do the critical values of linear correlation depend on sample size? 7EEacAwAAACAxOPs71C36KOSRcb3o0fT > I have been having an argument about this. > For this question, we're assuming linear regression is run on n data > points. We get a correlation of R^2, where R is the slope of the best > fit line with normalized (mean of 0, standard devitation of 1) > coordinates. Just say your sample correlation between X and Y is R. You want to test the hypothesis Ho: rho = 0. and you want to know > if you test it at some alpha level, how large must R be before Ho > can be rejected at the slpha level for a two-tailed test. The answer depends on the sample size n and alpha. The critical values of |R| is given in the table in the link you gave: > http://www.gifted.uconn.edu/siegle/research/Correlation/corrchrt.htm For example, at alpha = 0.05 and various df (n-2) your critical values > are df 20 40 60 80 100 |R|> .423 .304 .250 .217 .195 e.g., if n=102, you reject if |R| is greater than .195 > and if n= 20, you need a corr coefficient of |R|>.423 before you can > rej. This is WHY. If (X,Y) comes from a bivariate normal, then under the > null hyp. of rho = 0, the statistic R* sqrt((n-2)/(1-R*R)) has a > T distribution with (n-2) df. Thus, the null hypothesis is rejected if |R|* sqrt((n-2)/(1 - R*R)) > t(1-alpha/2;(n-2)). or equivalently, if |R| > t /sqrt((n-2) + t*t)) The right-hand side expression of the above, for various combinations > of alpha and d.f. (= n-2) are given in the web link. Exercise: verify any of the values in the table by the formula above > --------------------------------------------------------------------- Having given you the solution, I should caution that George Box said > in his (1978 JASA) paper it is better to get an approximate answer > to the right question than an exact answer to the wrong question. In the case of a regression, testing for the significance of R is > ALWAYS the WRONG QUESTION. John Tukey said something to the effect > that using R is sweep the data under the rug with a vengeance. A statistically significant R may be utterly useless if the sample > size is large, and the relation between X and Y may look like a > shot gun blast. For n=10,000 it takes only a correlation of .02 > for it to be statistically significant. Use prediction intervals and other means to look for the PRACTICAL > significance of any regression result. A final note is an easy and useless result for large samples. > If the sample size is large, the APPROXIMATE critical value for |R| > is Z/sqrt(n), because T --> Z, and (n-2) + Z*Z ---> n (approx). For example, for Z = 1.96 and n = 10,000, the critical value > is approx. 1.96/100 = 0.0196 or 0.02. I found two fairly obvious typos. The exact critical values is 1.9602/(9998 + 1.9602**2) = 0.01960. The denominator should be sqrt(9998 + 1.9602^2). If you're good at doing square roots of large numbers in your head > you can estimate the critical value of R for large samples to be > 2/n for alpha .05, and win a bar bet or impress some friends. 2/n should be 2/sqrt(n). -- Bob. > If you think that the 95% confidence critical value is always an R^2 of > 95%, you should choose the NO option. > If you think that the critical value of the R^2 correlation depends on > how many points n you have, like in this chart, then you should choose > the YES option. The critical values are listed in terms of |R|, BTW. Of > course |R| = sqrt(R^2). > http://www.gifted.uconn.edu/siegle/research/Correlation/corrchrt.htm > Choose YES or NO or INVALID QUESTION === Subject: Newton's quantity of matter WntVwQ0AAADl64T9c7l_8rwegqF5R8RR According to Newton, the quantity of matter in a body is the product of its bulk and density; either its weight-density, or mass-density: So that the quantity of matter in a cubic foot of water will be 62.4 lbf per cubic foot, and the quantity of matter in a cubic decimeter of water will be 1 kg per cubic decimeter. Don === Subject: Re: Newton's quantity of matter rCfV9AwAAAB6GOp4HP3aWrPYi4NqdB_x > According to Newton, the quantity of matter in a body is the product of > its bulk and density; either its weight-density, or mass-density: So > that the quantity of matter in a cubic foot of water will be 62.4 lbf > per cubic foot, and the quantity of matter in a cubic decimeter of > water will be 1 kg per cubic decimeter. Don But that is not the standard by which you define the mass of a body. It is only one of several (as you have observed) ways that we can conveniently measure that mass. Others include using the body's weight and the local acceleration of gravity, using an applied net force and the resultant acceleration, using the body in an oscillator and measuring the frequency, using measured kinetic energy and its velocity, using