mm-1409 === Subject: ignore my earlier post -- Ive solved it!!!! BUT now Im now stuck on a very important modification of the problem. Let T(n) be some nxn matrix. Let Z(n) = I(n) + T(n) T(n) . Let v(n) be the nx1 vector with all matrix elements equal to 1/n. Define the quadratic form q = v(n) T(n) Z(n)^(-1) T(n) v(n). What are necessary and sufficient conditions on the matrix T(n) such that q -> 0 as n -> infinity. (Hint: I believe it is true for ALL T(n), but I have trouble proving it.) Roddy === Subject: Re: ignore my earlier post -- Ive solved it!!!! BUT now >Im now stuck on a very important modification of the problem. >Let T(n) be some nxn matrix. Real, I assume. >Let Z(n) = I(n) + T(n) T(n) . >Let v(n) be the nx1 vector with all matrix elements equal to 1/n. >Define the quadratic form >q = v(n) T(n) Z(n)^(-1) T(n) v(n). >What are necessary and sufficient conditions on the matrix T(n) >such that >q -> 0 as n -> infinity. >(Hint: I believe it is true for ALL T(n), but I have trouble proving it.) Yes, its true. For convenience, Ill leave out all the (n)s. Write T = B R where B is the positive semidefinite square root of T T and R is orthogonal. Then T Z^(-1) T = R B (I+B^2)^(-1) B R so ||T Z^(-1) T|| <= 1 And then |q| <= ||v||^2 = 1/n Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: ignore my earlier post -- Ive solved it!!!! BUT now >Let T(n) be some nxn matrix. > Real, I assume. >Let Z(n) = I(n) + T(n) T(n) . >Let v(n) be the nx1 vector with all matrix elements equal to 1/n. >Define the quadratic form >q = v(n) T(n) Z(n)^(-1) T(n) v(n). >What are necessary and sufficient conditions on the matrix T(n) >such that >q -> 0 as n -> infinity. >(Hint: I believe it is true for ALL T(n), but I have trouble proving it.) > Yes, its true. > For convenience, Ill leave out all the (n)s. > Write T = B R where B is the positive semidefinite > square root of T T and R is orthogonal. Then > T Z^(-1) T = R B (I+B^2)^(-1) B R so > ||T Z^(-1) T|| <= 1 > And then |q| <= ||v||^2 = 1/n Rod > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada === Subject: Re: definition of definition Originator: harris@tcs.inf.tu-dresden.de (Mitchell Harris) >> Is it stipulative? ... >Mathematical Monthly has some interesting comments about the >role of stipulative definitions in mathematics: >Surprises from Mathematics Education Research: Student (Mis)use of >Mathematical Definitions >by Barbara S. Edwards and Michael B. Ward Mitch === Subject: Re: The origin of quaternions > Ive read about how quaternions were invented by Hamilton. What did he find > unsatisfactory about 3 dimensional numbers? > Was it specifically that they cant be used for representing rotations in 3d > space (Im not clear as to whether Hamilton had this purpose in mind at the > time?) Or is there some more fundamental ßaw making it impossible to > sensibly define +, * operators? > G.A. Plus is easy for three numbers (a,b,c) + (d,e,f,) = (a+d,b+e,c+f) is an example. Multiplication is harder depending on what you want it to do, for instance (a,b,c) *(d,e,f) = (0,0,0) is an example of a multiplication, but its kind of useless. Another one is (a,b,c)*(d,e,f)=(a*b,b*e,c*f), but while it seems more useful, you are basically just doing addition and multiplication in three unrelated spaces, there is no rhyme or reason for making a combined object. So since there are many ways to define + and *, Hamilton had to decide what kind of properties he wanted. Properties similar to rotations are nice properties, but yes then you need four numbers to get nice formulas. === Subject: Re: The origin of quaternions > Ive read about how quaternions were invented by Hamilton. What did he find > unsatisfactory about 3 dimensional numbers? > Was it specifically that they cant be used for representing rotations in 3d > space (Im not clear as to whether Hamilton had this purpose in mind at the > time?) Or is there some more fundamental ßaw making it impossible to > sensibly define +, * operators? .... Yes, there is. numbers on a plane. Mathematicians like generalizing things, so a natural question then was What about numbers represented in 3-dimensional space? This turned out to be difficult, and in particular Hamilton struggled with it for a long time before he reached the insight that 4 dimensions would work better than 3. Suppose you want to start with complex numbers x + iy and extend them to 3-dimensional numbers x + iy + jz (where x, y and z are real). Then the product ij must itself be some such number, say ij = a + ib + jc where a, b, c are some real numbers. Then i(ij) = i(a + ib + jc) = ia + (i^2)b + (ij)c = ia - b + (a + ib + jc)c = (ac - b) + i(a + bc) + j(c^2). But i(ij) = (i^2)j by the associative law = - j so (ac - b) + i(a + bc) + j(c^2) is the same number as - j. Equating the coefficients gives ac - b = 0 and a + bc = 0 and also c^2 = - 1, although c is supposed to be a real number. That may answer your question ... is there some more fundamental ßaw ...? Ken Pledger. === Subject: Re: The origin of quaternions >numbers on a plane. Mathematicians like generalizing things, so a >natural question then was What about numbers represented in >3-dimensional space? This turned out to be difficult, and in >particular Hamilton struggled with it for a long time before he reached >the insight that 4 dimensions would work better than 3. > Suppose you want to start with complex numbers x + iy and extend >them to 3-dimensional numbers x + iy + jz -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: The origin of quaternions My understanding was that he was studying vector analysis and looking at A divid B for two vectors. ( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). The consideration of A divid B , not as a vector but as an operation carrying a representative segment B into a coterminous representative segment of A, led Sir William Hamilton to the study of quaternions. The use of quaternions in a rotation of spherical triangle problem considerable simplified the math involved. >Ive read about how quaternions were invented by Hamilton. What did he find >unsatisfactory about 3 dimensional numbers? >Was it specifically that they cant be used for representing rotations in 3d >space (Im not clear as to whether Hamilton had this purpose in mind at the >time?) Or is there some more fundamental ßaw making it impossible to >sensibly define +, * operators? >G.A. -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : alternative email: rlbtftn@netscape.net URL : http://home.earthlink.net/~tftn === Subject: Re: The origin of quaternions > My understanding was that he was studying vector analysis > and looking at A divid B for two vectors. > ( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). > The consideration of A divid B , not as a vector but as an operation > carrying a representative segment B into > a coterminous representative segment of A, led Sir William Hamilton to > the study of quaternions..... This may be a misunderstanding. I remember that one of Hamiltons text-books does begin with the idea of dividing one vector by another, but that was written long after the original discovery/invention. Ken Pledger. === Subject: Re: The origin of quaternions - Hamiltons 1844 paper Perhaps the following is a relevant contribution to the newsgroup thread on the origin of quaternions. W.R.Hamiltons second paper on quaternions presented to the Royal Irish Academy: W.R.Hamilton: On quaternions, or on a new system of imaginaries in algebra; with some geometrical illustrations. Communicated November 11th 1844 to the Royal Irish Academy, Proc. Roy. Irish Acad. vol. 3 (1847), 1-16. If I remember well, it is in this paper that Hamilton tries to set up an algebra of proportionalities of spatial directions. In the horizontal plane one can say: North is to East as South is to West. In the meridional plane on can likewise say that Zenith is to South as Nadir is to North. These are geometrical rephrasings of properties of the complex number system. Hamilton did not succeed in obtaining a consistent algebra of spatial directions. Instead he found that when one more dimension to the three dimensions North-South, East-West, Zenith-Nadir, one indeed obtains a nice closed and consistent algebra of directions and their quotients. All this is actually a geometrical rephrasing of Hamiltons discovery (or invention?) of the quaternion algebra. H. knew that this fourth dimension can in no way be a spatial dimension. He imagined this dimension as laid out on a scale and coined the term scalar for this kind of numbers. H. was predominantly a phycisist and so identified this fourth dimension with time. To no avail, as would become apparent over the decades to come. In hindsight one may guess that Hamiltons motives for his quest were (1) to generalise the algebra of complex numbers to 3D, and to maintain in the process: (2) the concepts of quotient of directions and of quotient of vectors, and (3) the law of moduli of complex numbers, and of course (4) the often repeated questions at breakfast by his sons, aged six and eight in the autumn of 1843: Well, Papa, can you multiply triplets? (*) All this still does not answer the questions of exactly how H. got his brainwave to add a fourth dimension, and of why he got it then and there. My cherished speculation is that Hamiltons walk on October 16th 1843 from Dunsink to the city of Dublin helped to clear up his mind and open it up for a brainwave (a ßow of endorphin set in motion by enjoying a 5-mile walk in the cool autumn air). And then the brainwave will of course involve the subject closest at hand. (*) Se.87n ODonnell: William Rowan Hamilton - Portrait of a Prodigy. Boole Press Dublin, 1983, ISBN 0-906783-06-2 (hc) and 0-906783-15-1 (pbk) Johan E. Mebius >>My understanding was that he was studying vector analysis >> and looking at A divid B for two vectors. >>( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). >> The consideration of A divid B , not as a vector but as an operation >>carrying a representative segment B into >>a coterminous representative segment of A, led Sir William Hamilton to >>the study of quaternions..... >> > This may be a misunderstanding. I remember that one of Hamiltons >text-books does begin with the idea of dividing one vector by another, >but that was written long after the original discovery/invention. > Ken Pledger. === Subject: Re: The origin of quaternions - Hamiltons 1844 paper big snip > > In hindsight one may guess that Hamiltons motives for his quest were > (1) to generalise the algebra of complex numbers to 3D, and to maintain > in the process: (2) the concepts of quotient of directions and of > quotient of vectors, and (3) the law of moduli of complex numbers, and > of course (4) the often repeated questions at breakfast by his sons, > aged six and eight in the autumn of 1843: Well, Papa, can you multiply > triplets? (*) Well, I dont claim to be on the order of Hamilton but I do have a construction that extends complex properties to three dimensions and beyond. Please see polysigned numbers in sci.math. The solution is not orthogonal. It comes as a result the additive identity in higher signs. Complex numbers are equivalent to three-signed numbers. In four-signed numbers you can interpret them graphically as rays extending from the center of a tetrahedron to each corner. Labeling them -,+,*,# and enforcing the additive identity: - a + a * a # a = 0, where a is either an unsigned scalar or a four-signed value. Thence points in 3D space can be defined as sums in the four signs. The algebraic product rules follows: | - + * # ----------------- - | + * # - | + | * # - + | * | # - + * | # | - + * # This is simply a process of sign counting as the product of any two components are taken, wrapping the count at the highest sign. This obviously yields the rotational nature of the arithmetic. And so a general 3D product like: ( - a + b * c # d )( - e + f * g # h ) yields: + ae * af # ag - ah * be # bf - bg + bh # ce - cf + cg * ch - de + df * dg # dh This procedure in three-signed math yields the complex numbers. This procedure in two-signed math yields the real numbers. All of these can be looked at as number-line style algebra where each sign has a ray emanating from an origin. Taking the step of graphical interpretation or dimensional analysis shows that an n-signed system has dimension n - 1. This is easily seen right from the additive identity. All the commutative and associative properties that apply to the reals and complex numbers also work in higher signs, or higher dimensions. This approach is much simpler than queternions. -Tim > All this still does not answer the questions of exactly how H. got his > brainwave to add a fourth dimension, and of why he got it then and > there. My cherished speculation is that Hamiltons walk on October 16th > 1843 from Dunsink to the city of Dublin helped to clear up his mind and > open it up for a brainwave (a ßow of endorphin set in motion by > enjoying a 5-mile walk in the cool autumn air). And then the brainwave > will of course involve the subject closest at hand. > (*) Se.87n ODonnell: William Rowan Hamilton - Portrait of a Prodigy. > Boole Press Dublin, 1983, > ISBN 0-906783-06-2 (hc) and 0-906783-15-1 (pbk) > Johan E. Mebius > >>My understanding was that he was studying vector analysis >> and looking at A divid B for two vectors. >>( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). >> The consideration of A divid B , not as a vector but as an operation >>carrying a representative segment B into >>a coterminous representative segment of A, led Sir William Hamilton to >>the study of quaternions..... >> >> > This may be a misunderstanding. I remember that one of Hamiltons >text-books does begin with the idea of dividing one vector by another, >but that was written long after the original discovery/invention. > Ken Pledger. > === Subject: Re: The origin of quaternions - Hamiltons 1844 paper > big snip > > In hindsight one may guess that Hamiltons motives for his quest were > (1) to generalise the algebra of complex numbers to 3D, and to maintain > in the process: (2) the concepts of quotient of directions and of > quotient of vectors, and (3) the law of moduli of complex numbers, and > of course (4) the often repeated questions at breakfast by his sons, > aged six and eight in the autumn of 1843: Well, Papa, can you multiply > triplets? (*) > Well, I dont claim to be on the order of Hamilton but I do have a > construction that extends complex properties to three dimensions and > beyond. Please see polysigned numbers in sci.math. > The solution is not orthogonal. It comes as a result the additive > identity in higher signs. > Complex numbers are equivalent to three-signed numbers. > In four-signed numbers you can interpret them graphically as rays > extending from the center of a tetrahedron to each corner. Labeling > them -,+,*,# and enforcing the additive identity: > - a + a * a # a = 0, where a is either an unsigned scalar or a > four-signed value. > Thence points in 3D space can be defined as sums in the four signs. > The algebraic product rules follows: Funny, When I view this table in the original it is all screwed up, but now it looks good again. Put some spaces on the end just in case. > | - + * # > ----------------- > - | + * # - > | > + | * # - + > | > * | # - + * > | > # | - + * # > This is simply a process of sign counting as the product of any two > components are taken, wrapping the count at the highest sign. This > obviously yields the rotational nature of the arithmetic. > And so a general 3D product like: > ( - a + b * c # d )( - e + f * g # h ) > yields: > + ae * af # ag - ah * be # bf - bg + bh # ce - cf + cg * ch - de + df > * dg # dh > This procedure in three-signed math yields the complex numbers. > This procedure in two-signed math yields the real numbers. > All of these can be looked at as number-line style algebra where each > sign has a ray emanating from an origin. Taking the step of graphical > interpretation or dimensional analysis shows that an n-signed system > has dimension n - 1. This is easily seen right from the additive > identity. All the commutative and associative properties that apply to > the reals and complex numbers also work in higher signs, or higher > dimensions. > This approach is much simpler than queternions. > -Tim > All this still does not answer the questions of exactly how H. got his > brainwave to add a fourth dimension, and of why he got it then and > there. My cherished speculation is that Hamiltons walk on October 16th > 1843 from Dunsink to the city of Dublin helped to clear up his mind and > open it up for a brainwave (a ßow of endorphin set in motion by > enjoying a 5-mile walk in the cool autumn air). And then the brainwave > will of course involve the subject closest at hand. > > (*) Se.87n ODonnell: William Rowan Hamilton - Portrait of a Prodigy. > Boole Press Dublin, 1983, > ISBN 0-906783-06-2 (hc) and 0-906783-15-1 (pbk) > > Johan E. Mebius > > > > > > >>My understanding was that he was studying vector analysis >> and looking at A divid B for two vectors. >>( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). >> The consideration of A divid B , not as a vector but as an operation >>carrying a representative segment B into >>a coterminous representative segment of A, led Sir William Hamilton to >>the study of quaternions..... >> >> > > > This may be a misunderstanding. I remember that one of Hamiltons >text-books does begin with the idea of dividing one vector by another, >but that was written long after the original discovery/invention. > > Ken Pledger. > > === Subject: Re: The origin of quaternions > My understanding was that he was studying vector analysis > and looking at A divid B for two vectors. > ( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). > The consideration of A divid B , not as a vector but as an operation > carrying a representative segment B into > a coterminous representative segment of A, led Sir William Hamilton to > the study of quaternions. > The use of quaternions in a rotation of spherical triangle problem > considerable simplified the math involved. Interesting, Roger. So perhaps Hamilton was aware that he needed a system with a non-commutative multiplication from the outset? Shouldve narrowed his search a bit. I understand youre an aficionado of fractals. Quaternion fractals are very nice, but I feel that a dimensions being wasted. Did you ever wonder if you could define a 3d number system (I assume you only need +, * and magnitude operators) for drawing Julia sets etc? I dont have any idea if the impossibility of making a division ring implies that you couldnt create a system that makes nice Julia sets. cheers, G.A. === Subject: Re: The origin of quaternions Terry Gintz is the expert on quaternion and octonion iterative functions, but Ive done chaos differential equation solutions based on quaternions. The Euler rotation is what you are looking for ( based on so(3)/ SO(3)). It leaves out the fourth part. >>My understanding was that he was studying vector analysis >> and looking at A divid B for two vectors. >>( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). >> The consideration of A divid B , not as a vector but as an operation >>carrying a representative segment B into >>a coterminous representative segment of A, led Sir William Hamilton to >>the study of quaternions. >>The use of quaternions in a rotation of spherical triangle problem >>considerable simplified the math involved. >> >Interesting, Roger. So perhaps Hamilton was aware that he needed a system >with a non-commutative multiplication from the outset? Shouldve narrowed >his search a bit. >I understand youre an aficionado of fractals. Quaternion fractals are very >nice, but I feel that a dimensions being wasted. Did you ever wonder if >you could define a 3d number system (I assume you only need +, * and >magnitude operators) for drawing Julia sets etc? I dont have any idea if >the impossibility of making a division ring implies that you couldnt create >a system that makes nice Julia sets. >cheers, >G.A. -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : alternative email: rlbtftn@netscape.net URL : http://home.earthlink.net/~tftn === Subject: Re: The origin of quaternions its not clear to me that there really was 3d numbers, since a sufficient notation didnt exist for its arithmetic, other than *maybe* simple addition of vectors.... that may be why some folks get all het-up on grassmannians. if you look at Gibbs cross & inner products, they are directly adapted from Hamiltons terms; quaterions solve both siimultaneously. maybe this is part of the reason that Fermat went a further with linear algebra than did Descartes, but especially in the plane. you can see that most purely directional operations are done on the unit sphere, using quaternions with real part being zero; every two rotations can be achieved with one rotation; with two purely imaginary quaternions ... the product is, or is not neccesarily purely imaginary? > My understanding was that he was studying vector analysis > and looking at A divid B for two vectors. >Was it specifically that they cant be used for representing rotations in 3d >space (Im not clear as to whether Hamilton had this purpose in mind at the >time?) Or is there some more fundamental ßaw making it impossible to >sensibly define +, * operators? --Advice 0.05; Free, if wrong, again! http://tarpley.net/bush6.htm === Subject: Re: Invariant Galilean Transformations (FAQ) On All Laws >> Androcles is a well-known fool, and his connection with reality has >> been served no useful purpose when you treat him as if he was more >> intelligent than he actually is, as you have done above. It is interesting that she did not see fit to write explaining what specific objections to his points. Androcles === Subject: Re: Mathworld errors > I just got an email > inviting me to send him a list of errors on mathworld. > I havent saved such a list, thought Id mention this > in case anyone else has. >Recently, on sci.math, you mentioned: ><to believe, you might at least include a question mark or >something... >(Today must be your first visit to mathworld - its really full of >errors. I admit this is a good one.)>> >If you would be so kind as to send us a list, Eric and I will be >glad to fix them. I have kept lists of errors, and used to submit them to Eric every once in a while. But it eventually became apparent to me that I could spend a large part of my life doing nothing but correcting MathWorld errors... As it happens, the good one mentioned above was on my list of errors yet to be submitted, so thats one I can remove from the list. Just for the heck of it, I decided to take care of the last four errors on the list now. Well, the last one, which concerned an equation in the entry Triangle Wave, seems to have been taken care of already, so here are the remaining three: 1) In , the reference Whittaker, J. M. On the Cardinal Function of Interpolation Theory. Proc. Edinburgh Math. Soc. 2, 41-46, 1927. is slightly incorrect. (The mistake seems to have been copied from the reference in the paper by McNamee et al.) Instead of volume 2, it should be series 2, volume 1. 2) In absolute values are needed in the denominators. 3) In the correct spelling is casus irreducibilis, rather than ...lus. [A copy of this post has been sent to Ed Pegg.] David W. Cantrell === Subject: Re: Mathworld errors > I just got an email .... > 2) In > you meant absolute values are needed in the denominators for curvature or in the numerators for radius of curvature.. right? === Subject: Re: Mathworld errors > 2) In > you meant absolute values are needed in the denominators for curvature > or in the numerators for radius of curvature.. right? No. I meant what I said: In the entry for radius of curvature, absolute values are needed in the denominators. (Unless theres a usage which is different from the one Im familiar with, we want the radius of curvature to be nonnegative always.) I havent looked at the entry for curvature itself. David === Subject: Re: anti-cantorian probability theory and dart throwing > Suppose the reals are countable. :) > Then R = Q / Irr, where Irr is RQ. > If somebody throws a dart at a dartboard, > what is the probability that the x-coordinate > of the point where the dart lands is in Q (the rationals), > and why? > David Bernier I think the probability depends on the velocity and distance of the dart. On a massless spaceship moving at an acceleration of 1g and no air, then the dart would move in a parabola, so if you threw it with a rational velocity in the right direction from the right distand, then it would hit a rational point. In other circumstance it would be irrational. So the question is begged back towards you providing a PD of the phase space of the center of mass of the dart, and from that, we can kinematically transform the PD of the darts position at one time into a PD on the x coordinate of the dart later, and from that answer your probability question. === Subject: Re: anti-cantorian probability theory and dart throwing > Suppose the reals are countable. :) Starting from a false premise, anything that follows is worthless. If you want anyone to read the rest of your message, I suggest you not start off with a blatantly false premise. I suggest you avoid all mention of the reals. I suggest you devise some model that doesnt involve real numbers. === Subject: Re: anti-cantorian probability theory and dart throwing > Suppose the reals are countable. :) >> Then R = Q / Irr, where Irr is RQ. > So Irr = empty set , the mere fact that R is a countable set containing Q it does not follow that RQ is empty, but from the fact that R is both countable and uncountable you can conclude that Bertrand Russell is the Pope. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Functional Analysis: Equivalent of Taylor series for operators thank you foryour answer. I am interested by both questions: Let T be an operator between two Banach spaces (or Hilbert spaces), what are the conditions on T and on these spaces such that T admits a Taylor series expansion, and how to perform this expansion. I would appreciate any references about this. >Hi all, >is there an equivalent of Taylor series for operators between Banach >Spaces ?- if yes, what is the name of this equivalent, and what are >the conditions on this operators ? > I can think of at least two things the question might mean > (does an operator have a Taylor series/can we apply a Taylor > series to an operator). Try to be a little more specific about > the question... > ************************ > David C. Ullrich === Subject: Re: Functional Analysis: Equivalent of Taylor series for operators >Let T be an operator between two Banach spaces (or Hilbert spaces), >what are the conditions on T and on these spaces such that T admits a >Taylor series expansion, and how to perform this expansion. Taylors theorem for functions from a Banach space to a Banach space is Theorem 6 in Nelson, Topics in Dynamics I: Flows, Princeton U. Press 1969. