mm-4509 === Subject: Re: The sphere and hyperbolic geometry > Categorical means that it is uniquely determined by a set of axioms. So a theorem proved in one model is also a theorem in any other model. Ah, there's that word 'model' again! Time to resort to mathworld again. Is this valid terminology: all the models constitute a category? === Subject: Re: The sphere and hyperbolic geometry > Euclidean geometry, hyperbolic and elliptic geometry are categorical which means there is only one of them---except there can be several diferent representations or models. But in any of these geometries 2 points determine a line, so this also holds in any model you come up with. Maybe your geometry satisfies only some of the axioms needed to categorize it. > izzy >> 1. by categorical, do you mean for example that the surface of a 3- >> dim. sphere can have only one of the types of >> geometry defined on it, but this type may have several different >> models? I have no reason to argue against this. >> 2. My geometry - the surface of the sphere, all points and all circles >> - may indeed satisfy only some of the >> axioms listed for Euclidean geometry, and claimed (by Mathworld at >> least) to hold - the first four, specifically >> are at issue. And 2 points determining a line is definitely NOT >> satisfied by my model - in fact two points determine >> an infinite number of lines (circles) in my model. So you may be right > The model I've heard about for elliptic geometry (which gives > a 2-dimensional elliptic plane) starts with the surface > of a sphere and identifies antipodal points, those that > lie on a line through the center of the sphere. > I'm not sure what you mean by 'identifies' - certainly I can see that every point on the Northern hemisphere has an antipodal point - how do they figure in the geometry you're creating? We have the point in the Northern hemisphere - what does your geometry 'do' with the antipode? That way, the Northern hemisphere plus the equator > has a copy of every point. Also, from the North Pole, > the farthest points are all on the equator, and at > the same distance. > The latter sentence implies I think your antipodal geometry is essentially the Northern hemisphere. Two points on the equator which are antipodal are > identified. So from the North Pole to 0 lat., > 0 longitude, there seem to be two lines: - follow a meridian from North Pole to (0,0). > - follow a meridian from North Pole to 180 longitude, > zero latitude. > Yup, I can see both of those. > What are the geometry's lines? There's the rub! [...] Identifying means to consider to be the same. One way is to > consider all the diameters of the sphere: all the closed line segments > of length 2r that pass through the center of the sphere and with > end-points on the sphere, where r is the radius of the sphere. So the segment from North Pole to South Pole represents one > point. If two points are antipodal, either they are both on > the equator, or one is North and the other South of the > equator. So when we identify antipodal points, it's > enough to look only at the Northern hemisphere plus the > equator. If we have a diameter with one end North of > the equator, the other end is South of the equator. So on the closed hemisphere, points North of > the equator appear exactly once. For points > on the equator, the one at longitude 0 and > the one at longitude 180 are part of > a diameter, so they are identified. This happens for > all pairs of points on the equator separated by 180 > degrees. If we have two distinct diameters on the > sphere, then they lie in some plane. To go from > diameter A to diameter B, we take a path of > diameters in a non-obtuse sector -- one which subtends > an angle of 90 degrees or less. If the diameters are > separated by 10 degrees, or even 89 degrees, there is > just one shortest path of diameters. Unfortunately, > if the diameters are orthogonal, there are two paths > of diameters of length pi*r/2 . This contradicts > the axiom that two points determine a unique line segment > joining them (from euclidean geometry). We can use just the closed Northern hemisphere, but we need > to remember that to go from 10 degrees North and 180 longitude ( = A) > to 10 degrees North 0 longitude ( = B), the shortest route goes South > to the equator along longitude 180 until we are on the equator > at lat. = 0, long. = 180, and this is antipodal to > lat. = 0, long. = 0 (so they are the same point when we identify > antipodes). B is at 10 degrees North, 0 longitude. So next, > we go from lat.=0, long. = 0 to B. That means 20 degrees instead > of 160 degrees along a great circle which lies entirely in > the Northern hemisphere. > I'm confused about several things. First: I understand how you create a 'line', or a 'great circle' in this geometry, at least where the two points connected by the line are exactly opposite - they are on longitudes 180 degrees apart. But to create a line between two points which are NOT on lines of longitude 180 degrees apart, you have to go down to the equator from one point,move along the equator from the longitude of one to the longitude of the other, then zip across to the antipode, and up to the other point. Is this right? This collection of movements forms what the geometry calls a line? Second: It seems as tho you're saying that movement along the equator between two points is defined as taking the shortest distance, the distance where the degrees moved between the two points is < 90 deegrees, but when they're antipodes, there are two ways to go. But if the two points are at antipodes, then they're actually the same point, so the distance between them is 0. So there's NO line between them, they're the same point. That's as far as I can get so far. case. > I think this identification of antipodal points was in > Godel Escher Bach by Hofstadter. > I read that a long time ago and don't remember a single thing from it! > This is similar to the real projective space model here: This has a nice symmetry. What puzzles me is that > from what I understand, there are two segments from > the North Pole to lat. = 0 and long. = 0 ... > Sorry haven't had a chance to look at this yet. > David Bernier -- === Subject: Re: The sphere and hyperbolic geometry >> Euclidean geometry, hyperbolic and elliptic geometry are categorical which means there is only one of them---except there can be several diferent representations or models. But in any of these geometries 2 points determine a line, so this also holds in any model you come up with. Maybe your geometry satisfies only some of the axioms needed to categorize it. >> izzy > 1. by categorical, do you mean for example that the surface of a 3- > dim. sphere can have only one of the types of > geometry defined on it, but this type may have several different > models? I have no reason to argue against this. > 2. My geometry - the surface of the sphere, all points and all circles > - may indeed satisfy only some of the > axioms listed for Euclidean geometry, and claimed (by Mathworld at > least) to hold - the first four, specifically > are at issue. And 2 points determining a line is definitely NOT > satisfied by my model - in fact two points determine > an infinite number of lines (circles) in my model. So you may be right >> The model I've heard about for elliptic geometry (which gives >> a 2-dimensional elliptic plane) starts with the surface >> of a sphere and identifies antipodal points, those that >> lie on a line through the center of the sphere. > I'm not sure what you mean by 'identifies' - certainly I > can see that every point on the Northern hemisphere has an > antipodal point - how do they figure in the geometry > you're creating? We have the point in the Northern > hemisphere - what does your geometry 'do' with the > antipode? >> That way, the Northern hemisphere plus the equator >> has a copy of every point. Also, from the North Pole, >> the farthest points are all on the equator, and at >> the same distance. > The latter sentence implies I think your antipodal > geometry is essentially the Northern hemisphere. >> Two points on the equator which are antipodal are >> identified. So from the North Pole to 0 lat., >> 0 longitude, there seem to be two lines: >> - follow a meridian from North Pole to (0,0). >> - follow a meridian from North Pole to 180 longitude, >> zero latitude. > Yup, I can see both of those. > What are the geometry's lines? There's the rub! >> [...] > Identifying means to consider to be the same. One way is to >> consider all the diameters of the sphere: all the closed line segments >> of length 2r that pass through the center of the sphere and with >> end-points on the sphere, where r is the radius of the sphere. > So the segment from North Pole to South Pole represents one >> point. If two points are antipodal, either they are both on >> the equator, or one is North and the other South of the >> equator. So when we identify antipodal points, it's >> enough to look only at the Northern hemisphere plus the >> equator. If we have a diameter with one end North of >> the equator, the other end is South of the equator. > So on the closed hemisphere, points North of >> the equator appear exactly once. For points >> on the equator, the one at longitude 0 and >> the one at longitude 180 are part of >> a diameter, so they are identified. This happens for >> all pairs of points on the equator separated by 180 >> degrees. If we have two distinct diameters on the >> sphere, then they lie in some plane. To go from >> diameter A to diameter B, we take a path of >> diameters in a non-obtuse sector -- one which subtends >> an angle of 90 degrees or less. If the diameters are >> separated by 10 degrees, or even 89 degrees, there is >> just one shortest path of diameters. Unfortunately, >> if the diameters are orthogonal, there are two paths >> of diameters of length pi*r/2 . This contradicts >> the axiom that two points determine a unique line segment >> joining them (from euclidean geometry). > We can use just the closed Northern hemisphere, but we need >> to remember that to go from 10 degrees North and 180 longitude ( = A) >> to 10 degrees North 0 longitude ( = B), the shortest route goes South >> to the equator along longitude 180 until we are on the equator >> at lat. = 0, long. = 180, and this is antipodal to >> lat. = 0, long. = 0 (so they are the same point when we identify >> antipodes). B is at 10 degrees North, 0 longitude. So next, >> we go from lat.=0, long. = 0 to B. That means 20 degrees instead >> of 160 degrees along a great circle which lies entirely in >> the Northern hemisphere. I'm confused about several things. First: I understand how you create a 'line', or a 'great circle' in this > geometry, at least where the two points connected by > the line are exactly opposite - they are on longitudes > 180 degrees apart. But to create a line between two points which are NOT > on lines of longitude 180 degrees apart, you have > to go down to the equator from one point,move along the > equator from the longitude of one > to the longitude of the other, then zip across to > the antipode, and up to the other point. Is this > right? This collection of movements forms what > the geometry calls a line? Second: It seems as tho you're saying that movement along > the equator between two points is defined as > taking the shortest distance, the distance > where the degrees moved between the two points > is < 90 deegrees, but when they're > antipodes, there are two ways to go. But if the two points are at antipodes, > then they're actually the same point, > so the distance between them is 0. > So there's NO line between them, they're > the same point. That's as far as I can get so far. > case. > I think this identification of antipodal points was in >> Godel Escher Bach by Hofstadter. I read that a long time ago and don't > remember a single thing from it! > This is similar to the real projective space model here: > This has a nice symmetry. What puzzles me is that >> from what I understand, there are two segments from >> the North Pole to lat. = 0 and long. = 0 ... Sorry haven't had a chance to look at this yet. What surprises me is that they say: ``The two models represent different geometries about the hyperspherical model and the projective model, the one where antipodal points are identified or glued. Up to a scaling factor, there is just one hyperbolic plane. For the projective plane, the one where antipodal points are identified, cf.: < http://en.wikipedia.org/wiki/Real_projective_plane#Formal_construction > So, I was under the impression that there was just one kind of elliptic geometry, up to curvature constant... David Bernier -- === Subject: Re: The sphere and hyperbolic geometry <4772be00$0$28007$88260bb3@free.teranews.com> posting-account=Gg91ZwkAAAARyqtmT5dB5plbBnobCndP rv:1.8.1.11) Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Euclidean geometry, hyperbolic and elliptic geometry are categorical which means there is only one of them---except there can be several diferent representations or models. But in any of these geometries 2 points determine a line, so this also holds in any model you come up with. Maybe your geometry satisfies only some of the axioms needed to categorize it. > izzy >> 1. by categorical, do you mean for example that the surface of a 3- >> dim. sphere can have only one of the types of >> geometry defined on it, but this type may have several different >> models? I have no reason to argue against this. >> 2. My geometry - the surface of the sphere, all points and all circles >> - may indeed satisfy only some of the >> axioms listed for Euclidean geometry, and claimed (by Mathworld at >> least) to hold - the first four, specifically >> are at issue. And 2 points determining a line is definitely NOT >> satisfied by my model - in fact two points determine >> an infinite number of lines (circles) in my model. So you may be right > The model I've heard about for elliptic geometry (which gives > a 2-dimensional elliptic plane) starts with the surface > of a sphere and identifies antipodal points, those that > lie on a line through the center of the sphere. >> I'm not sure what you mean by 'identifies' - certainly I >> can see that every point on the Northern hemisphere has an >> antipodal point - how do they figure in the geometry >> you're creating? We have the point in the Northern >> hemisphere - what does your geometry 'do' with the >> antipode? > That way, the Northern hemisphere plus the equator > has a copy of every point. Also, from the North Pole, > the farthest points are all on the equator, and at > the same distance. >> The latter sentence implies I think your antipodal >> geometry is essentially the Northern hemisphere. > Two points on the equator which are antipodal are > identified. So from the North Pole to 0 lat., > 0 longitude, there seem to be two lines: > - follow a meridian from North Pole to (0,0). > - follow a meridian from North Pole to 180 longitude, > zero latitude. >> Yup, I can see both of those. >> What are the geometry's lines? There's the rub! > [...] Identifying means to consider to be the same. One way is to > consider all the diameters of the sphere: all the closed line segments > of length 2r that pass through the center of the sphere and with > end-points on the sphere, where r is the radius of the sphere. So the segment from North Pole to South Pole represents one > point. If two points are antipodal, either they are both on > the equator, or one is North and the other South of the > equator. So when we identify antipodal points, it's > enough to look only at the Northern hemisphere plus the > equator. If we have a diameter with one end North of > the equator, the other end is South of the equator. So on the closed hemisphere, points North of > the equator appear exactly once. For points > on the equator, the one at longitude 0 and > the one at longitude 180 are part of > a diameter, so they are identified. This happens for > all pairs of points on the equator separated by 180 > degrees. If we have two distinct diameters on the > sphere, then they lie in some plane. To go from > diameter A to diameter B, we take a path of > diameters in a non-obtuse sector -- one which subtends > an angle of 90 degrees or less. If the diameters are > separated by 10 degrees, or even 89 degrees, there is > just one shortest path of diameters. Unfortunately, > if the diameters are orthogonal, there are two paths > of diameters of length pi*r/2 . This contradicts > the axiom that two points determine a unique line segment > joining them (from euclidean geometry). We can use just the closed Northern hemisphere, but we need > to remember that to go from 10 degrees North and 180 longitude ( = A) > to 10 degrees North 0 longitude ( = B), the shortest route goes South > to the equator along longitude 180 until we are on the equator > at lat. = 0, long. = 180, and this is antipodal to > lat. = 0, long. = 0 (so they are the same point when we identify > antipodes). B is at 10 degrees North, 0 longitude. So next, > we go from lat.=0, long. = 0 to B. That means 20 degrees instead > of 160 degrees along a great circle which lies entirely in > the Northern hemisphere. > I'm confused about several things. > First: > I understand how you create a 'line', or a 'great circle' in this geometry, at least where the two points connected by the line are exactly opposite - they are on longitudes 180 degrees apart. > But to create a line between two points which are NOT on lines of longitude 180 degrees apart, you have to go down to the equator from one point,move along the equator from the longitude of one to the longitude of the other, then zip across to the antipode, and up to the other point. Is this right? This collection of movements forms what the geometry calls a line? > Second: > It seems as tho you're saying that movement along the equator between two points is defined as taking the shortest distance, the distance where the degrees moved between the two points is < 90 deegrees, but when they're antipodes, there are two ways to go. > But if the two points are at antipodes, then they're actually the same point, so the distance between them is 0. So there's NO line between them, they're the same point. > That's as far as I can get so far. case. I think this identification of antipodal points was in > Godel Escher Bach by Hofstadter. > I read that a long time ago and don't remember a single thing from it! This is similar to the real projective space model here: This has a nice symmetry. What puzzles me is that > from what I understand, there are two segments from > the North Pole to lat. = 0 and long. = 0 ... > Sorry haven't had a chance to look at this yet. What surprises me is that they say: > ``The two models represent different geometries > about the hyperspherical model and the projective model, > the one where antipodal points are identified or glued. Up to a scaling factor, there is just one hyperbolic plane. For the projective plane, the one where antipodal points > are identified, > cf.: > So, I was under the impression that there was just > one kind of elliptic geometry, up to curvature > constant... David Bernier -- Argle-bargle!* I have to get reading! Sorry for my dilatoriness. *You ever read 'Bored of the Rings' by the Harvard Lampoon? Came out in the late 60's? Hilarious if you're into that kind of collegiate humor. THere was one sentence where the text said something like 'a certain king' which was asterisked and the footnote said *either Argle-Bargle IV or someone else. Typical collegiate humor. I love it. === Subject: Re: Karzeddin's evolving conjectures ... (too much written matter and not possible to understand ? ) * * * Lets simplify my post to: > So the interesting will be, if for such > developed sentences: > t^n = a^n +b^n +2*n^u abtp > where natural n;t;a;b;p are of gcd =1 > and u>=2 You can find counterexamples ? * * * Ro-Bin Already for some n natural You could find quite a lot of counterexamples: I should remark again, that n is prime and n>5 === Subject: Re: Tetration again! Am 26.12.2007 20:54 schrieb mike3: > How actually the polynomials are built by hand, you may find at >> wikipedia or mathworld... if I recall right, in wikipedia the >> explanation was very elementar and instructive. What would I look up, though? For instance http://en.wikipedia.org/wiki/Polynomial_interpolation -- --- Gottfried Helms, Kassel === Subject: Re: Defining ln(D) o x^n ? On 26 d.8ec, 19:15, Stephen Montgomery-Smith Here ln: neperian logarithm , D or d/dx: derivative in x. >> We' ve got for operator I/D >> I/D o x^n = x^(n+1)/(n+1) (to a constant) >> Alain I have never seen log(D) used in the literature, but exp(tD) is very > common (it is the shift operator), as well as i sign(D) (it is the > Hilbert Transform). In general, to make sense of f(D), you need a quite sophisticated > background in mathematics. I think that the references to functional > calculus for sectorial operators is what you are needing. Stephen Bonjour Stephen, we may use expressions like : ln(D -I) o x^n and I/(I - D) o x^n this last one giving for x = 1 1 +n +n*(n -1)+n*(n-1)*(n-2)+...+n! So why not ln(D) Amiti.8e,Alain Generally if one is going to define f(T) for an operator T, and a > holomorphic function f, you usually need that f is holomorphic in a > neighborhood of the spectrum of T. Now D has spectrum equal to the > imaginary axis (actually it is an unbounded operator which complicates > things even more), and the log function is not analytic at zero. This > is why log(I-D) makes sense, because the spectrum of I-D is 1+imaginary > axis, and log definitely can be defined in a neighborhood of this. Now this doesn't mean that defining log(D) is impossible (although I > confess that I also don't know if it is possible), but one does have to > go deeper into the functional calculus than usual. (By the way, I tend to use log for ln - hope that doesn't cause too much > confusion.) Stephen If you use the multiplication-operator form of the Spectral Theorem it's pretty straightforward. The Fourier transform makes it quite explicit, turning D into multiplication by ix (with domain those functions f(x) for which f(x) and ix f(x) are both in L^2), and log(D) (for some branch of log) into multiplication by log(ix) (with domain those functions f(x) for which f(x) and log(ix) f(x) are both in L^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: Defining ln(D) o x^n ? On 26 d.8ec, 19:15, Stephen Montgomery-Smith We' ve got for operator I/D > I/D o x^n = x^(n+1)/(n+1) (to a constant) > Alain > I have never seen log(D) used in the literature, but exp(tD) is very >> common (it is the shift operator), as well as i sign(D) (it is the >> Hilbert Transform). > In general, to make sense of f(D), you need a quite sophisticated >> background in mathematics. I think that the references to functional >> calculus for sectorial operators is what you are needing. > Stephen > Bonjour Stephen, we may use expressions like : > ln(D -I) o x^n and I/(I - D) o x^n > this last one giving for x = 1 > 1 +n +n*(n -1)+n*(n-1)*(n-2)+...