mm-1060 === Subject: Re: The origin of quaternions - HamiltonÕs 1844 paper big snip In hindsight one may guess that HamiltonÕs 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 donÕt 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 HamiltonÕs 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 OÕDonnell: 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 HamiltonÕs >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 - HamiltonÕs 1844 paper > big snip In hindsight one may guess that HamiltonÕs 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 donÕt 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 HamiltonÕs 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 OÕDonnell: 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 HamiltonÕs >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: Vector invariants >I have 3 vectors A,B,C in the plane (with some origin O) >and like to construct invariants which are symmetric >(or antisymmetric) in A,B,C and independent of O, First of all, a function F(A,B,C) which is antisymmetric has the property that G(A,B,C) = ( F(A,B,C) )^2 is symmetric, so in some sense itÕs sufficient to find all possible symmetric functions. Try this: Writing A = (a1,a2), etc., it appears that you are looking for polynomials in six variables which are invariant under the diagonal action of S_3 (the symmetric group on three letters). These form a ring which is a finitely-generated module over a polynomial ring. Now, I was working on a similar problem just three years ago and asked in sci.math.symbolic about this sort of thing. The computations -- at least for n=3 ! -- are not too bad. Indeed, the invariant ring R can be viewed as containing the polynomial ring P on Z1=a1+b1+c1, Y1=a2+b2+c2, Z2=a1^2+b1^2+c1^2, Y2=a2^2+b2^2+c2^2, Z3=a1^3+b1^3+c1^3, Y3=a2^3+b2^3+c2^3 ; the whole invariant ring R is generated, as a ring, by these and three more invariants, W1 = a1*a2 + b1*b2 + c1*c2 , W2 = a1^2*a2 + b1^2*b2 + c1^2*c2 , W3 = a1*a2^2 + b1*b2^2 + c1*c2^2 , which are subject to certain relations (attached below). If you like, the whole invariant ring may be presented as a free module over P on six generators: 1, W1, W2, W3, and W4 = W1^2, W5 = W2*W3 ; the five relations show (respectively) how to express W1 W2, W1 W3, W1^3, W2^2, and W3^2 in terms of these generators. Now, there is a homomorphism phi from this invariant ring R to R[u,v] defined by sending a1 -> a1+u, a2 -> a2+v and likewise b1 -> b1+u, etc. When you say you want expressions which are independent of O, I take it that means you want the subring of invariants which are fixed by phi. I would have told you it was easy to compute this, except that in a thread several weeks ago I realized I didnÕt know how to compute invariants of just this type (e.g. phi(x,y) = (2x, 3y) on F[x,y]. The thread was called Curves invariant under a rational map.) But in this case life is not so complicated because we easily find phi(Z1)=Z1 + 3u, phi(Y1) = Y1 + 3v ; so itÕs not too hard to use example, phi(Z2) = Z2 + 2u Z1 + 3 u^2 and so we compute that Z2Õ = Z2 - Z1^2/3 is phi-invariant. Likewise Z3Õ = Z3 - Z1^2*Z2 + (2/9)*Z1^3 and similar expressions for the YÕs. Therefore our invariant subring R may be written F[ Z1, Y1, Z2Õ, Y2Õ, Z3Õ, Y3Õ ] on which it seems pretty clear that the portion preserved by phi is just F[ Z2Õ, Y2Õ, Z3Õ, Y3Õ ]. Likewise we may