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Functional Analysis: Equivalent of Taylor series for operators >>Let T be an operator between two Banach spaces (or Hilbert spaces), >>what are the conditions on T and on these spaces such that T admits a >>Taylor series expansion, and how to perform this expansion. >Taylors theorem for functions from a Banach space to a Banach space >is Theorem 6 in Nelson, Topics in Dynamics I: Flows, Princeton U. Press >1969. Huh. Can you give us a hint what sort of thing these are? I can only imagine two possibilities: Taylor series which really amount to series for the function of one variable f(t) = T(x + ty), or something like Taylor series in infinitely many variables, by analogy with Taylor series in R^n. >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia Vancouver, BC, Canada ************************ David C. Ullrich === Subject: Re: Functional Analysis: Equivalent of Taylor series for operators >>Taylors theorem for functions from a Banach space to a Banach space >>is Theorem 6 in Nelson, Topics in Dynamics I: Flows, Princeton U. Press >>1969. >Huh. Can you give us a hint what sort of thing these are? I can >only imagine two possibilities: Taylor series which really >amount to series for the function of one variable >f(t) = T(x + ty), or something like Taylor series in infinitely >many variables, by analogy with Taylor series in R^n. For a function f from an open subset U of Banach space E to Banach space F, the Frechet derivative (if it exists) is a function Df from U to L(E,F) (the Banach space of bounded linear operators E -> F) such that f(x+y) = f(x) + Df(x) y + o(||y||). The Frechet derivative of Df, if it exists, is then a function D^2 f from U to L(E,L(E,F)), which can be identified with the space L(E x E, F) of bounded bilinear forms from E x E to F. We write D^2 f(x) y^2 for the value of the form (D^2 f)(x) at (y,y). Similarly for any positive integer k, D^k f, if it exists, is a function from U to the bounded k-linear forms from E^k to F, and we write D^k f(x) y^k for (D^k f)(x)(y,...,y) with k ys. Then this version of Taylors theorem says that if f is C^k on U f(x+y) = f(x) + sum_{j=1}^k D^j f(x) y^j/j! + o(||y||^k) Of course if E is R^n and F is R, this can be thought of as a coordinate-free way of writing the multivariate Taylor series of f to kth order. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Functional Analysis: Equivalent of Taylor series for operators >Taylors theorem for functions from a Banach space to a Banach space >is Theorem 6 in Nelson, Topics in Dynamics I: Flows, Princeton U. Press >1969. >>Huh. Can you give us a hint what sort of thing these are? I can >>only imagine two possibilities: Taylor series which really >>amount to series for the function of one variable >>f(t) = T(x + ty), or something like Taylor series in infinitely >>many variables, by analogy with Taylor series in R^n. >For a function f from an open subset U of Banach space E to Banach space >F, the Frechet derivative (if it exists) is a function Df from U to >L(E,F) (the Banach space of bounded linear operators E -> F) such that >f(x+y) = f(x) + Df(x) y + o(||y||). The Frechet derivative of Df, if it >exists, is then a function D^2 f from U to L(E,L(E,F)), Realized a minute too late that this was probably what was >which can be >identified with the space L(E x E, F) of bounded bilinear forms from >E x E to F. We write D^2 f(x) y^2 for the value of the form (D^2 f)(x) >at (y,y). Similarly for any positive integer k, D^k f, if it exists, is a >function from U to the bounded k-linear forms from E^k to F, and we write >D^k f(x) y^k for (D^k f)(x)(y,...,y) with k ys. Then this >version of Taylors theorem says that if f is C^k on U >f(x+y) = f(x) + sum_{j=1}^k D^j f(x) y^j/j! + o(||y||^k) >Of course if E is R^n and F is R, this can be thought of as a >coordinate-free way of writing the multivariate Taylor series >of f to kth order. >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia Vancouver, BC, Canada ************************ David C. Ullrich === Subject: Re: Functional Analysis: Equivalent of Taylor series for operators >thank you foryour answer. >I am interested by both questions: >Let T be an operator between two Banach spaces (or Hilbert spaces), >what are the conditions on T and on these spaces such that T admits a >Taylor series expansion, and how to perform this expansion. The _other_ question is the topic of something called the functional calculus, which you can read about in many books on functional analysis. This question makes less sense to me offhand. A Taylor expansion for T would look something like T(x) = sum c_n x^n, but if x is in a Banach space theres no such thing as x^n. >I would appreciate any references about this. >>Hi all, >>is there an equivalent of Taylor series for operators between Banach >>Spaces ?- if yes, what is the name of this equivalent, and what are >>the conditions on this operators ? >> I can think of at least two things the question might mean >> (does an operator have a Taylor series/can we apply a Taylor >> series to an operator). Try to be a little more specific about >> the question... >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Re: Cantors Theory: Mathematical creationism > Given a closed set A, he [Cantor] considered the process of > removing the isolated points to get a new closed set A. > If so, one can consider the intersection of all of them, > A^(omega), which is of course also a closed set. Then A^(omega) > =A^(omega+1) may still be different from A^(omega). > Its natural, and not the result of theological training, to want to say > keep going to this process. Youre talking about a process which you repeat infinitely many times, and then you keep going from there. That is certainly a modified notion of process, which answers the following question you ask: > |Both Cantors theory and Creationism theory are pseudoscience. Both the > |Creationists and the Cantorians impose upon their disciples a world view > |in which people must modify their thinking to incorporate certain axioms > |handed down from higher authority, > Modify from what? The intuitions we gain from living and observing the world we live in. === Subject: Re: Cantors Theory: Mathematical creationism <41a556ab$15$fuzhry+tra$mr2ice@news.patriot.net> <41a96cd2$7$fuzhry+tra$mr2ice@news.patriot.net> at 06:03 PM, examachine@gmail.com (Eray Ozkural exa) said: >The description of Kabbalah you opposed to defined clearly the reason >how Kabbalah was related to Cantors theory. No. >Exactly in which way do you object to that? In that it is false. >Are you claiming that Kabbalah did not assert the existence of a >higher order of infinity, or other nonsense like that? The issue is not a higher order of infinity; the issue is cardinal numbers, defined in terms of 1-1 correspondences. That has nothing to do with Qaballah, or with the way that infinity[1] is used in Qaballah. But I am claiming that Qaballah is not concerned about infinity, but rather with concepts that do not translate well into English. >You may want to read the below excerpt from the wikipedia entry. Hardly a reliable source. Please quote something from Qaballah that refers to cardinal numbers. >Kabbalists speak of the first aspect of God as Ein Sof; this is >translated as the infinite, or that which has no limits. As I said, it doesnt translate well. And it isnt a cardinal number. >The sefirot mediate the interaction of the ultimate unknowable God >with the physical and spiritual world. K3wl. What does that have to do with cardinal numbers? >According to Kabbalah, the true essence of G-d is so transcendent >that it cannot be described, except with reference to what it is >not. This true essence of G-d is known as Ein Sof, which literally >means without end, Closer, but still not accurate. >Well, it may actually have something to do with their *reality*, not >that you can conceive them. Surely, you can think of fiction. Indeed, but I would never claim that Gandalf was a character in a Tale of Two Cities, nor that Sidney Carton was one of the nazgul. >I can conceive of a world, which has no causation, but that is not >our world. If I formalized my notion of such a world, would I have >made my fantasy real? Youre ducking the question; the specific issue is whether Cantors concept of cardinal numbers came from Qaballah, not whether either is really. If you formalize your fantasy and Joe down the block formalizes his, that does not mean that your fantasy derives from Joes. [1] It isnt, but I can see how someone might incorrectly translate various distinct phrases into the single English word infinity. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: number of serieses Hi! I would like to know the number of serieses consisting of n non-negative integers where the sum of a series is below a predefined positive integer, K. Is it possible to find it in a closed formula? === Subject: Re: number of serieses >I would like to know the number of serieses consisting of n non-negative >integers where the sum of a series is below a predefined positive integer, I hope you mean positive integers, because if you include 0 there are an infinite number of series. Look up Stirling numbers of the second kind. Is that what you want? http://mathworld.wolfram.com/ StirlingNumberoftheSecondKind.html >Is it possible to find it in a closed formula? Yes. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: number of serieses >>I would like to know the number of serieses consisting of n non-negative >>integers where the sum of a series is below a predefined positive integer, >>K. > I hope you mean positive integers, because if you include 0 there are an > infinite number of series. > Look up Stirling numbers of the second kind. Is that what you want? he is counting (0,0), (0,1), (0,2), (0,3), (1,0), (1,1), (1,2), (2, 0) (2,1) and (3,0). The answer in general is the binomial coefficient (n+K choose n). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Cantor reloaded We consider a matrix in the form: m_p1 m_p2 m_p3 m_p4 ... m_pi ... : : : : : m_1q: m_11 m_12 m_13 m_14 ... m_1i ... m_2q: m_21 m_22 m_23 m_24 ... m_2i ... m_3q: m_31 m_32 m_33 m_34 ... m_3i ... m_4q: m_41 ... . . . m_jq: m_j1 ... m_ji ... . . . The m_ji are numerals (for example 0 ... 9 in the decimal system). Every m_pi and m_jq should be associated to a real number betweeen 0 and 1 (as 0 . m_11 m_12 m_13 m_14 ...) . The matrix should be infinite in both directions (horizontal and vertical): while the irrational numbers are infinite in decimal places, the columns and the rows must be infinite. The matrix containes all real numbers. So this matrix is a equivalent to a Ôlist, in which the real numbers are numbered with real numbers. Like a list of natural numbers which ist numbered with natural numbers, this list of real numbers must be quadratic. This matrix (or list on other considerations) cant contain its anti-diagonal. Albrecht S. Storz === Subject: Re: Cantor reloaded > Do you think whether it is possible to put infinite things or objects, > especially mathematical objects, in a row? If the objects are infinite > like real numbers, is it also possible? And how do these things behave > in the infinite at the point at which the natural numbers end? These > questions are very silly, but they can lead to another understanding > of the things. > Why do you think, a matrix, filled with numerals of real numbers does > not exist? Of course you can make a list of all real numbers, by application of the well-ordering principle. However, the numbers of rows in your list must be equal to the cardinality of the real number system. So assume that you have such a list. > This matrix (or list on other considerations) cant contain its > anti-diagonal. This is where the problem lies. How can you be certain that such an antidiagonal exist. The construction should take the form of an argument by transfinite inducition. Correspond with each entry in a row an ordinal number. For finite ordinals one can take the procedure of Cantor, however if you have limit ordinals, things get complicated. Take a limit ordinal alpha. Assume that there exist a real number (probably more), different from every entry in a row gamma less then alpha. How can you be sure that there is a real number different from every ordinal number less or equal to alpha. If you can solve out above problem, your assertion is valid. But I think your problem can easily be falsified. Assume that the list of real numbers has an antidiagonal. This antidiagonal is the limit of a rational number. Just look at the construction of the antidiagonal. This limit must be a real number. (by definition, a rational number is the limit of a sequence of real numbers). Therefore the antidiagonal is a real number. By your assertion the antidiagonal is not in the list of real numbers, hence it cannot be a real number. Therefore, we have a contradiction. Either, the antidiagonal is in the list of real numbers or the list of real numbers does not have an antidiagonal. So if your assertion is true: the list of real numbers does not contain its anti-diagonal then the result follows: the antidiagonal of the list of real numbers does not exist. thomas *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Cantor reloaded > But I think your problem can easily be falsified. Assume that the list > of real numbers has an antidiagonal. This antidiagonal is the limit of > a rational number. Just look at the construction of the antidiagonal. > This limit must be a real number. (by definition, a rational number > is the limit of a sequence of real numbers). You dont mean this in the other way: a real number is the limit of a sequence of rational numbers? > Therefore the > antidiagonal is a real number. By your assertion the antidiagonal is > not in the list of real numbers, hence it cannot be a real number. > Therefore, we have a contradiction. Either, the antidiagonal is in > the list of real numbers or the list of real numbers does not have an > antidiagonal. > So if your assertion is true: the list of real numbers does not > contain its anti-diagonal then the result follows: the antidiagonal > of the list of real numbers does not exist. So you would argue in the same way: The conclusion of Cantors diagonal proof (on reals) is either: no (complete) list exists, or: the antidiagonal does not exists. ? Albrecht > thomas > *-----------------------* > www.GroupSrv.com > *-----------------------* === Subject: Re: Cantor reloaded >> Do you think whether it is possible to put infinite things or > objects, >> especially mathematical objects, in a row? If the objects are > infinite >> like real numbers, is it also possible? And how do these things > behave >> in the infinite at the point at which the natural numbers end? > These >> questions are very silly, but they can lead to another > understanding >> of the things. >> Why do you think, a matrix, filled with numerals of real numbers > does >> not exist? Of course you can make a list of all real numbers, by application of > the well-ordering principle. Why not just observe that the identity map from R to R is a list in the same sense? In the context of Cantors proof, a list is generally taken to be a map defined on the naturals, but if you are going to allow lists with uncountable domains, you may as well just choose R. > However, the numbers of rows in your list must be equal to the > cardinality of the real number system. So assume that you have such a > list. >> This matrix (or list on other considerations) cant contain its >> anti-diagonal. There is no such thing. The anti-diagonal is a function a: R^N -> R (*) such that, for each f in R^N (meaning f: N -> R is a list), there is a number x = a(f) in R such that x not in ran(f). If your list is the identity map on R, then it is a member of R^R, not R^N, and is not in the domain of a, as indicated by (*). Hence, the anti-diagonal of the identity on R (or of any mapping whose domain is not N) has no meaning. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Cantor reloaded at 03:08 PM, albstorz@gmx.de (albrecht) said: >The matrix containes all real numbers. No. >this list of real numbers must be quadratic. What do you mean by quadratic? >This matrix (or list on other considerations) cant contain its >anti-diagonal. Which tells you that it doesnt contain all real numbers. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Cantor reloaded The matrix containes all real numbers. > So this matrix is a equivalent to a Ôlist, in which the real numbers > are numbered with real numbers. Using Cantors diagonalization proof one can show that there must be a real number that is not in your list. Hence a matrix of all real numbers does not exist. [/quote]Like a list of natural numbers which ist numbered with natural numbers, this list of real numbers must be quadratic. [/quote] What do you mean with quadratic. So: The set consisting of the numbers {1,2,3,4} can be written as. 1 2 3 4 Is this set quadratic???? (whatever it may mean) thomas *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Cantor reloaded > The matrix containes all real numbers. > So this matrix is a equivalent to a Ôlist, in which the real > numbers > are numbered with real numbers. > Using Cantors diagonalization proof one can show that there must be a > real number that is not in your list. Hence a matrix of all real > numbers does not exist. Do you think whether it is possible to put infinite things or objects, especially mathematical objects, in a row? If the objects are infinite like real numbers, is it also possible? And how do these things behave in the infinite at the point at which the natural numbers end? These questions are very silly, but they can lead to another understanding of the things. Why do you think, a matrix, filled with numerals of real numbers does not exist? > [/quote]Like a list of natural numbers which ist numbered with > natural > numbers, this list of real numbers must be quadratic. > [/quote] > What do you mean with quadratic. So: The set consisting of the numbers > {1,2,3,4} can be written as. > 1 2 > 3 4 > Is this set quadratic???? (whatever it may mean) We consider a list or a matrix. If the list or the matrix isnt quadratic then the question whether an anti-diagonal number is contained or not is not so easy to answer. If the list or the matrix is quadratic, the diagonal and the anti-diagonal number is of the same kind as the numbers in rows and columns. Your matrix contained the diagonal number 1 4 and an anti-diagonal, for example 8 6. Both contained two digits such as your rows and columns. A. S. Storz > thomas > *-----------------------* > www.GroupSrv.com > *-----------------------* === Subject: Re: Cantor reloaded > We consider a matrix in the form: > m_p1 m_p2 m_p3 m_p4 ... m_pi ... > : : : : : > m_1q: m_11 m_12 m_13 m_14 ... m_1i ... > m_2q: m_21 m_22 m_23 m_24 ... m_2i ... > m_3q: m_31 m_32 m_33 m_34 ... m_3i ... > m_4q: m_41 ... > m_jq: m_j1 ... m_ji ... > The m_ji are numerals (for example 0 ... 9 in the decimal system). > Every m_pi and m_jq should be associated to a real number betweeen 0 > and 1 (as 0 . m_11 m_12 m_13 m_14 ...) . The matrix should be > infinite in both directions (horizontal and vertical): while the > irrational numbers are infinite in decimal places, the columns and the > rows must be infinite. > The matrix containes all real numbers. If the rows and columns of this matrix are indexed by the integers, then the claim that it can contain all real numbers requires proof. > So this matrix is a equivalent to a Ôlist, in which the real numbers > are numbered with real numbers. > Like a list of natural numbers which ist numbered with natural > numbers, this list of real numbers must be quadratic. > This matrix (or list on other considerations) cant contain its > anti-diagonal. Such a matrix can easily be reordered into the more standard form of list, from which as many anti-diagonals as one wishes. > Albrecht S. Storz === Subject: Re: Cantor reloaded > > We consider a matrix in the form: > > m_p1 m_p2 m_p3 m_p4 ... m_pi ... > : : : : : > m_1q: m_11 m_12 m_13 m_14 ... m_1i ... > m_2q: m_21 m_22 m_23 m_24 ... m_2i ... > m_3q: m_31 m_32 m_33 m_34 ... m_3i ... > m_4q: m_41 ... > . > . > . > m_jq: m_j1 ... m_ji ... > . > . > . > > The m_ji are numerals (for example 0 ... 9 in the decimal system). > Every m_pi and m_jq should be associated to a real number betweeen 0 > and 1 (as 0 . m_11 m_12 m_13 m_14 ...) . The matrix should be > infinite in both directions (horizontal and vertical): while the > irrational numbers are infinite in decimal places, the columns and the > rows must be infinite. > The matrix containes all real numbers. > If the rows and columns of this matrix are indexed by the integers, > then the claim that it can contain all real numbers requires proof. You are right. I should not use natural numbers as indices. But their function is only, to have labels to talk about. They are not necessary for the construction. I should label it in a better way with alpha, beta and so on. If we want indexing, can we use the Powerset of the natural numbers? > So this matrix is a equivalent to a Ôlist, in which the real numbers > are numbered with real numbers. > Like a list of natural numbers which ist numbered with natural > numbers, this list of real numbers must be quadratic. > > This matrix (or list on other considerations) cant contain its > anti-diagonal. > Such a matrix can easily be reordered into the more standard form of > list, from which as many anti-diagonals as one wishes. > I think this matrix could not be converted to a usual list. A list represents a bijection between a set X and the set of the natural numbers. I try to have a bijection between Y and R. > > Albrecht S. Storz === Subject: Re: Cantor reloaded posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs > The matrix containes all real numbers. What makes you think so? - Randy === Subject: Re: Cantor reloaded > The matrix containes all real numbers. > What makes you think so? > - Randy If the rows and the columns are filled with (different) irrational numbers without end, the matrix should represent all real numbers. Do you know the diagonal proof of G. Cantor? His proof consider every list of rational numbers. So I do for every matrix with real numbers. Also for a completely matrix with all reals. I dont need to have this matrix. But if this matrix exists it cant contain the anti-diagonal. Albrecht === Subject: Re: Cantor reloaded Correction: We consider a matrix in the form: m_palpha m_pbeta m_pgamma ....m_pi : : : : : m_alphaq: m_alphaalpha m_alphabeta m_alphagamma ... m_alphai ... m_betaq: m_betaalpha m_betabeta m_betagamm ... m_betai ... m_gammaq: m_gammaalpha m_gammabeta m_gammagamma ... m_gammai ... m_deltaq: m_deltaalpha ... . . . m_jq: m_jalpha ................... m_ji ... . . . The m_ji are numerals (for example 0 ... 9 in the decimal system). Every m_pi and m_jq should be associated to a real number betweeen 0 and 1 (as 0 . m_alphaalpha m_alphabeta m_alphagamma m_alphadelta ...). The indices must be taken out of a appropriate set (perhaps Powerset of |N?). But the indices are not necessary for the consideration. Essentially is only the matrix itself. The matrix should be infinite in both directions (horizontal and vertical): since the irrational numbers are infinite in decimal places, the columns and the rows must be infinite. The matrix is able to contain all real numbers. So this matrix is a equivalent to a Ôlist, in which the real numbers are numbered with real numbers (or with something with the same cardinality). This matrix (or list on other considerations) cant contain its anti-diagonal. Albrecht S. Storz === Subject: Re: Cantor reloaded posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs > The matrix containes all real numbers. > What makes you think so? > - Randy > If the rows and the columns are filled with (different) irrational > numbers without end, the matrix should represent all real numbers. Only if the irrational numbers are countable. > Do you know the diagonal proof of G. Cantor? His proof consider every > list of rational numbers. His proof says, suppose I have a list of real numbers and goes on to show that there exists a number not on the list. > So I do for every matrix with real numbers. Have you seen the proof of the countability of the rationals? It lays the rational numbers out in a matrix very much like yours, and shows that you can trace a path through that matrix that associates a natural number with every position in the matrix. That is, that the number of elements in the matrix is countable. You cant start by assuming the irrationals are countable. By Cantors method, as soon as you tell you youve got a matrix of irrationals, Ill find you one not on the list. You have infinitely many reals, but you still dont have ALL the reals. - Randy === Subject: Re: Cantor reloaded > The matrix containes all real numbers. > What makes you think so? > > - Randy > If the rows and the columns are filled with (different) irrational > numbers without end, the matrix should represent all real numbers. > Only if the irrational numbers are countable. Why? The matrix consist of the numerals of real numbers in rows. It has so many rows as there exists real numbers. So they are not enumerable. Why cant this matrix exist? If this matrix is thinkable it leads to the same conclusion like Cantors proof but for the impossibility of the bijection R --> R. > Do you know the diagonal proof of G. Cantor? His proof consider every > list of rational numbers. > His proof says, suppose I have a list of real numbers > and goes on to show that there exists a number not on > the list. > So I do for every matrix with real numbers. > Have you seen the proof of the countability of the > rationals? It lays the rational numbers out in a matrix > very much like yours, and shows that you can trace a > path through that matrix that associates a natural > number with every position in the matrix. That is, that > the number of elements in the matrix is countable. > You cant start by assuming the irrationals are countable. I dont assume like this. Albrecht > By Cantors method, as soon as you tell you youve got > a matrix of irrationals, Ill find you one not on the > list. You have infinitely many reals, but you still > dont have ALL the reals. > - Randy === Subject: topology.....--- hello........doctor~ if X is path-connected, then X is locally path-connected. -------------------------------------------- i think..... counter-example is {(x,y) | 0<= x <=1, y = x/n, n=1,2,....} this is path-connected, but this is not locally path-connected. um....is this right counter-example ?? === Subject: Re: topology.....--- > hello........doctor~ > if X is path-connected, then X is locally path-connected. > -------------------------------------------- > i think..... > counter-example is {(x,y) | 0<= x <=1, y = x/n, n=1,2,....} > this is path-connected, but this is not locally path-connected. > um....is this right counter-example ?? I dont believe so. what point doesnt have a path connected neighborhood? Given a point (x,x/n) one can certainly find some neighborhood that doesnt intersect the line y=x/(n-1) or y=x/(n+1). Perhaps you want to expand this set somewhat to make it non-locally path-connected. (Hint: what is the closure of this set?) === Subject: Re: topology.....--- > hello........doctor~ > if X is path-connected, then X is locally path-connected. > -------------------------------------------- > i think..... > counter-example is {(x,y) | 0<= x <=1, y = x/n, n=1,2,....} > this is path-connected, but this is not locally path-connected. > um....is this right counter-example ?? > I dont believe so. what point doesnt have a path connected > neighborhood? Given a point (x,x/n) one can certainly find some > neighborhood that doesnt intersect the line y=x/(n-1) or y=x/(n+1). > Perhaps you want to expand this set somewhat to make it non-locally > path-connected. (Hint: what is the closure of this set?) ------------------------------------------------------------- ----------- um........ Heres a compact Hausdorff (and hereditarily normal, but not metrizable) counterexample. Define the long circle using the following steps. Step 1: Let X be the set of all ordinals that are less than or equal to the first uncountable ordinal *:=omega_1 = aleph_1. Given the order topology, X is a non-metrizable compact Hausdorff space. Step 2: Form the quotient space Y := (X x [0,1])/~, where the points (a,1) are identified with (a+1,0), for all countable ordinals a. Y is a long interval, a connected, locally connected compact ordered space. Its almost what you want, but it fails to be path connected. We remedy that by identifying the first and last points. This space is now path connected and locally connected, but not locally path connected at the bad point with no countable neighborhood base. ------------------------------------------------------------- -------- ------------------------------------------------------------- -------- > I am searching for a subset of the Euclidean plane that is path > connected but not locally connected in *any* point. Consider {(1+ r e^{it})^2: 0 <= r < infinity, 0 < t < Pi, t rational} in the complex plane. ------------------------------------------------------------- -------- um......by Robert B. Israel. more complex........oh....my god. === Subject: Test statistic for a coin toss I want to form a statistical hypothesis built around the likelihood that a tossed coin will land on heads approximately 50% of the time. Could anyone help me get started? I mainly need help deciding which test statistic to use (Z, T, F, or chi squared). Any help would be great. === Subject: Re: Test statistic for a coin toss > I want to form a statistical hypothesis built around the likelihood that > a > tossed coin will land on heads approximately 50% of the time. Could > anyone > help me get started? I mainly need help deciding which test statistic to > use > (Z, T, F, or chi squared). You are interested in the binomial distribution, which can be approximated by a Gaussian distribution. Do you know of Bernoulli trials? -- Mostly economics: r c v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Help!! >Please help me to sove the Question︰ >S（ω）= 26+006ω/ω, ω>=5 >Graph S and state the process (domain , s, s) >Please re-mail: hanhaly@yahoo.com.tw >3Q very, very, very much!!!!!! I count five times now that youve posted this. If you were present in the room, would you just walk up and start yanking at a persons sleeve until they dropped what they were doing and answered you RIGHT NOW? Hmm .. since you demand an e-mail response, I guess a better analogy is until they left the room and answered you in private. Hint: Usenet is like God: it helps those who help themselves. Post _once_, and show what you yourself did to solve the problem. And since you post here, have the courtesy to read answers here. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: A079586 njas: The first %H was moved. %I A079586 %S A079586 3,3,5,9,8,8,5,6,6,6,2,4,3,1,7,7,5,5,3,1,7,2,0,1,1,3,0,2,9,1,8, 9,2,7,1, %T A079586 7,9,6,8,8,9,0,5,1,3,3,7,3,1,9,6,8,4,8,6,4,9,5,5,5,3,8,1,5,3,2, 5,1,3,0, %U A079586 3,1,8,9,9,6,6,8,3,3,8,3,6,1,5,4,1,6,2,1,6,4,5,6,7,9,0,0,8,7,2, 9,7,0,4 %N A079586 Decimal expansion of sum(k>=1,1/F(k)) where F(k) is the k-th Fibonacci number A000045(k). %H A079586 Eric Weissteins World of Mathematics, Recipr ocal Fibonacci Constant %H A079586 Eric Weissteins World of Mathematics, Fibonacci Number %F A079586 3.35988566624.... %Y A079586 Cf. A084119. %K A079586 cons,nonn %O A079586 1,1 -- Don Reble djr@nk.ca === Subject: Re: Function naming >function group as I described polynomials. I have some knowledge of math, >but I was always under the impression that a polynome was defined as a one >variable function, and that these were merely a special case of the >multivariable case. Probably has to do with my understanding ;-) (Please dont post upside down.) A good resource for definitions is http://mathworld.wolfram.com -- the definition at http://mathworld.wolfram.com/Polynomial.html says A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. I understand that this wouldnt help you when you didnt know the term, but when you do want a definition of a specific term (or facts about a term or concept) its a great starting point. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: JSH: Operator ambiguity, Escultura hey, then 0 = 5 - 5 = -5 - 5 = -10; you should turn that into a tutorial on Zero Divisors ... and watch out for those chaotic rounding-off errors! > Hey! I remember thinking this was really cool. I did it with 5 = -5, but --Advice 0.05; Free, if wrong, again! http://tarpley.net/bush6.htm === Subject: Re: JSH: Operator ambiguity, Escultura > You see, the ambiguity in the square root operator still remains, > despite the convention. Your insistence that it is impossible to define a function for sqrt(x), i.e. to restrict the possible solutions to a single value, invalidates your own prime count function and disqualifies you from further study of algebra. Analogous situations arise in evaluating periodic functions, in finding Ôprincipal values, in dealing with Ôbranch cuts and elsewhere. If you cannot understand how these issues are resolved, take up a different line of work. (Not sales, however. I doubt if you could sell manure to an organic farmer.) -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Operator ambiguity, Escultura > Assumption (b) is whats holding you back from learning math. Before you > can learn, you have to realize that you dont know. A Fundamental Truth. === Subject: Re: JSH: Operator ambiguity, Escultura > Assumption (b) is whats holding you back from learning math. Before you > can learn, you have to realize that you dont know. > A Fundamental Truth. Oh please, the poster never answered the mathematical issue raised but instead turned to personal attacks. You deleted out all of the context and made a reply that had nothing to do with the issue. For those who missed it, I used Esculturas example of i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction, to show that its naive to think that you can remove the ambiguity from the square root operator. Rather than give a cogent answer the sci.math poster Jim Ferry babbled about his childhood, and simply *claimed* as if thats all it takes that theres no problem. The regular sci.math poster Gib Bogle then came in to delete out all of the context and make his own reply. Thats how sci.mathers managed to paint me as a crank. They do this consistently, and sci.math readers cheer them on! The sci.math readership is remarkably stupid and gullible. James Harris === Subject: Re: JSH: Operator ambiguity, Escultura > > Assumption (b) is whats holding you back from learning math. Before you > can learn, you have to realize that you dont know. > > A Fundamental Truth. > Oh please, the poster never answered the mathematical issue raised but > instead turned to personal attacks. You deleted out all of the > context and made a reply that had nothing to do with the issue. > For those who missed it, I used Esculturas example of > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction, > to show that its naive to think that you can remove the ambiguity > from the square root operator. > Rather than give a cogent answer the sci.math poster Jim Ferry babbled > about his childhood, and simply *claimed* as if thats all it takes > that theres no problem. > The regular sci.math poster Gib Bogle then came in to delete out all > of the context and make his own reply. > Thats how sci.mathers managed to paint me as a crank. They do this > consistently, and sci.math readers cheer them on! Can you give me an example showing sci.math readers cheering them on? > The sci.math readership is remarkably stupid and gullible. Is it? How do you know? Brian Chandler http://imaginatorium.org === Subject: Re: JSH: Operator ambiguity, Escultura > > Assumption (b) is whats holding you back from learning math. Before you > can learn, you have to realize that you dont know. > > A Fundamental Truth. > > > Oh please, the poster never answered the mathematical issue raised but > instead turned to personal attacks. You deleted out all of the > context and made a reply that had nothing to do with the issue. > > For those who missed it, I used Esculturas example of > > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction, > > to show that its naive to think that you can remove the ambiguity > from the square root operator. > > Rather than give a cogent answer the sci.math poster Jim Ferry babbled > about his childhood, and simply *claimed* as if thats all it takes > that theres no problem. > > The regular sci.math poster Gib Bogle then came in to delete out all > of the context and make his own reply. > > Thats how sci.mathers managed to paint me as a crank. They do this > consistently, and sci.math readers cheer them on! > Can you give me an example showing sci.math readers cheering them on? Ive been posting for years. These people routinely get all kinds of support, which has convinced me that the sci.math readership is stupid. Just dig into the record, and youll find plenty of people getting on my case about me supposedly being wrong since all these people supposedly have counter examples, and such, versus very few, if any questioning posts asking if maybe these obsessive people replying to me werent full of it! > The sci.math readership is remarkably stupid and gullible. > Is it? How do you know? > Brian Chandler I was pissed off yesterday so basically I was ranting. I know, its better not to go around calling a lot of people stupid, but hey, if mathematics is explained to you in detail, where the conclusion is quite evident, and you listen to people who rather stupidly lie about the mathematics, and instead of properly questioning them, just swallow their b.s., then do you really think you get points? After all, eventually my research will come out on top, and what will the sci.math newsgroup have? A long record of rather amazing stupidity, where its most notable point, and what will probably define the newsgroup from now on, was the gang emailing to the Southwest Journal of Pure and Applied Mathematics to get my paper censored. Whether I call you stupid now or not, more than likely that and resisting my research, namecalling, and other childish behavior is ALL that the sci.math newsgroup will become known for. The sci.math newsgroup will have its place in history. James Harris === Subject: Re: JSH: Operator ambiguity, Escultura > A long record of rather amazing stupidity, where its most notable > point, and what will probably define the newsgroup from now on, was > the gang emailing to the Southwest Journal of Pure and Applied > Mathematics to get my paper censored. Who is in this gang, and how do you know they emailed SWJPA? Also, the word censored is a bit strong. Is criticism really censorship? As far as I can see, nobody here wanted your paper dropped. It would have been fun for us all if it were actually left published. It would have been a problem for SWJPA, but I for one would have enjoyed it if it were left up there. Perhaps your paper was dropped just because the editor realized that it was wrong and worthless. === Subject: Re: JSH: Operator ambiguity, Escultura ... > Oh please, the poster never answered the mathematical issue raised but > instead turned to personal attacks. You deleted out all of the > context and made a reply that had nothing to do with the issue. But some posters answered it. > For those who missed it, I used Esculturas example of > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction, The contradiction exists because the assumption is made that sqrt(a/b) = sqrt(a)/sqrt(b), which is false when in the complex numbers. Similarly log(a.b) = log(a) + log(b) is false when in the complex numbers. Read a bit about Riemann surfaces and you may understand. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Operator ambiguity, Escultura > Read a bit about Riemann surfaces and you may understand. He wont. -- --Tim Smith === Subject: Re: JSH: Operator ambiguity, Escultura > ... > i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction, This is a reductio proof that sqrt(x/y) <> sqrt(x)/sqrt(y) . This is a well-known and not very exciting fact. === Subject: Re: JSH: Operator ambiguity, Escultura 1 = 0 is a well known fact among insiders. There are hundreds of proofs for this fact. The last one was published by an Australian mathematician: Jamie Simpson, A New Proof that 1=0, JRM vol 32, Number 2, page 142 Richard Schorn > So, by the convention, sqrt(4) = 2, and thats good as, -2(-2) = 4, so > if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2, and 4 = 0, > which is not good. ^^^^ ??? === Subject: re:Determining number of roots of a polynomial I cant remember the details, but I remember having to use an algorithm involving Sturm sequences. The algorithm gave the number of real roots (which may include multiplicities) greater than some given number. Thus if you knew a lower bound for the roots, you could use it to get all real roots. Added note: Look up Sturm sequence using Google - it will get you the necessary information. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Dumbing down, dumbing up? >So whats going on here? There are constant barrages of anecdotes >about how stupid kids are these days. Arent more and more highschool >kids taking the Calculus AP exam or college level classes? > http://www.maa.org/features/dualenrollment.html >> More are, but fewer are taking real mathematics courses. >> The old-fashioned Euclidean geometry course, emphasizing >> proofs, was mathematics. So was induction in the old >> college algebra course. There was little additional >> mathematics in the old program until rigorous analysis >> or algebra was introduced. On this, nothing has changed. >just to get the terminology down correctly, mathematics means >mathematical subject matter with proofs, and not theory means not >real mathematics means pattern matching, plugging in formulas, >learning to apply symbolic algorithms? Plugging into formulas is like using a hammer or a typewriter; it is essentially training, rather than education. There is a huge difference here. The same holds for directly applying symbolic algorithms. Especially for the non-mathematician, the important parts are concepts and language. In my opinion, the use of variables is what freed post-Renaissance mathematics from the restraints which kept the Greeks from doing much more; one variable goes back to Diophantus (about 300), the systematic use of more than one for numbers to Viete, around 1600, functions to Euler, and the full modern use to those looking at the foundations around 1900 and later. But if this is extended to full linguistics, it becomes easy. The other point is concepts. This is much harder than it looks, but is very important. If one has the concepts and the language, one can translate between real world problems and formal problems, and the the full power of mathematics can be applied to the formal problems. A major stumbling block is that the same mathematical objects can have different interpretations. For the ordinary integers, the cardinal and ordinal concepts are quite different, and I believe that one of the problems with the new math was that the apparently easier, but really harder to work with, cardinal concept was chosen, rather than the apparently more complicated, but easier to develop, ordinal concept. There are other concepts which involve extensions to rational numbers, signed numbers, real and complex numbers, which involve introducing additional somewhat related concepts. So when I am asked what 1 means, how many answers should I give? It is quite possible to get an intuitive understanding of concepts, but learning mechanics does not seem to help. It is extremely doubtful if learning to add and multiply by rote helps at all in learning the underlying concepts, while the other does help, although it may leave the person asking why? The non-mathematician needs to understand what proofs are, and the need for them. The mathematician needs to be more familiar, and able to produce them sometimes. Learning to apply algorithms needs to be divided into its parts; formulate the problem as a mathematical problem, see if there are appropriate methods available, and then use them. The first part is language; for the second part, the formulation may be standard, in which case the method can be looked up, or it might have to be derived, or some other trick found. Sometimes the tricks can be found, but often it will take a good mathematician to see them. As for pattern matching, this is a research method, and cannot be taught. It can be demonstrated, and one can hope the student picks up some of it. Research consists in seeing the obvious, and research ability is essentially independent of background, alas. We need more good researchers, and they are not going to be produced by wasting their time going from specific to general. We do not have the time. >> All of the computation courses, given without any theory >> foundation, are not real mathematics courses. Honors >> students may still be taught axiom-theorem-proof mathematics, >> and computational procedures based on that theoretical >> background, but the proportion of college entrants who >> have had some rigorous mathematics has dropped from almost >> all to very few. >Isnt this also a trend in entry level undergrad calculus courses? >That is, more emphasis on doing calculus (computational) and less on >the reason why (proof based)? If we are going to do anything else, we will have to assume that the entering college students need to learn basics. >Is the trend also a reason for the introduction of bridge courses >(a proof methods class taken as pre-req for upper level undergrad math >classes) Understanding proofs belongs in elementary school. It has been taught to the upper half of fifth graders, and I think a less pedantic approach would even work for a larger set of third graders. The present proof methods courses are far too weak; a direct abstract algebra course might well work much better, as the students would not have too many examples available. Proof methods in analysis are much less effective. And it would not be a bad idea to teach the number systems by a modification of Landau, starting in first grade. First graders can handle those proofs, but it seems teachers cannot. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Uncle assAl: (SR) Lorentz t, x = Intervals I dont currently have time to discuss this right now, as much as I would like to, but I have a few question for you. 1) I read your clock analysis, do you think the HK experiment is possible. You said the counter is perfect in a clock (usually) but the oscillator is not, so is HK possible to perform and get accurate results? 2) I still have some difficult with your light example, because a component of the lights velocity is in the direction of ship, it seems like that should matter. In the other version (Garders) the light moves perpendicular to ship. Seems like this should matter, I cant quite put my finger on it though. Do you see what I am saying? Is there any validity to it? Peter === Subject: Re: Uncle assAl: (SR) Lorentz t, x = Intervals >I dont currently have time to discuss this right now, as much as I >would like to, but I have a few question for you. > 1) I read your clock analysis, do you think the HK experiment is > possible. You said the counter is perfect in a clock (usually) but the > oscillator is not, so is HK possible to perform and get accurate > results? No. > 2) I still have some difficult with your light example, because a > component of the lights velocity is in the direction of ship, it > seems like that should matter. In the other version (Garders) the > light moves perpendicular to ship. Seems like this should matter, I > cant quite put my finger on it though. Do you see what I am saying? No. Is there any > validity to it? No. > Peter === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) posting-account= y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g > This assumes that A is invertible but if > both A and D were not invertible Det(M)=0 Consider A = D = 0, B = C = I. I prefer a perturbation argument to overcome singularity of (for example) A. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) > This assumes that A is invertible but if > both A and D were not invertible Det(M)=0 > Consider A = D = 0, B = C = I. Yes, but you are still assuming that the matrices on a diagonal are invertible. You would then swap columns or choose a different multiplying matrix. In any case, it seems that you would have to assume that at least one matrix was non-singualar. > I prefer a perturbation argument to overcome singularity of (for > example) A. Yes, but I suppose you are going beyond elementary methods. I am not familiar with the textbook from which the excercise was taken. Knowing what topics had been covered might suggest how the problem should be solved using commutivity alone or with a peturbation argument if that had been covvered. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) > This assumes that A is invertible but if > both A and D were not invertible Det(M)=0 > Consider A = D = 0, B = C = I. Yes, but you are still assuming that the matrices on a diagonal are invertible. You would then swap columns or choose a different multiplying matrix. In any case, it seems that you would have to assume that at least one matrix was non-singualar. > I prefer a perturbation argument to overcome singularity of (for > example) A. Yes, but I suppose you are going beyond elementary methods. I am not familiar with the textbook from which the excercise was taken. Knowing what topics had been covered might suggest how the problem should be solved using commutivity alone or with a peturbation argument if that had been covvered. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) posting-account=JrQSJw0AAAA6UtXIbhl36iP_cLZ-hNeT Just to clarify what tools a reader of the textbook might reasonably bring to bear on the exercise, perturbation arguments and even invariant subspaces have not yet been introduced. Topics covered thus far include elementary row operations, vector spaces, bases, and the idea of dimension, linear transformations and their representation by matrices, dual spaces and dual bases, elementary polynomial algebra (ideals of polynomials and some simple results depending therefrom), and determinants of matrices. No canonical forms of any sort have been discussed, though similarity insofar as it relates to changes of bases has been covered. My suspicion, after reading through this thread, is that the solution theyre looking for involves something along the lines of first assuming that both A and D are invertible, proving the result based on something similar I.M. Davidsons line of reasoning, and then showing that it still holds even if A and/or D are both singular. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) >> This assumes that A is invertible but if >> both A and D were not invertible Det(M)=0 >> Consider A = D = 0, B = C = I. >Yes, but you are still assuming that >the matrices on a diagonal are invertible. >You would then swap columns or choose a >different multiplying matrix. >In any case, it seems that you would have to assume >that at least one matrix was non-singualar. >> I prefer a perturbation argument to overcome >singularity of (for >> example) A. >Yes, but I suppose you are going beyond >elementary methods. Not really. The determinant of a matrix being a polynomial in its elements is continuous. As the determinant of A+tI is a polynomial in t, it must be non-singular for all non-zero t which are sufficiently small, and hence A+tI is non-singular. Now pass to the limit as t -> 0. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) >> This assumes that A is invertible but if >> both A and D were not invertible Det(M)=0 >> Consider A = D = 0, B = C = I. >Yes, but you are still assuming that >the matrices on a diagonal are invertible. >You would then swap columns or choose a >different multiplying matrix. >In any case, it seems that you would have to assume >that at least one matrix was non-singualar. >> I prefer a perturbation argument to overcome > singularity of (for >> example) A. >Yes, but I suppose you are going beyond >elementary methods. > Not really. The determinant of a matrix being a > polynomial in its elements is continuous. As the > determinant of A+tI is a polynomial in t, it must > be non-singular for all non-zero t which are > sufficiently small, and hence A+tI is non-singular. > Now pass to the limit as t -> 0. Yes, I see what you mean. So are you also saying that the problem as stated cannot be solved without analysis, i.e without taking limits, unless you make an assumption about invertibilty ? === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) > This assumes that A is invertible but if > both A and D were not invertible Det(M)=0 > Consider A = D = 0, B = C = I. >>Yes, but you are still assuming that >>the matrices on a diagonal are invertible. >>You would then swap columns or choose a >>different multiplying matrix. >>In any case, it seems that you would have to assume >>that at least one matrix was non-singualar. > I prefer a perturbation argument to overcome >> singularity of (for > example) A. >>Yes, but I suppose you are going beyond >>elementary methods. >> Not really. The determinant of a matrix being a >> polynomial in its elements is continuous. As the >> determinant of A+tI is a polynomial in t, it must >> be non-singular for all non-zero t which are >> sufficiently small, and hence A+tI is non-singular. >> Now pass to the limit as t -> 0. >Yes, I see what you mean. >So are you also saying that the problem as stated >cannot be solved without analysis, i.e without taking limits, >unless you make an assumption about invertibilty ? No, you can get by with less. If one is working in matrices over a field with infinitely many elements, the determinant of A+tI will still be non-zero for all except a finite number of t. Thus, one will have equality of the determinants of the 2nx2n matrix and the nxn matrix for all except a finite number of t. However, both determinants are polynomials of degree n, so agreement at n+1 values of t guarantees that the equation wil hold for all. If the field does not have sufficiently many elements, go to the algebraic closure, which is infinite. The theorem holds there, but the determinants are not changed by the embedding. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) posting-account= y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g I.M. Davidson asked: > So are you also saying that the problem as stated > cannot be solved without analysis, i.e without taking limits, > unless you make an assumption about invertibilty ? This is a case where the analysis technique (continuity of polynomial functions) could be replaced by a formal algebraic technique, e.g. carrying out the computation: det( A + xI, B; C, D ) with indeterminate x as a polynomial in x. Note that since det( A + xI ) is a monic polynomial, its not a zero divisor in R[x], where R is a ring with unit containing the matrix coefficients. Since det( I, 0; 0, A + xI) = det( A + xI ), we have: det( A + xI, B; (A + xI)C, (A + xI)D ) = det(A + xI ) * det( A + xI, B; C, D ) Now subtract C times the top blocks of the matrix in the left-hand side determinant from the bottom blocks, which leaves the determinant unchanged (as a sequence of elementary row operations): det( A + xI, B; 0, (A + xI)D - BC) = det(A + xI ) * det( A + xI, B; C, D ) But the left-hand side is now block triangular, so: det(A + xI)*det( (A + xI)D - BC) = det(A + xI ) * det( A + xI, B; C, D ) and since det(A + xI) is not a zero divisor: det( (A + xI)D - BC) = det( A + xI, B; C, D ) Now the evaluation mapping x |--> 0 is a ring homomorphism from R[x] to R, so applying it to both sides: det(AD - BC) = det(A, B; C, D) === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) > I.M. Davidson asked: > So are you also saying that the problem as stated > cannot be solved without analysis, i.e without taking limits, > unless you make an assumption about invertibilty ? > This is a case where the analysis technique (continuity of > polynomial functions) could be replaced by a formal > algebraic technique, e.g. carrying out the computation: > det( A + xI, B; C, D ) > with indeterminate x as a polynomial in x. Note that since > det( A + xI ) is a monic polynomial, its not a zero divisor > in R[x], where R is a ring with unit containing the matrix > coefficients. > Since det( I, 0; 0, A + xI) = det( A + xI ), we have: > det( A + xI, B; (A + xI)C, (A + xI)D ) > = det(A + xI ) * det( A + xI, B; C, D ) > Now subtract C times the top blocks of the matrix > in the left-hand side determinant from the bottom > blocks, which leaves the determinant unchanged > (as a sequence of elementary row operations): > det( A + xI, B; 0, (A + xI)D - BC) > = det(A + xI ) * det( A + xI, B; C, D ) > But the left-hand side is now block triangular, so: > det(A + xI)*det( (A + xI)D - BC) > = det(A + xI ) * det( A + xI, B; C, D ) > and since det(A + xI) is not a zero divisor: > det( (A + xI)D - BC) > = det( A + xI, B; C, D ) > Now the evaluation mapping x |--> 0 is > a ring homomorphism from R[x] to R, > so applying it to both sides: > det(AD - BC) = det(A, B; C, D) Yes, that neatly meets the requirements of the excercise in a straightforward way. But if I was starting with the same background as the original poster says he had, I wouldnt have thought of this approach either. === Subject: Re: Determinants and commuting matrices (old Hoffman and Kunze Exercise) > The last exercise in section 5.4 of Hoffman and Kunzes Linear Algebra > asks the reader to prove that, given 4 commuting n x n matrices A, B, C, > and D, the determinant of the 2n x 2n matrix > A B > C D > is given by det(AD - BC). > Im feeling dense, but I dont see why this should be. Any insight would > be appreciated... If they all commute (including AD-BC), then they have the same invariant subspaces. Try to find a basis where the matrix [ A, B; C, D ] becomes block-diagonal. The determinant of a block diagonal matrix is the product of determinants of each block. Hope this helps. Igor === Subject: Course****R/S System: Advanced Programming, San Francisco***Seattle, December 20-21 R/S Advanced Programming course in Seattle on December 20-21 Please check out the full description and Agenda of the 2-day class on the website: www.xlsolutions-corp.com/Radv.htm And let us know if we should hold seats for you. Ask for group discount! Heres the outline: Course outline: - Overview of R/S fundamentals: Syntax and Semantics - Class and Inheritance in R/S-Plus - Concepts, Construction and good use of language objects - Coercion and efficiency - Object-oriented programming in R and S-Plus - Advanced manipulation tools: Parse, Deparse, Substitute, etc. - How to fully take advantage of Vectorization - Generic and Method Functions; S4 (S-Plus 6) - Search path, databases and frames Visibility - Working with large objects - Handling Properly Recursion and iterative calculations - Managing loops; For (S-Plus) and for() loops - Consequences of Lazy Evaluation - Efficient Code practices for large computations - Memory management and Resource monitoring - Writing R/S-Plus functions to call compiled code - Writing and debugging compiled code for R/S-Plus system - Connecting R/S-Plus to External Data Sources - Understanding the structure of model fitting functions in R/S-Plus - Designing and Packaging efficiently This course will also deal with lots of S-Plus efficiency issues and any special topics from participants is welcome. Please let us know if you and your colleagues are interested in this class to take advantage of group discount. Register now to secure your seat in this course! Elvis Miller, PhD Manager Training. XLSolutions Corporation 206 686 1578 www.xlsolutions-corp.com === Subject: Why is Q not locally compact? Every point in Q is compact, and every point is open as well because of discrete topology, so around every point x, there is a compact set (in this case it is x) which contains an open set (in this case x) containing x. Whats wrong with this reasoning? Isaac === Subject: Re: Why is Q not locally compact? >Every point in Q is compact, and every point is open as well because of >discrete topology, so around every point x, there is a compact set (in this >case it is x) which contains an open set (in this case x) containing x. >Whats wrong with this reasoning? The topology of Q is not the discrete topology. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Why is Q not locally compact? === Subject: Course ****R/S System: Advanced Programming, Seattle***December 20-21 R/S Advanced Programming course in Seattle on December 20-21 Please check out the full description and Agenda of the 2-day class on the website: www.xlsolutions-corp.com/Radv.htm And let us know if we should hold seats for you. Ask for group discount! Heres the outline: Course outline: - Overview of R/S fundamentals: Syntax and Semantics - Class and Inheritance in R/S-Plus - Concepts, Construction and good use of language objects - Coercion and efficiency - Object-oriented programming in R and S-Plus - Advanced manipulation tools: Parse, Deparse, Substitute, etc. - How to fully take advantage of Vectorization - Generic and Method Functions; S4 (S-Plus 6) - Search path, databases and frames Visibility - Working with large objects - Handling Properly Recursion and iterative calculations - Managing loops; For (S-Plus) and for() loops - Consequences of Lazy Evaluation - Efficient Code practices for large computations - Memory management and Resource monitoring - Writing R/S-Plus functions to call compiled code - Writing and debugging compiled code for R/S-Plus system - Connecting R/S-Plus to External Data Sources - Understanding the structure of model fitting functions in R/S-Plus - Designing and Packaging efficiently This course will also deal with lots of S-Plus efficiency issues and any special topics from participants is welcome. Please let us know if you and your colleagues are interested in this class to take advantage of group discount. Register now to secure your seat in this course! Elvis Miller, PhD Manager Training. XLSolutions Corporation 206 686 1578 www.xlsolutions-corp.com === Subject: JSH: Final exam In arguing with me posters continually push that x=0 is a special case. Well, heres yet another answer and I want you to pay careful attention to what the sci.mathers do now. Remember 49(300125x^3 - 18375 x^2 - 360 x + 22) = (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) where the as are the roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and I find out what the constant terms are by using x=0, but now Ill consider x = 7. Then I have a^3 + 3(342)a^2 - 49(816361) = 0 which is irreducible over Q, for the as and 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) where youll notice that 102040002 is coprime to 7. So 7 divides TWO and only TWO of the factors. How do you know? Because when x=0, two of the as equal 0, thats why. x = 0, is equivalent to x = 0 mod 7. Notice though that the cubic defining the as is still irreducible over Q, and irreducibility has no impact. However, in the ring of algebraic integers NONE of the as have 7 as a factor, and each has a non-unit factor in common with 7. You will get the same result with x = 7k, where k is an integer, including k=0, which is not a special case after all. Im curious to see if any of the sci.mathers wish to argue over this point, as, of course, I can add more detail. You see, the sci.math readership is stupid, and will believe just about anything if certain people say, especially if they argue with me. So those posters are used to saying stupid things which are mathematically incorrect. I find it interesting to see what they will try now. They are stupid, after all. So they will try. James Harris === Subject: Re: JSH: Final exam > In arguing with me posters continually push that x=0 is a special > case. > Well, heres yet another answer and I want you to pay careful > attention to what the sci.mathers do now. > Remember > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the as are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and I find out what the constant terms are by using x=0, but now Ill > consider x = 7. > Then I have > a^3 + 3(342)a^2 - 49(816361) = 0 > which is irreducible over Q, for the as and > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) > where youll notice that 102040002 is coprime to 7. > So 7 divides TWO and only TWO of the factors. How do you know? > Because when x=0, two of the as equal 0, thats why. Actually, thats not how you know. > x = 0, is equivalent to x = 0 mod 7. Nope. It turns out that it is an error on my part to claim that you can say anything about what happens with x = 0 mod 7 from x=0 with the as. > Notice though that the cubic defining the as is still irreducible > over Q, and irreducibility has no impact. > However, in the ring of algebraic integers NONE of the as have 7 as a > factor, and each has a non-unit factor in common with 7. > You will get the same result with x = 7k, where k is an integer, > including k=0, which is not a special case after all. > Im curious to see if any of the sci.mathers wish to argue over this > point, as, of course, I can add more detail. Nope. Cant do that as what I had was wrong. > You see, the sci.math readership is stupid, and will believe just > about anything if certain people say, especially if they argue with > me. So those posters are used to saying stupid things which are > mathematically incorrect. I find it interesting to see what they will > try now. > They are stupid, after all. So they will try. I was really ticked off yesterday and probably at least a little angry this morning and I notice that I talked about how stupid the sci.math readership is several times. Oh well. In any event, what I have in this thread was just wrong, as you cant say anything just from x= 0 mod 7. Now I can admit when Im wrong, and Ive noticed that when I get upset I tend to leap to conclusions even more so than normally! And I was just feeling really pissed off yesterday. James Harris === Subject: Re: JSH: Final exam > Now I can admit when Im wrong, and Ive noticed that when I get upset > I tend to leap to conclusions even more so than normally! > And I was just feeling really pissed off yesterday. projecting that weakness on everyone else. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Final exam ... > x = 0, is equivalent to x = 0 mod 7. > Nope. It turns out that it is an error on my part to claim that you > can say anything about what happens with x = 0 mod 7 from x=0 with the > as. > Here readers can get a good dose of pseudo-math from one the most > obnoxious posters on sci.math as this guy not only lies about basic > mathematics, as you can see in his post, but he copied from my Usenet > posts without my permission on to his own webpage, where he added in > negative commentary! of the five pages I have devoted to your attempts, only two are your you. I keep them in, because they clearly show your method of discussion, your proofs to show the lack of logic. Which also has not changed a bit. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Final exam > In arguing with me posters continually push that x=0 is a special > case. > Well, heres yet another answer and I want you to pay careful > attention to what the sci.mathers do now. > Remember > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the as are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and I find out what the constant terms are by using x=0, but now Ill > consider x = 7. > Then I have > a^3 + 3(342)a^2 - 49(816361) = 0 > which is irreducible over Q, for the as and > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) > where youll notice that 102040002 is coprime to 7. > So 7 divides TWO and only TWO of the factors. How do you know? You dont. Since the polynomial is irreducible, 7 divides either NONE a_1(7), a_2(7), or a_3(7), or ALL of them. Suppose A is a solution of a^3 + 3(342)a^2 - 49(816361) = 0 and A is divisible by 7 in the algebraic integers. That is, A = 7 * B, where B is an algebraic integer. Then B satisfies the equation 7 B^3 + 3(342) B^2 - 816361 = 0, which is irreducible and non-monic and primitive. None of the roots of this latter equation are algebraic integers. Therefore none of a_1(7), a_2(7), or a_3(7) are divisible by 7. However it is equally true that each of them share a nonunit factor with 7 in the ring of algebraic integers. > Because when x=0, two of the as equal 0, thats why. > x = 0, is equivalent to x = 0 mod 7. x = 0 is indeed a special case, because then the polynomial that the as satisfy is reducible. That is not true when x = 7 as you note above. > Notice though that the cubic defining the as is still irreducible > over Q, and irreducibility has no impact. Wrong. See above. > However, in the ring of algebraic integers NONE of the as have 7 as a > factor, and each has a non-unit factor in common with 7. True, as I just proved. But if that is true, how can you conclude that exactly TWO of (5 a_1(7) + 7), (5 a_2(7) + 7), and (5 a_3(7) + 7) are divisible by 7 ? [Of course you have slyly neglected to specify in what ring your original claims about divisibility were being made. The natural assumption is that it is the ring of algebraic integers. However, from what you say above, you evidently realize that this must not be true. This leads me to conclude that since you have not specified the ring, your statement is essentially vacuous.] [Note too here that the statement you just made is tantamount to conceding what we have said from the beginning about the main conclusion of ÔAdvanced Polynomial Factorization: i.e., the main conclusion is false. And it only took you a year and a half to get it.] > You will get the same result with x = 7k, where k is an integer, > including k=0, which is not a special case after all. > Im curious to see if any of the sci.mathers wish to argue over this > point, as, of course, I can add more detail. Go for it. > You see, the sci.math readership is stupid, and will believe just > about anything if certain people say, especially if they argue with > me. So those posters are used to saying stupid things which are > mathematically incorrect. I find it interesting to see what they will > try now. > They are stupid, after all. So they will try. It may turn out that you are wrong, as it has very often in the past and even now, as you are finally admitting regarding the conclusion of ÔAPF. In that case, who is stupid? Who passes the final exam ? Nora B. > James Harris === Subject: Re: JSH: Final exam > In arguing with me posters continually push that x=0 is a special > case. > Well, heres yet another answer and I want you to pay careful > attention to what the sci.mathers do now. > Remember > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the as are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and I find out what the constant terms are by using x=0, but now Ill > consider x = 7. > Then I have > a^3 + 3(342)a^2 - 49(816361) = 0 > which is irreducible over Q, for the as and > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) > where youll notice that 102040002 is coprime to 7. > So 7 divides TWO and only TWO of the factors. No. In this case, NONE of the factors are divisible by 7 in the algebraic integers. If any of 5a_i + 7 were divisible by 7, then for that factor, wed have to have a_i divisible by 7 so then b_i = a_i/7 would have to be an algebraic integer. Notice that since the as satisfy a^3 + 3*342*a^2 - 49*816361 = 0 then, rearranging and factoring 816361, a^3 = -3*342*a^2 + 49*7*13*8971 and dividing through by 7^3 we have (a^3)/(7^3) = -(3*342/7)((a^2)/(7^2)) + 13*8971 or, using b = a/7 we would have b^3 + (3*342/7)b^2 - 13*8971 = 0 so the bs all satisfy 7b^3 + 3*342b^2 - 7*13*8971 = 0 The left side is a non-monic irreducible polynomial, showing that b cannot be an algebraic integer, from which we conclude that none of the as are divisible by 7. Rick === Subject: Re: JSH: Final exam > In arguing with me posters continually push that x=0 is a special > case. Yup, I will argue. > Well, heres yet another answer and I want you to pay careful > attention to what the sci.mathers do now. > Remember > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the as are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and I find out what the constant terms are by using x=0, but now Ill > consider x = 7. > Then I have > a^3 + 3(342)a^2 - 49(816361) = 0 > which is irreducible over Q, for the as and > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) > where youll notice that 102040002 is coprime to 7. Yup. > So 7 divides TWO and only TWO of the factors. How do you know? The question is *why*? > Because when x=0, two of the as equal 0, thats why. Oh, not very convincing. When x = 0, sqrt(x) = 0, so sqrt(x) is divisible by 7? It would be true if the as were polynomials. They are not. > x = 0, is equivalent to x = 0 mod 7. Using equivalence classes mod 7 is only useful when you are doing additions and multiplications. With other functions it just does not work. You are actually saying that sqrt(7) should be 0 mod 7. Which means that sqrt(7) is divisible by 7, which means that 1/7 is an element of the ring you are in, and so 7 is a unit in that ring. > Notice though that the cubic defining the as is still irreducible > over Q, and irreducibility has no impact. Yup, and so the as are not polynomials and that has a terrific impact. > However, in the ring of algebraic integers NONE of the as have 7 as a > factor, and each has a non-unit factor in common with 7. And the non-unit factors multiply together to give a multiple of 49. So, what is the problem? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Final exam > In arguing with me posters continually push that x=0 is a special > case. > Yup, I will argue. Here readers can get a good dose of pseudo-math from one the most obnoxious posters on sci.math as this guy not only lies about basic mathematics, as you can see in his post, but he copied from my Usenet posts without my permission on to his own webpage, where he added in negative commentary! And refused to quit using my Usenet posts in violation of both the spirit and letter of international copyright law. And he lies about math. > Well, heres yet another answer and I want you to pay careful > attention to what the sci.mathers do now. > > Remember > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the as are the roots of > > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > > and I find out what the constant terms are by using x=0, but now Ill > consider x = 7. > > Then I have > a^3 + 3(342)a^2 - 49(816361) = 0 > which is irreducible over Q, for the as and > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) > where youll notice that 102040002 is coprime to 7. > Yup. > So 7 divides TWO and only TWO of the factors. How do you know? > The question is *why*? > Because when x=0, two of the as equal 0, thats why. Notice that you have 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) with 49 on the left side, and it has to divide through in some way. For two of the as you know they result from functions that equal 0, when x = 0 mod 7 as they equal 0, when x = 0 and you cant get any better than that! The third equals 3 when x = 0, so it is blocked from havinng factors in common with 7, when x has 7 itself as a factor, like when x = 7. Its basic. To believe otherwise you need to challenge algebra. > Oh, not very convincing. When x = 0, sqrt(x) = 0, so sqrt(x) is divisible > by 7? It would be true if the as were polynomials. They are not. Now notice, the poster cant deny the result when the as are polynomials as then you can physically SEE how it works, so he handles that right off the bat by asserting that whether or not they are polynomials matters. Also he tossed in a weird assertion that doesnt follow as theres no reason to suggest that sqrt(7) is divisible by 7. Hes just trying to throw up smoke as one other tactic is to confuse readers to the point that they just give up and trust that if someone is disagreeing with me, then I must be wrong. > x = 0, is equivalent to x = 0 mod 7. > Using equivalence classes mod 7 is only useful when you are doing > additions and multiplications. With other functions it just does So now congruences only matter with additions and multiplications, according to the sci.mather, so isnt it interesting what math this person is teaching you? But, I didnt necessarily phrase my own sentence well, as my point is that x=0 is equivalent to x = 0 mod 7, with respect to having factors of 7 so my using x = 0 mod 7, is a neat way to show that the result at x=0 is NOT a special case at all, as the constant terms work as Ive explained repeatedly, while the claims of people arguing with me fall ßat. If this werent such a HUGE issue with massive repercussions for mathematicians worldwide, then maybe these sci.math-ers might finally give up and admit the truth, but wait, Im giving them too much credit. It could be a minor issue with few repercussions and these people, from what Ive seen over the years would still lie to you. Lying is in their nature. > not work. You are actually saying that sqrt(7) should be 0 mod 7. No Im not. > Which means that sqrt(7) is divisible by 7, which means that 1/7 > is an element of the ring you are in, and so 7 is a unit in that ring. Desperation. > Notice though that the cubic defining the as is still irreducible > over Q, and irreducibility has no impact. > Yup, and so the as are not polynomials and that has a terrific impact. This poster hardly even tries, but why should he? These TACTICS WORK on sci.math, as Ive noted for years now. The sci.math readership is like some kind of religious cult, willing to believe just about anything as long as posters disagree with me. They are true believers on sci.math, and its freaky. > However, in the ring of algebraic integers NONE of the as have 7 as a > factor, and each has a non-unit factor in common with 7. > And the non-unit factors multiply together to give a multiple of 49. So, > what is the problem? Oh, hey, its all ok if it works out in the end, right? If you want the full picture you need to read my original post, and my other reply to the other sci.math-er. The story here is truly sad as the problem I found *was* just a historical one, where math professors today could simply point out that they followed what they were taught. But now as the days turn into months and the months move toward years, theres little reason to believe that such a huge result hasnt traveled through the math community, so what was a fascinating historical error, is now a case of modern fraud. Math professors who go out today to teach ßawed algebraic number theory to their students may be criminals and it may be possible to prosecute them under anti-fraud laws, but that sounds so removed from reality that they will probably continue without real fear. But, for those of you listening to them, the impact is very real. You are being taught false information, and being forced to do homework, and take tests to regurgitate that false information and God help you if you come back saying that its false based on what you learned on newsgroups from a guy called James Harris. I feel sorry for you, but if you look around you can see how often society does things that inside you feel are wrong, but people just go on about their business as thats what theyve learned is necessary. And many of you, unfortunately, will be like them, today, and the next day. So, in a great field, where truth is paramount, you will knowingly learn false things, and maybe even pride yourself on the grades you get from professors who are compromised. But youre living in a mirage... James Harris === Subject: Re: JSH: Final exam Your cubic polynomial in Ôa has coefficients which are functions of Ôx: Q(a,x) = a^3 + 3(-1 + 49x)a^2 - 49(2401x^3 - 147x^2 + 3x) Therefore its roots are functions of Ôx. Your polynomial in Ôx is: P(x) = 14706125x^2 - 900375x^2 - 17640x + 1078 which you have factored in terms of the roots of Q(a,x): P(x) = (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) You claim that since the roots of Q(a,x) evaluated at x = 0, are 0, 0, and 3 that P(0) = (0 + 7)(0 + 7)(5*3 + 7) = (7)(7)(22) = 1078 from which you somehow conclude that two of the terms in parentheses are divisible by 7 for any value of Ôx. Or have I misunderstood? If that is your claim, it is surely false. We have Q(a,x) @ x=0 : a^3 - 3a^2 Q(a,x) @ x=1 : a^3 + 144a^2 - 110593 Q(a,x) @ x=2 : a^3 + 291a^2 - 912674 Q(a,x) @ x=3 : a^3 + 438a^2 - 3112137 The roots of Q(a,x) are a_1(x), a_2(x) and a_3(x), so a_1(0) = 0 a_2(0) = 0 a_3(0) = 3 a_1(1) = -31.32993946314472877... a_2(1) = -138.2104344584112440... a_3(1) = 25.54037392155592731... a_1(2) = -63.31229381879031244... a_2(2) = -279.3003544412207553... a_3(2) = 51.61264826001106776... a_1(3) = -31.65614690939515622... a_2(3) = -420.3902390936727750... a_3(3) = 77.68498899747098938... Hence, Ôx = 0 is indeed a Ôspecial case from which you argue to a false generalization. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Final exam ... > So 7 divides TWO and only TWO of the factors. How do you know? > > The question is *why*? > > Because when x=0, two of the as equal 0, thats why. > > Notice that you have > 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7) > with 49 on the left side, and it has to divide through in some way. > For two of the as you know they result from functions that equal 0, > when > x = 0 mod 7 as they equal 0, when x = 0 This is wrong. They are 0 when x = 0, they are not 0 when x != 0, whether x = 0 mod 7 or not. Note that the as are the roots of: > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) for any a to be 0, 2401 x^3 - 147 x^2 + 3x must be 0. This is (with integer x) only the case when x = 0. Talking about false math... > The third equals 3 when x = 0, so it is blocked from havinng factors > in common with 7, when x has 7 itself as a factor, like when x = 7. This is also wrong. Because if x has 7 as a factor a3(x) does not necessarily have 7 as a factor. You are at least consistent in your errors. You assume that if 1. f(0) = 0 2. x = 0 mod 7 you can conclude that f(x) = 0 mod 7. That conclusion is false in general and holds only when f(x) is a polynomial in x. > Its basic. To believe otherwise you need to challenge algebra. Yup, it is basic, and to believe otherwise you need to challenge algebra. > Oh, not very convincing. When x = 0, sqrt(x) = 0, so sqrt(x) is divisible > by 7? It would be true if the as were polynomials. They are not. > Now notice, the poster cant deny the result when the as are > polynomials as then you can physically SEE how it works, so he handles > that right off the bat by asserting that whether or not they are > polynomials matters. > Also he tossed in a weird assertion that doesnt follow as theres no > reason to suggest that sqrt(7) is divisible by 7. Lets check your assertion above with sqrt(x) as function: 1. sqrt(0) = 0? yes. 2. 7 = 0 mod 7? yes. sqrt(7) = 0 mod 7? no. So your conclusion above is not a valid conclusion. It is your logic that suggests that sqrt(7) is divisible by 7. > x = 0, is equivalent to x = 0 mod 7. > > Using equivalence classes mod 7 is only useful when you are doing > additions and multiplications. With other functions it just does > So now congruences only matter with additions and multiplications, > according to the sci.mather, so isnt it interesting what math this > person is teaching you? What other operations do you think it works with? Not with the sqrt function. Also not with division. > x=0 is equivalent to x = 0 mod 7, with respect to having factors of 7 > so my using x = 0 mod 7, is a neat way to show that the result at x=0 > is NOT a special case at all, as the constant terms work as Ive > explained repeatedly, while the claims of people arguing with me fall > ßat. Can you please explain why your inference work *must* work for your function, but does not work for sqrt? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Final exam Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >You assume that if > 1. f(0) = 0 > 2. x = 0 mod 7 >you can conclude that > f(x) = 0 mod 7. >That conclusion is false in general and holds only when f(x) is a >polynomial in x. To be pedantic, there are other classes of function that it holds for. For example, it holds for all functions that are zero at all the integers. -- Richard === Subject: Re: JSH: Final exam >You assume that if > 1. f(0) = 0 > 2. x = 0 mod 7 >you can conclude that > f(x) = 0 mod 7. >That conclusion is false in general and holds only when f(x) is a >polynomial in x. > To be pedantic, there are other classes of function that it holds for. > For example, it holds for all functions that are zero at all the > integers. > -- Richard Yup. However, I *did* use it improperly and my example in my original post is misleading. I was wrong. It happens. I was kind of upset yesterday, and got emotional. James Harris === Subject: Re: JSH: Final exam >You assume that if > 1. f(0) = 0 > 2. x = 0 mod 7 >you can conclude that > f(x) = 0 mod 7. >That conclusion is false in general and holds only when f(x) is a >polynomial in x. >>To be pedantic, there are other classes of function that it holds for. >>For example, it holds for all functions that are zero at all the >>integers. >>-- Richard > Yup. However, I *did* use it improperly and my example in my original > post is misleading. I was wrong. > It happens. I was kind of upset yesterday, and got emotional. You are always emotional, and continually making errors. Why dont you save yourself some embarrassment and go away and learn the subject before proclaiming your mastery of it? === Subject: Re: JSH: Final exam > >> >You assume that if > 1. f(0) = 0 > 2. x = 0 mod 7 >you can conclude that > f(x) = 0 mod 7. >That conclusion is false in general and holds only when f(x) is a >polynomial in x. >> >>To be pedantic, there are other classes of function that it holds for. >>For example, it holds for all functions that are zero at all the >>integers. >> >>-- Richard > > > Yup. However, I *did* use it improperly and my example in my original > post is misleading. I was wrong. > > It happens. I was kind of upset yesterday, and got emotional. > You are always emotional, and continually making errors. Why dont you > save yourself some embarrassment and go away and learn the subject > before proclaiming your mastery of it? That doesnt make sense. I have a mathematical argument, which is correct in every detail. Problems arise when I try to find other arguments to explain what Ive already proven one way. Trying to prove in different ways helps my own understanding, and its fun. Besides, Ive discovered is a problem in what is commonly taught, so it hardly makes sense to try and learn from what is commonly taught! Mistakes happen. They are MUCH more likely when doing research that is breaking new ground, and when trying to understand something that hasnt been well-worked out before. Its what happens when youre in the lead, and dont have some book or reference to tell you the answer but have to figure it out yourself. I am in the lead position. Making mistakes isnt a problem. Its human. Not being able to admit it when you make mistakes IS a problem. James Harris === Subject: Re: JSH: Final exam >> >> >You assume that if >1. f(0) = 0 >2. x = 0 mod 7 >you can conclude that > f(x) = 0 mod 7. >That conclusion is false in general and holds only when f(x) is a >polynomial in x. >> >>To be pedantic, there are other classes of function that it holds for. >>For example, it holds for all functions that are zero at all the >>integers. >> >>-- Richard >Yup. However, I *did* use it improperly and my example in my original >post is misleading. I was wrong. >It happens. I was kind of upset yesterday, and got emotional. >>You are always emotional, and continually making errors. Why dont you >>save yourself some embarrassment and go away and learn the subject >>before proclaiming your mastery of it? > That doesnt make sense. I have a mathematical argument, which is > correct in every detail. Problems arise when I try to find other > arguments to explain what Ive already proven one way. Actually, what happens is enough complications disappear that you can see the error in your simplified explanations. What never happens is the logical step of realizing that the error in your simple cases *also* exists in your general argument. If your argument were correct, your simpler examples would be correct to. You admit they are not. Draw the natural conclusion. > Trying to prove in different ways helps my own understanding, and its > fun. But your method of attack is always the same. All you have shown is that your primary argument does not work, in general. > Besides, Ive discovered is a problem in what is commonly taught, so > it hardly makes sense to try and learn from what is commonly taught! > Mistakes happen. They are MUCH more likely when doing research that > is breaking new ground, and when trying to understand something that > hasnt been well-worked out before. > Its what happens when youre in the lead, and dont have some book or > reference to tell you the answer but have to figure it out yourself. The answers to a *lot* of your questions are in books. > I am in the lead position. > Making mistakes isnt a problem. Its human. Not being able to admit > it when you make mistakes IS a problem. When you find that you are making mistakes at a high rate, it is worth while to at least lose some of the arrogance. Actually, arrogance pisses people off, regardless of whether you are right or not. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Final exam > Yup. However, I *did* use it improperly and my example in my original > post is misleading. I was wrong. But you say things like There is no error. Then you say I was wrong. Do you wonder why everyone doubts you? > It happens. I was kind of upset yesterday, and got emotional. But you often say that you are just following the math, and things like emotion and feelings do not enter it. Which is it with you? === Subject: Re: Final exam > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the as are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and I find out what the constant terms are by using x=0, but now Ill > consider x = 7. Where do you get this silly arithmetic? What are you choosing this particular equation as your starting point. Does it have any significance? Where do you get this silly arithmetic? What are you choosing this particular equation as your starting point. Does it have any significance? Did it come to you in a Kekule style benzene dream? Did god give it to you? Is it special, or is it random? === Subject: Re: JSH: Final exam > In arguing with me posters continually push that x=0 is a special > case. > Well, heres yet another answer and I want you to pay careful > attention to what the sci.mathers do now. > Remember > 49(300125x^3 - 18375 x^2 - 360 x + 22) = When x = 0, the above is 1078, which is divisible by 11 When x = 1, the above is 13789188, which is not divisible by 11. Certainly there are at least *some* properties where x=0 leads to different results than x=1. Of course, x=1 could be the special case, but either way, the properties are not the same. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Final exam > In arguing with me posters continually push that x=0 is a special > case. > > Well, heres yet another answer and I want you to pay careful > attention to what the sci.mathers do now. > > Remember > > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > When x = 0, the above is 1078, which is divisible by 11 > When x = 1, the above is 13789188, which is not divisible by 11. > Certainly there are at least *some* properties where x=0 leads to > different results than x=1. > Of course, x=1 could be the special case, but either way, the properties > are not the same. Told you theyd fight. Consider also that the poster deleted out the relevant information, an old sci.math tactic, and now consider the facts 49(300125x^3 - 18375 x^2 - 360 x + 22) = (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) where the as are the roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and setting x=0, lets you see that two of the as go to 0, at that point as you have 49(300125(0)^3 - 18375 (0)^2 - 360 (0) + 22) = (5a_1(0) + 7)(5a_2(0) + 7)(5a_3(0) + 7) and a^3 + 3(-1 + 49(0))a^2 - 49(2401 (0)^3 - 147 (0)^2 + 3(0)) = 0 so a^3 - 3a^2 = 0, which is a^2(a - 3) = 0, so two of the as equal 0, while one equals, 3, at x=0. Since I have the correct mathematics I can endlessly explain it in different variations which is fun for me, but a headache for those who lie about my work as notice that I simply went to using x = 0 mod 7, and in my original post specifically used x = 7 to show that you get an irreducible over Q cubic for the as, which shows that in the ring of algebraic integers NONE of the as have 7 as a factor. But, from the result at x=0, you know that one of the as cannot have non-unit factors in common with 7 for x=0 mod 7, but they do in the ring of algebraic integers. It turns out what are non-unit factors in the ring of algebraic integers are actually unit factors that are not units in the ring of algebraic integers because they are not roots of some monic polynomial with integer coefficients. You see that definition focusing on roots of monic polynomials with integer coefficients gives a ring that has these quirks. You can figure out the quirks, if you trust algebra and use logic, but if youre a ßawed human being desperate to have an easy tool to prove things, then you will fight it. Much of the supposedly great accomplishments in algebraic number theory over the last hundred years plus are simply vapor. Now to believe the sci.mathers, you need to start fiddling with your understanding of congruence, and Im sure some of them will come forward to lie to you about it. Thats just the fun of the game I play on sci.math and this newsgroup. Whats not a game is that some of you probably will go to class today and some math professor is going to confidently teach you mathematical information that is wrong. When you go to class today, if your professor talks about algebraic number theory, or misuses Galois Theory (given what youve learned here) I want you to carefully notice how you feel. Hold on to that feeling so that you never forget it. James Harris === Subject: Re: JSH: Final exam > When you go to class today, if your professor talks about algebraic > number theory, or misuses Galois Theory (given what youve learned > here) I want you to carefully notice how you feel. If any students think about you in class at all (which I doubt), Im certain theyll be telling their professors something like, Wait till you hear what that idiot Harris said *this* time! -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Final exam >In arguing with me posters continually push that x=0 is a special >case. >Well, heres yet another answer and I want you to pay careful >attention to what the sci.mathers do now. >Remember >49(300125x^3 - 18375 x^2 - 360 x + 22) = >>When x = 0, the above is 1078, which is divisible by 11 >>When x = 1, the above is 13789188, which is not divisible by 11. >>Certainly there are at least *some* properties where x=0 leads to >>different results than x=1. >>Of course, x=1 could be the special case, but either way, the properties >>are not the same. > Told you theyd fight. > Consider also that the poster deleted out the relevant information, an > old sci.math tactic, and now consider the facts > 49(300125x^3 - 18375 x^2 - 360 x + 22) = > (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7) > where the as are the roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > and setting x=0, lets you see that two of the as go to 0, at that > point as you have > 49(300125(0)^3 - 18375 (0)^2 - 360 (0) + 22) = > (5a_1(0) + 7)(5a_2(0) + 7)(5a_3(0) + 7) > and > a^3 + 3(-1 + 49(0))a^2 - 49(2401 (0)^3 - 147 (0)^2 + 3(0)) = 0 > so a^3 - 3a^2 = 0, which is a^2(a - 3) = 0, so two of the as equal 0, > while one equals, 3, at x=0. > Since I have the correct mathematics I can endlessly explain it in > different variations which is fun for me, but a headache for those who > lie about my work as notice that I simply went to using x = 0 mod 7, > and in my original post specifically used x = 7 to show that you get > an irreducible over Q cubic for the as, which shows that in the ring > of algebraic integers NONE of the as have 7 as a factor. > But, from the result at x=0, you know that one of the as cannot have > non-unit factors in common with 7 for x=0 mod 7, but they do in the > ring of algebraic integers. And it doesnt occur to you that you are making an argument that 7 is in the same category of special cases as 0 for the purposes of dividing by 7? Many people have shown that your arguments are simply false when x=1. They didnt do anything fancy, just cranked out the numbers and showed they dont have the properties you claim they do. Did I change what was being talked about? Yes. Did you miss the point? Yes. The burden of proof is always on you to show that x=0 is *not* a special case. The best you have done is to argue that x=0 mod 7 has a special property. That is *still* a special case. -- Will Twentyman email: wtwentyman at copper dot net === Subject: JSH: Just playing You are stupid compared to me, but thats not a surprise. If you realized that early, would you have played with me? Instead Ive had years of fun, but now I need to move on to other games. So I now tell you outright that you are stupid, and show it to you with more mathematics. You will fight, and I will break you. I could have before, but then you wouldnt have played with me any more. James Harris === Subject: Re: JSH: Just playing > I can explain the algebra in detail to you, but if you simply choose > to claim its nonsense, then no progress can be made, right? > Good, start by explaining all the counterexamples I have > provided. You have quite a backlog. > As an easy first step explain why, with a(0) = 0 and w1(0) =1 > the constant terms of > (a(x) + 7) > and (a(x)/w1(x) + 7/w1(x)) > are different. > Now then, if you will accept algebra, and point out an area of > confusion then I can explain to you. But if you are simply going to > just deny then theres little point. > > Which is it? Will you follow algebra, and give a mathematical area > for me to explain, or are you just going to deny? > Which is it? Will you reply to the above and give mathematical > arguments, or are you just going to ignore? And the winner is: ignore. - William Hughes === Subject: Re: JSH: Just playing > You are stupid compared to me, but thats not a surprise. > If you realized that early, would you have played with me? > Instead Ive had years of fun, but now I need to move on to other > games. > So I now tell you outright that you are stupid, and show it to you > with more mathematics. > You will fight, and I will break you. > I could have before, but then you wouldnt have played with me any > more. And heres the runny-nosed little gutter-snipe again, shaking his fist at the rest of the kids and telling them through his tears that he doesnt want to be invited to their stinky old clubhouse, anyway. Most people outgrow that sort of thing eventually... -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Just playing > You are stupid compared to me, but thats not a surprise. > If you realized that early, would you have played with me? > Instead Ive had years of fun, but now I need to move on to other > games. > So I now tell you outright that you are stupid, and show it to you > with more mathematics. > You will fight, and I will break you. > I could have before, but then you wouldnt have played with me any > more. > James Harris Stop......Hammer time. === Subject: Re: JSH: Just playing !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ELIi $t^ VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> You are stupid compared to me, but thats not a surprise. >> If you realized that early, would you have played with me? >> Instead Ive had years of fun, but now I need to move on to other >> games. >> So I now tell you outright that you are stupid, and show it to you >> with more mathematics. >> You will fight, and I will break you. >> I could have before, but then you wouldnt have played with me any >> more. > Stop......Hammer time. I am wondering about the alcoholic content of this Hammer. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Just playing >You are stupid compared to me, but thats not a surprise. >If you realized that early, would you have played with me? >Instead Ive had years of fun, but now I need to move on to other >games. Facinating, the way we continually swing back and forth between the current youre just ants for mighty me to toy with and bitter complaints about how the awful treatment you get here has ruined your life. >So I now tell you outright that you are stupid, and show it to you >with more mathematics. >You will fight, and I will break you. >I could have before, but then you wouldnt have played with me any >more. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Just playing > You are stupid compared to me, but thats not a surprise. > If you realized that early, would you have played with me? > Instead Ive had years of fun, but now I need to move on to other > games. > So I now tell you outright that you are stupid, and show it to you > with more mathematics. > You will fight, and I will break you. > I could have before, but then you wouldnt have played with me any > more. Is this addressed to anyone in particular? Or everyone? Figuring there are tens or hundreds of thousand of sci.math readers, its statistically unlikely that you are smarter than ALL of them. Feynman had a remark along these lines in one of his popular essays. He pointed out that in a room full of people, he couldnt be sure he was the smartest person, but he was reasonably certain that he was smarter than the average of all the people. Personally, I dont consider myself smarter than you. Frankly, not even saner, just crazy in a different dimension. And not nearly as entertaining. But still, you dont quite understand what constitutes a mathematical proof; and you refuse to take the time to study enough abstract algebra to communicate with anyone. Youve clearly put a lot of time and energy into your work, and youve brought awareness of the algebraic integers to many of us who never knew about them before. You bring a lot of entertainment to this newsgroup. If youre about to take one of your periodic vacations from posting, I for one will miss you. See you next time around. === Subject: Re: JSH: Just playing > You are stupid compared to me, but thats not a surprise. > If you realized that early, would you have played with me? > Instead Ive had years of fun, but now I need to move on to other > games. Youve said this before. You came back. > So I now tell you outright that you are stupid, and show it to you > with more mathematics. > You will fight, and I will break you. Youve said this before as well, it didnt happen. > I could have before, but then you wouldnt have played with me any > more. Perhaps if you learned to relax you wouldnt say things like the above. Why do you feel this is a competition? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Just playing > You are stupid compared to me, but thats not a surprise. Narcissistic Personality Disorder A pervasive pattern of grandiosity (in fantasy or behavior), need for admiration, and lack of empathy, beginning by early adulthood and present in a variety of contexts, as indicated by five (or more) of the following: (1) has a grandiose sense of self-importance (e.g., exaggerates achievements and talents, expects to be recognized as superior without commensurate achievements) (2) is preoccupied with fantasies of unlimited success, power, brilliance, beauty, or ideal love (3) believes that he or she is special and unique and can only be understood by, or should associate with, other special or high-status people (or institutions) (4) requires excessive admiration (5) has a sense of entitlement, i.e., unreasonable expectations of especially favorable treatment or automatic compliance with his or her expectations (6) is interpersonally exploitative, i.e., takes advantage of others to achieve his or her own ends (7) lacks empathy: is unwilling to recognize or identify with the feelings and needs of others (8) is often envious of others or believes that others are envious of him or her (9) shows arrogant, haughty behaviors or attitudes Reprinted with permission from the Diagnostic and Statistical Manual of Mental Disorders, fourth Edition. Copyright 1994 American Psychiatric Association *** Perhaps instead of spewing your useless nonsense, you should study this disorder, go see a doctor and then drug and drink yourself to you pass out. Research into your delusional fits of grandeur would be a more appropriate vocation for you! Think about, eh! === Subject: Re: Just playing ßip_alpha@safebunch.com says... > You are stupid compared to me, but thats not a surprise. > Narcissistic Personality Disorder > A pervasive pattern of grandiosity (in fantasy or behavior), need for > admiration, and lack of empathy, beginning by early adulthood and present in > a variety of contexts, as indicated by five (or more) of the following: > (1) has a grandiose sense of self-importance (e.g., exaggerates achievements > and talents, expects to be recognized as superior without commensurate > achievements) James Harris nPhd. === Subject: finite maze solving algorithm I was wondering if anyone knows if all possible topologies of finite 2d mazes can be solved by a finite algorithm. For example, we know that all fully connected mazes can be solved by picking a wall and exhaustively following it. Can a general solution work for all mazes including the ones that are piecewise disconnected? If this is possible, is the general solution a solved problem? === Subject: Re: finite maze solving algorithm > I was wondering if anyone knows if all possible topologies of finite > 2d mazes can be solved by a finite algorithm. Yes, they can. > For example, we know > that all fully connected mazes can be solved by picking a wall and > exhaustively following it. Can a general solution work for all mazes > including the ones that are piecewise disconnected? If this is > possible, is the general solution a solved problem? Yes it is. Look for the Pledge Algorithm. It is well-described in the book Turtle Geometry, including a proof for how it solves every finite maze. The algorithm works by keeping track of the total amount of turning you do as you move throughout the maze. Using this purely local information (you dont need breadcrumbs, or to mark walls etc.) it is possible to escape. It is rather simple: 1. Initialize a counter to zero, and define an arbitrary direction to be Ônorth. 2. Move straight north until an obstacle is met. 3. Turn left and follow the obstacle. Keep track of Ôtotal turning and increment the counter by +1 for every full 360 degree turn clockwise, and -1 for every full 360 turn anti-clockwise. 4. Leave the obstacle when it is possible to move straight north and the counter reads zero. Goto step 2. Steven === Subject: Re: finite maze solving algorithm > I was wondering if anyone knows if all possible topologies of finite > 2d mazes can be solved by a finite algorithm. For example, we know > that all fully connected mazes can be solved by picking a wall and > exhaustively following it. Can a general solution work for all mazes > including the ones that are piecewise disconnected? If this is > possible, is the general solution a solved problem? Interesting. How are you defining a maze, here? Usually, I think of a maze as just a graph, which you can solve using something simple like depth-first search, regardless of whether it is planar. It is probably better to think in terms of connectedness of rooms than connectedness of walls. (I hope I am not being ignorant here, but you seem to be using topology and piecewise disconnected in strange ways. Obviously if the topological space is disconnected, there will be no path possible between points in different components - in the discrete topology, if youre not there, you cant get there.) HTH -- Kevin === Subject: Re: finite maze solving algorithm > I was wondering if anyone knows if all possible topologies of finite > 2d mazes can be solved by a finite algorithm. For example, we know > that all fully connected mazes can be solved by picking a wall and > exhaustively following it. Can a general solution work for all mazes > including the ones that are piecewise disconnected? If this is > possible, is the general solution a solved problem? > Interesting. How are you defining a maze, here? Usually, I think of a > maze as just a graph, which you can solve using something simple like > depth-first search, regardless of whether it is planar. It is probably > better to think in terms of connectedness of rooms than connectedness of > walls. > (I hope I am not being ignorant here, but you seem to be using > topology and piecewise disconnected in strange ways. Obviously if > the topological space is disconnected, there will be no path possible > between points in different components - in the discrete topology, if > youre not there, you cant get there.) > HTH I am probably using those terms in an unusual way, being entirely a layman. When I refer to piecewise disconnnected I mean the walls not the paths. I understand that an unconnected path can not be traversed. To look at this problem interms of paths (which I guess is the graph theory approach you refer to) you would need to posit a set of graphs that included loops, dead ends, routes that deadend in loops, and various structures of loops nested and sharing common branches. I wish usenet was friendlier to hand drawn diagrams embedded in text. That would make explaining this a lot easier. === Subject: Re: finite maze solving algorithm > I was wondering if anyone knows if all possible topologies of finite > 2d mazes can be solved by a finite algorithm. For example, we know > that all fully connected mazes can be solved by picking a wall and > exhaustively following it. Can a general solution work for all mazes > including the ones that are piecewise disconnected? If this is > possible, is the general solution a solved problem? A maze with n walls is homeomorphic to the n-times punctured plane. So the method of cuts used by Cauchy to derive a simply-connected domain can be applied. The cuts in this instance become barriers joining the walls in some sequence, the last one getting a final barrier going off to infinity. Then any two points in the maze can be joined by a path which is homotopically unique. to the same thing but with the barriers introduced on the run rather than in advance. === Subject: Re: finite maze solving algorithm >> I was wondering if anyone knows if all possible topologies of finite >> 2d mazes can be solved by a finite algorithm. For example, we know >> that all fully connected mazes can be solved by picking a wall and >> exhaustively following it. Can a general solution work for all mazes >> including the ones that are piecewise disconnected? If this is >> possible, is the general solution a solved problem? >Interesting. How are you defining a maze, here? Usually, I think of a >maze as just a graph, which you can solve using something simple like >depth-first search, regardless of whether it is planar. It is probably >better to think in terms of connectedness of rooms than connectedness of >walls. The basic problem with a depth-first search is the chance youll get into a loop going around and around a disconnected piece. The problem can be avoided by making a rule to never cross your own path. Or is there something more that Im missing? --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: finite maze solving algorithm > I was wondering if anyone knows if all possible topologies of finite > 2d mazes can be solved by a finite algorithm. For example, we know > that all fully connected mazes can be solved by picking a wall and > exhaustively following it. Can a general solution work for all mazes > including the ones that are piecewise disconnected? If this is > possible, is the general solution a solved problem? >>Interesting. How are you defining a maze, here? Usually, I think of a >>maze as just a graph, which you can solve using something simple like >>depth-first search, regardless of whether it is planar. It is probably >>better to think in terms of connectedness of rooms than connectedness of >>walls. > The basic problem with a depth-first search is the chance youll get into a > loop going around and around a disconnected piece. The problem can be > avoided by making a rule to never cross your own path. > Or is there something more that Im missing? If you keep going in circles, then you are not doing a depth-first search. Depth-first search will systematically explore *any* maze in its entirety. In depth-first-search, you keep traversing down previously untraversed paths until either you reach a place where all possible paths have already been traversed