+n! So why not ln(D) Amiti.8e,Alain >> Generally if one is going to define f(T) for an operator T, and a >> holomorphic function f, you usually need that f is holomorphic in a >> neighborhood of the spectrum of T. Now D has spectrum equal to the >> imaginary axis (actually it is an unbounded operator which complicates >> things even more), and the log function is not analytic at zero. This >> is why log(I-D) makes sense, because the spectrum of I-D is 1+imaginary >> axis, and log definitely can be defined in a neighborhood of this. > Now this doesn't mean that defining log(D) is impossible (although I >> confess that I also don't know if it is possible), but one does have to >> go deeper into the functional calculus than usual. > (By the way, I tend to use log for ln - hope that doesn't cause too much >> confusion.) > Stephen If you use the multiplication-operator form of the Spectral Theorem it's > pretty straightforward. The Fourier transform makes it quite explicit, > turning D into multiplication by ix (with domain those functions f(x) > for which f(x) and ix f(x) are both in L^2), and log(D) (for some branch > of log) into multiplication by log(ix) (with domain those functions f(x) > for which f(x) and log(ix) f(x) are both in L^2). Yes, your explanation is simpler than mine! === Subject: Re: Defining ln(D) o x^n ? > On 26 d.8ec, 19:15, Stephen Montgomery-Smith Here ln: neperian logarithm , D or d/dx: derivative in x. >> We' ve got for operator I/D >> I/D o x^n = x^(n+1)/(n+1) (to a constant) >> Alain I have never seen log(D) used in the literature, but exp(tD) is very > common (it is the shift operator), as well as i sign(D) (it is the > Hilbert Transform). In general, to make sense of f(D), you need a quite sophisticated > background in mathematics. I think that the references to functional > calculus for sectorial operators is what you are needing. Stephen >> Bonjour Stephen, > we may use expressions like : >> ln(D -I) o x^n and I/(I - D) o x^n >> this last one giving for x = 1 >> 1 +n +n*(n -1)+n*(n-1)*(n-2)+...+n! > So why not ln(D) > Amiti.8e,Alain > Generally if one is going to define f(T) for an operator T, and a > holomorphic function f, you usually need that f is holomorphic in a > neighborhood of the spectrum of T. Now D has spectrum equal to the > imaginary axis (actually it is an unbounded operator which > complicates things even more), and the log function is not analytic > at zero. This is why log(I-D) makes sense, because the spectrum of > I-D is 1+imaginary axis, and log definitely can be defined in a > neighborhood of this. Now this doesn't mean that defining log(D) is impossible (although I > confess that I also don't know if it is possible), but one does have > to go deeper into the functional calculus than usual. (By the way, I tend to use log for ln - hope that doesn't cause too > much confusion.) Stephen > If you use the multiplication-operator form of the Spectral Theorem it's >> pretty straightforward. The Fourier transform makes it quite explicit, >> turning D into multiplication by ix (with domain those functions f(x) >> for which f(x) and ix f(x) are both in L^2), and log(D) (for some branch >> of log) into multiplication by log(ix) (with domain those functions f(x) >> for which f(x) and log(ix) f(x) are both in L^2). Yes, your explanation is simpler than mine! And I think also provides something of an explanation to the original problem. So he wants to apply log(D) to x^n. Now the Fourier transform of x^n is something like the nth derivative of the dirac delta function. So you would multiply the dirac delta function by log(w), and then take its inverse Fourier transform. But this is differentiating the log function n times at the origin, and the singularity of log at 0 is bad enough that I just don't see how you are going to get any meaning out of this! === Subject: Re: Defining ln(D) o x^n ? <4bqdneFCMq7EPO_anZ2dnUVZ_sSlnZ2d@centurytel.net> posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) On 27 d.8ec, 02:31, Stephen Montgomery-Smith On 26 d.8ec, 19:15, Stephen Montgomery-Smith We' ve got for operator I/D > E E E E E E I/D Eo x^n = x^(n+1)/(n+1) E (to a constant) > Alain > I have never seen log(D) used in the literature, but exp(tD) is very >> common (it is the shift operator), as well as i sign(D) (it is the >> Hilbert Transform). > In general, to make sense of f(D), you need a quite sophisticated >> background in mathematics. EI think that the references to functional >> calculus for sectorial operators is what you are needing. > Stephen > Bonjour Stephen, we may use expressions like : > ln(D -I) o x^n Eand I/(I - D) o x^n > this last one giving for x = 1 > E E E E1 +n +n*(n -1)+n*(n-1)*(n-2)+...+n! So why not ln(D) Amiti.8e,Alain >> Generally if one is going to define f(T) for an operator T, and a >> holomorphic function f, you usually need that f is holomorphic in a >> neighborhood of the spectrum of T. ENow D has spectrum equal to the >> imaginary axis (actually it is an unbounded operator which >> complicates things even more), and the log function is not analytic >> at zero. EThis is why log(I-D) makes sense, because the spectrum of >> I-D is 1+imaginary axis, and log definitely can be defined in a >> neighborhood of this. > Now this doesn't mean that defining log(D) is impossible (although I >> confess that I also don't know if it is possible), but one does have >> to go deeper into the functional calculus than usual. > (By the way, I tend to use log for ln - hope that doesn't cause too >> much confusion.) > Stephen If you use the multiplication-operator form of the Spectral Theorem it's > pretty straightforward. EThe Fourier transform makes it quite explicit, > turning D into multiplication by ix (with domain those functions f(x) > for which f(x) and ix f(x) are both in L^2), and log(D) (for some branch > of log) into multiplication by log(ix) (with domain those functions f(x) > for which f(x) and log(ix) f(x) are both in L^2). > Yes, your explanation is simpler than mine! And I think also provides something of an explanation to the original > problem. So he wants to apply log(D) to x^n. ENow the Fourier transform of x^n is > something like the nth derivative of the dirac delta function. ESo you > would multiply the dirac delta function by log(w), and then take its > inverse Fourier transform. EBut this is differentiating the log function > n times at the origin, and the singularity of log at 0 is bad enough > that I just don't see how you are going to get any meaning out of this!- Masquer le texte des messages pr.8ec.8edents - - Afficher le texte des messages pr.8ec.8edents - Good morning, I am not sure we've got a way towards a useful workable solution, Alain === Subject: Vector Spaces as Manifolds I am reading John Lee's Riemannian Mflds. book, and in it, he refers to the standard manifold structure of real vector spaces(assumed finite-dim). I can think of giving charts to a f.d V.S V by using the isomorphism between V and IR^n , and using the coordinates, i.e: Choose a basis {v1,..,vn} for V . Then v in V has a rep. a_iv^i in this basis. Change of basis is linear, so coord. change is smooth. Then v has coords {a_i}, and a subset of V is open if its' coordinates form an open set in IR^n. Anyone know if this is the structure Lee is refering to? === Subject: Re: Vector Spaces as Manifolds > I am reading John Lee's Riemannian Mflds. book, > and in it, he refers to the standard manifold > structure of real vector spaces(assumed finite-dim). > I can think of giving charts to a f.d V.S V by using > the isomorphism between V and IR^n , and using the > coordinates, i.e: Choose a basis {v1,..,vn} for V . Then v in V > has a rep. a_iv^i in this basis. Change of basis > is linear, so coord. change is smooth. Then v has coords {a_i}, and a subset of V > is open if its' coordinates form an open set in > IR^n. Anyone know if this is the structure Lee is > refering to? Sort of, but is is more simple than that. Just fix some basis {v_1, ..., v_n} and consider the map from R^n onto V defined by (a_1,...,a_n) |-> a_1v_1 + ... a_nv_n. This is a chart. And that is all that you need; the set whose only element is this chart happens to be an atlas. There is no need to think about what would happen if you replace that basis by another one. Jose Carlos Santos === Subject: Re: Vector Spaces as Manifolds. Quick follow-up. > Anyone know if this is the structure Lee is refering to? Sort of, but is is more simple than that. Just fix > some basis > {v_1, ..., v_n} and consider the map from R^n onto V > defined by (a_1,...,a_n) |-> a_1v_1 + ... a_nv_n. This is a chart. And that is all that you need; the > set whose only > element is this chart happens to be an atlas. There > is no need to > think about what would happen if you replace that > basis by another > one. > Jose Carlos Santos A follow-up, please: Don't we need to check 2nd countability, Hausdorffness, etc. Does it follow immediately? === Subject: Re: Fermat's Last Theorem > It seems that there may be new developments from that > another planet math pages at the following links: http://www.mathpages.com/home/kmath367.htm > http://www.mathpages.com/home/kmath230.htm > I hope they would interest you Bassam Karzeddin > Message was edited by: bassam king karzeddin I used to visit these links shortly after Your post. These developments are very old indeed. I've then start to write, but some fault used to I'll do it again: Your and mine developments are common in some point to so called Abel-Formulas. These formulas were known much earlier then norwegian mathematician Abel used to sketch them, but with his name they come to mathematical C.V. (nowadays at least in russian sources) also for X;Y;Z not divided by exponent n: 1) X+Y =(z1)^n 2) (X^n +Y^n)/(X+Y) = (z2)^n 3) Z-X = (y1)^n 4) (Z^n -X^n)/(Z-X) = (y2)^n 5) Z-Y = (x1)^n 6) (Z^n -Y^n)/(Z-Y) = (x2)^n Now for one of X;Y;Z; divided by n (1;2) or (3;4) or (5;6) lines should be replaced by: I) for Z/n : 1) X+Y = n^(nu-1) (z1)^n 2) (X^n +Y^n)/(X+Y) = n(z2)^n II) for Y/n : 3) Z-X = n^(nu-1) (y1)^n 4) (Z^n -X^n)/(Z-X) = n(y2)^n III) for X/n : 5) Z-Y = n^(nu-1) (x1)^n 6)(Z^n -Y^n)/(Z-Y) = n(x2)^n Where u is some natural number. (further You'll see u>=2) Now once we'll combine it with input TAB: Z=T+A+B; X=T+B; Y=T+A we'll see interesting relations: A = Z-X = (y1)^n = a^n B = Z-Y = (x1)^n = b^n 2T+A+B = X+Y = (z1)^n = t^n but for: Y/n ; A = n^(nu-1) a^n X/n ; B = n^(nu-1) b^n Z/n ; 2T+A+B = n^(nu-1) t^n after executing input TAB: (T+A)^n +(T+B)^n = (T+A+B)^n T^n = nAB(2T+A+B)Ext where Ext = F[n-3](T,A,B) function of (n-3)degree of T;A;B; for n=3 Ext = 1 for n=5 Ext = 2T^2 +2(A+B)T + A^2 +AB +B^2 etc. and etc. more and more complicated (I have already from 1995 year combinatoric form of T^n = nAB F[n-2](T,A,B) once Ext = F[n-3](T,A,B) is much more complicated for to express in general combinatoric coefficients, but for the most important absolute therm Abst = F(A,B) it is possible ) This Ext(ension) for X;Y;Z not divided by n will be: Ext = n^(nu-1) p^n or once one of X;Y;Z divided by n: will be: Ext = p^n where p is some ideal natural number. and so on we can confirm: T = n^u abtp Existence of such number is the first step for to find counterexample or solution for FLT. After solving Ext=p^n or Ext= n^(nu-1) p^n resists only X+Y = t^n : also 2*n^u abtp + a^n +b^n = t^n ..............(1) or for Y/n 2*n^u abtp +n^(nu-1) a^n +b^n = t^n ...........(2) or X+Y = n^(nu-1) t^n for Z/n : so 2*n^u abtp a^n +b^n =n^(nu-1) t^n ..........(3) If such p was true solution for Ext so also we'll find solution for X+Y=t^n. Once already we now after prof. A.Wiles proof, that FLT is true so also I can be sure, that for such proper forms as eq. eq. (1),(2),(3) quasi or anybody else can not find any counterexample ! Nowadays I hope for to deliver some special retrieved configuration of possible factors for the 1-st case of FLT: eq.(1). See my new topic: Slightly different approach for 1-st case of FLT Ro-Bin P.S. Some like see things less complicated but there is no way for to escape to some flat FLT... Happy New Year 2008 !!! === Subject: A tale of human fortitude posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) I recently posted the following in another group (a group for teachers of high school calculus) and decided that it may be of interest here as well. I recently came across something that I thought would be of interest to this group. Before electronic computers (roughly, before the early 1940s), many people spent enormous amounts of time computing mathematical tables. Most texts at and below the level of calculus before the mid 1980s included abbreviated tables for student use (although computers were crunching these numbers from the mid 1940s onward, it wasn't until widespread use of calculators that these tables in math texts began to be unnecessary), and I imagine many (most?) of the people in this group have had some experience with them (probably trig. and logarithm tables) when growing up. Also, if you ever browsed through the library while you were in college, and again I suspect most of those in this group did, you probably came across a number of huge and old volumes that were quite a bit more extensive and advanced than the ones found at the back of textbooks. By the way, I highly recommend David Alan Grier's recent (2005) book When Computers Were Human if you want to read more about this part of mathematical history. The main point I want to make is that people literally spent years of 12+ hour days on these calculations. With this in mind, I offer what is one of the most compelling tales of perseverance in the face of disaster I've encountered in quite some time: T. H. Gronwall, Review of [...] by Keiichi Hayashi, Bulletin of the American Mathematical Society 32 (1926), 716. The work under review is far more elaborate and complete than any other tables of this kind. In the main part of Table I, x runs from 0.100 to 2.999 by intervals of 0.001, and the following columns are found on two opposite pages: x, phi(= x converted into degrees, minutes, and seconds), sin x, cos x, tan x, arcsin x, arccos x, arctan x (left page); x, e^x, e^(-x), sh x, ch x, th x, arg sh x, arg ch x (part of the table only), arg th x (right page). At the bottom of each page there is an additional table [...] The number of decimal places carried varies with the function tabulated from twelve for sin x and cos x to seven for arg sin x. [...] Table II gives sin(x*pi/2) and cos(x*pi/2) from x = 0.000 to 0.500 to ten places, and table II [...] Last paragraph of the review --> One cannot but admire the author's fortitude in carrying on his work in the face of disaster. As he tells us in the preface, the manuscript of over a thousand pages was completed toward the end of 1923, only to be destroyed in a fire that consumed the engineering building of the University of Kyushu. The author immediately began his work anew, and in another two years completed the manuscript of the present tables which will certainly remain the standard of their kind for many years to come. http://www.ams.org/journals/bull/1926-32-06/home.html http://tinyurl.com/349dhf [URL for .pdf file] Dave L. Renfro === Subject: Re: A tale of human fortitude posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 2.0.50727),gzip(gfe),gzip(gfe) > I recently posted the following in another group > (a group for teachers of high school calculus) > and decided that it may be of interest here as well. I recently came across something that I thought > would be of interest to this group. Before > electronic computers (roughly, before the early > 1940s), many people spent enormous amounts of > time computing mathematical tables. Most texts > at and below the level of calculus before the > mid 1980s included abbreviated tables for student > use (although computers were crunching these > numbers from the mid 1940s onward, it wasn't > until widespread use of calculators that these > tables in math texts began to be unnecessary), > and I imagine many (most?) of the people in this > group have had some experience with them (probably > trig. and logarithm tables) when growing up. > Also, if you ever browsed through the library > while you were in college, and again I suspect > most of those in this group did, you probably > came across a number of huge and old volumes > that were quite a bit more extensive and advanced > than the ones found at the back of textbooks. By the way, I highly recommend David Alan Grier's > recent (2005) book When Computers Were Human if > you want to read more about this part of mathematical > history. The main point I want to make is that people > literally spent years of 12+ hour days on > these calculations. With this in mind, I offer > what is one of the most compelling tales of > perseverance in the face of disaster I've > encountered in quite some time: T. H. Gronwall, Review of [...] by Keiichi Hayashi, > Bulletin of the American Mathematical Society > 32 (1926), 716. The work under review is far more elaborate and > complete than any other tables of this kind. In > the main part of Table I, x runs from 0.100 to > 2.999 by intervals of 0.001, and the following > columns are found on two opposite pages: x, > phi(= x converted into degrees, minutes, and > seconds), sin x, cos x, tan x, arcsin x, arccos x, > arctan x (left page); x, e^x, e^(-x), sh x, ch x, > th x, arg sh x, arg ch x (part of the table only), > arg th x (right page). At the bottom of each page > there is an additional table [...] The number of > decimal places carried varies with the function > tabulated from twelve for sin x and cos x to seven > for arg sin x. [...] Table II gives sin(x*pi/2) > and cos(x*pi/2) from x = 0.000 to 0.500 to ten > places, and table II [...] Last paragraph of the review -- > One cannot but admire the author's fortitude in > carrying on his work in the face of disaster. As > he tells us in the preface, the manuscript of over > a thousand pages was completed toward the end of > 1923, only to be destroyed in a fire that consumed > the engineering building of the University of Kyushu. > The author immediately began his work anew, and in > another two years completed the manuscript of the > present tables which will certainly remain the > standard of their kind for many years to come. http://www.ams.org/journals/bull/1926-32-06/home.htmlhttp://tinyurl.com/349d h f[URL for .pdf file] Dave L. Renfro Jesus Christ, didn't he make Xerox backup copies? ;-) === Subject: Re: A tale of human fortitude posting-account=iBgNeAoAAADRhzuSC4Ai7MUeMmxtwlM7 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) On Dec 26, 7:03 pm, mensana...@aol.compost I recently came across something that I thought would be of interest to this group. Before electronic computers (roughly, before the early 1940s), many people spent enormous amounts of time computing mathematical tables. Most texts at and below the level of calculus before the mid 1980s included abbreviated tables for student use (although computers were crunching these numbers from the mid 1940s onward, it wasn't until widespread use of calculators that these tables in math texts began to be unnecessary), and I imagine many (most?) of the people in this group have had some experience with them (probably trig. and logarithm tables) when growing up. Also, if you ever browsed through the library while you were in college, and again I suspect most of those in this group did, you probably came across a number of huge and old volumes that were quite a bit more extensive and advanced than the ones found at the back of textbooks. > By the way, I highly recommend David Alan Grier's recent (2005) book When Computers Were Human if you want to read more about this part of mathematical history. > The main point I want to make is that people literally spent years of 12+ hour days on these calculations. With this in mind, I offer what is one of the most compelling tales of perseverance in the face of disaster I've encountered in quite some time: > T. H. Gronwall, Review of [...] by Keiichi Hayashi, Bulletin of the American Mathematical Society 32 (1926), 716. > The work under review is far more elaborate and complete than any other tables of this kind. In the main part of Table I, x runs from 0.100 to 2.999 by intervals of 0.001, and the following columns are found on two opposite pages: x, phi(= x converted into degrees, minutes, and seconds), sin x, cos x, tan x, arcsin x, arccos x, arctan x (left page); x, e^x, e^(-x), sh x, ch x, th x, arg sh x, arg ch x (part of the table only), arg th x (right page). At the bottom of each page there is an additional table [...] The number of decimal places carried varies with the function tabulated from twelve for sin x and cos x to seven for arg sin x. [...] Table II gives sin(x*pi/2) and cos(x*pi/2) from x = 0.000 to 0.500 to ten places, and table II [...] > Last paragraph of the review -- One cannot but admire the author's fortitude in carrying on his work in the face of disaster. As he tells us in the preface, the manuscript of over a thousand pages was completed toward the end of 1923, only to be destroyed in a fire that consumed the engineering building of the University of Kyushu. The author immediately began his work anew, and in another two years completed the manuscript of the present tables which will certainly remain the standard of their kind for many years to come. >http://www.ams.org/journals/bull/1926-32-06/home.htmlhttp://tinyurl.c...