replace our module generators by phi-invariant ones: W1Õ = W1 - Z1*Y1/3, W2Õ = W2 - (Y1*Z2/3)-2*(Z1*W1)/3 + 2*(Y1*Z1^2)/9, W3Õ = W3 - (Y2*Z1/3)-2*(Y1*W1)/3 + 2*(Z1*Y1^2)/9, so that the whole S_3 - invariant ring R is now described as: the rank-6 free module over F[ Z1, Y1, Z2Õ, Y2Õ, Z3Õ, Y3Õ] spanned by 1, W1Õ, W2Õ, W3Õ, W4Õ = (W1Õ)^2, and W5=(W2Õ)(W3Õ); the action of phi is easily given since all but two generators are phi-invariant. ItÕs obvious (isnÕt it?) that if S is any ring and phi : S[Z1,Y1] --> S[Z1,Y1,u,v] is the map which is the identity on S but sends phi(Z1) = Z1+u, phi(Y1)=Y1+v, then the subring fixed by phi is precisely S. In our case, that means (drumroll...) The invariants you are looking for are the polynomials in : Z2= (a1^2+b1^2+c1^2) - (b1*a1+c1*a1+b1*c1) Y2= (a2^2+b2^2+c2^2) - (b2*a2+c2*a2+b2*c2) Z3= 2*(a1^3+b1^3+c1^3) + 12*a1*b1*c1 - 3*(c1*b1^2+c1^2*b1+c1*a1^2+b1*a1^2+a1*c1^2+a1*b1^2) Y3= 2*(a2^3+b2^3+c2^3) + 12*a2*b2*c2 - 3*(c2*b2^2+c2^2*b2+c2*a2^2+b2*a2^2+a2*c2^2+a2*b2^2) W1= 2*(a1*a2+b1*b2+c1*c2) - (a1*b2+a1*c2+b2*c1+b1*c2+a2*c1+a2*b1) W2= 2*(a1^2*a2+b1^2*b2+c1^2*c2) + 4*(b1*a1*c2+c1*a1*b2+b1*c1*a2) -(a2*c1^2+a2*b1^2+c2*b1^2+c1^2*b2+a1^2*b2+a1^2*c2) -2*(b1*a1*b2+c1*a1*a2+c1*a1*c2+b1*c1*b2+b1*c1*c2+b1*a1*a2) W3= 2*(a1*a2^2+b1*b2^2+c1*c2^2) + 4*(c1*a2*b2+b1*a2*c2+a1*b2*c2) -( c2^2*a1+b2^2*a1+b2^2*c1+c2^2*b1+a2^2*b1+a2^2*c1) -2*( a2*b2*b1+c2*a1*a2+a2*c2*c1+c2*b2*b1+b2*c1*c2+b2*a1*a2) (I scaled the generators to get rid of fractions.) Just an example, one of the obvious invariants is the square of the area of the triangle formed by the endpoints of the vectors, a quantity which is known (Heron) to be expressible as a polynomial in the coordinates. It turns out to be (1/3)( 4 Z2 Y2 - (W1)^2 ) . Something related to the perimeter ought to be in there too. I will let you find another five or so geometric quantities which are clearly both polynomial and invariant so that there is a completely geometric description of all the invariants. (Note that we donÕt expect most of them to be _rotation_ - invariant.) You said something about taking cross products but IÕm going to ignore that because if you REALLY meant to do that, then IÕd have to repeat all these computations within the ring F[a1,a2,a3,b1...,c3]. That will be uglier. You go on to ask, >Same with 4 vectors. Any suggestions? which would force me to work out Sym(4) invariants in a polynomial ring on 12 generators; if you really want that, itÕs going to cost you... dave Here are the five generators among the Zs,Ys, and Ws: -6*W1*W2 + Z1^3*Y2-2*Z1^2*Y1*W1-Z1^2*W3+Z1*Y1^2*Z2+2*Z1*Y1*W2 -4*Z1*Z2*Y2+4*Z1*W1^2-Y1^2*Z3+3*Z2*W3+3*Y2*Z3, -6*W1*W3 + Z1^2*Y1*Y2-Z1^2*Y3-2*Z1*Y1^2*W1+2*Z1*Y1*W3+Y1^3*Z2 -Y1^2*W2-4*Y1*Z2*Y2+4*Y1*W1^2+3*Z2*Y3+3*Y2*W2, -3*W1^3 + Z1^3*Y1*Y2+Z1^3*Y3-2*Z1^2*Y1^2*W1-Z1^2*Y1*W3-3*Z1^2*Y2*W1 +Z1*Y1^3*Z2-Z1*Y1^2*W2-Z1*Y1*Z2*Y2+7*Z1*Y1*W1^2-6*Z1*Z2*Y3 +6*Z1*Y2*W2+Y1^3*Z3-3*Y1^2*Z2*W1+6*Y1*Z2*W3-6*Y1*Y2*Z3 +3*Z2*Y2*W1+9*Z3*Y3-9*W2*W3, -6*W2^2 + Z1^4*Y2-2*Z1^3*Y1*W1+Z1^2*Y1^2*Z2-4*Z1^2*Z2*Y2+3*Z1^2*W1^2 +2*Z1*Y1*Z2*W1-2*Z1*Z2*W3+2*Z1*Y2*Z3-Y1^2*Z2^2+4*Y1*Z2*W2 -4*Y1*Z3*W1+Z2^2*Y2-Z2*W1^2+6*Z3*W3, -6*W3^2 + Z1^2*Y1^2*Y2-Z1^2*Y2^2-2*Z1*Y1^3*W1+2*Z1*Y1*Y2*W1+4*Z1*Y2*W3 -4*Z1*Y3*W1+Y1^4*Z2-4*Y1^2*Z2*Y2+3*Y1^2*W1^2+2*Y1*Z2*Y3 -2*Y1*Y2*W2+Z2*Y2^2-Y2*W1^2+6*Y3*W2 === Subject: Re: Vector invariants a sentence. (Well, at least I do understand almost every