or you reach a dead end. In these situations, you backtrack along the path youve just traversed until reach a place where there is an unexplored corridor, and then you head down that corridor. To find your way through an actual maze (i.e. not one written on paper), you need some way of marking passages. A 19th century algorithm for traversing a maze is as follows. Bring along a pile of pennies to mark passages as you enter and leave junctions. During the algorithm, passages without pennies havent been traversed yet, those with one penny have been traversed exactly once, and those with two pennies have been traversed and then backtracked along in the opposite direction. Enter the maze. When you encounter a junction that you havent seen before (no passageways with pennies), then choose a passage at random (remember to drop a penny in the passage you just left and in the passage you just entered). If you hit a dead end, turn around and go back. If you are walking down a passage for the first time and you encounter a junction youve seen before, then turn around and head back (remember to drop two pennies, one for entering the junction and one for leaving). If you are walking down a passage for the second time and encounter a junction, then take a new passage if there is one, otherwise take an old passage (marked with one penny). When you reach the solution, passages marked exactly once will indicate a direct path back to the start. If there is no solution, then you will end up back at the start with all passages marked twice. === Subject: Re: finite maze solving algorithm >> I was wondering if anyone knows if all possible topologies of finite >> 2d mazes can be solved by a finite algorithm. For example, we know >> that all fully connected mazes can be solved by picking a wall and >> exhaustively following it. Can a general solution work for all mazes >> including the ones that are piecewise disconnected? If this is >> possible, is the general solution a solved problem? >Interesting. How are you defining a maze, here? Usually, I think of a >maze as just a graph, which you can solve using something simple like >depth-first search, regardless of whether it is planar. It is probably >better to think in terms of connectedness of rooms than connectedness of >walls. > The basic problem with a depth-first search is the chance youll get into a > loop going around and around a disconnected piece. The problem can be > avoided by making a rule to never cross your own path. > Or is there something more that Im missing? > --Keith Lewis klewis {at} mitre.org > The above may not (yet) represent the opinions of my employer. I dont think that rule would be sufficient, How would it handle the case where a path deadended in a lopp. If you could never cross your path again youd be stuck. Certainly there are lots of graphs that can not be traced if you dont allow, jumping, retracing or crossing? Maybe I am misunderstanding your suggestion. === Subject: Re: finite maze solving algorithm >>I was wondering if anyone knows if all possible topologies of finite >>2d mazes can be solved by a finite algorithm. For example, we know >>that all fully connected mazes can be solved by picking a wall and >>exhaustively following it. Can a general solution work for all mazes >>including the ones that are piecewise disconnected? If this is >>possible, is the general solution a solved problem? >Interesting. How are you defining a maze, here? Usually, I think of a >maze as just a graph, which you can solve using something simple like >depth-first search, regardless of whether it is planar. It is probably >better to think in terms of connectedness of rooms than connectedness of >walls. >>The basic problem with a depth-first search is the chance youll get into a >>loop going around and around a disconnected piece. The problem can be >>avoided by making a rule to never cross your own path. >>Or is there something more that Im missing? >>--Keith Lewis klewis {at} mitre.org >>The above may not (yet) represent the opinions of my employer. > I dont think that rule would be sufficient, How would it handle the > case where a path deadended in a lopp. If you could never cross your > path again youd be stuck. Certainly there are lots of graphs that can > not be traced if you dont allow, jumping, retracing or crossing? > Maybe I am misunderstanding your suggestion. OK, in laymans terms, assume you have a long rope and a loaf of bread. Tie one end of the rope to a hitch outside, and keep it reasonably taut. Whenever you enter a new room, drop a bread crumb. Now, determine which adjacent rooms have bread in them. If all adjacent rooms have bread, follow the rope back to the last room. If any of them have no bread, go in that room, drop a crumb, check the adjacent rooms for crumbs, and so on. You are then guaranteed to visit every room in the maze, assuming the rooms are all connected somehow. HTH -- Kevin === Subject: Re: finite maze solving algorithm >I was wondering if anyone knows if all possible topologies of finite >2d mazes can be solved by a finite algorithm. For example, we know >that all fully connected mazes can be solved by picking a wall and >exhaustively following it. Can a general solution work for all mazes >including the ones that are piecewise disconnected? If this is >possible, is the general solution a solved problem? >>Interesting. How are you defining a maze, here? Usually, I think of a >>maze as just a graph, which you can solve using something simple like >>depth-first search, regardless of whether it is planar. It is probably >>better to think in terms of connectedness of rooms than connectedness of >>walls. > The basic problem with a depth-first search is the chance youll get into a > loop going around and around a disconnected piece. The problem can be > avoided by making a rule to never cross your own path. > Or is there something more that Im missing? > --Keith Lewis klewis {at} mitre.org > The above may not (yet) represent the opinions of my employer. I was under the impression that rule was part of DFS on arbitrary graphs, though it isnt needed for trees. Otherwise cycles are a problem. === Subject: Re: JSH: Being me > I know, many of you probably think its horrible being me, with all > these people calling me name, putting up nasty webpages, and spending > so much time talking bad about me. I have to be honest, if being you is horrible, those are not the causes I would have listed for it. You receive very little name-calling, the websites mainly say you are obstinately wrong. Ive seen people treated far worse in the public schools. Youve got nothing on some of the politicians out there. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Being me >I know, many of you probably think its horrible being me, with all >these people calling me name, putting up nasty webpages, and spending >so much time talking bad about me. Why in the world would you think anyone cares? And just out of curiosity, who are you adressing here? This happens every once in a while, you seem to be talking to all your other readers, the ones who are not part of the vast conspiracy. What makes you think that they exist? >Some of you are in your own little world and maybe still think Im >wrong, while some of you realize that Im right and STILL wouldnt >want to be me considering how much opposition my results have faced >and are likely to face. Like, what gives you the idea that there are people out there who think youre right? Nobody ever _says_ they think youre right. >But hey, its actually fun! >Like I havent just done arguing with people on math or even just math >research as I have an open source project on SourceForge, and I have >made some friends (believe it or not) in a few Internet communities. That _is_ hard to believe. Why do you keep coming back here? >I *thought* I could use the Internet in a groundbreaking way to >introduce major results but here I am WAITING ON A JOURNAL, when I >said in the past that I wouldnt even use journals! Youve also said in the past that the internet didnt matter, all that mattered was the fact that you were about to be published in a journal. Given just about anything youve ever said, youve said something directly contraditory in the past. ************************ David C. Ullrich === Subject: Re: JSH: Being me posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK > Like, what gives you the idea that there are people out there > who think youre right? Nobody ever _says_ they think youre > right. > Wasnt there that guy from the genius club? What ever happened to him? I dont recall Quinn Tyler-Jackson ever saying Harris was right, he objected to how Harris was treated. I dont recall A. Beckwith ever saying Harris was right, he just tried to steal the credit for his paper. === Subject: Re: JSH: Being me > Like, what gives you the idea that there are people out there > who think youre right? Nobody ever _says_ they think youre > right. > Just think. > Sci.math has *lots* of readers. What are the odds that theyre *all* > wrong about Jamess work? Well, I think sci.math readers are stupid. Besides, I do remind you that Im just here goofing off, while I wait. It makes me no never mind to play with you people and see just what I can see. I trace out your neural pathways this way. Like, tomorrow I will come to see who replies to this post, and read information bounced around in their heads. I do this enough that I can build a map, and a model, then I simply test the model. If it fails to respond as you would, then I test again, until it responds as you do. So I build a world and then I can simply test that world, moving the pieces in it as I see fit. And then I know how you will move, as I see fit. I am mostly done. James Harris === Subject: Re: JSH: Being me > Well, I think sci.math readers are stupid. > Besides, I do remind you that Im just here goofing off, while I wait. > It makes me no never mind to play with you people and see just what I > can see. > I trace out your neural pathways this way. > Like, tomorrow I will come to see who replies to this post, and read > information bounced around in their heads. > I do this enough that I can build a map, and a model, then I simply > test the model. > If it fails to respond as you would, then I test again, until it > responds as you do. > So I build a world and then I can simply test that world, moving the > pieces in it as I see fit. And then I know how you will move, as I > see fit. > I am mostly done. > James Harris It appears you have managed to elevate your delusions of grandeur to delusions of divinity. Keep it up. There are special places for you -- even here on earth! -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Being me > > Like, what gives you the idea that there are people out there > who think youre right? Nobody ever _says_ they think youre > right. > > Just think. > > Sci.math has *lots* of readers. What are the odds that theyre *all* > wrong about Jamess work? > Well, I think sci.math readers are stupid. You read sci.math, right? === Subject: Re: JSH: Being me > Frightening thought. Must be hell in there. Dirk Vdm === Subject: Re: topology.....+++ by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAU32mj11249; >hello......doctor~ >let X be a topolotical space and A, B be subsets of X with B in A. >We give the subspace topology on A. >(a) if B is dense in A and A is dense in X, >then show that B is dense in X. >(b) Given an example of a dense proper subset of Q, the set of >all rational number. >------------------------------------------------------------ >i can do (a). >but i cant find the example of (b). >i only know that Q is dense subset R. >so, i need your adivice. >thank you very much for your advice. Unless Im missing something, (b) seems trivial. Just leave out one point from Q. The rest is a dense proper subset of Q. === Subject: Unstoppable Force vs Immovable Object by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAU32nX11264; What happens when an unstoppable force collides with a immovable object? This question definitely lies within the realm of the theoretical but there is a definite solution that one may establish if such an event could happen. Just use basic properties of physics! The Normal Force of the unstoppable force is equal to infinity ( F=oo ). Then you take the Strong Force that makes any object solid- the reason you do not fall to the center of the earth- which is a set number and therefore less then infinity. Now Normal Force exceeds Strong Force and the unstoppable force simply passes through the unmovable object. I may be in error but I have been thinking about this and come to this solution. If you have another expiation then write it, this has been plaguing my mind for awhile. === Subject: Re: Unstoppable Force vs Immovable Object > What happens when an irresistable force collides with a immovable object? There is an inconceivable concussion! . === Subject: Re: Unstoppable Force vs Immovable Object > What happens when an unstoppable force collides with a immovable > object? The same thing that happens when you go further North than the North Pole. This is a silly, purely verbal claptrap question. Bob Kolker === Subject: Re: Unstoppable Force vs Immovable Object > What happens when an unstoppable force collides with a immovable object? This question definitely lies within the realm of the theoretical but there is a definite solution that one may establish if such an event could happen. Just use basic properties of physics! The Normal Force of the unstoppable force is equal to infinity ( F=oo ). Then you take the Strong Force that makes any object solid- the reason you do not fall to the center of the earth- which is a set number and therefore less then infinity. Now Normal Force exceeds Strong Force and the unstoppable force simply passes through the unmovable object. > I may be in error but I have been thinking about this and come to this solution. If you have another expiation then write it, this has been plaguing my mind for awhile. It is logically impossible for an unstoppable force to meet an immovable object. When a force meets an object, of logical necessity either the object moves or fails to move. So, of logical necessity, either the object isnt immovable or the force isnt unstoppable. === Subject: Re: Unstoppable Force vs Immovable Object > It is logically impossible for an unstoppable force to meet an > immovable object. When a force meets an object, of logical necessity > either the object moves or fails to move. So, of logical necessity, > either the object isnt immovable or the force isnt unstoppable. Any finite mass on which a net force is exerted will accelerate in the direction of the net force. Newtons Second Law. Bob Kolker === Subject: Re: Unstoppable Force vs Immovable Object > > It is logically impossible for an unstoppable force to meet an > immovable object. When a force meets an object, of logical necessity > either the object moves or fails to move. So, of logical necessity, > either the object isnt immovable or the force isnt unstoppable. > Any finite mass on which a net force is exerted will accelerate in the > direction of the net force. Newtons Second Law. > Bob Kolker Yes, I know. In that case, its also physically impossible that there should be an immovable object. I was ignoring the laws of physics and going for logic alone. === Subject: Re: Unstoppable Force vs Immovable Object > Yes, I know. In that case, its also physically impossible that there > should be an immovable object. I was ignoring the laws of physics and > going for logic alone. Logic, as such, has very little to say about physical forces. As I said you are just playing a word game. Bob Kolker === Subject: Re: Unstoppable Force vs Immovable Object > Yes, I know. In that case, its also physically impossible that there > should be an immovable object. I was ignoring the laws of physics and > going for logic alone. > Logic, as such, has very little to say about physical forces. As I said > you are just playing a word game. > Bob Kolker Why? Whats wrong with the argument? === Subject: Re: Unstoppable Force vs Immovable Object > Why? Whats wrong with the argument? Force has a physical meaning, not a logical meaning. A force is some kind of action that changes the momentum of a body. Bob Kolker === Subject: Re: Unstoppable Force vs Immovable Object > > Why? Whats wrong with the argument? > Force has a physical meaning, not a logical meaning. A force is some > kind of action that changes the momentum of a body. > Bob Kolker I dont think youve identified a problem with my reasoning. === Subject: Re: Unstoppable Force vs Immovable Object > I dont think youve identified a problem with my reasoning. Your reasoning addresses a physical concept in a non-physical way. As I said, you are playing word games. Bob Kolker === Subject: Re: Unstoppable Force vs Immovable Object > > I dont think youve identified a problem with my reasoning. > Your reasoning addresses a physical concept in a non-physical way. As > I said, you are playing word games. > Bob Kolker Whats that supposed to mean, I address a physical concept in a non-physical way? I conclude that its logically impossible for an unstoppable force to meet an immovable object. Do you disagree? Do you think its logically possible? === Subject: Re: Unstoppable Force vs Immovable Object > Whats that supposed to mean, I address a physical concept in a > non-physical way? > I conclude that its logically impossible for an unstoppable force to > meet an immovable object. Do you disagree? Do you think its logically > possible? If you consider what the word force means you would conclude there is no unstopable (i.e. infinite) force in the physical universe. Likewise, an immovable object would have infinite momentum (physically impossible). In either case you are talking about things which not only dont exist, but cant exist. Bob Kolker === Subject: Re: Unstoppable Force vs Immovable Object > What happens when an unstoppable force collides with a immovable > object? They married, gave birth to me and then divorced shortly afterward having caused sufficient trouble. Bart === Subject: Re: Unstoppable Force vs Immovable Object > What happens when an unstoppable force collides with a immovable object? You get a tedious riddle. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Unstoppable Force vs Immovable Object > What happens when an unstoppable force collides with a immovable > object? This question definitely lies within the realm of the > theoretical but there is a definite solution that one may establish > if such an event could happen. Just use basic properties of physics! > The Normal Force of the unstoppable force is equal to infinity > ( F=oo ). Then you take the Strong Force that makes any object solid- > the reason you do not fall to the center of the earth- which is a set > number and therefore less then infinity. Now Normal Force exceeds > Strong Force and the unstoppable force simply passes through the > unmovable object. Then its not really unmovable, is it? In the real world, there are no such things as unstoppable forces or immovable objects. So invoking physics to solve this problem is inappropriately mixing the hypothetical world of the puzzle with the real world. In the hypothetical world of the puzzle, if there is such a thing as an unstoppable force, then there cannot exist such a thing as an immovable object, and vice versa. This puzzle is just as trivial as saying Joe is taller than Jim, and Jim is taller than Joe. Which one is taller? The given conditions are contradictory. --Mark === Subject: Re: Unstoppable Force vs Immovable Object > What happens when an unstoppable force collides with a immovable > object? This question definitely lies within the realm of the > theoretical but there is a definite solution that one may establish > if such an event could happen. Just use basic properties of physics! > The Normal Force of the unstoppable force is equal to infinity > ( F=oo ). Then you take the Strong Force that makes any object solid- > the reason you do not fall to the center of the earth- which is a set > number and therefore less then infinity. Now Normal Force exceeds > Strong Force and the unstoppable force simply passes through the > unmovable object. > I may be in error but I have been thinking about this and come to > this solution. If you have another expiation then write it, this has > been plaguing my mind for awhile. I have two responses, both with the same impact. Ill preface it all by acknowledging that some of what Ive written may seem a bit harsh, but I sincerely mean no offense. I just think youre being a bit self-indulgent, and (if this has *really* been plaguing your mind) perhaps you have too much time on your hands. Have you considered taking up masturbation? First (quickndirty): This is nearly as stupid as the dungheap that has been erected by JSH over the past 8 years. Get a life. Second (on the technical side): The strong force is not the reason objects are solid. It is the reason atomic nuclei hold together, but at the distances atoms relate to one another, the strong interaction is ineffective. If it were the strong force at play, youd see a lot more nuclear fusion than is common today. Probably more than youd like to see, if you get my drift. You should read about the fundamental forces. Next, the irresistable force/immovable object conundrum is nothing more than a classical paradox, in which mutually-contradictory entities are proposed, after which one attempts to proclaim ones own smarty- pantsitude. Its a game of infantile logic grappling with semantics. As they say: Philosophy is the wading pool of the mind Is that what you mean to do with your life? Why not solve the existence of Evil in a world created by a Perfect God? What about the nature of human hope and the problem of personal mortality? Well, maybe a problem that isnt just us gnashing our teeth: how about the problem of human suffering? Immense wealth and abysmal poverty? Getting JSH to pull his head out of his own ass? The point is that if you are an intelligent person, you are doing yourself a monumental disservice by wasting your limited time in this lifecycle on semantic quibbling. If this problem is truly plaguing your mind, then you need to get off it, and do something that matters. If youre a dimbulb, then it doesnt matter what the hell you do, just as long as you stay out of the way of others. Suit yourself, of course, but dont expect everyone to be wowed by your savoir-faire at the bar when youre explaining it over a beer. Dale. === Subject: Re: Unstoppable Force vs Immovable Object >What happens when an unstoppable force collides with a immovable object? This >question definitely lies within the realm of the theoretical but there is a >definite solution that one may establish if such an event could happen. Just >use basic properties of physics! The Normal Force of the unstoppable force is >equal to infinity ( F=oo ). Then you take the Strong Force that makes >any object solid- the reason you do not fall to the center of the earth- >which is a set number and therefore less then infinity. Now Normal Force >exceeds Strong Force and the unstoppable force simply passes through the >unmovable object. >I may be in error but I have been thinking about this and come to this >solution. If you have another expiation then write it, this has been plaguing >my mind for awhile. f/a = pressure = p1 and p2 When considering this situation you have to consider the area that the force is applied which is the pressure. ( And assume they both use the same level of hardness and material, for the unstoppable force and unmovable object) unstoppable force | p1 v p2 f/a--->|O|<---f/a ( The object has to have ^ the balancing f/a) | Immovable Object If the area of the immovable object comes into contact with the same area of the unstoppable force and they remain in balance, then the result is just zero movement. p1=p2, p1 - p2 = 0, the result is like no force applied, and they just sit there. If the unstoppable force has a smaller area than the Immovable object, then even though the immovable object never moves, it could be penetrated by the unstoppable force. If the hardness level changes in regard to the specific material used, this could also change things. If the unstoppable force is made of a brittle applied material as compared to the unmovable object, the unstoppable force would fall apart, and the unmovable object just stays in place unchanged. This can also be applied to self-defense logic. Even though a person may be unmovable, a knife could still penetrate that person, unless that person uses a shield that can counter the applied knife, like a knife proof vest to prevent penetration. Once the unstoppable force of the knife reaches the immovable-impenetrable object, they just sit there and nothing happens. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: Unstoppable Force vs Immovable Object >>What happens when an unstoppable force collides with a immovable object? >This >>question definitely lies within the realm of the theoretical but there is a >>definite solution that one may establish if such an event could happen. Just >>use basic properties of physics! The Normal Force of the unstoppable force >>equal to infinity ( F=oo ). Then you take the Strong Force that makes >>any object solid- the reason you do not fall to the center of the earth- >>which is a set number and therefore less then infinity. Now Normal Force >>exceeds Strong Force and the unstoppable force simply passes through the >>unmovable object. >>I may be in error but I have been thinking about this and come to this >>solution. If you have another expiation then write it, this has been >plaguing >>my mind for awhile. > > f/a = pressure = p1 and p2 > When considering this situation you have to consider the area that the >force >is applied which is the pressure. ( And assume they both use the same level >hardness and material, for the unstoppable force and unmovable object) >unstoppable force > | > p1 v p2 >f/a--->|O|<---f/a ( The object has to have > ^ the balancing f/a) > | > Immovable > Object > If the area of the immovable object comes into contact with the same area >the unstoppable force and they remain in balance, then the result is just >zero >movement. p1=p2, p1 - p2 = 0, the result is like no force applied, and they >just sit there. > If the unstoppable force has a smaller area than the Immovable object, then >even though the immovable object never moves, it could be penetrated by the >unstoppable force. If the hardness level changes in regard to the specific >material used, this could also change things. If the unstoppable force is >made >of a brittle applied material as compared to the unmovable object, the >unstoppable force would fall apart, and the unmovable object just stays in >place unchanged. > This can also be applied to self-defense logic. Even though a person may >unmovable, a knife could still penetrate that person, unless that person uses >shield that can counter the applied knife, like a knife proof vest to prevent >penetration. Once the unstoppable force of the knife reaches the >immovable-impenetrable object, they just sit there and nothing happens. Another interesting thought about this is, you have heard it said, those that live by the sword will die by the sword, but those that live by the shield will continue to live by the shield. A shield doesnt attack anyone or hurt anyone. The best offense is a good defense. A person that did happen to attack by a sword would get tired trying to hurt that person, while the other person with the shield just sits there and takes it easy. And another thought .... The best way to win a nuclear war, is to have a good defensive shield and have a good high energy EM pulse directed beam shield that would shut down the electronics used in the nuclear weapon, before it even had a chance to explode. I would say that these timers and switches used in nuclear bombs are electronics and IC circuits. If a strong enough EM pulse beam is directed to the missle, it should shut down, even the guidence system wouldnt work. >Smarts Alt. Physics News Group >http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >S. Enterprize (Science Journal) >http://smart1234.s-enterprize.com/ Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: Unstoppable Force vs Immovable Object > Another interesting thought about this is, you have heard it said, those > that live by the sword will die by the sword, but those that live by the shield > will continue to live by the shield. A shield doesnt attack anyone or hurt > anyone. Hogwash. A shield can be a useful offensive weapon. Perhaps you should crosspost to one of the SCA groups where Im sure lots of people would be willing to point out just how wrong you are. The best offense is a good defense. A person that did happen to attack > by a sword would get tired trying to hurt that person, while the other person > with the shield just sits there and takes it easy. Hogwash^2. Rick === Subject: Re: simple math question. > In my book, > They resolve that > SQRT(56) is equal to 2*SQRT(14). > Correct, but what method would you use to come to that conclusion. > crzzy1. > sqrt(56) = sqrt (2*2*2*7) > = sqrt (2*2) * sqrt (2*7) > = 2 * sqrt (14) > General approach: > To simplfy sqrt(n), express n as a product of prime factors. Wherever some > prime factor appears two or more times (like 2 in this case), move out an > even number of them under their own square root and simplify. > For example, sqrt (2 * 3^7 * 5^2 * 17^3) - whatever that number is - will be > = sqrt(3^6 * 5^2 * 17^2) * sqrt(2 * 3 * 17) > = 3^3 * 5 * 17 * sqrt(102) Excellent explanation. crzzy1. === Subject: Re: Partial products > It is possible that the following problem has been solved already by > someone, but I couldnt find any refence with Google. > Given is a set of n=2^m numbers {x(0),...x(n-1)}. It is required > to compute all partial products P(0) to P(n-1), such that each > includes n-1 numbers from the set: > P(0) = 1 *x(1)*...