[UR L for .pdf file] > Dave L. Renfro Jesus Christ, didn't he make Xerox backup copies? ;-) Maybe he kept the backup copies in the same building? === Subject: Re: Comprehensive Solution Manual for Textbooks posting-account=rLOz6QoAAAAmvEIbrGZd27QhtZqovu5R rv:1.8.1.11) Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) === Subject: Re: Comprehensive Solution Manual for Textbooks 5.1),gzip(gfe),gzip(gfe) contenueing that how aboutryour your exa? are u study very well? please do properly! by by GOOD with u! === Subject: Re: easy proof? I am a beginner in the subject of proofs. I have > the following > question: Prove that for any natural number n, either n is > a prime or a perfect > square , or n divides (n-1)! Kind regads The statement is literally correct because or in > a mathematical statement means that at least one of the > alternatives is true. But it might be better to state it as follows, since it > gives more information For any natural number n greater than 1, either n > is a prime or the square of a prime, or n divides (n-1)!. For example, 36 is a perfect square, but it also > divides 35!. I added greater than 1 because the statement is trivially > true for 1. The proof depends on the unique factorization > theorem: any natural number greater than 1 is a product of distinct > prime powers and that factorization is unique except for the order of the > factors. A number will divide another if each of those prime power > factors does. You just have to check whether or not each of those prime > power factors is less than n. > 9 | 8! even though 9 is the square of a prime. 3|3, > 3|6 No extra > numbers > 25 | 24! ; 5|5, 5|10, 5|15, 5|20 Two extra numbers > 121 | 120! there are many 11's to spare in 120!. > and so on any other square let S = n^2 once we'd like S|(n^2 -1)! so in the (n^2 -1)! we'll have at least following factors: n; 2n; 3n; 4n; ... till (n-1)n; also only n=2 S=4 not divide (4-1)! = 3! = 1*2*3 = 6 More interesting will be to find such power of n , that n^p not divide (n^p -1)! ? It looks, that for powers from n=3 it works even better. 2^3 = 8 ; 7! = 1*2*3*4*5*6*7 = (2^4)*(3^2)*5*7 Ro-Bin === Subject: Re: Lorenz ODE and analyticity > < http://planetmath.org/encyclopedia/LorenzEquation.html > there is an attracting set which is an ellipsoid if the > parameters sigma, tau and beta are all positive. sigma = 10, tau = 28 and beta = 8/3. They mention > the initial condition (x_0, y_0, z_0) = (3, 15, 1). We know that, if a solution exists for t in [0, T], > then x(t), y(t) and z(t) are C^1 in (0, T). > I think it follows that x(t), y(t) and z(t) are C^oo > in (0, T). Is it true that x, y and z are real-analytic > in (0, T)? And if they are real-analytic, can anything > be said about the radii of convergence at some time > t_0 in (0, T) ? The solutions of a DE system where the right sides are analytic are always analytic in any region of the complex plane where the solution exists. The only question is for what (complex) values of t the solution goes off to infinity. The simplest a priori bounds are as follows. Consider the DE system dW/dz = F(W) where W = (w_1(z),...,w_n(z)), and F = (f_1,...,f_n) is analytic and bounded in the region ||W - W_0|| < beta (where ||W|| = sqrt(sum_j |w_j|^2)) with ||F(W)|| < mu there. Then the system with initial condition W(z_0) = W_0 has a unique analytic solution in |z - z_0| < beta/mu. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Lorenz ODE and analyticity > < http://planetmath.org/encyclopedia/LorenzEquation.html > there is an attracting set which is an ellipsoid if the >> parameters sigma, tau and beta are all positive. > sigma = 10, tau = 28 and beta = 8/3. They mention >> the initial condition (x_0, y_0, z_0) = (3, 15, 1). > We know that, if a solution exists for t in [0, T], >> then x(t), y(t) and z(t) are C^1 in (0, T). >> I think it follows that x(t), y(t) and z(t) are C^oo >> in (0, T). Is it true that x, y and z are real-analytic >> in (0, T)? And if they are real-analytic, can anything >> be said about the radii of convergence at some time >> t_0 in (0, T) ? The solutions of a DE system where the right sides are > analytic are always analytic in any region of the complex plane > where the solution exists. The only question is for what (complex) > values of t the solution goes off to infinity. The simplest a priori > bounds are as follows. Consider the DE system dW/dz = F(W) where W = (w_1(z),...,w_n(z)), and F = (f_1,...,f_n) is analytic > and bounded in the region ||W - W_0|| < beta (where > ||W|| = sqrt(sum_j |w_j|^2)) with ||F(W)|| < mu there. Then > the system with initial condition W(z_0) = W_0 has a unique > analytic solution in |z - z_0| < beta/mu. axis, but with a finite radius of convergence, is provided by: dw/dz = - w^2 , with initial condition w(0) = 1. For z in [0, oo), we have: w(z) = 1/(z+1) and dw/dz = - 1/(1+z)^2 . Both w(z) and dw/dz have poles at z = -1 . If I remember correctly, this means the Taylor series for w(z) diverges for |z| > 1. I wonder what happens if we try to solve the Lorenz system for t<0 . David Bernier -- === Subject: Re: Lorenz ODE and analyticity > < http://planetmath.org/encyclopedia/LorenzEquation.html > there is an attracting set which is an ellipsoid if the > parameters sigma, tau and beta are all positive. sigma = 10, tau = 28 and beta = 8/3. They mention > the initial condition (x_0, y_0, z_0) = (3, 15, 1). We know that, if a solution exists for t in [0, T], > then x(t), y(t) and z(t) are C^1 in (0, T). > I think it follows that x(t), y(t) and z(t) are C^oo > in (0, T). Is it true that x, y and z are real-analytic > in (0, T)? And if they are real-analytic, can anything > be said about the radii of convergence at some time > t_0 in (0, T) ? The solutions of a DE system where the right sides are analytic are always analytic in any region of the complex plane where the solution exists. The only question is for what (complex) values of t the solution goes off to infinity. The simplest a priori bounds are as follows. Consider the DE system dW/dz = F(W) where W = (w_1(z),...,w_n(z)), and F = (f_1,...,f_n) is analytic and bounded in the region ||W - W_0|| < beta (where ||W|| = sqrt(sum_j |w_j|^2)) with ||F(W)|| < mu there. Then the system with initial condition W(z_0) = W_0 has a unique analytic solution in |z - z_0| < beta/mu. axis, but with a finite radius of convergence, is provided by: dw/dz = - w^2 , with initial condition w(0) = 1. For z in [0, oo), we have: w(z) = 1/(z+1) and dw/dz = - 1/(1+z)^2 . > Both w(z) and dw/dz have poles at z = -1 . If I remember correctly, this > means the Taylor series for w(z) diverges for |z| > 1. I wonder what happens if we try to solve the Lorenz system for t<0 . By looking at a series solution of the Lorenz system (with the parameter values and initial conditions you mentioned), it seems to me like the radius of convergence is somewhere between 0.1 and 0.15. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Lorenz ODE and analyticity >> < http://planetmath.org/encyclopedia/LorenzEquation.html > there is an attracting set which is an ellipsoid if the >> parameters sigma, tau and beta are all positive. >> sigma = 10, tau = 28 and beta = 8/3. They mention >> the initial condition (x_0, y_0, z_0) = (3, 15, 1). >> We know that, if a solution exists for t in [0, T], >> then x(t), y(t) and z(t) are C^1 in (0, T). >> I think it follows that x(t), y(t) and z(t) are C^oo >> in (0, T). Is it true that x, y and z are real-analytic >> in (0, T)? And if they are real-analytic, can anything >> be said about the radii of convergence at some time >> t_0 in (0, T) ? The solutions of a DE system where the right sides are > analytic are always analytic in any region of the complex plane > where the solution exists. The only question is for what (complex) > values of t the solution goes off to infinity. The simplest a priori > bounds are as follows. Consider the DE system dW/dz = F(W) where W = (w_1(z),...,w_n(z)), and F = (f_1,...,f_n) is analytic > and bounded in the region ||W - W_0|| < beta (where > ||W|| = sqrt(sum_j |w_j|^2)) with ||F(W)|| < mu there. Then > the system with initial condition W(z_0) = W_0 has a unique > analytic solution in |z - z_0| < beta/mu. axis, but with a finite radius of convergence, is provided by: dw/dz = - w^2 , with initial condition w(0) = 1. For z in [0, oo), we have: w(z) = 1/(z+1) and dw/dz = - 1/(1+z)^2 . Both w(z) and dw/dz have poles at z = -1 . If I remember correctly, this means the Taylor series for w(z) diverges for |z| > 1. I wonder what happens if we try to solve the Lorenz system for t<0 . By looking at a series solution of the Lorenz system (with the parameter > values and initial conditions you mentioned), it seems to me like the > radius > of convergence is somewhere between 0.1 and 0.15. On further investigation using numerical solution methods, it seems to me that the closest singularities to 0 are at approximately -.16244 (+/-) 0.06791 i. So the radius of convergence should be approximately 0.17606. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: online proof of Fubini's theorem posting-account=eCCYhwkAAABFc2vcqsxfXKC3sQXDmAFh 5.1),gzip(gfe),gzip(gfe) spider-dtc-te09.proxy.aol.com[CDBC7089] (Prism/1.2.1), HTTP/1.1 cache-dtc-ad05.proxy.aol.com[CDBC74C7] (Traffic-Server/6.1.5 [uScM]) > I am looking for a rigorous proof of Fubini's theorem online. ?Could > anyone help me? ?(I'm already aware of several excellent proofs in > texts such as Rudin's Real and Complex Analysis etc. ?However, I don't > have easy access to maths books at present, or much of a budget for > buying.) > Paul Epstein I assume you did some obvious Google searches and didn't find anything. Those searches probably wouldn't find the proof in the free online book by Shlomo Sternberg at http://www.math.harvard.edu/~shlomo/docs/Real Variables.pdf I found Sternberg's book at http://www.geocities.com/alex stef/mylist.html#FuncAn which is a list of free online math books. If you Google around you can probably find lots of other such free book lists. Happy holidays, Eric === Subject: Re: online proof of Fubini's theorem posting-account=fVOpuAkAAAB0gOUkQMH0DG_KdwTVgKXP CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > I am looking for a rigorous proof of Fubini's theorem online. ?Could anyone help me? ?(I'm already aware of several excellent proofs in texts such as Rudin's Real and Complex Analysis etc. ?However, I don't have easy access to maths books at present, or much of a budget for buying.) Paul Epstein I assume you did some obvious Google searches and didn't find > anything. Those searches probably wouldn't find the proof in the free > online book by Shlomo Sternberg athttp://www.math.harvard.edu/~shlomo/docs/Real Variables.pdf I found Sternberg's book athttp://www.geocities.com/alex stef/mylist.html#FuncAn > which is a list of free online math books. If you Google around you > can probably find lots of other such free book lists. Happy holidays, Eric although I'm familiar with Allan Hatcher's free online book on algebraic topology which has had wonderful reviews. Google is a wonderful resource but it does have its limitations. If encounter text that says something like we will not provide a proof of Fubini's theorem here ... However, a human who saw another person solve such difficult problems in artificial intelligence. Happy holidays to you as well. Paul Epstein === Subject: Confusing Christmas Present posting-account=eyodQwoAAABPdJRS9cyjWeOaXdI6SKnG Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) Hey all, Got a cool partial's book for Christmas, and had a few questions that weren't explained very well in it. The first has to do with potential, with =full double partial of U PDE: U(x,y)=0 with (x,y) in the Domain D:({(x,y): x>0, 0 Hey all, > Got a cool partial's book for Christmas, and had a few questions that > weren't explained very well in it. The first has to do with potential, with =C4=3Dfull double partial of U PDE: =C4U(x,y)=3D0 with (x,y) in the Domain D:({(x,y): x>0, 0 u(0,y)=3D f (y), Finally, u(x,y) is bounded.... It might help if you used ordinary ASCII text instead of those special characters that my newsreader has troubles with. My guess is that the PDE is laplacian u = 0 and the domain is {(x,y): x > 0, 0 < y < pi}. > Part a) > The exercise asks to determine all the solutions in the form > u(x,y)=3DX(x) sin( N*y), where X(x) is a function of x, and N is a > positive? integer. Hint: substitute u(x,y) = X(x) sin(N*y) into the PDE, and divide by sin(N*y). You should get an ODE that you can solve. What solutions of that ODE are bounded for x > 0? > Part b) > Next, we're asked to assume that f has a piecewise continuous. > Determine the Fourier series representation of the above BVP (Boundary > Value Problem). You want to take the Fourier series f(y) = sum_N b_N sin(N y). Then if X_N(x) is the bounded solution of the ODE above with X(0) = 1, your solution will be u(x,y) = sum_N b_N X_N(x) sin(N y). -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: 5 Reasons why Godels incompleteness theorem invalid posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG > The Autralian philosopher Colin Leslie Dean points out Godels theorems are > invalid for 5 reasons: [...] > No real need to show invalidity. > Just show inconsequentialness. > Look at what all we've achieved in spite of it. Indeed. It appears that all those statements which exist but cannot be proved true or false are mostly of the form this sentence refers to itself in a way that may produce an apparent paradox. I think we've gotten along quite well without all of those rather esoteric, and mostly useless, statements. > But - does it prevent some really high levels of achievement in the > far future? There is something to be said about the formalizations of inconsistency and incompleteness. But again, I doubt that they have any impact on any theorems of significance. > These are mysteries we shall learn when we die - Catholic nuns from > 3 rd grade. Or not. === Subject: Re: 5 Reasons why Godels incompleteness theorem invalid The Autralian philosopher Colin Leslie Dean points out Godels theorems are invalid for 5 reasons: [...] > No real need to show invalidity. Just show inconsequentialness. Look at what all we've achieved in spite of it. Indeed. It appears that all those statements which exist > but cannot be proved true or false are mostly of the > form this sentence refers to itself in a way that may > produce an apparent paradox. I think we've gotten > along quite well without all of those rather esoteric, > and mostly useless, statements. This is a common misunderstanding. Every Godel sentence does have an interpretation of the form this statement cannot be proven. But the statement is also a statement in the language of number theory, and has an interpretation purely about numbers, without any reference to statements or provability. A Godel statement states a property of natural numbers which cannot be proven or disproven in that formal system. However, the Godel statement about numbers is so horrendously complex that it could perhaps be argued that no human being would ever care whether it is provable or not. Personally, I think the principle that all such formal systems are incomplete is critically important, independent of whether such pathological statements are useful or not. --Mark === Subject: Re: 5 Reasons why Godels incompleteness theorem invalid posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG > The Autralian philosopher Colin Leslie Dean points out Godels > theorems are invalid for 5 reasons: [...] > No real need to show invalidity. > Just show inconsequentialness. > Look at what all we've achieved in spite of it. > Indeed. It appears that all those statements which exist > but cannot be proved true or false are mostly of the > form this sentence refers to itself in a way that may > produce an apparent paradox. I think we've gotten > along quite well without all of those rather esoteric, > and mostly useless, statements. > This is a common misunderstanding. Every Godel sentence does have an > interpretation of the form this statement cannot be proven. But the > statement is also a statement in the language of number theory, and has > an interpretation purely about numbers, without any reference to > statements or provability. A Godel statement states a property of > natural numbers which cannot be proven or disproven in that formal > system. Just to be clear, I am well aware of this. I was merely mentioning the layman's high-level non-numeric interpretation of that property. > However, the Godel statement about numbers is so horrendously complex > that it could perhaps be argued that no human being would ever care > whether it is provable or not. Saying that there are numeric statements that are unprovable could lead one to deduce that these statements are simple regular theorems, instead of, as you say, horrendously complex numeric encodings of self-referential logic. And in fact, many people, especially many philosophers, have assumed exactly that, and used that assumption to make all sorts of pathological claims about the impossibility of logic, artificial intelligence, and even natural intelligence. The bookshelves are filled with books based on misunderstandings of Goedel's result. Hence the importance of specifying exactly what it was that Goedel was proving. > Personally, I think the principle that > all such formal systems are incomplete is critically important, > independent of whether such pathological statements are useful or not. Yes, that conclusion is important, saying a lot about formal systems and their limitations. And it is just as commonly misinterpreted by many, including those who are mathematically adept. E.g., we can never know everything in Mathematics (which is true) therefore Mathematics is a fatally flawed system (which is a non sequitur). Or even, the universe is based on fundamental mathematical principles (a misstatement) therefore, because of Geodel's theorem, the universe is inherently inconsistent and flawed at the deepest level (another non sequitur). === Subject: Re: 5 Reasons why Godels incompleteness theorem invalid Furst day widda brane? > The Autralian philosopher Colin Leslie Dean points out Godels theorems are -- === Subject: Re: 5 Reasons why Godels incompleteness theorem invalid posting-account=n4TzyQkAAADLWxrRHqyiUZ-1SZdOB4vv Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) The Autralian philosopher Colin Leslie Dean points out Godels theorems are invalid for 5 reasons: he uses the axiom ofreducibility- which is invalid, he uses the axiom of choice, he constructs impredicative statements - which are invalid ,he miss uses the theory of types, he falls into 3 paradoxes > http://gamahucherpress.yellowgum.com/books/philosophy/GODEL5.pdf > G.85DEL'S INCOMPLETENESS THEOREM. ENDS IN ABSURDITY OR MEANINGLESSNESS G.85DEL IS A COMPLETE FAILURE AS HE ENDS IN UTTER MEANINGLESSNESS CASE STUDY IN THE MEANINGLESSNESS OF ALL VIEWS By COLIN LESLIE DEAN B.SC, B.A, B.LITT (HONS), M.A, B,LITT (HONS), M.A, M.A (PSYCHOANALYTIC STUDIES), MASTER OF PSYCHOANALYTIC STUDIES, GRAD CERT (LITERARY STUDIES) 2007 > A case study ...... No real need to show invalidity. Just show inconsequentialness. Look at what all we've achieved in spite of it. It's like a dent in a car's fender - you can still use the car for all > sorts of wonderful things. But - does it prevent some really high levels of achievement in the > far future? > Or it might enable some achievements. Isn't it , that the structure and order given with counting of natural numbers and addition/subtraction/ Multiplication/division just highlighted these numbers, which can only be divided by itself or by 1 without rest, like 7 can not be divided by 2, 3 ... What about these? They are chaotic, all kind of endeavours to find order and structure in them, failed up to now. So there is chaos. But what mathematicians did? They took them as primary, as prime numbers, and found, that every natural has a unique decomposition into these numbers. When You want structure and order of a level a bit above finite combinatorics, this will create chaos out of it, this not as a law, and sometimes one can create nice things out of it. It's not a bumb in the fender, it's a major feature of many axiomatic systems. Some decennia ago, this fantastic new technic of creating energy was develeoped, and we were assured by mathematical calculations, that it is nearly foolproof - one big failure in a million years or so.