measured momentum and velocity, measuring total energy and momentum in an inertial frame, and so on. PD === Subject: Newton's quantity of matter WntVwQ0AAADl64T9c7l_8rwegqF5R8RR According to Newton, the quantity of matter in a body is the product of its bulk and density; either its weight-density, or mass-density: So that the quantity of matter in a cubic foot of water will be 62.4 lbf per cubic foot, and the quantity of matter in a cubic decimeter of water will be 1 kg per cubic decimeter. Don === Subject: Re: factorization in Z_3 LPDiaQ0AAACsJLMUu95_O2ZUX0DZOvYQ Er... But it is a homework problem. We should not use Mathematica. === Subject: Re: Dubya can play chess? firewolf@juliewaters.com says... > I have no idea whether Dean would have done better or worse than Kerry, > but how in the world is Dean a nut case? He's so crazy that he thinks that it's a good idea for a politician to > tell the truth sometimes. What a whacko! Well he *says* he thinks that. But isn't this the guy who had his gubernatorial records sealed, then offered to release them if Bush did, was backfooted when the Bush campaign team pointed out that Bush's had already been released, and resigned from the contest shortly afterwards? - Gerry Quinn === Subject: Re: Russell's Paradox > While I was reading about Russell's paradox I concluded I do not > understand the basic terms of set theory. Russell's paradox concerns whether a set A is a member of itself, i.e., A > in A. Now my confusion comes from the idea that a set A has elements (members), > and those elements may be individual sets, but how can the set A ever be > an element of itself? > Some informal examples may help. Consider the set of all things that are not cats. Intuitively, it is NOT itself a cat, and should be an element of itself. The set of cats, however, is NOT a cat and is therefore NOT an element of itself. More abstractly, consider the possible existence of the set of all sets. Or the set of everything. Intuitively, it seems that both should be elements of themselves. Dan Download my DC Proof software at http://www.dcproof.com === Subject: Re: Mathematical cartoons (related to simplification) > I agree. I will work on this system to make it more >> sophisticated. Unfortunately, I am a klutz when it comes to nice GUI, >> so, any thoughts on usability will be appreciated. My suggestion would be to create a smallish JavaScript and put a button > beside the start image. If butten pressed, show next step. This way you > can think about the problem before you see the answer. Who knows. Maybe > the one looking at it will actually learn something, too. ;-) JS code can be found scattered all over the 'net, so you don't have to > do it yourself. Perhaps the animation should be a little smoother, like fading from one > img to the next. Otherwise, nice idea! :-) javascript can control display of a GIF cartoon, I will definitely add such buttons. If not, I can make regular links. i === Subject: Re: Mathematical cartoons (related to simplification) Sny74g0AAADy66iGh6ZMdSlIFta_KAXh > math formula as an animation/cartoon. It is targeted towards children. Example is at > http://www.algebra.com/services/rendering/regression.mpl > The idea is that it will be used for my free tutoring system. Any > tutor explaining a math problem or writing a lesson, could easily > write a simple expression that would draw a math cartoon. It is already > working, actually. > As I said, it already works. Some feedback will be appreciated. This will, eventually, be a part of the simplifier that I am writing, > but right now it will be used by humans solving problems manually and > wanting their solution to be fun and good looking. I think it is pretty neat. More interaction would be good though: since people read pages at different speed, who knows what that animated GIF is showing when they finally reach it. Also each display may not last the right amount of time... it runs so fast that many people, particularly the beginners such an application would target might not get to see what is going on before the display changes. I agree.. It'd be nice to let the user control when it starts and how fast it goes. It might also be useful to you to have your system be able to draw a red arrow to the box and supply a short english sentence in a bubble describing which principles are going to be applied in this step to the boxed-in text.. or how multiple elements will be combined, etc. -Mysid === Subject: Re: Mathematical cartoons (related to simplification) >> math formula as an animation/cartoon. It is targeted towards children. >> Example is at >> http://www.algebra.com/services/rendering/regression.mpl > The idea is that it will be used for my free tutoring system. Any >> tutor