term :-) Yes, I think I do need rotational invariance too. -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn === Subject: Re: Vector invariants >Yes, I think I do need rotational invariance too. Oh, then the problem is trivial! There are natural one-to-one correspondences between these categories of thing: * lists of three vectors, up to permutations, translations, & rotations * sets of three points, up to translations, & rotations * congruence classes of triangles * sets of three positive lengths (subject to triangle inequality) * solutions to cubic polynomials (of some restricted type) In other words, all the invariants of the type youÕre looking for will be naturally built from the three symmetric functions of the three distances between the ends of the vectors. You can use the squares of the distances just as well; they can be expressed as polynomials in the coordinates. In other words, try using just these three invariants (and polynomials in them): -2*a1^2+2*a1*b1-2*b1^2-2*a2^2+2*a2*b2-2*b2^2+2*a1*c1-2*c1^2 +2*a2*c2-2*c2^2+2*c1*b1+2*c2*b2 (-a2*b1+a1*b2+a2*c1-b2*c1-a1*c2+b1*c2)^2 (a1^2-2*a1*b1+b1^2+a2^2-2*a2*b2+b2^2)*(a1^2-2*a1*c1+c1^2+a2^2 -2*a2*c2+c2^2) *(c1^2-2*c1*b1+b1^2+c2^2-2*c2*b2+b2^2) > a sentence. dave === Subject: congruence classes What does it mean to have a negative congruence class mod n? For example, the congruence class, [-4], mod 16. = [12] ? Why? === Subject: Re: congruence classes > What does it mean to have a negative congruence class mod n? > For example, > the congruence class, [-4], mod 16. > = [12] ? Why? [-4]_16 = [12]_16 because -4 = 12 (mod 16) because 16 | -4-12 === Subject: Re: congruence classes Because the number in the brackets is just a representative member of a set: ..., -20, -4, 12, 28, 44, ... And it doesnÕt really matter which number we choose to be the representative, since anyone will uniquely determine the congruence class it is in. -Darren > What does it mean to have a negative congruence class mod n? > For example, > the congruence class, [-4], mod 16. > = [12] ? Why? === Subject: 6th degree equations Hi. Can one produce a solution to the general sixth-degree equation x^6 + a5 x^5 + a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0 = 0 with a function mag_n(z), defined such that mag_n(z)^n + mag_n(z) = z? (mag means magic)? I know you can express the quinticÕs solution in terms of mag_5(z), but could you express the 6th degree equationÕs solution using mag_5(z) and mag_6(z)? -- We should have a town named Alderaan someday. No, seriously. LetÕs put it on the table. === Subject: Re: 6th degree equations > Can one produce a solution to the general sixth-degree > equation > x^6 + a5 x^5 + a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0 > = 0 > with a function mag_n(z), defined such that > mag_n(z)^n + mag_n(z) = z? (mag means magic)? > I know you can express the quinticÕs solution in terms of > mag_5(z), but could you express the 6th degree > equationÕs solution using mag_5(z) and mag_6(z)? You canÕt. However, using so called Tschirnhausen transformation you can show that the solution of an arbitrary equation of degree n can be reduced to the solution of equations of degrees =< 3 and one equation of the form: X^n + b_4*X^{n-4} + b_5*X^{n-5} + ... + b_n = 0. This is the Bring-Jerard theorem. In particular, the solution of an equation of degree 6 can be obtained from