( So now we can be sure, that the next two millions years nothing will happen, as we had at least two accidents already. Only imagine, the mathematicians, who can look into the future, they can predict an eclipse of the moon in two thousand years from now on, find a rigid order and structue, which absolutely rules everything. This would be fate for us, not be bypassed and leaving no choices. Who fears these imperfections? Only the ones, who want to imply their structure without dispute to be followed. With friendly greetings Hero === Subject: Jacobi Method for finding eigenvalues of symmetric matrix. (I posted this on sci.math.research, but it didn't cross-post properly!) I have been looking at the Jacobi Method for finding eigenvalues of a converges with quadratic order when the eigenvalues are distinct. What is known about the speed of convergence when the eigenvalues are not necessarily distinct? P.S. I only learned this method the day before yesterday. It really is a very cool algorithm! === Subject: Re: PUZZLE. Do you ever cut apple in half? posting-account=TV2szgkAAACrA1vyuh8IN_0zzgzcwogw 5.1),gzip(gfe),gzip(gfe) Don S. McDonald Hacking Butter (late NW C..ollins ca. Nov 1976, MenZed.) Part II apple peel. > PUZZLE. EDo you ever cut apple in half? > 1/ Please cut up an apple presently. Suppose that a > freshly cut piece// of apple may turn brownish after 1 > hr. If I wish to eat a 1/4 apple now and I have a > sharp knife handy, how can I make 2 plane cuts into > the apple (assume spherical) while exposing the > minimum // new surface liable to browning? e.g. pi. r^2, > wedge. sure, cut out a wedge, same SA..surface area ....... no r^2 needed not a radial wedge. what angle and depth? > 2/ (..) a pitched span roof with 2 > inclined sides meeting Bevel Cant Ridge? (right angle, > greater or less.) Angle, A. Place 2 plane cuts Notch// > meeting .., with the ridge, so as to obtain the > maximum volume // of a block with new cut surface = 1 sq > met = 1 m^3. what is the question? a cartesian variation of question 1/? rectangular coords. e.g. toffee apple meniscus. dip apple maximum bouyancy per surface area?? > 3/ A Platonic solid (regular n-hedron) may often have > 3 plane faces meeting at each vertex. Please insert > 1, 2 or 3 plane cuts in a platonic butter block so as > to cut off a small 1 cm^3 (cu. cm.) piece of butter > with minimum new surface. typically a very hard problem especially for all of them various symmetry. the 1976 puzzle asked 2 plane cuts out of a cube. which don approximated with hp-25 programmable calculator. e.g. v1 v2 v3. square roots. attempt with BASIC. iterate. Don. McDonald 27.12.2007. nz.general === Subject: Re: two counterfeit coins > We are given 12 coins, 10 of which are of the same weight and 2 of which are lighter but mutually of the same weight. > we also have a two arm balance at our disposal. How can we detect the two lighter coins in at most four weighings? > To do it in five weighings is easy. > C. Wildhagen > Rotterdam > The Netherlands I read a book, whose name I forget, that suggested that you try to maximize the entropy with each weighing. So suppose you decide to start weighing 4 coins against 4. You calculate the probabilities p,q,r of the balance going to the left, the right, and staying still. Then calculate the entropy -(p log p + q log q + r log r). You do this for all the other possible experiments (3 against 3, 5 against 5, etc), see for which one the entropy is maximized! If you are doing the standard one coin which is heavier or lighter, then it is clear that the best choice is 4 against 4, when the entropy is log(3). However in your case, presumably the calculations will be a little more delicate! === Subject: Re: two counterfeit coins We are given 12 coins, 10 of which are of the same weight and 2 of which are lighter but mutually of the same weight. we also have a two arm balance at our disposal. How can we detect the two lighter coins in at most four weighings? To do it in five weighings is easy. > C. Wildhagen Rotterdam The Netherlands > I read a book, whose name I forget, that suggested that you try to > maximize the entropy with each weighing. So suppose you decide to start > weighing 4 coins against 4. You calculate the probabilities p,q,r of > the balance going to the left, the right, and staying still. Then > calculate the entropy -(p log p + q log q + r log r). You do this for > all the other possible experiments (3 against 3, 5 against 5, etc), see > for which one the entropy is maximized! Good idea. If you weigh k against k, then r = ((k choose 0)^2 (12-2k choose 2) + (k choose 1)^2 (12-2k choose 0)) /(12 choose 2) = 1 - 23/66 k + k^2/22 and p = q = (1-r)/2. The entropy is maximized at k=4. In fact, with k=4 we have p = q = r = 1/3. So let's say the first weighing is coins 1,2,3,4 against 5,6,7,8. I'll write (Li), (Ri) or (Ei) to mean left down, right down or equal on the i'th weighing. The second weighing can be coins 1,5,9 against coins 2,6,10. Now we have the following cases: (L1,L2): Third weighing is 7,11 against 8,12. The possibilities for light coins are then: (L1L2L3): (6,8), (6,12), (8,10). Decide by weighing # 10 against # 12. (L1L2R3): (6,7), (6,11), (7,10). Decide by weighing # 10 against # 11. (L1L2E3): (6,10). No fourth weighing needed. (L1,E2): Third weighing is # 5 against # 7. The possibilities are then: (L1E2L3): (7,8), (7,11), (7,12). Decide by weighing #8 against #11. (L1E2E3): (6,9), (8,11), (8,12). Decide by weighing #6 against #11. (L1E2R3): (5,6), (5,10). Decide by weighing #6 against #10. (L1,R2): Third weighing is #7 against #8. The possibilities are then: (L1R2L3): (5,8), (8,9). Decide by weighing #5 against #9. (L1R2E3): (5,9), (5,11), (5,12). Decide by weighing #9 against #11. (L1R2R3): (5,7), (7,9). Decide by weighing #5 against #9. (E1,L2): Third weighing is #2 against #3. The possibilities are then: (E1L2L3): (3,6). No fourth weighing. (E1L2E3): (4,6), (10,11), (10,12). Decide by weighing #4 against #11. (E1L2R3): (2,6), (2,7), (2,8). Decide by weighing #6 against #7. (E1,E2): Third weighing is 1,2 against 3,7. The possibilities are then: (E1E2L3): (3,7), (3,8), (4,7). Decide by weighing #4 against #8. (E1E2E3): (4,8), (9,10), (11,12}. Decide by weighing #4 against #9. (E1E2R3): (1,6), (2,5). Decide by weighing #2 against #5. (E1,R2): Third weighing is 1,3 against 4,5. The possibilities are then (E1R2L3): (2,5), (4,7), (4,8). Decide by weighing #2 against #7. (E1R2E3): (9,10), (11,12). Decide by weighing #9 against #11. (E1R2R3): (1,6), (3,7), (3,8). Decide by weighing #1 against #7. (R1...) similar to (L1...) using symmetry (interchange (1,2,3,4) with (5,6,7,8)). -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: two counterfeit coins > Stephen Montgomery-Smith We are given 12 coins, 10 of which are of the > same weight and 2 of which > are lighter but mutually of the same weight. > we also have a two arm balance at our disposal. > How can we detect the two > lighter coins in at most four weighings? > To do it in five weighings is easy. C. Wildhagen > Rotterdam > The Netherlands I read a book, whose name I forget, that suggested > that you try to maximize the entropy with each weighing. So > suppose you decide to start weighing 4 coins against 4. You calculate the > probabilities p,q,r of the balance going to the left, the right, and > staying still. Then calculate the entropy -(p log p + q log q + r log > r). You do this for all the other possible experiments (3 against 3, 5 > against 5, etc), see for which one the entropy is maximized! Good idea. If you weigh k against k, then > r = ((k choose 0)^2 (12-2k choose 2) + (k choose 1)^2 > (12-2k choose 0)) > /(12 choose 2) = 1 - 23/66 k + k^2/22 > and p = q = (1-r)/2. The entropy is maximized at > k=4. In fact, with k=4 > we have p = q = r = 1/3. So let's say the first > weighing is coins 1,2,3,4 > against 5,6,7,8. I'll write (Li), (Ri) or (Ei) to > mean left down, right down > or equal on the i'th weighing. The second weighing can be coins 1,5,9 against coins > 2,6,10. Now we have the > following cases: (L1,L2): Third weighing is 7,11 against 8,12. The > possibilities for light > coins are then: > (L1L2L3): (6,8), (6,12), (8,10). Decide by weighing > # 10 against # 12. > (L1L2R3): (6,7), (6,11), (7,10). Decide by weighing > # 10 against # 11. > (L1L2E3): (6,10). No fourth weighing needed. (L1,E2): Third weighing is # 5 against # 7. The > possibilities are then: > (L1E2L3): (7,8), (7,11), (7,12). Decide by weighing > #8 against #11. > (L1E2E3): (6,9), (8,11), (8,12). Decide by weighing > #6 against #11. > (L1E2R3): (5,6), (5,10). Decide by weighing #6 > against #10. (L1,R2): Third weighing is #7 against #8. The > possibilities are then: > (L1R2L3): (5,8), (8,9). Decide by weighing #5 > against #9. > (L1R2E3): (5,9), (5,11), (5,12). Decide by weighing > #9 against #11. > (L1R2R3): (5,7), (7,9). Decide by weighing #5 > against #9. (E1,L2): Third weighing is #2 against #3. The > possibilities are then: > (E1L2L3): (3,6). No fourth weighing. > (E1L2E3): (4,6), (10,11), (10,12). Decide by > weighing #4 against #11. > (E1L2R3): (2,6), (2,7), (2,8). Decide by weighing #6 > against #7. (E1,E2): Third weighing is 1,2 against 3,7. The > possibilities are then: > (E1E2L3): (3,7), (3,8), (4,7). Decide by weighing #4 > against #8. > (E1E2E3): (4,8), (9,10), (11,12}. Decide by weighing > #4 against #9. > (E1E2R3): (1,6), (2,5). Decide by weighing #2 > against #5. (E1,R2): Third weighing is 1,3 against 4,5. The > possibilities are then > (E1R2L3): (2,5), (4,7), (4,8). Decide by weighing #2 > against #7. > (E1R2E3): (9,10), (11,12). Decide by weighing #9 > against #11. > (E1R2R3): (1,6), (3,7), (3,8). Decide by weighing #1 > against #7. > (R1...) similar to (L1...) using symmetry > (interchange (1,2,3,4) with (5,6,7,8)). > -- > Robert Israel > israel@math.MyUniversitysInitials.ca > Department of Mathematics > http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, > BC, Canada Actually, the composition of the three weighings can be chosen a priori, e.g. 1. {1,2,3,4} vs. {5,6,7,8} 2. {1,2,3,5} vs. {4,9,10,11} 3. {1,6,9,12} vs. {2,5,7,10} It's enough to write a table with all possible results in order to detect the desired coin. It seems that the composition of the three weighings {a,b,c,d} vs. {x,y,z,t} can be chosen almost at random. Valeriu Anisiu === Subject: Re: two counterfeit coins Stephen Montgomery-Smith We are given 12 coins, 10 of which are of the same weight and 2 of which are lighter but mutually of the same weight. we also have a two arm balance at our > disposal. How can we detect the two lighter coins in at most four weighings? To do it in five weighings is easy. > C. Wildhagen Rotterdam The Netherlands > Actually, the composition of the three weighings > can be chosen a priori, e.g. > 1. {1,2,3,4} vs. {5,6,7,8} > 2. {1,2,3,5} vs. {4,9,10,11} > 3. {1,6,9,12} vs. {2,5,7,10} It's enough to write a table with all possible > results > in order to detect the desired coin. It seems that the composition of the three weighings > {a,b,c,d} vs. {x,y,z,t} can be chosen almost at > random. Valeriu Anisiu Sorry, the presented solution is for a single false coin and I just saw that there are two :-) I did not check if the apriori weighings works in this case. V. A. === Subject: Re: two counterfeit coins <22887370.1198758723645.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=ogMREwkAAAC5xUr8sg7heGtsvzzF18LA Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Stephen Montgomery-Smith We are given 12 coins, 10 of which are of the > same weight and 2 of which > are lighter but mutually of the same weight. > we also have a two arm balance at our disposal. > How can we detect the two > lighter coins in at most four weighings? > To do it in five weighings is easy. > C. Wildhagen > Rotterdam > The Netherlands > Actually, the composition of the three weighings can be chosen a priori, e.g. 1. {1,2,3,4} vs. {5,6,7,8} 2. {1,2,3,5} vs. {4,9,10,11} 3. {1,6,9,12} vs. {2,5,7,10} > It's enough to write a table with all possible results in order to detect the desired coin. > It seems that the composition of the three weighings {a,b,c,d} vs. {x,y,z,t} can be chosen almost at random. > Valeriu Anisiu Sorry, the presented solution is for a single false coin > and I just saw that there are two :-) > I did not check if the apriori weighings > works in this case. V. A. with 2 light coins, there are 12 C 2 = 66 possible combinations of light coins. Three weighings (with apriori settings) can only distinguish 27 cases. Therefore it is not sufficient for this problem. Theoretically, there can be an apriori list of 4 weighings (since they have 81 possible cases), but I haven't worked on this problem yet. === Subject: Re: The Virgin Birth of Points <6sspm3hlc9rd8m0t626te27hmv9bn79q0f@4ax.com> posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG > Not is a unary logical operator. So, by definition, not > cannot stand by itself as a well-formed logical statement. > Your form of argument is defective. I say X is true since alternatives > to X are false. You say X is false because X isn't Y and you assume Y > to be true. I said that not is a logical operator, and you say that my argument is defective. So you must be saying that not is not a logical operator, but something else. Alternatives to not are other logical operators. Or are you saying that's not the case? > You're interested in Y only because assumptions of truth can be false. If by assumptions of truth you mean axioms, no, that's not the case. Mathematical axioms are always true. If by assumptions of truth you mean definitions, no, that's not the case either. Definitions are not true or false. === Subject: Re: The Virgin Birth of Points >> Not is a unary logical operator. So, by definition, not >> cannot stand by itself as a well-formed logical statement. > Your form of argument is defective. I say X is true since alternatives >> to X are false. You say X is false because X isn't Y and you assume Y >> to be true. I said that not is a logical operator, and you say that >my argument is defective. So you must be saying that >not is not a logical operator, but something else. By your own words right above you don't just say that not is a logical operator. You say that not is a unary logical operator. Consequently my comment is rather obviously directed at the idea not is not unary not that it is not logical. >Alternatives to not are other logical operators. >Or are you saying that's not the case? I'm saying not is not unary just as differences are not unary and are taken between things and can stand by itself as a well formed logical statement. >> You're interested in Y only because assumptions of truth can be false. If by assumptions of truth you mean axioms, no, that's not >the case. Mathematical axioms are always true. Incorrect by your own words. You say elsewhere: An axiom is an assumed truth. Period. Assumed truth and always true are different predicates. >If by assumptions of truth you mean definitions, no, that's >not the case either. Definitions are not true or false. Well definitions are always definitions. Doesn't make them true or false. Contradictions between predicates in definitions make them false as in squircles are square circles. ~v~~ === Subject: Re: The Virgin Birth of Points <11m7n3p4a3em8bj4ud3199fibooah69g47@4ax.com> posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG > Not is a unary logical operator. So, by definition, not > cannot stand by itself as a well-formed logical statement. > Your form of argument is defective. I say X is true since alternatives > to X are false. You say X is false because X isn't Y and you assume Y > to be true. > I said that not is a logical operator, and you say that > my argument is defective. So you must be saying that > not is not a logical operator, but something else. > By your own words right above you don't just say that not is a > logical operator. You say that not is a unary logical operator. > Consequently my comment is rather obviously directed at the idea not > is not unary not that it is not logical. If not is not a unary operator, is it a binary operator? Or perhaps it's a logical value, or maybe a free variable, since logical statements are composed of operators, logical values (true and false), and variables, and a few other kinds of convenience punctuation. > Alternatives to not are other logical operators. > Or are you saying that's not the case? > I'm saying not is not unary just as differences are not unary and > are taken between things and can stand by itself as a well formed > logical statement. The only thing that can stand by itself in a logical statement is a simple value, such as true. Not is not a value, it's an operator, and therefore cannot stand by itself. Differences are values, not operators. They are the result of applying a binary difference operator to two arguments (operands). The statement d = a - b has a value, d, equated to the result of a binary operator - taking two arguments, a and b. I can't see how not = (whatever) can be a well-formed sentence. What is not equal to as a value? It is equal to false? > You're interested in Y only because assumptions of truth can be false. > If by assumptions of truth you mean axioms, no, that's not > the case. Mathematical axioms are always true. > Incorrect by your own words. You say elsewhere: > An axiom is an assumed truth. Period. > Assumed truth and always true are different predicates. Let's just expand that, then, to: Axioms are statements of truth, and are always true, and are therefore always assumed to be true (obviously). If an axiom is assumed to be false, it is not an axiom. In other words, a false statement cannot be an axiom. Hence axioms cannot be false. You seem to be hung up on the assumed part. > If by assumptions of truth you mean definitions, no, that's > not the case either. Definitions are not true or false. > Well definitions are always definitions. Doesn't make them true or > false. Contradictions between predicates in definitions make them > false as in squircles are square circles. Contradicting predicates makes a definition void or vacuous, but it does not make it false. Definitions are never false or true. Squircles are square circles is a vacuous definition; there are no geometric objects that meet the definition of squircle. Yet the definition itself is neither true nor false; it just is. Consider a Goldbach non-integer is an integer greater than 2 that is not the sum of two primes. We don't know if there are any integers meeting this definition, but that does not make the definition false, or true, or unknown, or whatever. The definition itself is completely valid. === Subject: #492 pseudosphere deconstructed to join together with its given sphere and make a Euclidean Cube ; new textbook: Mathematical-Physics (AP-adic primer) for students of age 6 onwards <12348601.1198409709803.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=fsC03QkAAAAwkSNcSEKmlcR-W_HNitEd Gecko/20021120 Netscape/7.01,gzip(gfe),gzip(gfe) If not for the fact that the Pseudosphere is clipped at one end > whereas the > Equator on the Pseudosphere is a perfect line then a Construction > proof > would render the above conjecture that a Cube is formed. But since > there > is a clipped Pseudosphere, the calculus has to be involved But anyone can visualize in their mind's eye how the Pseudosphere + > Sphere > when cut and joined form either a Cube or a rectangle that is almost a > cube > (clipped end of pseudosphere) Just picture the profile of the Pseudosphere crosssection: | > | > | > | > |___-------____ > So one can easily visualize that such a crosssection profile would > fit a circle crosssection of the sphere. So where every sphere > crosssection is > fitted with 4 Pseudosphere crosssections and where the end result is > either > a Cube or a Rectangle that is a near Cube, depending on the clipping > of the Pseudosphere. So, has anyone proven that above conjecture? > Now actually one can go the opposite route and take a fresh cube and delicately cut its 8 corners into that of 8 sections of the Pseudosphere and then assemble those into the Pseudosphere. 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: #493 pseudosphere deconstructed to join together with its given sphere and make a Euclidean Cube ; new textbook: Mathematical-Physics (AP-adic primer) for students of age 6 onwards <12348601.1198409709803.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=fsC03QkAAAAwkSNcSEKmlcR-W_HNitEd Gecko/20021120 Netscape/7.01,gzip(gfe),gzip(gfe) Now actually one can go the opposite route and take a fresh cube and > delicately > cut its 8 corners into that of 8 sections of the Pseudosphere and then > assemble > those into the Pseudosphere. Now I am not 100 percent sure if the above works. As I said many times before, that the mind's vision in NonEuclidean geometry is often deceived and these constructions in the mind's eye are tricky. What I do see if I tried to assemble those freshly cut 8 corners, that they form 4 fins or 4 ridges if I tried to construct the pseudosphere. So it appears that once I have the 8 corners cut from the cube, that I then have to make at least 8 more cuts to cut those 8 corners into 16 slices. So I envision assembling those 16 slices where no ridges are formed and have a pseudosphere. But unless I actually can construct such a contraption with my own hands can I really believe in the procedure. For one aspect is telling me that maybe this is impossible, in the aspect that these corners may still retain positive curvature of the sphere removed and that you cannot end up with negative curvature from that of positive curvature. But an arguement against that is the cube from whence this started is zero curvature and we carved out a sphere which is different curvature. But unless I can actually hold the project in my hands and carve it out and then reassemble into a pseudosphere can I for sure be certain. And perhaps this maybe the idea for a classroom toy in mathematics to illustrate the three curvatures of geometry of positive and negative and zero. Much like that other toy where you have to get all the different color squares lined up (what is the name? Rubriks cube or something like that?). Only this cube is taken apart (precut lines) and then reassembled into a Pseudosphere and a Sphere. Now if someone can confirm that my mind's eye is correct in the above deconstruction and to how many cuts required, or how many pieces (is it 16 pieces), then I would accept that my intuition is correct and not playing tricks on me. 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: #494 pseudosphere deconstructed to join together with its given sphere and make a Euclidean Cube ; new textbook: Mathematical-Physics (AP-adic primer) for students of age 6 onwards <12348601.1198409709803.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=fsC03QkAAAAwkSNcSEKmlcR-W_HNitEd Gecko/20021120 Netscape/7.01,gzip(gfe),gzip(gfe) Maybe the answer is that a deconstruction of those cutaways works only it is not a finite reconstruction of say 16 parts or 32 parts but an infinity of parts of every line is deconstructed and then reassembled to form the Pseudosphere for it sure looks to me like those lines of a Pseudosphere are of the same form as the 8 corners cutaway of a cube with a sphere inside. So maybe the answer is a line by line deconstruction and then a line by line reassemblage to form the pseudosphere. There is no point in me continuing with the pseudosphere until I can resolve that issue. Lwal, what do you say? Can those 8 corners be deconstructed and then reassembled to form a Pseudosphere? 