explaining a math problem or writing a lesson, could easily >> write a simple expression that would draw a math cartoon. It is already >> working, actually. >> As I said, it already works. Some feedback will be appreciated. >> This will, eventually, be a part of the simplifier that I am writing, >> but right now it will be used by humans solving problems manually and >> wanting their solution to be fun and good looking. I think it is pretty neat. More interaction would be good though: > since people read pages at different speed, who knows what that > animated GIF is showing when they finally reach it. The speed is actually settable and is set by me, right now it is 2 seconds between views, and 5 seconds after the cartoon ends. > Also each display may not last the right amount of time... > it runs so fast that many people, particularly the beginners > such an application would target might not get to see what is > going on before the display changes. I agree.. It'd be nice to let the user control when it starts > and how fast it goes. What I want to do, and it should not be difficult, is to present the cartoon in several ways. One that you have seen, one more would be to just dump toe expressions consecutively, an yet one more would be a presentation where the user has to click next. It's not hard to do given the engine that I have. It might also be useful to you to have your system be able to > draw a red arrow to the box and supply a short english sentence in > a bubble describing which principles are going to be applied > in this step to the boxed-in text.. or how multiple elements will be > combined, etc. Yep. It is a good idea, however I have to be careful there as it would require rewriting of my parser. I do want to do it though, someday. i === Subject: Re: mathematician salaries >Mark has no use for so many memory registers. Mark cannot keep track >of so many layers of numbers in his head. Who goes around doing such >complex calculations while walking around with a calculator except >nerds? Most normal people sit down after dinner with a cup of tea in >front of a computer. Then, after exchanging a few friendly emails and >googling, they call up the Excel and crank up the colorful pie charts >they need for tomorrow's presentation. So what. Who cares how many registers *you* need in your work. You are not the arbiter of the type of mathematics or calculations which other people should use. That is not your job, and it is an extreme conceit of yours that you think that you have the right to dictate to others what calculators or mathematics they should, or should not, have or use. The type of calculations that you use is not indicative of the type of calculations that everybody uses in their professions. Other people use more complicated mathematics in their daily work lives. For instance, my father, who was a Civil Engineer, was the person who bought the HP67 calculator with the Reverse Polish Notation and the 26 memory registers to use in his work, as well as using the statistical packages which came with the HP67 calculator. Without decent calculations, how safe would the structures and foundations of buildings and bridge be? Think about it. Life would be far more dangerous if the calcultions, which were used for the work, was not up to the high standard of calculations required, as any buildings or bridges etc., which were built on the basis of inferior calculation, could be much more prone to collapse. Navigators in the Navy use much more complicated mathematics than you ever would, and they would have benefited much more from a calculator with Reverse Polish Notation than calculators with a = button. On the sort of mathematics used by professionals in general, if nobody used the sort of mathematics that you appear to have little time for, then we would be far less technologically advanced (maybe back to the level of the nineteenth century, or even earlier), and you might have had to have found yourself a more constructive job than marketing. >If the old calcultors are so superior, why did they die off? Not even >the Russians or Chinese try to copy them, and they like math as much as >anyone. This presupposes that good products survive and bad products get discontinued. This supposition is so much against personal experience that one has to wonder at the intelligence of anybody who makes such a supposition in the first place. Quality has degenerated in so many things in today's world, and not just in calculators. If your assumption that superior products survive were true, then the profiferation of McDonalds around the world would never have happened. In many places, McDonalds is actually replacing far superior restaurants and cafes, and the general public often get