*x(n-1) > P(1) =x(0)* 1 *...*x(n-1) > ... > P(n-1)=x(0)*x(1)*...* 1 > The only operation permited is multiplication. I need an efficient > algorithm to generate these partial products using minimum number > of multiplications. > My use of the above is in a factorization algorithm. > A.D. I hoped that this problem would attract some attention from the readers of this group. In the meantime I made some progress on solving it. Obviously, naive direct method requires n*(n-1) multiplication operations. Desired approach will pre-compute some values and combine them to obtain the products P(i) using minimum number of operations. The pre-computed values can be stored in a 2-dimensional matrix. Then partial products P(i) can be obtained with minimum number of additional operations by cleverly combining the matrix elements. Let pre-computed value be stored in a matrix d(m,2^m) in the following manner: for i=0 to n-1 d(0,i)=x(i) next i Ô for j=1 to m-1 k=2^(m-j)-1 for i=0 to k d(j,i)=d(j-1,2*i)*d(j-1,2*i+1) next i next j The filling up of the matrix d(j,i) needs n-2 operations. The following step will combine the elements from the matrix into the products P(i). For example take m=4, n=16. One gets the following: P(0)=d(0,1)*d(1,1)*d(2,1)*d(3,1) which uses 3 operations. However P(1) needs only 1 operation if d(0,1) only is replaced in the above product with d(0,0). It can be shown that P(2) will need 2 operations, P(3) again 1 operation, P(4) 3 operations, and so on ... After counting all the operations I found that the total is 3*(16-2) including those needed to calculate d(j,i) products. It looks that optimum method needs 3*(n-2) operations, versus n*(n-1) operation needed by direct inefficient method. For n=2^7 the difference is quite significant, 378 versus 16256 operations! What I still need is an algorithm to combine the elements from matrix d(m,2^m) to obtain P(i)s. Any suggestion is appreciated. A.D. === Subject: Re: Partial products This best way to deal with this is the Forward-Backward Algorithm. we construct to arrays: 1st array is: x(0) x(0)*x(1) x(0)*x(1)*x(2), ......... 2st array is: x(n-1) x(n-1)*x(n-2) x(n-1)*x(n-2)*x(n-3), ........... It takes O(n) to construct these 2 arrays. every partial product is the product from 2 number in 1st and 2st array. and it takes constant time to locate the 2 numbers, so the whole process takes O(n). BTW, what are you going to make use of this algorithm? I would like to know it. === Subject: Re: Proposed definition for comparing the sizes of two sets > There are definitions of relative set sizing besides cardinality, for > example, there is a definition of set sizing that a proper superset > of a set is larger than the set Are you referring to the partial order you get by using the subset predicate? I.e. the question is whether S1 is a subset of S2? Unfortunately that method leaves almost all pairs of sets uncomparable. For example consider two sets {a, b} and {a, c}. Neither is a subset of the other, so the two sets are uncomparable by the subset method, even though almost anyone can see that each has two elements so they ought to be considered the same size. The whole idea of cardinality is to clarify what the common definition means for finite sets in a way that directly applies in the same way to non-finite sets. Your proposal of using the subset property doesnt even give the right answer for finite sets, so it fails at the desired task. > in the consideration of number theory, the set of even integers, as > an example, has half the asymptotic density of the integers, within > the integers. Do you know the difference between a set, which has no properties other than what are elements and what arent elements, and a more complicated structure that has a set plus some additional operators defined on the elements of the set? The set of integers, and the set of even integers, have no such thing as asymptotic density when all you can consider is what elements are in the sets and what elements arent. Asymptotic density can be defined only when you have some kind of metric defined between elements of the sets. If you use the real metric, whereby the distance between any two integers is the absolute value of their arithmetic difference, and as with any metric space you can define neighborhoods of various radii, and you can define various kids of limits with respect to that metric, such as the limit as the size of the ball from some fixed point grows larger without limit, then you can define asymptotic density in terms of those metric-space-related terms. But if you use a different metric on the very same set, such as the 2-adic metric, you get a completely different result. So obviously asymptotic density isnt a property of the set itself, but only a property of one or another specific metric space using the set as elements in the space. > The asymptotic density is rather useful Only in a metric space. In just set theory its not even defineable (unless you use set theory to define a metric space of course, like the way set theory is used to construct the natural numbers, from which you can construct the integers, from which you can construct the rationals, from which you can define a metric, etc.). === Subject: Re: Proposed definition for comparing the sizes of two sets > There are definitions of relative set sizing besides cardinality, for > example, there is a definition of set sizing that a proper superset > of a set is larger than the set > Are you referring to the partial order you get by using the subset predicate? > I.e. the question is whether S1 is a subset of S2? > Unfortunately that method leaves almost all pairs of sets uncomparable. > For example consider two sets {a, b} and {a, c}. Neither is a subset of > the other, so the two sets are uncomparable by the subset method, even > though almost anyone can see that each has two elements so they ought > to be considered the same size. The whole idea of cardinality is to > clarify what the common definition means for finite sets in a way that > directly applies in the same way to non-finite sets. Your proposal of > using the subset property doesnt even give the right answer for > finite sets, so it fails at the desired task. > in the consideration of number theory, the set of even integers, as > an example, has half the asymptotic density of the integers, within > the integers. > Do you know the difference between a set, which has no properties other > than what are elements and what arent elements, and a more complicated > structure that has a set plus some additional operators defined on the > elements of the set? The set of integers, and the set of even integers, > have no such thing as asymptotic density when all you can consider is > what elements are in the sets and what elements arent. Asymptotic > density can be defined only when you have some kind of metric defined > between elements of the sets. If you use the real metric, whereby the > distance between any two integers is the absolute value of their > arithmetic difference, and as with any metric space you can define > neighborhoods of various radii, and you can define various kids of > limits with respect to that metric, such as the limit as the size of > the ball from some fixed point grows larger without limit, then you can > define asymptotic density in terms of those metric-space-related terms. > But if you use a different metric on the very same set, such as the > 2-adic metric, you get a completely different result. So obviously > asymptotic density isnt a property of the set itself, but only a > property of one or another specific metric space using the set as > elements in the space. > The asymptotic density is rather useful > Only in a metric space. In just set theory its not even defineable > (unless you use set theory to define a metric space of course, like the > way set theory is used to construct the natural numbers, from which you > can construct the integers, from which you can construct the rationals, > from which you can define a metric, etc.). Do you have any infinite sets? Please name an infinite set besides the integers or reals. The counting numbers, the natural integers, are very useful for counting these other sets of things. As each set contains only unique elements and there is an ordering relation on each of those sets of things, it is simple to see why each is numbered individually. When the sets are disjoint, disjoint, or partially disjoint, then there exists a proper superset containing each element of both. Youll notice that the proper superset is larger than the set. (Youll notice that was ignored.) You might see why you could apply that to any pair of sets, thus that it is universally applicable. About density and metrics, the metric is a great thing. Im interested in this measure theory. For example, I consider the sigma algebra in relation to my rootsets and decorated ordinals, in the theory with ubiquitous ordinals where the ordinals are Z, the integers, in the complete, concrete, and consistent theory. I browse the MathWorld definition of the sigma algebra and wonder what he means when Eric has sequences in the statement. In terms of the sets where they are defined to be containing the, for example, integers, they are then those integers. The notion of counting in any form, enumerating each, for example via a choice function or well ordering and induction, reßects back upon the completely intuitive natural counting numbers. Thats especially so in the finite case. Besides the metrics there is also the intertwined notion of probability distributions. Not about that, the set of even integers within the integers, is defined by the integral modulus. Cardinality has nothing to say about asymptotic density except no opinion. Are half the infinite binary sequences normal or through restricted sequence element interchange convertible to the canonical sequence with equal zero and one density, one third or two thirds? The infinitesimal predates the cardinal, in a sense being the classical. The cardinal is in a sense a modern red herring. The set of all sets is its own powerset. The infinite is not necessarily simple nor intuitive except that it is: half of the integers are even. Bijections exist between the integers and even integers, and the particular one f(x)=2x, a plain straight line, shows that there are twice as many integers as even integers, in the integers or superset of the integers. You have some good points there, but in comparing the relative sizes or infinite sets there is sometimes a reason that has to do with solving a real world problem instead of ßights of fancy about meaningless escapisms from foundational foundations. Asymptotic density is a useful notion that runs right back to one plus one equals two. Ken, 2 + 2 = 4. To Scandinavians herring is a way of life, to Americans its a Monty Python sketch. I saw this the other day, its haunting, yet funny: http://www.khaaan.com/ . It gets more haunting and less funny. There is an implied point set topology and metric space, from nothing, wherefrom all is implied. Ross Finlayson === Subject: Re: Proposed definition for comparing the sizes of two sets > Do you have any infinite sets? Please state what you mean when you use the word have. The ordinary meaning, synonym with own, isnt applicable, because sets are abstract mathematical ideas which cant be owned by any one person. > Please name an infinite set ... Please state what you mean when you use the word name. The ordinary meaning, meaning to assign a name to an object, for example naming a baby, doesnt seem useful in this context. So theres this infinite set, and you want me to name it Fred?? > The counting numbers, the natural integers, are very useful for > counting these other sets of things. Are you talking about using natural numbers (positive integers) as context-sensitive names for individual elements of sets, for example in the context of the set of rational numbers you can call 1/3 #1 and you can call 4/7 #2 and you can call 0 #3 and you can call -5000 #4 etc., assigning a different natural-number label to each element in the set? Or are you talking about cardinality of finite subsets of some context-establishing set, for example the rationals, whereby the subset consisting of {1/3, 4/7} would be counted as size 2, and the subset consisting of {-5000, 0, 22/7, 1/2, 1/3} would be counted as size 5? > As each set contains only unique elements I have no idea what you mean by a unique element. Every thing is unique, so of course every element of any set is a unique thing hence a unique element. Do you actually mean anything by what you said? > and there is an ordering relation on each of those sets of things, That is such a gross understatment that its grossly misleading hence basically a lie. Not only is there one ordering relation on a set, but for any set containing at least two elements there are at least two different ordering relation on the set, and for any set containing at least three elements there are at least six different ordering relations on the set, and for the integers there are an uncountable infinity (aleph_null factorial, which equals aleph_one) of different ordering relations. > completely intuitive natural counting numbers. The natural numbers arent completely intuitive. Only the numbers from one up to about four or five are completely intuitive for humans, where they can just glance at a visual image showing that many similar objects in any random orientation and immediately know intuitively, without needing to count them, how many there are. Some birds can intuitively recognize cardinality up to about seven, beating humans by a couple, which is very useful for detecting if any eggs have been removed from the nest or added to the nest. If objects are in standard partterns, such as pips on a die-face or half-domino, then we can recognize them up to nine, but put those same pips in random pattern and suddenly the problem gets much more difficult. > the set of even integers within the integers, is defined by the > integral modulus. Wrong! If you treat the integers as *nothing* except a set, no order relation, no arithmetic properties, and the individual integers arent defined in terms of something else such as cardinality of sets whereby you can use that definition to generate arithmetic properties, there is no way whatsoever to define which are even and which arent even except by an infinitely long list enumerating each and every even integer (or alternately by enumerating the complement set which are odd). Here are four examples of sets of integers: (1) Recursive definition: The empty set, and any set containing exactly one element which is an integer. Thus {} {{}} {{{}}} etc. are the consecutive integers. Even can be defined recursively like this: N is even iff N = {M} and M is not even. This works because integers arent just abstract elements, but are actually constructed via set theory in a way such that even can be defined from that. (2) Arithmetic definition: Start with Peanos postulates, in particular the successor function S, with natural numbers defined recursively as 1, and any S(N) where N is an integer. Even can be defined recursively like this: N is even iff N = S(M) where M is not even. (3) Explicit listing of just a few integers because each one is listed separately and I dont have an infinite amount of time to type them all: {apat, isa, lima, delawa, tatlo}. Unless you know that Ive used Tagalog names for those five integers, and unless you actually know what those five Tagalog words mean, and unless Im actually using the corect Tagalog names instead of shufßing them, you cant figure out which of those are odd and which are even in my set of integers. If I tell you that apat and delawa are even, and the other three arent even, would you believe me? If I told you something else instead, would you believe me? On what basis could you decide for yourself which are even and which arent per my definition unless I simply tell you the answer and promise not to change my definition to pull a trick on you? (4) In all the above, I had some sort of name for each integer. But suppose I dont have any names at all. Here are a bunch of integers, each displayed as an asterisk: * * * * * * * * * * * * * * * * * * * * * * * * ** * * * * * * * * Now suppose that I claimed that was a viewport into a small section of my integers, that really there are an infinite number of them, and they are not displayed in any particular order that would make sense to you, but I promise I do have a grand design whereby that pattern you see above is one 3-by-62 portion of the overall design. Now how would you decide which of those asterisks are even numbers and which arent? No matter how you guess which are even and which arent, I could legitimately claim you are 100% wrong. If you get two guesses, I could legitimately claim you are 50% wrong, or worse, in each case. Suppose I told you an algorithm for deciding in each position whether one of the integers is there or that position is empty, not just for that 3-by-62 viewport but for the entire grand pattern. How would you define even vs. not-even for all the asterisks? Well because they are located in a lattice, you could perhaps make up a definition that is based on the location within the pattern. It wouldnt match my definition, but at least you *could* define what is meant by even on that set of integers. But that would not be defining even on the elements themselves, rather youd be definining even based on their position within some sort of structue, in this case a regular lattice. But suppose they werent in any lattice structure, but just abstract objects that you could get somehow, not in any particular order, not in any structure, no way to examine them internally, the only predicates you have are: (a) For any item, either that item is in the set (of integers) or not; (b) For any two items in the set, they are either the same element or not the same. With only those two predicates, theres no way to define even rigorously. Suppose in some object-oriented language, for example Java, I provided for you a playpen whereby you could execute commands interactively. I provide for you the following functions/methods: (1) static getInteger() ==> an integer object (2) integerObject.toString() ==> SomeInteger (every integer prints the same) (3) integerObject.hashCode() ==> 0 (every integer has the same hashcode too!) (4) integerObject.equals(integer) ==> true if same object, false otherwise By calling getInteger() from time time, getting several such integerObjects, can you decide just from calls to the above API which of them are even and which arent? No, the best you can do is make arbitrary decisions, which wouldnt be consistent from one run of the demo to another. Can you define a predicate: boolean isEven() which is well defined, using *only* calls to the API Ive specified above? No, you cant. The best you can do is randomly decide for each new integerObject whether its even or odd, and keep a cache of such decisions so you dont contradict yourself later, but thats not a definition, thats a random sampler, and again, just like the manual demo, youd get different results with each run of the program, so your function/method doesnt *define* a function, it merely generates a new random sample each time its run. By the way, with *all* the integers, not just the positive integers, even if you know the total ordering, thats still not enough to define what is even and what isnt even. The best you can do is divide the ordered set into two subsets, alternating, and then pick one at random to be even. And its even worse: If you have a playpen where you can call an API which gives you random integers and tells for any two integers whether they are equal and if not which is the smaller of the two, since you dont know whether two integers are adjacent or not you cant in a finite number of steps determine the two alternating sets. > The set of all sets ... Theres no such thing. If you understood proof-by-contradiction I could present you a very simple proof to that fact, but you dont seem to understand even that simple aspect of mathematical logic so it would be a waste of my time to present it to you. > half of the integers are even. As just a set, 99% of them are even too. > Bijections exist between the integers and even integers, and the > particular one f(x)=2x, a plain straight line, shows that there are > twice as many integers as even integers, in the integers or superset of > the integers. It shows no such thing!! The bijection shows there are exactly the same number of integers as even integers. For every integer theres a corresponding even integer, and vice versa. What you said is equivalent to saying theres a bijection between the fingers on my left hand and the fingers on my right hahd, which shows there are twice as many fingers on my left hand as on my right hand. === Subject: Re: Proposed definition for comparing the sizes of two sets <41AAD2D3.301BBFC3@tiki-lounge.com> at 07:03 PM, rem642b@Yahoo.Com (tinyurl.com/uh3t) said: >Theres no such thing. There are set theories that include a set of all sets. They dont, however, satisfy, e.g., the axiom of foundation. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Proposed definition for comparing the sizes of two sets > Do you have any infinite sets? > Please state what you mean when you use the word have. The ordinary > meaning, synonym with own, isnt applicable, because sets are > abstract mathematical ideas which cant be owned by any one person. > Please name an infinite set ... > Please state what you mean when you use the word name. The ordinary > meaning, meaning to assign a name to an object, for example naming a > baby, doesnt seem useful in this context. So theres this infinite > set, and you want me to name it Fred?? > The counting numbers, the natural integers, are very useful for > counting these other sets of things. > Are you talking about using natural numbers (positive integers) as > context-sensitive names for individual elements of sets, for example in > the context of the set of rational numbers you can call 1/3 #1 and you > can call 4/7 #2 and you can call 0 #3 and you can call -5000 #4 etc., > assigning a different natural-number label to each element in the set? > Or are you talking about cardinality of finite subsets of some > context-establishing set, for example the rationals, whereby the subset > consisting of {1/3, 4/7} would be counted as size 2, and the subset > consisting of {-5000, 0, 22/7, 1/2, 1/3} would be counted as size 5? Identify an infinite set. > As each set contains only unique elements > I have no idea what you mean by a unique element. Every thing is > unique, so of course every element of any set is a unique thing hence a > unique element. Do you actually mean anything by what you said? Yes, I tend to be sufficiently exact. > and there is an ordering relation on each of those sets of things, > That is such a gross understatment that its grossly misleading hence > basically a lie. Not only is there one ordering relation on a set, but > for any set containing at least two elements there are at least two > different ordering relation on the set, and for any set containing at > least three elements there are at least six different ordering > relations on the set, and for the integers there are an uncountable > infinity (aleph_null factorial, which equals aleph_one) of different > ordering relations. Its definitely not a lie. There are obviously many ordering relations, pick one. > completely intuitive natural counting numbers. > The natural numbers arent completely intuitive. Only the numbers from > one up to about four or five are completely intuitive for humans, where > they can just glance at a visual image showing that many similar > objects in any random orientation and immediately know intuitively, > without needing to count them, how many there are. Some birds can > intuitively recognize cardinality up to about seven, beating humans by > a couple, which is very useful for detecting if any eggs have been > removed from the nest or added to the nest. If objects are in standard > partterns, such as pips on a die-face or half-domino, then we can > recognize them up to nine, but put those same pips in random pattern > and suddenly the problem gets much more difficult. How many states are in the union? How many continents are on the planet? How many stars are in the sky? > the set of even integers within the integers, is defined by the > integral modulus. > Wrong! If you treat the integers as *nothing* except a set, no order > relation, no arithmetic properties, and the individual integers arent > defined in terms of something else such as cardinality of sets whereby > you can use that definition to generate arithmetic properties, there is > no way whatsoever to define which are even and which arent even except > by an infinitely long list enumerating each and every even integer (or > alternately by enumerating the complement set which are odd). > Here are four examples of sets of integers: > (1) Recursive definition: The empty set, and any set containing exactly > one element which is an integer. Thus {} {{}} {{{}}} etc. are the > consecutive integers. Even can be defined recursively like this: N is > even iff N = {M} and M is not even. This works because integers arent > just abstract elements, but are actually constructed via set theory in > a way such that even can be defined from that. > (2) Arithmetic definition: Start with Peanos postulates, in particular > the successor function S, with natural numbers defined recursively as > 1, and any S(N) where N is an integer. Even can be defined recursively > like this: N is even iff N = S(M) where M is not even. OK. The powerset is the successor is the order type. > (3) Explicit listing of just a few integers because each one is listed > separately and I dont have an infinite amount of time to type them > all: {apat, isa, lima, delawa, tatlo}. Unless you know that Ive used > Tagalog names for those five integers, and unless you actually know > what those five Tagalog words mean, and unless Im actually using the > corect Tagalog names instead of shufßing them, you cant figure out > which of