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: #495 constructing a sphere out of successive smaller bands and a pseudosphere from the same bands ; new textbook: Mathematical-Physics (AP-adic primer) for students of age 6 onwards posting-account=fsC03QkAAAAwkSNcSEKmlcR-W_HNitEd Gecko/20021120 Netscape/7.01,gzip(gfe),gzip(gfe) Now here is an interesting way of constructing the Pseudosphere. Wikipedia calls it the Partial Pseudosphere: http://en.wikipedia.org/wiki/Pseudosphere I place before me a number of ring bands that are successively smaller than the largest and these bands are very thin of say 2mm thickness. Now if I place the ring bands successively stacked one on top of another I simulate the globes latitudes and thus build a sphere. However, if I stack the rings so that they are not successively contiguous so that they stack on top of one another but instead where they need propping-up because they would collapse since they would fall inside one another in successive additions of bands, well, I am constructing a pseudosphere as shown in that wikipedia photo of a partial pseudosphere. So here I have found a physical link between taking successive sized circular bands or loops and if they are laying on top of one another successively I form a sphere, if they are nested successively I form a partial pseudosphere. This is very curious, for it suggests that the difference between Hyperbolic geometry and Elliptic geometry is the mere physical difference that the successive bands of latitude are contiguous and do not nest inside one another, whereas in Hyperbolic geometry the successive bands do nest inside one another. Very, very curious. Perhaps that will be a new definition that distinguishes Hyperbolic from Elliptic geometry instead of the definition that one has all triangle angle sums less than 180 whereas the other has all triangles greater than 180 degrees. But perhaps both of these definitions come from one another in that if you visualize Euclidean Geometry Sieve as a 180 degree triangle that a Hyperbolic triangle will nest inside whereas the Elliptic triangle cannot penetrate through the Euclidean and thus stacks up outwards. 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: EUROMEDIA 2008, April 9-11, 2008, FEUP-University of Porto, Portugal - 2nd Call for Papers posting-account=N5wzPwoAAACfM7xTOME2HKP_WjJviHXd Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) ---------------------------------------------------------------------------- -------------------------------------------------------------- (Apologies for cross-posting) EUROMEDIA 2008, APRIL 9-11, 2008, FEUP-UNIVERSITY OF PORTO, PORTUGAL http://www.eurosis.org/cms/?q=taxonomy/term/100 We would appreciate if you could distribute this information by your colleagues and co-workers. ---------------------------------------------------------------------------- -------------------------------------------------------------- 2ND Call for Papers EUROMEDIA 2008 14TH ANNUAL MULTIMEDIA CONFERENCE WEBSITE http://www.eurosis.org/cms/?q=taxonomy/term/100 UNIVERSITY OF PORTO PORTUGAL APRIL 9-11, 2008 Organized by ETI Sponsored by EUROSIS Delft University of Technology Belgacom TTVI Ghent University DG-INFSO this is a reminder that the abstract submission deadline for EUROMEDIA 2008 is next January 20. As a reminder also please find enclosed the list of the conference topics: [] WEBTEC * Internet Viewers and Programs * Visual Programming Languages * 3D Web Programs * Video and Audio Streaming on the Web * AI on the Web * Software for Web-based Business Applications [] MEDIATEC * Multimedia Techniques and Telecom * Multimedia Authoring Tools and Software * Multimedia Building Blocks [] COMTEC * Telecommunications Technologies * Networks * Network Security * Mobile Communications * TV Technology * QoS [] APTEC * Telematics Consumer Applications * Cooperative Consumer Application * Tele-Education * Integrated Enterprise Software and Groupware * Telemedicine * Ubiquitous Computing Applications * Embedded Systems and Future Product Market Combinations * Multimodal communication [] ETEC * Tele-X and E-Commerce * Knowledge Management and E-Mobility Special Tracks: [] Knowledge Management and e-Mobility [] In-Car Applications [] Facial Recognition or Non Verbal Communication [] Virtual Reality 2.0 Applications Special Workshops: [] Medical Imaging Systems In recent years, extensive research has been performed to develop more and more efficient and powerful medical imaging systems. Such systems are crucial for medical specialists, allowing a deeper analysis and to understand what is going inside the human body, and therefore they play an essential role for adequate medical diagnosis and treatments. To accomplish efficient and powerful medical imaging systems, many research works have being done in many domains, like the ones related with medical image devices, signal processing, image processing and analysis, biomechanical simulation and data visualization. The main goal of the Workshop Medical Imaging Systems is to bring together researchers involved in the related domains, in order to set the major lines of development for the near future. Therefore, the proposed Workshop will consist of researchers representing various fields related to Medical Devices, Signal Processing, Computational Vision, Computer Graphics, Computational Mechanics, Scientific Visualization, Mathematics and Medical Imaging. The Workshop endeavors to contribute to obtain better solutions for more efficient and powerful medical imaging systems, and attempts to establish a bridge between clinicians and researchers from these diverse fields. The proposed Workshop will cover topics related with medical imaging systems, such as: * image acquisition * signal processing * image processing and analysis * modelling and simulation * computer aided diagnosis * surgery, therapy, and treatment * computational bioimaging and visualization * software development * virtual reality * telemedicine systems and their applications [] D-TV Scope: In contrast to traditional TV DTV is a telecommunication system for broadcasting and receiving moving pictures and sound by means of digital signals. It uses digitally compressed modulation data, which requires decoding by a specially designed television set or a standard receiver (set-top box) as well as via PC-TV based on ADLS (VDSL). Digital Television has several advantages regarding technically as well as content and programme related aspects. One of the most significant is the use of a smaller channel bandwidth. This frees up space for more digital channels, other non-television services such as pay-multimedia services and the generation of new revenue models based on interactive advertisement and marketing features. Furthermore there are special services such as multicasting (more than one programme on the same channel), electronic programme guides and interactivity providing a wider range of application. Accordingly within the last year a growing up of specialised digital TV channels (Special Interest Channels) has to be observed European wide. This new growing up stream is mainly characterised by the phenomena of merging television (TV) and information technology (IT) know how, methods and techniques. In this context the workshop targets to show up the development in both directions: the technically aspects as well as the content and programme issues. It provides a presentation and discussion platform for exchanging experiences made within the establishment as well as the maintenance of digital special interest channels. Main focus will be to point out the interdependencies and the mutual influences of technically streams and content respectively service related issues. Topics: * Broadband Television * Interactive Television * IPTV - Web-TV * DVB-H, DVB-T, DVB-S, DVB-C * MHP & Interactive Television * Set-Top-Boxes * Full Digital Production Chain Management * Digital-Video-Journalism * DMB (Digital Multimedia Broadcasting) * FreeTV & PayTV * Interactive Video * Interactive Advertising Formats * Event Oriented Interview Formats * Event Driven Program Schemes & Formats * Mobile Content * Cross-Format Content & Knowledge Pooling * Cross-Media Content Production & Distribution * Indexing and Retrieval of Digital Content * Semantically Enrichment of Digital Content * Security & Digital Rights Management (DRM) * Watermarking & Copyrights & Protection * Cross-platform productions on TV Media-Convergence * Peer-to-peer Grid TV Applications For more information on the workshops see: http://www.eurosis.org/cms/index.php?q=node/476 Conference Committee General Conference Chair Jo.8bo Manuel R. S. Tavares, FEUP, University of Porto, Porto, Portugal Renato Natal Jorge, FEUP, University of Porto, Porto, Portugal WEBTEC Programme Committee Sameh Abdel-Naby, University of Trento, Trento, Italy Dr. Paul Dowland, University of Plymouth, Plymouth, United Kingdom Dipl.-Inf. Steffen Harneit, CUTEC-Institute GmbH, Clausthal-Zellerfeld, Germany Ass. Prof. Qingping Lin, Nanyang Technological University, Singapore Prof. Dr. J.9arn Loviscach, Hochschule Bremen, University of Applied Sciences, Bremen, Germany Assoc. Prof. Wenji Mao, Institute of Automation, Chinese Academy of Sciences, Beijing, P.R. China Lorenzo Motta, Ansaldo Segnalamento Ferroviaro s.p.a. Genova, Italy Dr. H. Joachim Nern, Aspasia Knowledge Systems, Dusseldorf, Germany Dr. Carlos E. Palau, Universidad Politecnica de Valencia, Valencia, Spain Prof. Paola Salomoni, Universita di Bologna, Bologna, Italy Dr. Elpida Tzafestas, National Technical University of Athens, Athens, Greece Dr. Matthew Warren, Deakin University Geelong, Victoria, Australia MEDIATEC Programme Committee Assoc. Prof. Vincent Charvillat, IRIT-ENSEEIHT, Toulouse cedex, France Dr. Fernando Boronat Segui, Universidad Politecnica de Valencia, Gran de Gandia, Spain Ignazio Infantino, ICAR-CNR, Palermo (PA), Italy PhD. Jens Mueller-Iden, University of M.9fnster, M.9fnster, Germany Dr. Ana Pajares, Universidad Politecnica de Valencia, Valencia, Spain Jehan Francois Paris, University of Houston, Houston, USA Prof. Marco Roccetti, Universita' di Bologna, Bologna, Italy Dr. Leon Rothkrantz, Delft University of Technology, Delft, The Netherlands Dr. Leonid Smalov, Coventry University, Coventry, United Kingdom Prof. Rik Van de Walle, Ghent University, Ghent, Belgium COMTEC Programme Committee Prof. Dr. Marwan Al-Akaidi, De Montfort University, Leicester, United Kingdom Boguslaw Butrylo, Bialystok Technical University, Bialystok, Poland Dr. Nathan Clarke, University of Plymouth, Plymouth, United Kingdom Dr. Steven Furnell, University of Plymouth, Plymouth, United Kingdom Prof. Chris Guy, The University of Reading, Reading, United Kingdom PhD Mohammad Riaz Moghal, Ali Ahmad Shah-University College of Engineering and Technology, Mirpur, Pakistan PhD Roberto Montemanni, IDSIA, Manno-Lugano, Switzerland Maria Papadaki, University of Plymouth, Plymouth, United Kingdom Ph. D. Oryal Tanir, Bell Canada, Montreal, Canada Ass. Prof. Vassilis Triantafillou, ?echnological Educational Institution of Messolonghi Applied, Greece APTEC Programme Committee Prof. Dr. J. Broeckhove, RUCA-UA, Antwerp, Belgium Dr. Juan Carlos Guerri Cebollada, Universidad Politecnica de Valencia, Valencia, Spain Hatice Gunes, University of Technology, Sydney (UTS), NSW Australia Carsten Magerkurth, SAP RESEARCH, St. Gallen, Switzerland Dr.ir. Johan Opsommer, Belgacom - BUS, Brussels, Belgium Prof. Matthias Rauterberg, Eindhoven University of Technology, Eindhoven, The Netherlands. Francisco Reinaldo, FEUP, University of Porto, Porto, Portugal Zasriati Azla Sabot, University College Jubail, Jubail Industrial City, Saudi Arabia Prof. Jeanne Schreurs. Hasselt University, Diepenbeek, Belgium Ass. Prof. Ramiro Vel.87zquez, Universidad Panamericana, Aguascalientes, Mexico Dr. Charles van der Mast, Delft University of Technology, Delft, The Netherlands E-TEC Programme Committee Dr.Steven Furnell, University of Plymouth, Plymouth, United Kingdom Dr. Paul Dowland, University of Plymouth, Plymouth, United Kingdom Knowledge Management and E-Mobility Prof. Ricardo Chalmeta, Universidad Jaume I, Castellon, Spain Prof. Dr.-Ing. Stephan Kassel, University of Applied Sciences Zwickau, Germany Workshops Medical Imaging Systems General Chair Jo.8bo Manuel R. S. Tavares, FEUP, University of Porto, Porto, Portugal General Co-Chair Renato Natal Jorge, FEUP, University of Porto, Porto, Portugal International Programme Committee Alberto De Santis, Universit.88 degli Studi di Roma La Sapienza, Italy Arrate Mu.96oz Barrutia, University of Navarra, Spain Behnam Heidari, University College Dublin, Ireland Bernard Gosselin, Faculte Polytechnique de Mons, Belgium Chandrajit Bajaj, University of Texas, USA Christos E. Constantinou, Stanford University School of Medicine, USA Daniela Iacoviello, Universit.88 degli Studi di Roma La Sapienza, Italy Dinggang Shen, University of Pennsylvania, USA Djemel Ziou, University of Sherbrooke, Canada Gerald Schaefer Aston University, United Kingdom Jo.8bo Krug Noronha, Dr. Krug Noronha Clinic, Portugal Jo.8bo Manuel R. S. Tavares, Faculty of Engineering of University of Porto, Portugal Jo.8bo Paulo Costeira, Instituto Superior T.8ecnico, Portugal Jorge M. G. Barbosa, Faculty of Engineering of University of Porto, Portugal Lyuba Alboul, Sheffield Hallam University, United Kingdom Manuel Gonz.87lez Hidalgo, Balearic Islands University, Spain Maria Elizete Kunkel, Universit.8at Ulm, Germany M.87rio Forjaz Secca, Universidade Nova de Lisboa, Portugal Miguel Angel L.97pez, Faculty University of Ciego de Avila, Cuba Miguel Velhote Correia, Faculty of Engineering of University of Porto, Portugal Patrick Dubois, Institut de Technologie M.8edicale, France Reneta Barneva, State University of New York, USA Renato M. Natal Jorge, Faculty of Engineering of University of Porto, Portugal Sabina Tangaro, University of Bari, Italy Valentin Brimkov, State University of New York, USA Yongjie Zhan, Carnegie Mellon University, USA D-TV Workshop Dr. Hans-Joachim Nern, TTVI, Germany CORRESPONDENCE ADDRESS Philippe Geril Ghent University Faculty of Engineering Dept. of Industrial Management Technologiepark 803 B-9062 Ghent-Zwijnaarde, Belgium Tel: +32 9 2645509 Fax: + 32 9 2645824 Email: philippe.geril@eurosis.org -- Philippe Geril Tel: +32.9.264.55.09 EUROSIS -ETI Fax: +32.9.264.58.25 Ghent University E-mail: philippe.geril@eurosis.org Dept.of Industrial Mgmt. E-mail: pgeril@yahoo.co.uk Technologiepark 903 URL: http://www.eurosis.org Campus Ardoyen B-9052 Ghent-Zwijnaarde Belgium **************************************************************************** ******* * Your Scientific information site on * * Computer Simulation - Concurrent Engineering - Multimedia- Games * * WWW.EUROSIS.ORG * **************************************************************************** ******* === Subject: Re: Is i^i transcendental? Nntp-Posting-Host: hera.cwi.nl ... Yes, the sets of solutions are not equal. Does that mean the equation 1^(1/4) = (1^2)^(1/8) is incorrect? If yes, where did we go wrong? Ok, it may be incorrect. > i = -1^(1/2) = ((-1)^2)^1/4 = 1^(1/4) which is not correct, because i > is only one of the roots for 1^(1/4). And it is only one of the roots of (-1)^(1/2). > But then, in general, if a^(p/q) and b^(m/n) have one root in common > where a,b,p,q,m,n are all positive integers, then can't we say a^(p/q) > = b^(m/n)? Only as long as you consider real answers only, and only those that are >= 0. But doing this with non-real answers is simply wrong. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Analysis with sum.. posting-account=z1ZA6AoAAACEgXDaRRTJFG5d4vJvYyOY SV1),gzip(gfe),gzip(gfe) Hello sir~ n is positive integer. n + (n-1).2 + (n-2).2^2 + ... + 2.2^(n-2) + 2^(n-1) = ? --------------------------------------------- Maybe... sum{k=0 to n-1} (n-k).2^k = sum{k=0 to n-1} n.2^k - sum{k=0 to n-1} k.2^k sum{k=0 to n-1} k.2^k ?? How can you calculate this ? === Subject: Re: Analysis with sum.. posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > Hello sir~ n is positive integer. n + (n-1).2 + (n-2).2^2 + ... + 2.2^(n-2) + 2^(n-1) = ? --------------------------------------------- > Maybe... sum{k=0 to n-1} (n-k).2^k = sum{k=0 to n-1} n.2^k E- Esum{k=0 to n-1} k.2^k sum{k=0 to n-1} k.2^k ?? How can you calculate this ? S = sum{k=0 to n-1} k.x^k S/x = sum{k=0 to n-1} k.x^(k-1) = d/dx(sum{k=0 to n-1} x^k) ... === Subject: Re: Analysis with sum.. posting-account=z1ZA6AoAAACEgXDaRRTJFG5d4vJvYyOY SV1),gzip(gfe),gzip(gfe) > Hello sir~ > n is positive integer. > n + (n-1).2 + (n-2).2^2 + ... + 2.2^(n-2) + 2^(n-1) = ? > --------------------------------------------- Maybe... sum{k=0 to n-1} (n-k).2^k > = sum{k=0 to n-1} n.2^k - sum{k=0 to n-1} k.2^k > sum{k=0 to n-1} k.2^k ?? How can you calculate this ? S = sum{k=0 to n-1} k.x^k S/x = sum{k=0 to n-1} k.x^(k-1) > = d/dx(sum{k=0 to n-1} x^k) ... Yes, good idea. d/dx[1 + x + x^2 + ... + x^(n-1)] = d/dx[{(x^n)-1}/(x-1)] ==> 1 + 2x + ... + (n-1).x^(n-2) = [{n.x^(n-1).(x-1)} - {(x^n)-1}]/ (x-1)^2 ==> x*[1 + 2x + ... + (n-1).x^(n-2)] = x*[[{n.x^(n-1).(x-1)} - {(x^n)-1}]/(x-1)^2] If x = 2, then sum{k=0 to n-1} k.2^k = n.2^n - 2^(n+1) + 2. so, sum{k=0 to n-1} n.2^k - sum{k=0 to n-1} k.2^k = n.2^n - n - [n.2^n - 2^(n+1) + 2] = 2^(n+1) - n - 2. === Subject: Re: Analysis with sum.. posting-account=a6woBRAAAADpNFZJBA7ZBx35zXaKmaP4 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Hello sir~ n is positive integer. n + (n-1).2 + (n-2).2^2 + ... + 2.2^(n-2) + 2^(n-1) = ? --------------------------------------------- > Maybe... sum{k=0 to n-1} (n-k).2^k = sum{k=0 to n-1} n.2^k - sum{k=0 to n-1} k.2^k sum{k=0 to n-1} k.2^k ?? How can you calculate this ? Try expressing it as a double summation, with the inner sum being a geometric series. === === === Subject: Welcome to my webside that provides help in English Study. posting-account=fichWgoAAADR2tEgXerph8SNcScYbuNK TencentTraveler ),gzip(gfe),gzip(gfe) The address is http://englearn.zhan.cn.yahoo.com === Subject: Re: Frechet Differentiablity giving me problems. I imagine the way to show this is to use the chaine rule. Ignoring the absolute value the derivative of Ef-> (f)^p is p|f|^{p-1} by the binomail theorem without any problem. The problem seems to be showing that the derivative of f -> |f| is sign(f). Here's my issue: |f+g| - |f| - sign(f) g = [sign(f+g)-sign(f)][f+g] So I'm integrating over the region where sign(f+g) and sign(f) differ. Now if I take g supported outside the support of f, this integral will be the order of the integral of |g| and the limit won't go to zero. Now I Edo realize the denominator is the L^p norm of g, but since g can be arbitary I don't see that this will help. I really appreciate everyone's help. If there is something easier please let me know. On Dec 22, 10:13Epm, The World Wide Wade I'm studying for a PhD general exam and am a bit caught up on the topic of frechet differentiability. Here is specifically the problem I'm trying to work out: Consider A_p : L^p [0,1] -> L^1 [0,1] defined by A_p f=|f|^p. For what p is A_p (once and twice) strongly differentiable, and in these cases what are the derivatives? I would generally post my progress as good-faith evidence that I'm not trying to get the group to do my homework, but my problem seems to be very fundamental. In, say, the standard case of differentiation, the derivative is defined as a limit, which one can try to compute. Here I need to _find_ a linear operator such that ||A_p(f+g)-A_p(f)-Tg ||/ || g|| -> 0. EThis seems difficult in that the binomial expansion of (f +g)^p is messy, and there are the absolute values further complicating matters. The more I think about this, unless I'm missing something, this > doesn't appear true. Since, as I stated above the limit in the case > that the support is on different sets, is the l^1 norm of g over the > l^1 norm of g which tends to 1 not zero. Any help please? I didn't > state this in the original formulation but we may assume that the > domain is [0,1], although this doesn't get around the issues I > mentioned. You already said the domain was [0,1]; see your own post. And why did newsreader should have a reply feature that quotes all previous text automatically. Try it. I did not understand what you you meant with Ignoring the absolute value the derivative of Ef-> (f)^p is p|f|^{p-1} by the binomail theorem without any problem above. Certainly f -> |f|, going from L^p to L^p, is not going to be differentiable at certain f's (like f = 0). Back to the original problem. Assume p > 1. We want int ||f+g|^p - |f|^p - p|f|^(p-1)sgn(f)*g| = o(||g||_p) as ||g||_p -> 0. On the set where |f|<= |g|, the integrand is dominated by a constant times |g|^p. And int |g|^p/||g||_p -> 0. On the set where 0 < |g| <= |f| (we can ignore the set where g = 0), factor out |g| to get |g| * |(|f+g|^p - |f|^p)/g - p|f|^(p-1)sgn(f)|. We can assume g = g_n and g_n -> 0 a.e. Call the factor on the right h_n; clearly h_n -> 0 a.e. Apply Holder to get the integral of the above <= ||g_n||_p*||h_n||_q, with q = p/(p-1). We have |h_n|^q <= constant time |f|^p on this set for all n; apply Lebesgue's dominated convergence theorem. === Subject: Re: A question in module theory! <5tfc35F1bsu78U1@mid.individual.net> posting-account=tUxNdAoAAACNcm0ILHRuAqJoxjIfbQUy 5.1),gzip(gfe),gzip(gfe) Hi! I dared to solve this problem two days but no gain. So I put this problem in this group hoping to get some clue. Let S be the multiplicative set of commutative ring with unity R and > J is an injective R-module. > Then, J/S is an injective R/S-module. (here, R/S is a quotient ring > of R) > You mean localization with respect to S, not an actual quotient ring, right? > quasi and as a second hint: localization of exact sequences remain exact- [Micro]uA AO.bd .bcuaa - - [Micro]uA AO.bd a - To quasi: Right! Localization is with respect to S. To Axel: Your hint is right! However, finging exact sequence of R- module from exact R/S-module sequence is too dificult. Namely, Our goal is to show, given exact R/S-module sequence 0->J/S->A- >B->0, it is indeed split sequence. To use J is injective module, we want to get R-module exact sequence 0->J->A->B->0 but I didn't find such sequence induced from the given one. === Subject: Re: A beginner's guide to forcing posting-account=klgwcwoAAACz-mmR4MZRhJc8QJZqsKqZ 98),gzip(gfe),gzip(gfe) > P.S. -- messages in this thread are not showing up > in my newsreader. EDon't know why. I downloaded this previous post of mine as soon as I had posted it (from Google). Now it's gone off my newsreader, along with all the rest in this topic. Weird. -- hz === Subject: Re: contradiction method of proof > Would you please show me or give me some clues for solving the > following proof question: Prove that if x is a positive real number, then x/(x+1) < (x+1)/(x+2). Since 0 < x < x + 1 < x + 2. x/(x+1) < (x+1)/(x+2) iff x(x + 2) < (x + 1)^2 === Subject: Re: contradiction method of proof > Would you please show me or give me some clues for solving the >> following proof question: > Prove that if x is a positive real number, then x/(x+1) < (x+1)/(x+2). One method, in discovery mode, is