no say in the matter. Furniture used to be made to last for years (even centuries). Now, furniture is made with a built-in obsolescence so that you need to get more in a few years time. The same thing is true of electronic equipment, such as televisions and videorecorders. So your question: If the old calcultors are so superior, why did they die off is a complete non sequitir, since it is based on a false assumption. >IPOD, now, that's progress! That is only in the opinion of somebody whose tastes run in that particular direction. ----- === Subject: Re: mathematician salaries Well, Mark likes to answer questions that indirectly? This is not relevant to my question. In fact, I just wanted to raise a pair of seemingly contradictory statements made by Mark in a short time and see how Mark will answer it (Bear me to follow Mark's way of speaking): Mark does equity valuation but doesn't use calculator!! Perhaps Mark could do all pricing in mind -- hard to believe. > Are you working for an i bank? Mark Demers works for EquityValue Investments. You can look the > EquityValue Google group for more information. EquityValue Investments > is a private equity firm. First, this means that Mark's firm is > private. Second, don't confuse Mark's firm with those big ugly > infamous i banks, who just take your money, churn your portfolio, and > eat up a 1% management fee every year. Mark actual CREATES wealth for his clients the old fashion way --- by > finding good investments for them, and buying and HOLDING while the > eggs hatch. > === Subject: Re: mathematician salaries BhnVAA0AAADpcidP-5AvYEnIArz5kqjP > Mark does equity valuation but doesn't use calculator!! > Perhaps Mark could do all pricing in mind -- hard to believe. Has Aldar C-F Chan ever heard of a program called Excel? Mark bets he has not. === Subject: Re: mathematician salaries <59rla1pboopn89srfel6et39jcsjd7ib63@4ax.com> BhnVAA0AAADpcidP-5AvYEnIArz5kqjP The rank of the matrix is the number of independent vectors formed by its rows or columns. The rank of the columns equals the rank of the rows. The rank must be less than or equal to the number of rows and the number of columns. Mark knows that very well. Check this out: Suppose you're given 2 known vectors, v and z, both in Rn. Let 1 be the vector in Rn with all elements 1. Suppose v'1 = 1 and v'z = 0. How do you construct v? Just another one of the many portoflio optimization puzzles Marky enjoy Wyoming or Manhattan while he peruses the financial pages. Marky Mark === Subject: Re: mathematician salaries BhnVAA0AAADpcidP-5AvYEnIArz5kqjP > This shows how stupid you really are! You paid $300 for a calculator!!! This shows Mark how stupid YOU are. You actually pay for your calculators??? Marky's office manager buys all the calculators around here. Mark just finds them magically appearing in his desk supply drawer. === Subject: Re: mathematician salaries <59rla1pboopn89srfel6et39jcsjd7ib63@4ax.com> BhnVAA0AAADpcidP-5AvYEnIArz5kqjP The rank of the matrix is the number of independent vectors formed by its rows or columns. The rank of the columns equals the rank of the rows. The rank must be less than or equal to the number of rows and the number of columns. Mark knows that very well. Check this out: Suppose you're given 2 known vectors, v and z, both in Rn. Let 1 be the vector in Rn with all elements 1. Suppose v'1 = 1 and v'z = 0. How do you construct v? Just another one of the many portoflio optimization puzzles Marky enjoy Wyoming or Manhattan while he peruses the financial pages. Cheerio and toodle looo! === Subject: Re: mathematician salaries <59rla1pboopn89srfel6et39jcsjd7ib63@4ax.com> NUqquQ0AAACZwMApkWIjE--Er3vJbAtG Check this out: Suppose you're given 2 known vectors, v and z, both in > Rn. Let 1 be the > vector in Rn with all elements 1. Suppose v'1 = 1 and v'z = 0. How do > you construct v? v is given, so I wouldn't bother constructing it. Mike === Subject: Re: mathematician salaries BhnVAA0AAADpcidP-5AvYEnIArz5kqjP Are you working for an i bank? Mark Demers works for EquityValue Investments. You can look the EquityValue Google group for more information. EquityValue Investments is a private equity firm. First, this means that Mark's firm is private. Second, don't confuse Mark's firm with those big ugly infamous i banks, who just take your money, churn your portfolio, and eat up a 1% management fee every year. Mark actual CREATES wealth for his clients the old fashion way --- by finding good investments for them, and buying and HOLDING while the eggs hatch. === Subject: Re: Cantor and the binary tree >> Is here a bijection possible between uncountable sets? >> Sometimes. >> There exist bijections between R and C, for instance. But >> not between R and P(R). >> Can you enumerate