the solution of equations of degree =< 3 and one equation: X^6 + b_4*X^2 + b_5*X + b_6. For details see e.g. A. Mostowski, M. Stark, Introduction to higher algebra, Pergamon Press, 1964, pp. 366-370. Pawel Gladki === Subject: Re: 6th degree equations What kind of function would work? Any known functions? > Can one produce a solution to the general sixth-degree > equation > x^6 + a5 x^5 + a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0 > = 0 > with a function mag_n(z), defined such that > mag_n(z)^n + mag_n(z) = z? (mag means magic)? > I know you can express the quinticÕs solution in terms of > mag_5(z), but could you express the 6th degree > equationÕs solution using mag_5(z) and mag_6(z)? > You canÕt. However, using so called Tschirnhausen transformation you can > show that the solution of an arbitrary equation of degree n can be > reduced to the solution of equations of degrees =< 3 and one equation of > the form: > X^n + b_4*X^{n-4} + b_5*X^{n-5} + ... + b_n = 0. > This is the Bring-Jerard theorem. In particular, the solution of an > equation of degree 6 can be obtained from the solution of equations of > degree =< 3 and one equation: > X^6 + b_4*X^2 + b_5*X + b_6. > For details see e.g. A. Mostowski, M. Stark, Introduction to higher > algebra, Pergamon Press, 1964, pp. 366-370. > Pawel Gladki === Subject: Re: 6th degree equations >Can one produce a solution to the general sixth-degree >equation >x^6 + a5 x^5 + a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0 > = 0 >with a function mag_n(z), defined such that >mag_n(z)^n + mag_n(z) = z? (mag means magic)? >I know you can express the quinticÕs solution in terms of >mag_5(z), but could you express the 6th degree >equationÕs solution using mag_5(z) and mag_6(z)? >>You canÕt. However, using so called Tschirnhausen transformation you can >>show that the solution of an arbitrary equation of degree n can be >>reduced to the solution of equations of degrees =< 3 and one equation of >>the form: >>X^n + b_4*X^{n-4} + b_5*X^{n-5} + ... + b_n = 0. >>This is the Bring-Jerard theorem. In particular, the solution of an >>equation of degree 6 can be obtained from the solution of equations of >>degree =< 3 and one equation: >>X^6 + b_4*X^2 + b_5*X + b_6. >>For details see e.g. A. Mostowski, M. Stark, Introduction to higher >>algebra, Pergamon Press, 1964, pp. 366-370. >What kind of function would work? Any known >functions? That depends... The general sextic equation can be solved using so called Kamp.8e de F.8eriet functions (which are some kind of generalization of hypergeometric functions). For details, see e.g. P. Appell, J. Kamp.8e de F.8eriet, Fonctions hyperg.8eom.8etriques et hypersph.8eriques: polynomes dÕHermite, Gauthier-Villars, 1926. For arbitrary equations of any degree it is possible to express solutions in terms of the Mellin integrals. Pawel Gladki === Subject: Re: 6th degree equations >Can one produce a solution to the general sixth-degree >equation x^6 + a5 x^5 + a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0 > = 0 with a function mag_n(z), defined such that >mag_n(z)^n + mag_n(z) = z? (mag means magic)? I know you can express the quinticÕs solution in terms of >mag_5(z), but could you express the 6th degree >equationÕs solution using mag_5(z) and mag_6(z)? >>You canÕt. However, using so called Tschirnhausen transformation you can >>show that the