to multiply both sides by the least > common denominator (which is guaranteed to be positive -- why?). At > each step, make sure the new inequality is equivalent to the old one. > When you reach a truism, you can make a proof by just reversing the > steps. Another method (but the same at heart) is to simplify RHS - LHS and > show that it's always positive. An entirely different method is to note that the inequality to be > proved has the form f(x) < f(x+1) where f(x) = x/(x+1). Then show, > using calculus, that f is increasing on the interval (0,infinity). More geometrically insightful: (x+1)/(x+2) is the Farey mediant of x/(x+1) < 1/1 so it necessarily lies strictly between them. For more on Farey mediants follow this link to my prior post --Bill Dubuque === Subject: Re: contradiction method of proof posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) An entirely different method is to note that the inequality to be proved has the form f(x) < f(x+1) where f(x) = x/(x+1). Then show, using calculus, that f is increasing on the interval (0,infinity). No need for calculus ... > -- > Angus Rodgers > Contains mild peril *************************************************************8 You were the one that proposed the good idea of looking at the expression f(x) < f(x+1), with f(x) = x/(x+1), and asked Do you see anything about the function f that helps? (No need for a proof by contradiction, by the way.) Then you said to quasi that no calculus required, so what did you mean then by seeing anything about f that helps? The basic algebraic methods to solve the inequation have already been described here, but what else than using calculus and showing that f' > 0 always and thus f is ascending, can be used here with the function f(x)? Tonio === Subject: Re: contradiction method of proof > An entirely different method is to note that the inequality to be proved has the form f(x) < f(x+1) where f(x) = x/(x+1). Then show, using calculus, that f is increasing on the interval (0,infinity). No need for calculus ... As quasi had already given two methods not involving calculus (which you snipped), your comment is a bit redundant. === Subject: Re: contradiction method of proof >As quasi had already given two methods not involving calculus >(which you snipped), your comment is a bit redundant. I was trying to hint specifically that one does not need to use calculus to prove that f(x) = 1 - 1/(x + 1) is increasing - and it simply never occurred to me that my comment was ambiguous. (Hinting helpfully must be harder than I thought!) -- Angus Rodgers Contains mild peril === Subject: Re: contradiction method of proof >As quasi had already given two methods not involving calculus >>(which you snipped), your comment is a bit redundant. I was trying to hint specifically that one does not need to use >calculus to prove that f(x) = 1 - 1/(x + 1) is increasing - and >it simply never occurred to me that my comment was ambiguous. >(Hinting helpfully must be harder than I thought!) That's a fine method. In fact, it gives more insight as to the why than the other methods. quasi === Subject: Algebra with sum... Hello sir~ n is positive integer. n + (n-1).2 + (n-2).2^2 + ... + 2.2^(n-2) + 2^(n-1) = ? --------------------------------------------- Maybe... sum{k=0 to n-1} (n-k).2^k = sum{k=0 to n-1} n.2^k - sum{k=0 to n-1} k.2^k sum{k=0 to n-1} k.2^k ?? How can you calculate this ? === Subject: Re: Algebra with sum... >n is positive integer. n + (n-1).2 + (n-2).2^2 + ... + 2.2^(n-2) + 2^(n-1) = ? --------------------------------------------- >Maybe... sum{k=0 to n-1} (n-k).2^k = sum{k=0 to n-1} n.2^k - sum{k=0 to n-1} k.2^k >sum{k=0 to n-1} k.2^k ?? How can you calculate this ? Start with the familiar formula for the geometric series n-1 --- k 1 - x^n > x = ------- [1] --- 1 - x k=0 Differentiate to get n-1 --- k-1 1 + (n-1) x^n - n x^{n-1} > k x = ------------------------- [2] --- (1-x)^2 k=0 Multiply [1] by n and [2] by x, then subtract to get n-1 --- k n - (n+1) x + x^{n+1} > (n-k) x = --------------------- [3] --- (1 - x)^2 k=0 For the original question, set x = 2 to get n-1 --- k n+1 > (n-k) 2 = 2 - n - 2 [4] --- k=0 Rob Johnson take out the trash before replying === Subject: Re: Algebra with sum... posting-account=MCw_jgoAAACQIHCmmNFnEMO1eaBkLS8n Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) NetApp/6.0.5) > Hello sir~ n is positive integer. n + (n-1).2 + (n-2).2^2 + ... + 2.2^(n-2) + 2^(n-1) = ? --------------------------------------------- > Maybe... sum{k=0 to n-1} (n-k).2^k = sum{k=0 to n-1} n.2^k - sum{k=0 to n-1} k.2^k sum{k=0 to n-1} k.2^k ?? How can you calculate this ? S(x) = x^0 + x^1 + x^2 + ... + x^k = (x^(k+1) - 1)/(x-1) Thus, xS'(x) = 1*x^1 + 2*x^2 + ... + k*x^k = x*(1 + k*x^(k+1) -(x^k)*(k+1))/ ((x-1)^2) Putting x = 2 gives, sum{k=0 to n-1}k.2^k = 2*(1 + (n-1)*2^n - (2^(n-1))*n)) Reetesh Mukul === Subject: Re: Algebra with sum... posting-account=MCw_jgoAAACQIHCmmNFnEMO1eaBkLS8n Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) NetApp/6.0.5) > Hello sir~ > n is positive integer. > n + (n-1).2 + (n-2).2^2 + ... + 2.2^(n-2) + 2^(n-1) = ? > --------------------------------------------- Maybe... sum{k=0 to n-1} (n-k).2^k > = sum{k=0 to n-1} n.2^k - sum{k=0 to n-1} k.2^k > sum{k=0 to n-1} k.2^k ?? How can you calculate this ? S(x) = x^0 + x^1 + x^2 + ... + x^k = (x^(k+1) - 1)/(x-1) > Thus, > xS'(x) = 1*x^1 + 2*x^2 + ... + k*x^k = x*(1 + k*x^(k+1) -(x^k)*(k+1))/ > ((x-1)^2) Putting x = 2 gives, sum{k=0 to n-1}k.2^k = 2*(1 + (n-1)*2^n - (2^(n-1))*n)) Reetesh Mukul Or Better, 2*(1 + (n-2)*2^(n-1)) === Subject: Re: Algebra with sum... > Hello sir~ n is positive integer. n + (n-1).2 + (n-2).2^2 + ... + 2.2^(n-2) + 2^(n-1) = ? --------------------------------------------- > Maybe... sum{k=0 to n-1} (n-k).2^k = sum{k=0 to n-1} n.2^k - sum{k=0 to n-1} k.2^k > sum{k=0 to n-1} k.2^k ?? How can you calculate this ? Sorry. Analysis with sum... === Subject: PageRank Explained - Keeping SEO Math Simple posting-account=5Hx47goAAADQG8G8UkplmcgrtL2O-Rcm SLCC1; .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) PageRank Explained - Keeping SEO Simple What is Google's PageRank? If you have ever done any reading about search engine optimization or were just curious how you can get your site to the top of the Google search engine results, understanding PageRank is vital. I'm going to introduce you to the basics of PageRank and also provide a brief discussion on how much you should really worry about PageRank if you are running a website or Internet business. Google's founders, Larry Page and Sergey Brin, invented PageRank and it forms the basis for how Google works. Google didn't become the best search engine in the world by chance, it became the best search engine because it provided the best results. PageRank is in fact the technology that gave Google its competitor-killing edge, a way to greatly improve the accuracy and validity of a search response to a user query. In essence PageRank provides a means to determine the value of a website for any given search term or keyword phrase. This value is determined by how websites link together with the more popular (and theoretically better) sites receiving more links. It's these incoming links that help the site have a high PageRank value and thus display higher up in search results. Let's read how Google explains their PageRank system: PageRank relies on the uniquely democratic nature of the web by using its vast link structure as an indicator of an individual page's value. Google interprets a link from page A to page B as a vote, by page A, for page B. But, Google looks at more than the sheer volume of votes, or links a page receives; it also analyzes the page that casts the vote. Votes cast by pages that are themselves important weigh more heavily and help to make other pages important. Important, high-quality sites receive a higher PageRank, which Google remembers each time it conducts a search. Of course, important pages mean nothing to you if they don't match your query. So, Google combines PageRank with sophisticated text-matching techniques to find pages that are both important and relevant to your search. Google goes far beyond the number of times a term appears on a page and examines all aspects of the page's content (and the content of the pages linking to it) to determine if it's a good match for your query. The key rule to understand is that it is a combination of variables that determine how well your site performs in Google. These are the most important variables to worry about: Incoming links to your site. The relevancy (to your site's theme) of the pages linking to your site and the PageRank of these pages. The keywords that other sites use to link to your site. The keywords on your website in particular in places like page titles and headlines. Some of those factors you can control, others you can manipulate but not directly control. The important thing to understand regarding PageRank is that all those variables will determine how high your site shows up in search engine results. PageRank is the name for the technology that ranks sites and includes all those variables and many more. PageRank Numbers - The Little Green Bar If you install the Google Toolbar into your browser you can choose to switch on the PageRank display (it's in the options). This will make a little green bar appear above web pages you visit. The green bar represents the PageRank of the page you are viewing in your browser. The ranking starts at 0 (no ranking) up to 10, the highest ranking and can be blanked out completely if the page has been banned from Google. If you don't want to use the toolbar you can try this free PageRank lookup tool to find the ranking for any web address. Google created quite a storm when it launched its green PageRank bar. Webmasters became obsessed with methods to increase their PageRank and high PageRank sites started selling text links for hundreds of dollars. A link from a high PageRank page, from a PageRank 7, 8, 9 or 10, has been known to make lower PageRank pages increase a full number, even two if the incoming link is from a PageRank 10, and there is no doubt it is good for search engine rankings. The problem with PageRank being displayed in a little green bar is that it is very hard to really gauge how valuable a ranking is. The Google PageRank technology is complex containing many variables, some of which I mentioned above, and to interpret a number from 0-10, especially when only Google really knows how it works, is difficult. Worse still, the visible representation, the green bar that the public can see, only changes on a quarterly basis, while the real PageRank of a page changes on a daily basis. Most of the time you are looking at a very outdated ranking value. PageRank paranoia is an issue that every webmaster may fall victim to. There are rumours that Google will be changing the PageRank system because they are not happy with how it is being manipulated and interpreted. As a rule of thumb, watch the green bar with interest but don't take it too seriously or spend too much time trying to force it to increase (staring and yelling at it will do you no good, trust me on that one). The Randomness Of PageRank Search engine optimization experts actively track PageRank and investigate things like page backlinks to try and work out what the top search engine ranked sites are doing right so they can replicate and then surpass them in the rankings. This is a very good strategy for any person running a web business looking to improve their search ranking. There is no need to reinvent the wheel - copy what works and do it slightly better than the competition. This is all good in theory, but unfortunately there is a good amount of randomness in PageRank and search engine results. Google of course would argue that it's not randomness and their PageRank system is merely using algorithms that we don't understand, and no doubt that is true, but for the human webmaster trying to get traffic, PageRank and Google can be baffling sometimes. There are instances of high PageRanked sites having little to no backlinks. Given that incoming links are one of the most important variables used in PageRank calculations you have to scratch your head and wonder how a site with no links could have a big green bar. To Backlinks - for details) is another phenomenon that search engine experts often choose to ignore rather than try and evaluate. as well, with your sites jumping high into search results in places where you wouldn't expect it. The only consistency is randomness but there is logic that can be followed and smart search engine optimization practices that when implemented well will work. Just don't expect it to work precisely how or when you want it to. What You Should Know And Do About PageRank This advice I offer from experience as an avid PageRank chaser and search engine optimizer. The key to gaining PageRank is to ignore it and focus on the variables that control it. Having people link to your site has always been a good thing and PageRank was in fact a result of this. Don't get confused with the order of things, first came the Internet and links and then came PageRank. Focus on amassing quality incoming links from quality sites relevant to your site. This practice will naturally improve your PageRank and also increase the amount of visitors coming to your site. Don't get bogged down chasing links from only high PageRank sites or waste energy adding links from just any site willing to link to you. Do things naturally and your site will grow naturally. with this. Keywords play a crucial role in bringing the right type of traffic to your site but you should never spend half an hour in front of a computer trying to come up with the perfect title for your and easily develop good keywords without spending hours and hours tweaking every little phrase and heading. See what your competitors do If you build a good website with good content, always keep in mind your important keywords and proactively work every day to earn and create new backlinks to your site you will improve your PageRank. The best sites with the highest PageRank never worry about PageRank, they simply keep churning out content that people love to link to. This is a strategy that every webmaster and Internet entrepreneur should emulate for success online. Webmaster http://www.123movingcompany.com Internet Business Coach === === === Subject: help for Number Theory text-books posting-account=33KaEgkAAAA9tz8WICNABjrkyMKXFbGS Gecko/20070309 Firefox/2.0.0.3,gzip(gfe),gzip(gfe) My friend KRR Gandhi, a number theory enthusiast from India wishes to get the following used books at as low a cost as possible in his further research. Can you please help him? ( Expert in Narasimham Fibonacci and Lucas Numbers Author: Koshy, Thomas Publisher: John Wiley & Sons Inc ISBN: 0471399698 Fibonacci Numbers and Their Applications Author: Bergum, Gerald E. (Edt) Publisher: Kluwer Academic Print on Demand ISBN: 902772234X === === === === === Subject: Emil Post works - where can I find them? Hi I wanted to read The two-valued iterative systems in mathematical logic, but all I could find on Internet was 2-pages review on JSTOR. Do you know any other site of this kind? === Subject: Re: Emil Post works - where can I find them? posting-account=-eQqtQoAAACZVM-kNEsOn3k7GSvoJoS4 MathPlayer 2.0; .NET CLR 1.1.4322; InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Hi I wanted to read The two-valued iterative systems in mathematical logic, > but all I could find on Internet was 2-pages review on JSTOR. Do you know > any other site of this kind? > HTH === Subject: Re: Emil Post works - where can I find them? === Subject: Re: Emil Post works - where can I find them? posting-account=iBgNeAoAAADRhzuSC4Ai7MUeMmxtwlM7 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) Meta-learning. === Subject: Ring Problem for the Holiday Season Ok..this might be quite an easy problem, I haven't tried solving myself but I think it would be interesting to post the problem here and as we solve the problem I will add more conditions to form some newer problem to make it interesting. The first question I am thinking is something like this: 1. Find an example of two nontrivial unitary commutative rings (I'll name these CRings from now on, this is the way most people denote this category) say A and B such that a. A is a subring of B b. There are two nonzero elements b and b' in B such that i) There is an element a in A such that ab in A{0} ii) For any element a' in A such that ab in A{0} we have the property a'b = a'b' in A{0} iii) b and b' are not equal! I might post an example later on if one pops in my mind. PS: For those who like names.. If A is a semiprime ring then the ring A[b] is called a rational extension of A (for obvious reasons). Jose Capco === Subject: Re: Ring Problem for the Holiday Season posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) Ok..this might be quite an easy problem, I haven't tried solving myself but I think it would be interesting to post the problem here and as we solve the problem I will add more conditions to form some newer problem to make it interesting. The first question I am thinking is something like this: 1. Find an example of two nontrivial unitary commutative rings (I'll name these CRings from now on, this is the way most people denote this category) say A and B such that a. A is a subring of B > b. There are two nonzero elements b and b' in B such that i) There is an element a in A such that ab in A{0} > ii) For any element a' in A such that ab in A{0} we have the property a'b = a'b' in A{0} > iii) b and b' are not equal! > Note that I_b := { a in A | ab in A } is an ideal in A You want b in BA such that I_b is non-zero and (b-b')*I_b = 0. Let A = Z/4Z, B = A[1/2] = A[X]/(X-2). Then I_{1/2} = 2A. Since 2*2A=0, let b' = b+2 More explicitly, we have b = 1/2, b' = 2 + 1/2, a = 2 and necessarily a'= 2. > I might post an example later on if one pops in my mind. PS: For those who like names.. If A is a semiprime ring then the ring A[b] is called a rational extension of A (for obvious reasons). Jose Capco === Subject: Re: Ring Problem for the Holiday Season posting-account=hhC2JwoAAAAQt9ZcdRPKAFNCTwZjbe1M 1.1.4322),gzip(gfe),gzip(gfe) Ok..this might be quite an easy problem, I haven't tried solving myself but I think it would be interesting to post the problem here and as we solve the problem I will add more conditions to form some newer problem to make it interesting. The first question I am thinking is something like this: 1. Find an example of two nontrivial unitary commutative rings (I'll name these CRings from now on, this is the way most people denote this category) say A and B such that a. A is a subring of B > b. There are two nonzero elements b and b' in B such that E Ei) There is an element a in A such that ab in EA{0} > E Eii) For any element a' in A such that ab in A{0} we have the property Did you mean a'b in A{0}? > a'b = a'b' in A{0} > E Eiii) b and b' are not equal! What's wrong with this: B = {000, 001, 010, 011, 100, 101, 110, 111}, the set of all binary vectors of length 3, with componentwise addition mod 2; A = {000, 011, 100, 111}, b = 110, b' = 111? === Subject: Re: Ring Problem for the Holiday Season posting-account=hhC2JwoAAAAQt9ZcdRPKAFNCTwZjbe1M 1.1.4322),gzip(gfe),gzip(gfe) Ok..this might be quite an easy problem, I haven't tried solving myself but I think it would be interesting to post the problem here and as we solve the problem I will add more conditions to form some newer problem to make it interesting. > The first question I am thinking is something like this: > 1. Find an example of two nontrivial unitary commutative rings (I'll name these CRings from now on, this is the way most people denote this category) say A and B such that > a. A is a subring of B b. There are two nonzero elements b and b' in B such that > E Ei) There is an element a in A such that ab in EA{0} E Eii) For any element a' in A such that ab in A{0} we have the property Did you mean a'b in A{0}? > a'b = a'b' in A{0} E Eiii) b and b' are not equal! What's wrong with this: B = {000, 001, 010, 011, 100, 101, 110, 111}, the set of all binary > vectors of length 3, with componentwise addition mod 2; A = {000, 011, 100, 111}, b = 110, b' = 111? Oops, I overlooked your second message. My example is no good because b' is in A. Also I neglected to define multiplication. All right, what about this: B is the set of all binary vectors of length 4, with componentwise addition and multiplication mod 4; A = {0000, 0111, 1000, 1111}, b = 1100, b' = 1110. === Subject: Re: Ring Problem for the Holiday Season One more thing to add.. Of course I want b and b' not to be an element of A.. otherwise this is a trivial question.. === Subject: is this feasible?? hi please check out the design @ NASA tech contest and comment on it. http://www.createthefuturecontest.com/pages/view/entriesdetail.html?entryID= 798 === Subject: Re: is this feasible?? posting-account=HZYXOQoAAAB0CZtsRCtABgys4tHYIT8J 3.2.0; .NET CLR 1.1.4322; InfoPath.2),gzip(gfe),gzip(gfe) > hi please check out the design @ NASA tech contest and comment on it. http://www.createthefuturecontest.com/pages/view/entriesdetail.html?e... > It might work but it is far from being the best solution. The best solution is a gyroscope. This can be also be used to store energy and brake regeneratively. Also with a gyroscope you can simply sit in your vehicle as you do with a car. See http://ianparker.g3z.com/DriverlessCar.htm You will see these features. This is also conceived of as a vehicle which can travel along monorail type tracks. If you are in a sitting position you can have a teardrop and you can travel (in a controlled environment) greenly at 200km/h or so. You will note the squiggles at the top. This will provide you with a translation in Arabic should you want it. The lack of a mass transit system in Damascus prompted this. This would provide a state of the art system. It will drive itself on monorail type tracks, or it could be driven in an uncontrolled environment. You mention NASA. Could it be driven on Mars? I don't see why not, although on Mars you would probably want a unicycle which was stabalized in both directions. - Ian Parker === === === === === === === === === === === === === Subject: lambertW2 z= lambertW2(z)* exp( exp (lambertW2(z)) ) in other words lambertW2(z) is the inverse of z*exp(exp(z)) a nice function ... === Subject: Re: lambertW2 > z= lambertW2(z)* exp( exp (lambertW2(z)) ) > in other words lambertW2(z) is the inverse of z*exp(exp(z)) No it's not. The principal branch of the inverse of z*exp(exp(z)) is real-valued for example in [0,+oo). W(2,x) is complex valued there. The inverse of z*exp(exp(z)) cannot be expressed in terms of the Lambert W function. > a nice function ... -- I.N. Galidakis === Subject: proof involving primes posting-account=CcB_dwoAAADW9dap2h4IHtjSfIr3uklw Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) How can I prove that there an infinite number of primes under the assumption that every natural number greater than 1 has a prime divisor? === Subject: Re: proof involving primes How can I prove that there an infinite number of primes under the > assumption that every natural number greater than 1 has a prime divisor? HINT NN+N has more primes than N --Bill Dubuque === Subject: Re: proof involving primes posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) How can I prove that there an infinite number of primes under the > assumption that every natural number greater than 1 has a prime > divisor? > ************************************************************ Let p1,p2,...,pn prime numbers ==> N:= p1*p2*...