R <--> R? >> I can construct a bijection from R to R, if that is what >> you are asking. >> Here's one: f(x) = x. And I can construct a bijection from nodes to paths: 0. > / > 0 1 > / / > 0 1 0 1 > / / / / > ................... Everywhere a path branches off, it gets the node. The other one keeps >that node which already was mapped on it before. I need not knwow which >path gets which node, because all are of similar value. And so on in >infinity. That is not a bijection. If you compare, '000...' with '010...', you would map '000...' to '0' and '010...' to '01'. If you compare '000...' with '011...', you would map '000...' to '0' and '011...' to '01'. QED. Martin === Subject: Re: Now how did we end up with this genius for President? <42AA5C26.74766C81@Hovnanian.com> fp02fQwAAAAuxt4f6SfRn5n70zto2rPY This is a math forum. Because you insist on keeping this totally worthless and off topic thread open, I'm going to do a little math problem. Suppose Bush started drinking when he was 18 and quit when he was 40. If he only got drunk twice per week on the weekends, he spent 2*52*22=2288 days of his life drunk. Now 2288/365=6.26849315068493150684931506849315. Ergo, your man, Bush, may have spent approximately 6 years and 3 months of his life drunk. BTW, I'm glad he quit drinking. I suggest that this be the last post on this subject in sci.math. === Subject: Re: Now how did we end up with this genius for President? >As opposed to Dan Rather, Newsweek, NYT etc? Yup! >jt Dan Rather reported, correctly, that your vicious gay little fuhrer Bush deserted from his military obligation. Newsweek reported, correctly, that desecration of the Koran was indeed among the many vile insults Bush has caused to be heaped among the innocents in his concentration camps. He doesn't have the videos of that sent to him, though: he's into the ones with the child rapes and tortures. The NYT has had the integrity to admit and make reparations for its mistakes, which is more than your corrupt little god Bush will ever learn enough ethics to do. === Subject: Re: Now how did we end up with this genius for President? >...work >for the enemy. Your gods the Bush family have been doing that for generations. === Subject: Re: Now how did we end up with this genius for President? <04q5a1952fcnhsk6tqagutvibl01pkqrht@4ax.com> <42a45254$10$fuzhry+tra$mr2ice@news.patriot.net> V-lr_w0AAACtoSR1HsJcjzqp_3sT1WfE Gore is a robot. And clinton in the future of history will be seen for what he is: a socialist endeavering to be CEO of the United Nations. What he did will come out. === Subject: Re: An new and very interesting intellectual game! <42a16b48$2@clarion.carno.net.au> Z1jBAg0AAACsf5GcKoo6NNfQcrFfV7aO >> --------------------- >> The third problem: > During the excavations in the Andes, archeologists discovered an >> ancient (Incan?) wall with many drawings on it. At first the scientists >> couldn't understand why there would be so many pictures that are very >> similar to each other. But later they saw a llama and came up with an >> explanation. In one minute repeat their discovery. >> Speculation: A stable for llamas. > No. Think about animation. A succession of frames of a walking llama. :-) I got it in less than a minute, but I got to the thread three days late. > )-; > Close but no cigar. The pictures were indeed frames. But frames to be watched by people riding a llama! Sort of an ancient drive-in movie theater to be watched from the privacy of your own llama. BTW, the pictures were of religious sacrifice rituals. === Subject: Re: Math posers. Re: An new and very interesting intellectual game! > OK. Two more questions. Not from What?Where?When? but from my > childhood. Still good. 1. The president is looking out of the window and sees the president's > father's son play chess against president's son's father. What's going > on? 2. A monk was accused of breaking celibacy and spending his nights > visiting local women. A commissioner came to the monk's room to > investigate the accusations. The monk denied ever leaving his room at > night. The commissioner left. Three days later the Commission received > a letter from the monk accusing the commissioner of stealing his > candlestick. Three days alter another such letter arrived. The > Commission threw the monk out of the monastery (defrocked him?). Why > would they be so petty? > 2) The comissioner told them that if the monk slept in his own bed, he'd have found the candlestick by now. -- ------------------------ Mark Jeffrey Tilford tilford@ugcs.caltech.edu === Subject: Re: Math posers. Re: An new and very interesting intellectual game! Z1jBAg0AAACsf5GcKoo6NNfQcrFfV7aO Nice try. But too late. > OK. Two more questions. Not from What?Where?When? but from my > childhood. Still