those are odd and which are even in my set of integers. > If I tell you that apat and delawa are even, and the other three arent > even, would you believe me? If I told you something else instead, would > you believe me? On what basis could you decide for yourself which are > even and which arent per my definition unless I simply tell you the > answer and promise not to change my definition to pull a trick on you? Do they mean the same thing as {1, 2, 3, 4, 5}? > (4) In all the above, I had some sort of name for each integer. But > suppose I dont have any names at all. Here are a bunch of integers, > each displayed as an asterisk: > * * * * * * * * * * > * * * * * * * * * * * * > * * ** * * * * * * * * > Now suppose that I claimed that was a viewport into a small section of > my integers, that really there are an infinite number of them, and they > are not displayed in any particular order that would make sense to you, > but I promise I do have a grand design whereby that pattern you see > above is one 3-by-62 portion of the overall design. Now how would you > decide which of those asterisks are even numbers and which arent? I dont care. I wouldnt. > No matter how you guess which are even and which arent, I could > legitimately claim you are 100% wrong. If you get two guesses, I could > legitimately claim you are 50% wrong, or worse, in each case. Suppose I > told you an algorithm for deciding in each position whether one of the > integers is there or that position is empty, not just for that 3-by-62 > viewport but for the entire grand pattern. How would you define even > vs. not-even for all the asterisks? Well because they are located in a > lattice, you could perhaps make up a definition that is based on the > location within the pattern. It wouldnt match my definition, but at > least you *could* define what is meant by even on that set of > integers. But that would not be defining even on the elements > themselves, rather youd be definining even based on their position > within some sort of structue, in this case a regular lattice. But > suppose they werent in any lattice structure, but just abstract > objects that you could get somehow, not in any particular order, not in > any structure, no way to examine them internally, the only predicates > you have are: (a) For any item, either that item is in the set (of > integers) or not; (b) For any two items in the set, they are either the > same element or not the same. With only those two predicates, theres > no way to define even rigorously. > Suppose in some object-oriented language, for example Java, I provided > for you a playpen whereby you could execute commands interactively. I > provide for you the following functions/methods: > (1) static getInteger() ==> an integer object > (2) integerObject.toString() ==> SomeInteger (every integer prints the same) > (3) integerObject.hashCode() ==> 0 (every integer has the same hashcode too!) > (4) integerObject.equals(integer) ==> true if same object, false otherwise > By calling getInteger() from time time, getting several such > integerObjects, can you decide just from calls to the above API which > of them are even and which arent? No, the best you can do is make > arbitrary decisions, which wouldnt be consistent from one run of the > demo to another. An integer has an integer value. If you have integerObject.add(IntegerObject i) returning the sum, then you should be able to tell which are even. > Can you define a predicate: > boolean isEven() > which is well defined, using *only* calls to the API Ive specified above? > No, you cant. The best you can do is randomly decide for each new > integerObject whether its even or odd, and keep a cache of such > decisions so you dont contradict yourself later, but thats not a > definition, thats a random sampler, and again, just like the manual > demo, youd get different results with each run of the program, so your > function/method doesnt *define* a function, it merely generates a new > random sample each time its run. Thats irrelevant. Half of the integers are even. > By the way, with *all* the integers, not just the positive integers, > even if you know the total ordering, thats still not enough to define > what is even and what isnt even. The best you can do is divide the > ordered set into two subsets, alternating, and then pick one at random > to be even. And its even worse: If you have a playpen where you can > call an API which gives you random integers and tells for any two > integers whether they are equal and if not which is the smaller of the > two, since you dont know whether two integers are adjacent or not you > cant in a finite number of steps determine the two alternating sets. > The set of all sets ... > Theres no such thing. No, there is. Obviously enough there is not in ZF. > If you understood proof-by-contradiction I could > present you a very simple proof to that fact, but you dont seem to > understand even that simple aspect of mathematical logic so it would be > a waste of my time to present it to you. No, youre wrong. Several theories including my own have sets of all sets. > half of the integers are even. > As just a set, 99% of them are even too. No, the integers have integer values. > Bijections exist between the integers and even integers, and the > particular one f(x)=2x, a plain straight line, shows that there are > twice as many integers as even integers, in the integers or superset of > the integers. > It shows no such thing!! The bijection shows there are exactly the same > number of integers as even integers. For every integer theres a > corresponding even integer, and vice versa. Youre wrong, it shows exactly that thing. > What you said is equivalent to saying theres a bijection between the > fingers on my left hand and the fingers on my right hahd, which shows > there are twice as many fingers on my left hand as on my right hand. No, it doesnt. Look at any function from the integers to the integers of the form y=mx+b, a straight line, for integer m. The range has asymptotic density of 1/m in the integers. As it is so for any straight line function, except for arguably m=0, where that bijection exists for the domain of the integers, it shows that the range has an asyptotic density, which is a useful comparison of sets sizes, in this case the sets of the domain and range, comparing the integers to a subset of the integers and illustrating why the subset comprises half of the integers, or generally 1/m. Do you not see how terribly, horribly wrong youve been about all this? Half of the integers are even, true or false? Its true. If its false then through contradiction the integer is neither even nor odd. Besides your caffeinated integers, an integer is even or odd. Anyways, the key point to consider is that the proper subset definition of sizing is universally applicable. Also, when you talk about sets of only integers, not labels but integers, then all structural aspects of the integers hold true in the comparison of collections of them. Ross === Subject: Re: Proposed definition for comparing the sizes of two sets at 08:57 PM, dchris@netcom.ca (Dan Christensen) said: >Proposed Definition: A set Y is said to be larger than a set X iff >there exists no function mapping X onto all of Y. (i.e. there is no >surjection from X to Y) >Is this a workable definition that covers all cases? If so, is it >widely used? Are you assuming the Axiom of Choice, or some equivalent? If not, you have to deal with incomparable pairs of sets. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Proposed definition for comparing the sizes of two sets > at 08:57 PM, dchris@netcom.ca (Dan Christensen) said: >Proposed Definition: A set Y is said to be larger than a set X iff >there exists no function mapping X onto all of Y. (i.e. there is no >surjection from X to Y) >Is this a workable definition that covers all cases? If so, is it >widely used? > Are you assuming the Axiom of Choice, or some equivalent? If not, you > have to deal with incomparable pairs of sets. I dont currently have AC built into my program, but you can introduce it as a premise at the beginning of any proof. I am planning to make it a true axiom, as well as building in some defintions for cardinal numbers in a future release. Some details need to be worked out. Dan Download DC Proof 1.0 at http://www.dcproof.com === Subject: Re: Proposed definition for comparing the sizes of two sets >> Proposed Definition: A set Y is said to be larger than a set X iff there >> exists no function mapping X onto all of Y. (i.e. there is no surjection >> from X to Y) >This is no different that what is already being used. >Because magnitudes of sets are linear ordered > X < Y iff not Y <= X >(you may prefer to read X < Y as |X| < |Y|) Linear ordering of the magnitudes is equivalent to the axiom of choice. There are lots of models without it, and the definition is never a good one without it. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Proposed definition for comparing the sizes of two sets >Proposed Definition: A set Y is said to be larger than a set X iff there >exists no function mapping X onto all of Y. (i.e. there is no surjection >from X to Y) >Is this a workable definition that covers all cases? If so, is it widely >used? This is not a workable definition. Without at least some form of AC, one can have too sets, each of which comes out as larger than the other. Consider a non-wellorderable set X and a well-orderable set Y which is not smaller than or equal to (in the usual sense) P(X). If there is a non-wellorderable set, this must exist. Now a surjection of X onto Y is equivalent to an injection of Y into P(X), which we have assumed does not hold. And a surjection of a well-orderable set onto any set gives a well-ordering of that set. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? A > Certainly, if you integrate by parts, you would choose > int(f(t).d(t-T)) as the integrated bit to yield f(T), but when > I try this, I get 0!...... > int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... > with f(t).d(t-T) as V and e^(-sT) as U ..... > f(T).e^(-st) - int(e^(-st)/-s . f(T))..... > f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... > f(T).e^(-sT) - f(T).e^(-sT).... > 0. lets integrate by parts correctly: L[x(t).delta(t-T)] = int_0^infinity exp(-st).x(t).delta(t-T).dt using int_a^b udv = (uv)|_a^b - int_a^b v.du as our formulation of parts, choose: u = exp(-st) => du = -s.exp(-st)dt dv = x(t).delta(t-T) dt => v = int_0^t x(u) delta(u-T)du = 0 if t < T, or x(T) if t >= T so, L[x(t).delta(t-T)] = [exp(-st).v(t)]_0^inf - int_0^inf -s.exp(-st).v(t)dt = 0 + s.int_T^inf x(T)exp(-st)dt = [-x(T).exp(-st)]_T^inf = x(T).exp(-sT) hopefully that is satisfactory for you, I think the sloppiness is warranted. Richard === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? hey Beanie, i have proposal for you. would you promise to shut up and go away (or, at least, just shut up) if i show you where your ßaw is in your integration by parts? its amazing why you dont just accept the simple proof (applying the definition of the dirac impulse function) that +inf integral{ x(t) d(t-T) e^(-st) dt} ( d(t) = dirac impulse function ) -inf +inf = integral{ [x(t) e^(-st)] d(t-T) dt} -inf +inf = integral{ x1(t) d(t-T) dt} ( where x1(t) = x(t) e^(-st) ) -inf +inf = integral{ x1(u+T) d(u) du} ( substitute u = t-T ) -inf +inf = integral{ x2(u) d(u) du} ( where x2(t) = x1(t+T) ) -inf = x2(0) ( by definition of the dirac impulse ) = x1(T) = x(T) e^(-sT) now thats how normal people without mental illness will evaluate that bilateral Laplace Transform integral. (people *with* mental illness may want to use a less direct and more difficult method so that they will more likely screw up and get the wrong answer.) for some reason Beanie wants to evaluate it by parts and he insists he gets a different answer (zero). setting aside the lesson of Occams Razor ( http://en.wikipedia.org/wiki/Occam%27s_razor the simplest explanation should be used), the answer *if* you integrate by parts *correctly* (which Beanie cant seem to do correctly), the integral turns out to be the same. Laplace{ x(t) d(t-T) } = x(T) e^(-sT) not zero as Beanie claims Beanie, if you promise to behave yourself and stop trolling this newsgroup, ill show your ßaw in your integration by parts. both integrating by parts and the straight forward integration comes out with the same answer (as if any of us suspected otherwise). but im gonna use the *textbook* definition of integrating by parts: b | b b integral{ u dv} = (uv)| - integral{ v du} a | a a (Beanie, youll need a mono-spaced font to read this ASCII math well.) Beanie wants (if he agrees to conventional notation) u = e^(-st) and dv = x(t) d(t-T) dt Beanie, if you agree to those (very reasonable) terms, i will show you (and everyone else reading it) the ßaw in your derivation that erroneously says the result is zero. r b-j > with as follows..... > You need to do int(+/-inf)(f(t).d(t-T).e^(-st)) to > determine your result. > You simply cannot say that f(t).d(t-T) yields f(T) unless > you do so under an integral. This arises from the fundamental > properties of the Diracian Delta function . > Certainly, if you integrate by parts, you would choose > int(f(t).d(t-T)) as the integrated bit to yield f(T), but when > I try this, I get 0!...... thats because your knowledge and insight is far less than you think and youre also quite mentally ill. (perhaps only the latter is true if youre the master troll Jerry makes you out to be, but both are true if youre simply the obnoxious crackpot and jerk that is more ostensible.) > int(UV) = U.int(V) - int[dU.int(V)] giving..... > int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... > with f(t).d(t-T) as V and e^(-sT) as U ..... > f(T).e^(-st) - int(e^(-st)/-s . f(T))..... > f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... > f(T).e^(-sT) - f(T).e^(-sT).... > 0. > What I seek is a sound mathematical proof of the claim > that f(t).d(t-T) gives rise to a spectrum contribution > of f(T).e^-(st), and claiming that f(t).d(t-T) equals > f(T) is not sound. We use the properties of the Diracian > Delta Function in so many other aspects of signal > processing that it is just not right to pull-a-fast-one. >> Now you can evaluate f(t) at T, and use that in the laplace transform like: >> integral(from 0 to inf) of ( f(T).e^-(st) ) dt >> if f(T) is a constant you can pull it out the front. === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? Because of your childish behaviour quoted below, something of value to contribute to the discussion, then please present it in a style more appropriate to an international forum. > hey Beanie, > i have proposal for you. would you promise to shut up and go away (or, at > least, just shut up) === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? are you afraid to see your mistake explicitly exposed? cluck, cluck r b-j > Because of your childish behaviour quoted below, > something of value to contribute to the discussion, > then please present it in a style more appropriate to > an international forum. >> hey Beanie, >> i have proposal for you. would you promise to shut up and go away (or, at >> least, just shut up) === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? I am not afraid of anything, be that anything you, or the truth. It is simply that I refuse to deal with you when you indulge yourself in infantile outbursts. Shame on you. You should know better, including not indulging in silly yah-boo-sucks tirades as in your latest rant below. You do nothing for your reputation by so behaving. > are you afraid to see your mistake explicitly exposed? > cluck, cluck > Because of your childish behaviour quoted below, > something of value to contribute to the discussion, > then please present it in a style more appropriate to > an international forum. === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? > What I seek is a sound mathematical proof of the claim > that f(t).d(t-T) gives rise to a spectrum contribution > of f(T).e^-(st) For what its worth, tongue firmly in cheek, and with apologies to David Ulrich, who did it properly: You seem to have a problem with the Dirac Delta definition. Anyway, presumably you are happy with the Laplace transform L[f(t).delta(t-T)] = int_0^infinity f(t).delta(t-T).exp(-st) dt Maybe the notion that the Delta function is, loosely speaking, zero nearly everywhere is sort-of vaguely familiar to you, so you might be convinced that something like this is true: L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(t).delta(t-T).exp(-st)dt And maybe considering f(t) and exp(-st) might get you ruminating on their continuity, and a daring insight like this may leap out: L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(T).delta(t-T).exp(-sT)dt Forgive me if Im getting carried away, but isnt the Riemann Integral linear? I think it might be, so wouldnt that mean that L[f(t).delta(t-T)] = lim M->0 f(T).exp(-sT) int_(T-M)^(T+M) delta(t-T)dt I really think that I might be onto something here, but I need to work out how to integrate that pesky Dirac Delta ... any takers? === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? >> What I seek is a sound mathematical proof of the claim >> that f(t).d(t-T) gives rise to a spectrum contribution >> of f(T).e^-(st) > For what its worth, tongue firmly in cheek, and with apologies to David > Ulrich, who did it properly: > You seem to have a problem with the Dirac Delta definition. > Anyway, presumably you are happy with the Laplace transform > L[f(t).delta(t-T)] = int_0^infinity f(t).delta(t-T).exp(-st) dt > Maybe the notion that the Delta function is, loosely speaking, zero nearly > everywhere is sort-of vaguely familiar to you, so you might be convinced > that something like this is true: > L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(t).delta(t-T).exp(-st)dt > And maybe considering f(t) and exp(-st) might get you ruminating on their > continuity, and a daring insight like this may leap out: > L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(T).delta(t-T).exp(-sT)dt > Forgive me if Im getting carried away, but isnt the Riemann Integral > linear? I think it might be, so wouldnt that mean that > L[f(t).delta(t-T)] = lim M->0 f(T).exp(-sT) int_(T-M)^(T+M) delta(t-T)dt > I really think that I might be onto something here, but I need to work out > how to integrate that pesky Dirac Delta ... any takers? Perhaps youre just kidding, or I misunderstand something. Isnt the Dirac Delta *defined* such that the int_-inf^+inf(delta(t-T)dt = 1.0??? And since the Dirac delta functions value is *defined* as zero everywhere except when t-T = 0, then the lim M->0 int_(T-M)^(T+M) delta(t-T)dt is also equal to 1.0 ??? The partial integral from -inf to (T-M) as lim M->0 remains 0.0 as well as the partial integral from (T+M) to +inf remains 0.0. So the difference between lim M->0 int_(T-M)^(T+M) delta(t-T)dt and int_-inf^+inf delta(t-T)dt is zero (because M never reaches exactly 0). daestrom === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? <30uk8bF359mi8U1@uni-berlin.de> <41aa856b$0$20379$afc38c87@news.optusnet.com.au> <31075eF3628reU3@uni-berlin.de> <1t6rd.40278$AL5.20715@twister.nyroc.rr.com> posting-account=LVhRgA0AAAC_z_tVq31kKK9RKztdjt01 >Perhaps youre just kidding, or I misunderstand something. Just kidding :) > Isnt the Dirac >Delta *defined* such that the int_-inf^+inf(delta(t-T)dt = 1.0??? And since >the Dirac delta functions value is *defined* as zero everywhere except when >t-T = 0, then the lim M->0 int_(T-M)^(T+M) delta(t-T)dt is also equal to 1.0 >??? absolutely :) === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? In your second equation, how come youve still got t in your expression of the Diracian, but youve replaced it by T in the other two terms? It would seem to me that if a principle is to be applied, then it must be applied everywhere, so that you must end up with d(T -T) = d(0). > L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(t).delta(t-T).exp(-st)dt > And maybe considering f(t) and exp(-st) might get you ruminating on > their continuity, and a daring insight like this may leap out: > L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(T).delta(t-T).exp(-sT)dt === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? <30uk8bF359mi8U1@uni-berlin.de> <41aa856b$0$20379$afc38c87@news.optusnet.com.au> <31075eF3628reU3@uni-berlin.de> <312sn9F35si4mU18@uni-berlin.de> posting-account=LVhRgA0AAAC_z_tVq31kKK9RKztdjt01 > L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(t).delta(t-T).exp(-st)dt > And maybe considering f(t) and exp(-st) might get you ruminating on > their continuity, and a daring insight like this may leap out: > L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) f(T).delta(t-T).exp(-sT)dt > In your second equation, how come youve > still got t in your expression of the Diracian, but > youve replaced it by T in the other two terms? > It would seem to me that if a principle is to be applied, then > it must be applied everywhere, so that you must end up > with d(T -T) = d(0). Perhaps you missed the daring insight ... since were considering continuous functions, if you make the interval arbitrarily small, exp(-st) and f(t) approach constant values exp(-sT) and f(T) on that interval. As I said, its not rigorous, I would never submit something like that as a proof, but you should get the idea in a handwavy manner why David Ullrichs proof was correct. Richard === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? >> L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) >> f(t).delta(t-T).exp(-st)dt >> And maybe considering f(t) and exp(-st) might get you ruminating on >> their continuity, and a daring insight like this may leap out: >> L[f(t).delta(t-T)] = lim M->0 int_(T-M)^(T+M) >> f(T).delta(t-T).exp(-sT)dt > In your second equation, how come youve still got t in your > expression of the Diracian, but youve replaced it by T in the other > two terms? > It would seem to me that if a principle is to be applied, then it > must be applied everywhere, so that you must end up with d(T -T) = > d(0). Not true - what I was getting at in a handwavy manner is that if you make the interval arbitrarily small, because they are continuous, f(t) and exp(-st) approach constant values f(T) and exp(-sT). delta(t) is not continuous, so you cant apply the same principle to this. Anyway, just in case you misunderstood, this is *not* rigorous, and I would not consider this a proof. I was hoping in a handwavy manner to convince you that David Ullrichs initial proof was correct in the first place. Richard === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? >> What I seek is a sound mathematical proof of the claim >> that f(t).d(t-T) gives rise to a spectrum contribution >> of f(T).e^-(st) ... > I really think that I might be onto something here, but I need to work > out how to integrate that pesky Dirac Delta ... any takers? its actually pretty straight-forward to do it right (its on the other post i just made). Beanie prefers to use integration-by-parts, and then makes a mistake (he actually doesnt do it right from the beginning) and comes out with zero so then he claims that all accumulated knowledge since Heaviside and Nyquist about it is in error. dont fall for his bullshit. he really doesnt know what hes typing about. r b-j === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? >>I really think that I might be onto something here, but I need to work >>out how to integrate that pesky Dirac Delta ... any takers? > its actually pretty straight-forward to do it right (its on the other post > i just made). Beanie prefers to use integration-by-parts, and then makes a > mistake (he actually doesnt do it right from the beginning) and comes out > with zero so then he claims that all accumulated knowledge since Heaviside > and Nyquist about it is in error. dont fall for his bullshit. he really > doesnt know what hes typing about. > r b-j Disclaimer, my previous post was not to be taken very seriously - the maths was intentionally hand-wavy and tongue-in-cheek. David Ullrich did it the standard way further up. I just integrated it by parts for Mr Bean, so he is hopefully now satisfied ... Richard === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? > I just integrated it by parts for Mr Bean, so he is hopefully now > satisfied ... but shucks! you gave it away! for free! i wanted to extract a promise (that the asshole would probably not keep) before telling him what he did wrong. oh well, its public knowledge anyway. r b-j === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? <30uk8bF359mi8U1@uni-berlin.de> <41aa856b$0$20379$afc38c87@news.optusnet.com.au> <31075eF3628reU3@uni-berlin.de> <1VRqd.13100$3U4.257649@news02.tsnz.net> posting-account=LVhRgA0AAAC_z_tVq31kKK9RKztdjt01 Sorry, I didnt see your post in time - and then I realised that I posted it in the wrong place with a couple of errors and resubmitted. ah well Richard === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? > For what its worth, tongue firmly in cheek, and with apologies to David > Ulrich, who did it properly: Even more apologies to David Ullrich for spelling his name wrong === Subject: Re: Derivation Of The Spectrom Due To f(t).d(t - T)????? >> For what its worth, tongue firmly in cheek, and with apologies to David >> Ulrich, who did it properly: >Even more apologies to David Ullrich for spelling his name wrong No problem - I assumed you were talking about someone else, wondered why I didnt see any posts from him... (the spelling with one l is much more common.) ************************ David C. Ullrich === Subject: Definition of Z > Some people want to ask: Why is the definition of Z fixed? What makes > it fixed? What is it fixed *relative* to? What does this definition > *necessarily* depend on? > These are philosophical questions, something I believe you are not > good at. > This is just silliness, not philosophical questions. Had you > understood the usual elementary definition, you could have explained > to your misguided friend why the representation of the integers in > base n, for any n, is finite. Fine. You think these are not philosophical questions, but silliness. Thats okay. My friend was misguided, and I didnt learn anything about the elementary definitions in the secondary school about sets and integers, or later in my life. Thats quite nice. I wonder then, perhaps it is not right to question these things? If mathematics is so well defined, and so beyond doubt, why is there a need for philosophy of mathematics at all? Apparently none. Not any more. We only need history of philosophy of mathematics, you know all the silly questions like Russells paradox, until it was finally settled forever by modern set theory. Its just silliness, all of it. Since foundations of mathematics is rock solid today, then I think all those philosophers and mathematicians who work on this silly stuff can pack up and go on a really long vacation. -- Eray Ozkural === Subject: Re: Definition of Z > why is there a need for philosophy > of mathematics at all? Apparently none. Not any more. Good, now that youve finally grasped that, perhaps you can pack your bags and go home. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Definition of Z > If mathematics is so > well defined, and so beyond doubt, why is there a need for philosophy > of mathematics at all? Apparently none. Your comments are based on a misapprehension, not as to the significance of the philosophy of mathematics, but regarding the quality of your own contributions. === Subject: Nonlocally compact groups, representations, applications What are some examples of nonlocally compact groups whose representations are important for either physics or other areas? Still wondering about harmonic analysis on these guys. In compact groups, representations are very useful because of Peter-Weyl. But could it be the case that we dont need to think about representations of non-compact groups when we try to do harmonic analysis on non-compact groups? Could it be the case that something totally different than representations is needed to study this? Isaac