*pn + 1 is either a prime number or divisible by a prime number different from p1,...,pn ==> in any case, you get a prime different from the n above, and this is true for any n different primes you take. Tonio === Subject: best one posting-account=ieyfCwoAAAAGZHDgkXOd2bhOI7BXVajk Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR.3.0.04131.06; MEGAUPLOAD 2.0),gzip(gfe),gzip(gfe) test this one for free : http://www.AWSurveys.com/HomeMain.cfm?RefID=rezanassimi === Subject: Re: Question re: accumulation points of a set >...because x-d would be the new sup of S. More or less. If x is an upper bound for S, x is not an element of S, and (x-d,x) contains no point of S then x-d is an upper bound for S. Hence x is not the _least_ upper bound. You can't conlude that x-d is the sup of S unless you specify that we took d as large as possible... And thank you for pointing out my either-OR mistake. > As noted in my first post, this is an introductory analysis book. > There is no definition for adherent in the book. The definition >given in the book is: >Let S be a set of real numbers. A real number A is an accumulation > point of S iff every neighborhood of A contains infinitely many > points of S > In addition, the book provides the following as a lemma: >Let S be a set of real numbers. Then A is an accumulation point > of S iff each neighborhood of A contains a member of S different > from A > Ok, that gives you an easy way to do the problem. >> Say x = sup(S) and x is not an element of S. You >> need to show that x is an accumulation point of S. >> So by the lemma it's enough to show that if d > 0 >> then the interval (x-d, x) contains a point of S. > Suppose that on the other hand d > 0 and (x-d, x) >> does not contain any element of S. That gives a >> contradiction: x cannot be the sup of S, because... > I am working on my own through Gaughan's introductory > analysis book, > and I'm stuck on question 22 of chapter 1 on > Sequences, which reads as > follows: > - Let S be a nonempty set of real numbers that is > bounded from above > (below) and let x = sup S (inf S). Prove that either > x belongs to S or > x is an accumulation point of S. > Assuming that S is bounded from above and that S is > infinite, isn't it > the case that if x is a member of S, it would still > be an accumulation > point of the set, since every neighborhood of x would > contain an > infinite number of elements of S (to the left of x)? > The question is asking to prove an 'either A or B' > statement, which > means only one of them can be true. Is there > something I'm missing > here?? I'm I reading it wrongly? > I think you're just not reading the definitions carefully enough. >> First, the 'or' is to be taken in the inclusive sense. >> So in your example, x can satisfy both conditions. > Second, I'd suggest rereading the definition of accumulation (limit) point, and then possibly comparing it to the definition of adherent point. >> Perhaps you're confusing the two. > Let me ask you this: >> Suppose the author had used the word adherent instead of accumulation. >> Would a proof still be possible? > ************************ > David C. Ullrich ************************ David C. Ullrich === Subject: Re: Question re: accumulation points of a set <11171626.1198665197588.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=FIpO7goAAAALHHUK_A5_cLG4lDv5-_LG Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) On Dec 28, 12:37 am, David C. Ullrich x is not an element of S, and (x-d,x) contains > no point of S then x-d is an upper bound for S. > Hence x is not the _least_ upper bound. You can't conlude that x-d is the sup of S unless > you specify that we took d as large as possible... Point noted, thank you. As you said, we've contradicted the fact that x is the least upper bound, thus the interval (x-d,x) must contain at least one member of S. => x is an accumulation point of S (by the given lemma) === Subject: Re: Question re: accumulation points of a set <11171626.1198665197588.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=FIpO7goAAAALHHUK_A5_cLG4lDv5-_LG Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > On Dec 28, 12:37 am, David C. Ullrich You can't conlude that x-d is the sup of S unless you specify that we took d as large as possible... Point noted, thank you. As you said, we've contradicted the > fact that x is the least upper bound, thus the interval (x-d,x) > must contain at least one member of S. > Should add here: so any neighborhood (x-d, x+d) for d >0 must contain at least one element of S > => x is an accumulation point of S (by the given lemma) === Subject: Re: Trig equations <2r01n3d797o2ai62954blsbg9spd3o77g0@4ax.com> Cc: deepkdeb@yahoo.com posting-account=iJfBPwgAAACaBAH7DreA6VfAOYvL5VDz 1.1.4322),gzip(gfe),gzip(gfe) > Consider the following two equations under the given conditions. > cos kD = (E^1/2)^k E E E E E E E E(1) > sin kD = (F^1/2)^k E E E E E E E E E(2) > Conditions: E and F are rational but not perfect squares. Both cos D and sin D are irrational. E, F < 1. Let E^1/2 = cos G and F^1/2 = sin G. Odd k > 3 0 < D, G < pi/2 > Then from (1) one gets cos kD = (cosG)^k E E (1.1) and from (2) one gets sin kD = (sin G)^k E E E E(2.1) > Assertion: There exists some k for which G = D > One helpful example: Let sin D = 1/3^(1/2) then sin 5D = (sin D)^5 > Any comment about the correctness of the assertion will be gratefully appreciated. Wow, you really didn't read my prior reply. Your hypothesis can be shown to be unsatisfiable using the same > argument that I used to invalidate the hypothesis of your previously > posted problem. Here's the argument again ... Given information (I'll use only what I need to get a contradiction): E Esin(k*D) = (F^(1/2))^k E Ecos(k*D) = (E^(1/2))^k E EE,F are positive rationals E Ek is a positive integer, k > 2 My claim: The above conditions are unsatisfiable. proof: E sin^2(k*D) + cos^2(k*D) = 1 => E^k + F^k = 1, contrary to Fermat's Last Theorem. Hence, since your hypothesis can't be satisfied, your assertion is > meaningless. > You don't have to use FLT to solve a problem which is so elementary in nature. Don't be silly. With a false hypothesis, you can prove anything. Hence > your assertion, while technically true, is meaningless. > Do you have a solution of the following equation? Sin(kD) = (sin D)^k Ewhere odd k > 5 and 0 < D < pi/2 For all positive integers k (odd or even), the equation sin(kD) = > sin^k(D) has at least one real solution with 0 < D < Pi/2. For k>9, the equation probably can't be solved in radicals, but you > can always solve numerically. quasi- Hide quoted text - - Show quoted text - I will greatly appreciate if you would kindly give me the solutions of the following two equations. Otherwise, kindly refer me to some appropriate literature.(D = ?, E = ?) sin kD =(sin D)^k (1) cos kE = (cos E)^k (2) Conditions: odd k > 5, 0 < D, E < pi/2 === Subject: Algebra with order pq and center. Hello sir~ Let G be a group of order pq, where p and q are distinct primes. Prove that center Z(G) cannot be a proper non-trivial subgroup of G. --------------------------------------- Don't worry. Let's go... 1) Lemma) G/Z(G) is cyclic then G is Abelian. If we suppose that Z(G) is a proper non-trivial subgroup then its order is either p or q and so the order of G/Z(G) is either q or p. so cyclic. By lemma, G is abelian and so Z(G) = G, a contradiction. 2) Class equation) |G| = |Z(G)| + sum{i=1 to n}|G : C(a_i)| Suppose |Z(G)| = q and p < q and x not in Z(G). Then the centralizer C(x) of every x has more than q elements. Because, Z(G) subset C(x) and Z(G) =/= C(x). Because, If Z(G) = C(x), then x in Z(G). contradiction. Since |C(x)| divides |G|, |C(x)| = pq. so, |Z(G)| + sum{i=1 to n}|G : C(a_i)| > |G|. contradiction. Now suppose |Z(G)| = p and p < q and x not in Z(G). Then the centralizer C(x) of every x has more than p elements, and its order divides pq, hence it has q or pq. If |C(x)| = pq , contradiction as before. If |C(x)| = q , the order of every non-trivial element in C(x) is q. Since Z(G) subset C(x), the order of every non-trivial element in Z(G) is p. so, contradiction. === === === === === Subject: Re: Fermat's Last Theorem simple proof impossible? > On Nov 6, 6:30 am, Roman B. Binder > question. Hint : 350 years of the best > mathematical mionds were barely > enough to find a proof. That doesn't indicate much. Think of it > this way: Maybe math isn't sexy enough for the US to get involved in it > very much. And therefore there aren't enough brilliant mathematicians to > tackle > that problem. E.g. soccer isn't very popular in the US. > Therefore > only mediocre nations like Brasil or Italy win the world cup. On > the > other hand - if soccer would appeal more to the average > US-American > the > USA would have an all-star team that could pummel all other nations > 10 to nil. Think of it this way ;) -- Thomas Nordhaus > Even more, think about the difference between > impossible > event and event with probabilty 0. > Fernando. There is difference in used words ... But on the other hand, could You see some chances from classified solutions for equation: ax^(p-1) + by^(p-1) + cz^(p-1) = 0 where p prime numbers >=3 and x;y;z of gcd=1 ? ... Chears Ro-Bin > Perhaps I was very diplomatic. I meant that the > commonly used > argument that thousand of mathematicians have > failed before, > is irrelevant. Talent and erudition are not the > same > things, think > for example about Lord Kelvin and the Wright > Brothers. > If one of my students says he had found an > elementary > proof > of the LFT I dont think he/she is wright or not, > I > want to see > what and how he/she has reasoned, it is something > exiting. > But I have an amateur spirit which is a necessary > and > not sufficient > condition for being a good professional. > By contrast, compare this with some belligerent > behaviours. > Fernando. .. Again so much words ? I used to ask only for solutions of: ax^(p-1) + by^(p-1) + cz^(p-1) = 0 for x;y;z; of gcd = 1 and p>=3 and prime .. Ro-Bin- Hide quoted text - - Show quoted text - Ha ha! The crank Bassam loves you now. Ha ha! My previous link to: a(x)^(n-1) +b(y)^(n-1) = c(z^(n-1) really doesn't give something significant. Ha ha ha ah ! But beginning with pseudo-FLT for n=3: t^3 = a^3 + b^3 +2abt for natural numbers ? What was pointed out by Mr.Bassam King Karzeddin: (I used to sketch 2 proofs how it didn't) Then analysing 1-st case of FLT from substituted equation as: t^n = a^n + b^n + 2*n^u abtp ..................(*) I've come to some certain results. counterexamples for eq. like eq.(*) because it is true substitute of FLT. Where to stop or where to continue so you have the chance to argue... Ro-Bin Ha h === === === === === Subject: Re: approximating pi posting-account=G_G-iQoAAAB08LNQidt_LsMkopmIb4ZS Gecko/20060111 Firefox/1.5.0.1 Mnenhy/0.7.3.0,gzip(gfe),gzip(gfe) > k,a,b,c,d,E,F are positive integers. C_n_2n is a binomium. sum n=1 -> oo (n^k * 2^n)/ C_n_2n = a*pi + b. lim k-> oo b/a = pi sum n=1 -> oo n^k/ C_n_2n = c*2*pi*sqrt(3) + d. lim k-> oo d/c = 2*pi*sqrt(3) sum n=1 -> oo (n^k * 3^n)/ C_n_2n = E*pi*sqrt(3)/2 + F. lim k-> oo F/E = pi*sqrt(3)/2 nice hmm marry xmas tommy1729 PI / 4 = 1 - 1/3 + 1/5 - 1/7 - ... + 1/(2n -1) - ..... n = 1, 2, 3, ... nice AND simple! Bill J PS Hoppy New Year! === Subject: Re: approximating pi > k,a,b,c,d,E,F are positive integers. C_n_2n is a binomium. sum n=1 -> oo (n^k * 2^n)/ C_n_2n = a*pi + b. Not quite, but close... > Erm, how do you know that there are > positive integers a and b with this > property? Using Maple, it seems that S(k) = sum(n^k*2^n/binomial(2*n,n), n=1..infinity) = a(k)*pi/2 + b(k) for integers a(k) and b(k): 1+(1/2)*Pi, 3+Pi, 11+(7/2)*Pi, 55+(35/2)*Pi, 355+113*Pi, 2807+(1787/2)*Pi, 26259+(16717/2)*Pi, 283623+90280*Pi, 3473315+(2211181/2)*Pi, 47552791+(30273047/2)*Pi, 719718067+229093376*Pi, ... It seems that a(k) is OEIS sequence A014307 but b(k) is not in the OEIS. The exponential generating function for S(k) is G(t) = sum(exp(t*n)*2^n/binomial(2*n,n), n=1..infinity) = e^t/(2 - e^t) + e^(t/2) arcsin(sqrt(e^t/2))/(sqrt(2) (1-e^t/2)^(3/2)) Thus G^(k)(0) = S(k). The pi's come from the fact that arcsin(1/sqrt(2)) = pi/4. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: approximating pi k,a,b,c,d,E,F are positive integers. C_n_2n is a binomium. sum n=1 -> oo (n^k * 2^n)/ C_n_2n = a*pi + b. Not quite, but close... > Erm, how do you know that there are >> positive integers a and b with this >> property? Using Maple, it seems that >S(k) = sum(n^k*2^n/binomial(2*n,n), n=1..infinity) = a(k)*pi/2 + b(k) >for integers a(k) and b(k): 1+(1/2)*Pi, 3+Pi, 11+(7/2)*Pi, 55+(35/2)*Pi, 355+113*Pi, 2807+(1787/2)*Pi, >26259+(16717/2)*Pi, 283623+90280*Pi, 3473315+(2211181/2)*Pi, >47552791+(30273047/2)*Pi, 719718067+229093376*Pi, ... It seems that a(k) is OEIS sequence A014307 >G(t) = sum(exp(t*n)*2^n/binomial(2*n,n), n=1..infinity) > = e^t/(2 - e^t) + e^(t/2) arcsin(sqrt(e^t/2))/(sqrt(2) (1-e^t/2)^(3/2)) >Thus G^(k)(0) = S(k). The pi's come from the fact that >arcsin(1/sqrt(2)) = pi/4. I figured it was something like that. Not having gone through the above carefully, it's not clear to me whether this actually gives an algorithm for finding a(k) and b(k) for a given k. (Without knowing an approximate value for pi, of course). ??? ************************ David C. Ullrich === Subject: Re: approximating pi k,a,b,c,d,E,F are positive integers. C_n_2n is a binomium. sum n=1 -> oo (n^k * 2^n)/ C_n_2n = a*pi + b. Not quite, but close... > Erm, how do you know that there are > positive integers a and b with this > property? Using Maple, it seems that S(k) = sum(n^k*2^n/binomial(2*n,n), n=1..infinity) = a(k)*pi/2 + b(k) for integers a(k) and b(k): 1+(1/2)*Pi, 3+Pi, 11+(7/2)*Pi, 55+(35/2)*Pi, 355+113*Pi, 2807+(1787/2)*Pi, 26259+(16717/2)*Pi, 283623+90280*Pi, 3473315+(2211181/2)*Pi, 47552791+(30273047/2)*Pi, 719718067+229093376*Pi, ... It seems that a(k) is OEIS sequence A014307 but b(k) is not in the OEIS. The exponential generating function for S(k) is G(t) = sum(exp(t*n)*2^n/binomial(2*n,n), n=1..infinity) = e^t/(2 - e^t) + e^(t/2) arcsin(sqrt(e^t/2))/(sqrt(2) (1-e^t/2)^(3/2)) Thus G^(k)(0) = S(k). The pi's come from the fact that arcsin(1/sqrt(2)) = pi/4. I figured it was something like that. Not having gone through > the above carefully, it's not clear to me whether this actually > gives an algorithm for finding a(k) and b(k) for a given k. > (Without knowing an approximate value for pi, of course). Sure it does. You just ask your favourite CAS to compute the k'th derivative of G(t), and evaluate at t=0. For example, using Maple to get a(30) and b(30): > G:= exp(t)/(2-exp(t)) +exp(t/2)*arcsin(sqrt(exp(t)/2))/sqrt(2)/(1-exp(t)/2)^(3/2); eval(diff(G,t$30),t=0); 85950144383076253408132013000868398677/2*Pi +135010171084427194623890031993168567507 -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === === === Subject: Re: !!! TO WHOM IT MAY CONCERN !!! >Moderation would remove the spontaneity and almost chat like >atmosphere enjoyed by the group. I can post a question and get a >response within minutes. Exactly! And that strongly argues in favor of not messing with the structure of sci.math. quasi === Subject: Re: !!! TO WHOM IT MAY CONCERN !!! posting-account=UmxzHAoAAADiGn5q_OAJNdqryOZkrQSe 1.1.4322),gzip(gfe),gzip(gfe) IMO what we need, at least for GG users, is some way for trusted >sci.math as soon as it appears -- or prevent it from ever appearing. > Why should Google Groups be the main focus? I mention it only because it's what I use, and I have little idea how >other newsreaders work or how applicable my suggestions might be to >them. > It's not the newsreaders -- it's the news servers , of which Google is only one of them. There are thousands of news servers worldwide which get the same posts as Google, essentially simultaneously. Any posts deleted from Google Groups would still be on all the other servers, hence most usenet users would still see all the junk. > There is no way to delete a post from all usenet servers, Right, I wasn't sure about that. > thus the benefit would only be to Google Groups users. > Also, even on Google Groups, what makes you think Google would allow administrators to delete other user's posts? They already do for non-Usenet groups. See But in any case, it's besides the point. If it only works for Google Groups, there's no benefit for the majority of sci.math readers. > quasi- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - At the bottom of sci.math 1st page, it reads: In response to legal complaints we received, we have removed some messages. If you wish, you may read the legal complaints. (link) When you access the link, you'll find: Notice Unavailable. DMCA Complaint to Google. The notice is not available So it appears that someone is already taking some action or counter action, as the case may be!! Monir === Subject: Re: !!! TO WHOM IT MAY CONCERN !!! posting-account=-gYNIQoAAAD-MKGbMY_-Jpj1EqcOgRWx 3.0.2; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) IMO what we need, at least for GG users, is some way for trusted sci.math as soon as it appears -- or prevent it from ever appearing. Why should Google Groups be the main focus? > I mention it only because it's what I use, and I have little idea how other newsreaders work or how applicable my suggestions might be to them. It's not the newsreaders -- it's the news servers , of which Google > is only one of them. There are thousands of news servers worldwide > which get the same posts as Google, essentially simultaneously. Any > posts deleted from Google Groups would still be on all the other > servers, hence most usenet users would still see all the junk. There is no way to delete a post from all usenet servers, > Right, I wasn't sure about that. thus the > benefit would only be to Google Groups users. Also, even on Google Groups, what makes you think Google would allow > administrators to delete other user's posts? > They already do for non-Usenet groups. See > But in any case, it's > besides the point. If it only works for Google Groups, there's no > benefit for the majority of sci.math readers. quasi- Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text - At the bottom of sci.math 1st page, it reads: > In response to legal complaints we received, we have removed some > messages. EIf you wish, you may read the legal complaints. (link) > When you access the link, you'll find: > Notice Unavailable. > EDMCA Complaint to Google. > EThe notice is not available > So it appears that someone is already taking some action or counter > action, as the case may be!! I think that this is a slightly different issue though. This is for messages where there is a potential legal problem with the content, and, presumably, Google's lawyers advise that it would be best for them to get rid of the message(s). Someone trying to sell legal products at sci.math wouldn't qualify, and I'm not sure that Google will still get through. === Subject: Re: !!! TO WHOM IT MAY CONCERN !!! posting-account=UmxzHAoAAADiGn5q_OAJNdqryOZkrQSe 1.1.4322),gzip(gfe),gzip(gfe) IMO what we need, at least for GG users, is some way for trusted >sci.math as soon as it appears -- or prevent it from ever appearing. > Why should Google Groups be the main focus? I mention it only because it's what I use, and I have little idea how >other newsreaders work or how applicable my suggestions might be to >them. > It's not the newsreaders -- it's the news servers , of which Google is only one of them. There are thousands of news servers worldwide which get the same posts as Google, essentially simultaneously. Any posts deleted from Google Groups would still be on all the other servers, hence most usenet users would still see all the junk. > There is no way to delete a post from all usenet servers, Right, I wasn't sure about that. > thus the benefit would only be to Google Groups users. > Also, even on Google Groups, what makes you think Google would allow administrators to delete other user's posts? They already do for non-Usenet groups. See But in any case, it's besides the point. If it only works for Google Groups, there's no benefit for the majority of sci.math readers. > quasi- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - > At the bottom of sci.math 1st page, it reads: In response to legal complaints we received, we have removed some messages. EIf you wish, you may read the legal complaints. (link) When you access the link, you'll find: Notice Unavailable. EDMCA Complaint to Google. EThe notice is not available So it appears that someone is already taking some action or counter action, as the case may be!! I think that this is a slightly different issue though. This is for > messages where there is a potential legal problem with the content, > and, presumably, Google's lawyers advise that it would be best for > them to get rid of the message(s). Someone trying to sell legal > products at sci.math wouldn't qualify, and I'm not sure that Google > will still get through.