good. > 1. The president is looking out of the window and sees the president's > father's son play chess against president's son's father. What's going > on? > 2. A monk was accused of breaking celibacy and spending his nights > visiting local women. A commissioner came to the monk's room to > investigate the accusations. The monk denied ever leaving his room at > night. The commissioner left. Three days later the Commission received > a letter from the monk accusing the commissioner of stealing his > candlestick. Three days alter another such letter arrived. The > Commission threw the monk out of the monastery (defrocked him?). Why > would they be so petty? > > 2) The comissioner told them that if the monk slept in his own bed, he'd > have found the candlestick by now. > -- > ------------------------ > Mark Jeffrey Tilford > tilford@ugcs.caltech.edu === Subject: Re: Math posers. Re: An new and very interesting intellectual game! Z1jBAg0AAACsf5GcKoo6NNfQcrFfV7aO Answers: > OK. Two more questions. Not from What?Where?When? but from my > childhood. Still good. 1. The president is looking out of the window and sees the president's > father's son play chess against president's son's father. What's going > on? > Putitng aside ideas about triple windows/mirrors and various forms of incest, the natural and simplest answer is: The president sees her brother play chess with her husband. A couple of people got it. 2. A monk was accused of breaking celibacy and spending his nights > visiting local women. A commissioner came to the monk's room to > investigate the accusations. The monk denied ever leaving his room at > night. The commissioner left. Three days later the Commission received > a letter from the monk accusing the commissioner of stealing his > candlestick. Three days alter another such letter arrived. The > Commission threw the monk out of the monastery (defrocked him?). Why > would they be so petty? > The comissioner had hidden the candlestick in the monk's bed. Since the monk kept on complaining about the loss for many days, this is proof that he never slept in his own bed. Nobody got it. New question: In the good old times, young Austrian military officers were taught good dining table manners using two books. Name these books. === Subject: Re: Math posers. Re: An new and very interesting intellectual game! >> 2. A monk was accused of breaking celibacy and spending his nights >> visiting local women. A commissioner came to the monk's room to >> investigate the accusations. The monk denied ever leaving his room at >> night. The commissioner left. Three days later the Commission received a >> letter from the monk accusing the commissioner of stealing his >> candlestick. Three days alter another such letter arrived. The >> Commission threw the monk out of the monastery (defrocked him?). Why >> would they be so petty? > The comissioner had hidden the candlestick in the monk's bed. Since the > monk kept on complaining about the loss for many days, this is proof that > he never slept in his own bed. Nobody got it. And what am I, chopped liver? Heard this one (or its like) before: Fubegyl orsber gur svefg yrggre jnf frag, gur pbzzvffvbare fgbyr gur zbax'f pnaqyrfgvpx. Fubegyl nsgre gur svefg yrggre jnf frag, gur pbzzvffvbare cynprq gur pnaqyrfgvpx ba gur orq va gur zbax'f ebbz. ObVariant: (uhfonaq pbzrf ubzr sebz svfuvat gevc) Uhfonaq: Ubarl, jul qvqa'g lbh cnpx zl wnpxrg yvxr V nfxrq? Jvsr: V chg vg va lbhe gnpxyr obk. === Subject: Re: Math posers. Re: An new and very interesting intellectual game! Z1jBAg0AAACsf5GcKoo6NNfQcrFfV7aO >> 2. A monk was accused of breaking celibacy and spending his nights >> visiting local women. A commissioner came to the monk's room to >> investigate the accusations. The monk denied ever leaving his room at >> night. The commissioner left. Three days later the Commission received a >> letter from the monk accusing the commissioner of stealing his >> candlestick. Three days alter another such letter arrived. The >> Commission threw the monk out of the monastery (defrocked him?). Why >> would they be so petty? > The comissioner had hidden the candlestick in the monk's bed. Since the > monk kept on complaining about the loss for many days, this is proof that > he never slept in his own bed. > Nobody got it. And what am I, chopped liver? > Heard this one (or its like) before: Fubegyl orsber gur svefg yrggre jnf frag, gur pbzzvffvbare fgbyr gur > zbax'f pnaqyrfgvpx. Fubegyl nsgre gur svefg yrggre jnf frag, gur > pbzzvffvbare cynprq gur pnaqyrfgvpx ba gur orq va gur zbax'f ebbz. ObVariant: (uhfonaq pbzrf ubzr sebz svfuvat gevc) > Uhfonaq: Ubarl, jul qvqa'g lbh cnpx zl wnpxrg yvxr V nfxrq? > Jvsr: V chg vg va lbhe gnpxyr obk. > Da, no gde kandeliabra?