- Hide quoted text - - Show quoted text - You're probably right! Monir === Subject: Re: !!! TO WHOM IT MAY CONCERN !!! In response to legal complaints we received, we have removed some > messages. If you wish, you may read the legal complaints. (link) > When you access the link, you'll find: > Notice Unavailable. > DMCA Complaint to Google. > The notice is not available What's DMCA mean? Is that all that's in the link? Was the complain to Google explicitly or was it made to all servers that provide sci.math such as Math Forum? > So it appears that someone is already taking some action or counter > action, as the case may be!! > Is the action local to Google or is it global to all providers of sci.math? === Subject: Re: !!! TO WHOM IT MAY CONCERN !!! posting-account=UmxzHAoAAADiGn5q_OAJNdqryOZkrQSe 1.1.4322),gzip(gfe),gzip(gfe) At the bottom of sci.math 1st page, it reads: In response to legal complaints we received, we have removed some messages. EIf you wish, you may read the legal complaints. (link) When you access the link, you'll find: Notice Unavailable. EDMCA Complaint to Google. EThe notice is not available What's DMCA mean? EIs that all that's in the link? Was the complain to Google explicitly or was it > made to all servers that provide sci.math such > as Math Forum? > So it appears that someone is already taking some action or counter action, as the case may be!! Is the action local to Google or is it global to all providers of > sci.math? Basically that was all to the link, with sub-links to something like Chilling Effects, Cleaninghouse, etc. which I didn't have the curiosity to access! According to a colleague of mine, DMCA stands for Digital Millennium Copyright Act passed by the US Congress years ago to protect ISPs from liabilities for the illegal activities by their users. Whether it covers the issue at hand, I don't know! Monir === Subject: Re: !!! TO WHOM IT MAY CONCERN !!! >According to a colleague of mine, DMCA stands for Digital Millennium >Copyright Act passed by the US Congress years ago to protect ISPs >from liabilities for the illegal activities by their users. >Whether it covers the issue at hand, I don't know! The key word is Copyright. It covers infringement of copyright. That's all it covers. quasi === Subject: Re: !!! TO WHOM IT MAY CONCERN !!! >Is the action local to Google or is it global to all providers of >sci.math? Of course it's local. Almost every ISP in the world runs a news server. Some get an indirect feed from a larger server, others are directly part of the main swarm of usenet servers. Thus, the providers of sci.math are spread all over the world. There are thousands of them, and the list of them is not even static. DMCA (Digital Millennium Copyright Act) is a US law, hence not directly applicable to non-US providers. Even for the US, a company claiming infringement of their copyright would have to contact each provider individually. Trying to impose some form of censorship on usenet as a whole is practically impossible, but given the reality of media censorship and self-censorship, you should be thankful for that. Global control of usenet content would be a major loss of true free speech. The choice is simple. Either create a moderated group or live with a quasi === Subject: Re: #485 what is infinite and what is finite in geometry; there are not infinitely many polygons on sphere; new textbook: Mathematical-Physics (AP-adic primer) for students of age 6 onwards <12348601.1198409709803.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG > That means in Euclidean geometry, lines do not go to infinity since > the leftward portion is always finite. > This is, of course, the same old argument that in order for there to > be infinitely many natural numbers, there must exist an infinite > natural number, rewritten geometrically. > As usual your comments seem well-informed and apropos. Hey I hope you > might describe further the notions about that if there are infinite > integers then infinite integers. (oo + 1 = oo <=> N E N.) He meant, of course, that AP's statement is the same old *incorrect* argument. === Subject: Re: #485 what is infinite and what is finite in geometry; there are not infinitely many polygons on sphere; new textbook: Mathematical-Physics (AP-adic primer) for students of age 6 onwards <12348601.1198409709803.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=fsC03QkAAAAwkSNcSEKmlcR-W_HNitEd Gecko/20021120 Netscape/7.01,gzip(gfe),gzip(gfe) He meant, of course, that AP's statement is the same old *incorrect* > argument. When the above fool gave his rendition of Euclid Infinitude of Primes, he was not even intelligent enough to use the definition of prime in his offering. Such fools are no judge of anyone elses mathematics. === Subject: Re: A question on Reimann (ref Prime Obsession by John Derbyshire) posting-account=nKQXlgkAAABFcwCQitpPqM7oKmh-_khh Gecko/20071128 Camino/1.5.4,gzip(gfe),gzip(gfe) > I'm reading Prime Obsession by John Derbyshire, on the Riemann > Hypothesis. > And I've reached a line that I don't think I understand > Obviously > Log 1 , log 2, log 3 increases from 0 to just over 1 > and 1^2, 2^2, 3^2 increases from 1 to 9 > So it is true for x^2 but is it true for, for example x^(0.002) ? Is that what he means? Tony Another way to see that log(x)/x^c gets small (for small positive c) for large, assuming log(x)/x gets arbitrarily small as x gets big, is this: log(x)/x^c = (1/c)*(c*log(x))/(x^c) = (1/c)*log(x^c)/(x^c) Since x^c gets large for any fixed positive c (though more slowly for smaller c), log(x^c)/(x^c) gets small. Multiplying by (1/c) does not affect the getting smallness. This is informal, but can easily be made rigorous: f(x) gets small means that for any a > 0 there is a b (b is usually a function of a) such that abs(f(x)) < a for x > b; f(x) gets large means that for any a > 0 there is a b (b is usually a function of a) such that abs(f(x)) > a for x > b. === Subject: Re: A question on Reimann (ref Prime Obsession by John Derbyshire) posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > Obviously > Log 1 , log 2, log 3 increases from 0 to just over 1 > and 1^2, 2^2, 3^2 increases from 1 to 9 > So it is true for x^2 but is it true for, for example x^(0.002) ? Is that what he means? I'm jumping in late, but here's one way to look at things that might help. Instead of looking at an increasing sequence such as 1, 2, 3, ..., consider the sequence 10, 100, 1000, ..., 10^n, ... (where n = 1, 2, 3, ...). In the case of x^(0.002) = x^(1/500), we get outputs that increase at an exponential rate (much faster than geometrically, which means that each succeeding term is a fixed constant times the previous term): 10^(1/50), 10^(1/5), 10^2, 10^20, 10^200, ... Note that x^(0.002) had no significant effect in slowing the growth rate. We inputed a sequence that grows at a simple exponential rate and the output is a sequence that grows at a simple exponential rate. [Simple exponential rate means constant base to a linearly increasing exponent. Thus, 10^(n^2) for n = 1, 2, 3, ... is not simple exponential growth rate.] However, in the case of log(x), we get outputs that increase arithmetically (much slower than geometrically): 1, 2, 3, 4, 5, ... Note: In the case of the natural logarithm, ln(x), you still get an arithmetic growth rate -- d, 2d, 3d, ... where d = ln(10). Dave L. Renfro === Subject: On hyper-exponential numbers posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) Those interested in literature involving iterated exponentiation (tetration) will want to take note of the following: A. Cunningham, On residues of hyper-even numbers Brit. Assoc. Rep. York 7 (1906), 485-486. http://www.emis.de/cgi-bin/JFM-item?37.0226.02 http://tinyurl.com/2onr7a [.pdf file of JFM review] Revue Sem. Publ. Math. review (in English) is at http://tinyurl.com/348by2 H. J. Woodall, On hyper-exponential numbers, Sphinx, Bruxelles 7 (1937), 18-26. http://www.emis.de/cgi-bin/JFM-item?63.0107.02 http://tinyurl.com/ytcout [.pdf file of JFM review] For more appearances of this notion, see: Dave L. Renfro === Subject: Stupid Journalists Make Stupid Economic Forecasts As usual, I am unceasingly amazed at the vapidity of the news media. For several weeks now, since the advent of the sub-prime mortgage collapse, we have been listening to these hacks discuss the issue of whether a recession will occur. The mistake those fools repeatedly make is to attempt to predict the future. The fact is that there are economic problems happening right now, and in the future, we can we can look forward to more of exactly the same as what is happening now. It is going to take a little time to clean up the mortgage mess, and during that process of cleaning it up, foreclosures will happen and write-offs will occur, with the losses and human pain that correspond to them. This is what we have to put up with, right now, and for the foreseeable future, not the nonexistent fantasy threat of whether a recession will happen in the unspecified near future. I am also extremely annoyed by economists constant assessment of a 50/50 chance of the possibility that a recession will arrive. Since even stupid people, like journalists, know that economics is a quantitative science, this number is presented as though it had been calculated by some mathematical formula, when if fact, it has absolutely no mathematical basis at all. It is just the intuitive impression of whatever economist happens to be speaking at the moment, and even stupid people like journalists should know that intuitive impressions are not a mathematical science. In fact, when somebody says 50/50, it should be common knowledge that the phrase is a code word for I don't know. Whatever is being discussed might happen or it might not, with equal probability. Again, this is something that stupid reporters should know. Finally, on the topic of a 50/50 chance of a recession, reporters should look at this number in a slightly different way than a probability that something might or might not happen. They should interpret it as an assessment of the degree to which it will happen. In other words, right now, we are experiencing 50% of a recession and 50% of a good economy. That's a recipe for the stagnation that is currently happening. However, reporters continually make the mistake of trying to predict the future, and in the process, do nothing but make themselves look stupid. Maybe they should take a math class, so they can learn to understand how numbers work. === Subject: Re: pdf solutions manual posting-account=1GVcMgoAAACscFtGWsruD3cqC45_vX5q .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.2),gzip(gfe),gzip(gfe) > GetSolution Team we have a lot ofsolutionsmanual in low cast > to get solution manual you want please send message to > getsolut...@hotmail.com getsolution(at)hotmail.com PS: These are part of oursolutions, if the solution you want isn't on > the list, do not give up, just contact with us if you want any book not justsolutionsjust contact with us you can see our Eweb site:www.solutions.freeweb7.com this is new list Ein 19/12/2007 E<> 7ed Grant R. Fowles, George > L. Cassiday > Ecomputer networks Andrew S. Tanenbaum 4th edition > EDesign with Operational Amplifiers and Analog Integrated Circuits, > 3rd edt. by F > EFundamental of Physics 7 edsolutionsmanual of Halliday and Resnick > EInstructor's Manual: Im Experiments with Economic Principles E By > Bergstrom > EMechanicsof FluidsSolutionsManual E8ed EBy John Ward-Smith > EStatistical Digital Signal Processing and Modeling -SOLUTIONSMANUAL > By Monson H. Hayes > Advanced Engineering Mathematics, 9th Edition EBy Erwin Kreyszig > Advanced Macroeconomics,SolutionsManual ERomer > Advanced Modern Engineering Mathematics, 3rd Edt by Glyn James - > solutionmanuel > APPLIED MATHEMATICS AND MODELLING FOR CHEMICAL ENGINEERS-8th Edition > (solution manual) EBy Erwin Kreyszig > T. Thornton, Jerry B. Marion > Computational Techniques forFluidDynamics: ASolutionsManual By > Karkenahalli Srinivas, Clive A. J. Fletcher > Corporate Finance-6th Edition by Stephen A. Ross , Randolph W. > Westerfield , Jeffrey Jaffe > Corporate Finance-7th Edition by Stephen A. Ross , Randolph W. > Westerfield , Jeffrey Jaffe > DATABASE MANAGEMENT SYSTEMS 3rd Edition by Ramakrishnan, Gehrke, > Derstad, Seliko, Zhu- Solution Manual > Design and Analysis of ExperimentsSolutionsManual 6th edition > Digital and Analog Communication Systems -SOLUTIONSMANUAL EBy Leon > W. Couch, Leon W. > Digital Integrated Circuits by Rabaey 2nd edt. - solution manuel .8bi > .82.8b.87 > Econometric Analysis [ONLY theSolutionsManual to the 6th Edition By > William H. Greene > Electric Circuits 8th edition Nilsson Riedel > Electrical Engineering: Principles and Applications 3ed Allan R. > Hambley > Electronic Devices-6th Edition by Thomas L. Floyd Solution Manual > Engineering Economy - Leland Blank & Anthony Tarquin 6th Edition > EngineeringFluidMechanics, 7th Edition - StudentSolutionsManual > Engineering Mathematics, 4th edt. by John Bird - solution manue > Engineering Mathematics, Fourth Edition By John Bird > Finite Element Method: Volume 1, The Basis (Finite Element Method Ser) > 5ed > Fundamentals of Chemical Reaction Engineering -SolutionsManual EBy > Mark E. E. Davis, Robert J. J. Davis, > Fundamentals of Electric Circuits 2nd by Alexander Sadiku > Fundamentals of Engineering Electromagnetics--Cheng > Fundamentals of Organic Chemistry, 5E, Study Guide andSolutions > Manual EBy T. W. Graham Solomons > Fundamentals of Physics, 7th Edition - Instructor'sSOLUTIONSMANUAL > By Halliday, Resnick and Walker > Fundamentals of Probability 3/e (SolutionsManual EBy Saeed > Ghahramani > Fundamentals of Probability With stochastic processes E3/e (Solutions > Manual ) By Saeed Ghahramani > Fundamentals of Thermal-FluidSciences Solution Manual E2ed By Yunus > A. Cengel, Robert H. Turner, > Fundamentals of Thermodynamics SOLUTION MANUAL 6ed By Richard E. > Sonntag, Claus Borgnakke, Gordon J. Van Wylen, > hemical Engineering:Solutionsfor Volumes 2 and 3 by coulson > 2002-12-11 > In Experiments with Economic Principles, Instructor's Manual By > Bergstrom > Instructor Manual to An Introduction to Thermodynamics and StatisticalMechanics2ed E By Keith Stowe > Instructor Manual to Introduction to Solid State Physics Eighth > Edition EBy Charles Kittel > Instructor Manual to Introductory Quantum Optics EBy Christopher Gerry > and Peter Knight > Instructor Manual to SEMICONDUCTOR DEVICES EPhysics and Technology > Second Edition E By S.M.Sze > InstructorSolutionsManual to accompany Boyce Elementary Differential > Equations and Boundary Value Problem 8ed by Charles W. Haines, William > E. Boyce, > Instructor's Manual for Solving ODEs with MATLAB EBy L. F. Shampine, > I. Gladwell, S. Thompson > Instructor'sSolutionsfor: Design of Analog CMOS Integrated Circuits > Instructor'sSolutionsManual for Atkins' Physical Chemistry, 7/e > byP.W. Atkins, & J. de Paula > Introduction to Chemical Engineering Thermodynamics E7 ed(solution > manual) EBy J.M. Smith, Hendrick C Van Ness, > Introduction to Heat Transfer - 013391061X Solution's Manual EBy > peyman pourmoghaddam , Vedat S. Arpaci, > Introduction to Heat Transfer 4th Edition SOLUTION MANUAL EBy Frank P. > Incropera, David P. DeWitt, > Introduction to Probability by Dimitri P. Bertsekas - solution > Introduction to QuantumMechanics(Second Edition) -SolutionsManual > By David J. Griffiths > Introductory Econometrics: A Modern Approach 2ed Jeffrey Wooldridge > Modern Digital and Analog Communications Systems - B P LathiSolutions > Manual 3ed > modern electronic communication 8e - gary miller beasley - > solutions > New Perspectives on Computer Concepts-8th Edition Solution Manual > nvestment Analysis and Portfolio Management-7th Edition Frank K. > Reilly, Keith C. Brown > Physical Chemistry (Instructor'sSolutionsManual) By Peter Atkins & > Julio de Paula > Selected Answers-Basic Engineering Circuit Analysis-7th Ed. by J. > David Irwin > Shigley's Mechanical Engineering Design 8th Ed - Solution Manual By by > Richard Budynas, J. Keith Nisbett > SoilMechanicsSolutionsManual (2nd Edition) By William Powrie > SoilMechanicsSolutionsManual (2nd Edition) By William Powrie > Solution Manual for Numerical Solution of Partial Differential > Equations: An Introduction 2ed E By K. W. Morton, D. F. Mayers > Solution Manual Nonlinear Programming 2nd Edition by Dimitri P. > Bertsekas > Solution Manual-Microelectronics-Digital and Analog Circuits and > Systems by MillmanSolutionsManual for: Materials Science and Engineering: An > Introduction 6E EBy William D. CallisterSolutionsManual for: System Dynamics 3rd Ed EBy Katsuhiko Ogata > Statistical Digital Signal Processing and Modeling -SOLUTIONS > MANUAL E By Monson H. Hayes > the Guide to Energy Management 5ed E2005-12 EBy Klaus-Dieter E. > Pawlik > Undergraduate EconometricsSolutionsManual - Hill, Judge and > Griffiths > Fundamentals of Semiconductor Devices - Anderson > Solution Manual Statistical Digital Signal Processing & Modeling By > MonsonH.rar > electronic devices 6ed+electron flow version-4th edition by thomas > l.floyd.pdf > Engineering Circuit Analysis 7ed by Hayt > Electric Circuits, Nillson, 7th edition > Electric Circuits, Nillson, 8th edition > Calculus Single Variable 4ed chapter 1 to 11 Hughes-Hallett, Gleason, > McCallum, et al. > Calculus Third Editon By Strauss, Bradley and Smith > Fourier and Laplace Transform - Antwoorden > Field and Wave Electromagnetics, 2nd edition, Cheng > EFundamentals of Chemical Reaction Engineering -SolutionsManual E By > Mark E. E. Davis, Robert J. J. Davis,SolutionsManual for Equilibrium and Non-Equilibrium Statistical > Thermodynamics EBy Michel Le Bellac > Introduction To Electric Circuits 6th Ed [SolutionsManual] By R. C. > Dorf and J. A. Svoboda > Harcourt Mathematics 12 Geometry and Discrete MathematicsSolutions > Manual EBy McGraw-Hill Mechanicsof FluidsSolutionsManual Eby John Ward-Smith 8 ed > E EiranChemical Reaction Engineering, 3rd Edition Solution Manual EBy > Octave Levenspiel > Advanced Engineering Mathematics by Erwin Kreyszig E8edsolutions > manual > Advanced Macroeconomics,SolutionsManual E 3ed > advanced engineering mathematics 8ed - erwin kreyszig - > solutions manual.rar.htmlt.txt > Analysis and Design of Analog Integrated Circuits (4th Edition) Gray, > Hurst, Lewis and Meyer > E > AnalyticalMechanics:1SolutionsManual EGrant R. Fowles, George L. > Cassiday, 7 ed > ANTENNAS FOR ALL APPLICATIONS, THIRD EDITION.txt > c++ deteil E3ed > CMOS VLSI Design 3e - David Harris H E Weste.rar.html > Communication Systems 4Ed - A Bruce CarlsonSolutionsManualt.txt > communication systems engineering by poakis > Computer Organization and Design, Revised Printing, 3rd EditionSolutionsManual By David A. Patterson, John L. Hennessy, > Control Systems Engineering by Nise 4ed > Device Electronics for Integrated CircuitsSolutionsManual E3ed > Digital Communications Fundamentals and Applications 2E > 0130847887.pdft.txt > Digital Signal Processing - Proakis & Manolakis -SolutionsManual 3ed > Discrete-Time Signal Processing 2nd Edition, 1999-02 > Electronic Circuit Analysis and Design 2nd edt. by Donald A. > Neamen E.txt > Elementary Differential Equations and Boundary Value Problems , 8th > Edition .txt > Elementary Differential Equations and).txt > Elementary Principles of Chemical ProcessesSolutionsManual E3 ed EBy > Richard M.Felder > Engineering Circuit Analysis 6Ed - HaytSolutionsManual.pdf > Engineering Electromagnetics -Hayt (2001) > EngineeringFluidMechanics, 7th Edition - By Clayton T. Crowe, > Donald EngineeringMechanics, Dynamics 5/e -Solutionsmanual EBy J. L. > Meriam, L. G. Kraige, > Fundamentals of Digital Logic with Verilog Design bt.txt > Fundamentals of Logic Design 5Ed.txt > Fundamentals of Machine Component Design 3ed -SolutionsManual By > Robert C. Juvinall, Kurt > Fundamentals of Physics, 7th Edition - Instructor'sSOLUTIONSMANUAL > Greene Econometric Analysis (5th t.txt > Heat Transfer and EThermodynamics.txt > IntroductionFluidMechanics, 6Th Edition Solutio.txt > Introduction to Algorithms, 2nd Edition > Introduction to VLSI Circuits and Systems (2001 draft) - John P > Uyemura -SolutionsManual > Introductory Quantum Optics by Knight and Gerry.txt > Jackson s Classical Electrodynamics.zip.html.txt > James Stewart - Calculus 5Th Ed - Complete Instructor 'sSolutions > Manual.txt > linear systems and signals - b p lathi solutions manual 194t.txt > Ljungqvist-Sargent.-.Recursive.Macroeconomic.Theory.-.Solutions.Manual. > Materiala Science and Engineering An Introduction.txt > Mathematical Methods for Physics and Engineering A Comprehensive Guide > 3ed > Microeconomic Theory - -SolutionsManual for Mas-Colell .txt > Microeconomic Theory - -SolutionsManual for Mas-Colell > microwav rf ....txt > Modern QuantumMechanics- J. J. Sakurai - Solution.txt > Physical Chemistry (Instructor'sSolutionsManual) EPeter Atkins & > Julio de Paula 7ed > Power System Analysis Solution Manual Jo.txt > Power Systems Analysis -SolutionsManual EJohn Grainger, William D. > Stevenson > Principles of Electronic Materials and Devices,SolutionsManual ONLY > Safa O. Kasap E2ed > Probability and Statistics for Engineering and the Sciences [Solutions > Manual] E7 ed > Probability and Stochastic Processes A Friendly Introduction for > Electrical and Computer Engineers > Recursive Methods in Economic Dynamics By Claudio Irigoyen ,Esteban > RF circuit Design Theory and Application by Ludwig bretchko - > solution manuel.rar.html.txt > Safko,< Semiconductor Device Fundamentals 1st edt. by Robert F > semiconductor physics and devices 3ed neamen > getsolut...@hotmail.com.txt > Signals and Systems 2e Oppenheim.txtSolutionsManual for the Guide to Energy Management 5 ed E2005-12 > Klaus-Dieter E. PawlikSolutionsManual to accompany Corporate Finance EBy ROSS E6 editionSolutionsof Engineering Electromagnetics -Hayt (2001).rar > solutions Electric Machinery and Power System Fundamentals > Thermodynamics: An Engineering Approach Eby: Yunus A. Cengel > Wireless Communications: Principles and Practice, 2nd edition > theodore rappaportsolutionsmanual My name is Sadie and I a trying to fing a soutions manual for Robert A. Granger's 'Fluid Mechanics'. This book was published by Dover Publications in New York and Copyrighted in 85 & 95 by Robert A. Granger. I could not find this book through your website, so any help you could offer would be great. Sadie Barnett P.S Can you print out E-Books?