solution of an arbitrary equation of degree n can be >>reduced to the solution of equations of degrees =< 3 and one equation of >>the form: >>X^n + b_4*X^{n-4} + b_5*X^{n-5} + ... + b_n = 0. >>This is the Bring-Jerard theorem. In particular, the solution of an >>equation of degree 6 can be obtained from the solution of equations of >>degree =< 3 and one equation: >>X^6 + b_4*X^2 + b_5*X + b_6. >>For details see e.g. A. Mostowski, M. Stark, Introduction to higher >>algebra, Pergamon Press, 1964, pp. 366-370. > >What kind of function would work? Any known > >functions? > That depends... The general sextic equation can be solved using so > called Kamp.8e de F.8eriet functions (which are some kind of generalization > of hypergeometric functions). For details, see e.g. P. Appell, J. Kamp.8e > de F.8eriet, Fonctions hyperg.8eom.8etriques et hypersph.8eriques: polynomes > dÕHermite, Gauthier-Villars, 1926. Is there an English translation? > For arbitrary equations of any degree it is possible to express > solutions in terms of the Mellin integrals. > Pawel Gladki === Subject: Re: 6th degree equations >>The general sextic equation can be solved using so >>called Kamp.8e de F.8eriet functions (which are some kind of generalization >>of hypergeometric functions). For details, see e.g. P. Appell, J. Kamp.8e >>de F.8eriet, Fonctions hyperg.8eom.8etriques et hypersph.8eriques: polynomes >>dÕHermite, Gauthier-Villars, 1926. > Is there an English translation? Not to my knowledge... Pawel Gladki === Subject: Re: 6th degree equations > That depends... The general sextic equation can be solved using so > called Kamp.8e de F.8eriet functions (which are some kind of generalization > of hypergeometric functions). For details, see e.g. P. Appell, J. Kamp.8e > de F.8eriet, Fonctions hyperg.8eom.8etriques et hypersph.8eriques: polynomes > dÕHermite, Gauthier-Villars, 1926. > For arbitrary equations of any degree it is possible to express > solutions in terms of the Mellin integrals. Interesting! Do you have any reference? Wilbert === Subject: Re: 6th degree equations >>For arbitrary equations of any degree it is possible to express >>solutions in terms of the Mellin integrals. > Interesting! Do you have any reference? H. Mellin, Ein allgemeiner Satz .9fber algebraische Gleichungen, Ann. Soc. Fennicae, (A) 7, Nr. 8, 44 S. (1915) Pawel Gladki === Subject: Re: 6th degree equations > What kind of function would work? Any known > functions? I believe that the closed form solutions of any degree polynomial can be expressed in terms of Theta functions. > Can one produce a solution to the general sixth-degree > equation x^6 + a5 x^5 + a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0 > = 0 with a function mag_n(z), defined such that > mag_n(z)^n + mag_n(z) = z? (mag means magic)? I know you can express the quinticÕs solution in terms of > mag_5(z), but could you express the 6th degree > equationÕs solution using mag_5(z) and mag_6(z)? > You canÕt. However, using so called Tschirnhausen transformation you can > show that the solution of an arbitrary equation of degree n can be > reduced to the solution of equations of degrees =< 3 and one equation of > the form: > X^n + b_4*X^{n-4} + b_5*X^{n-5} + ... + b_n = 0. > This is the Bring-Jerard theorem. In particular, the solution of an > equation of degree 6 can be obtained from the solution of equations of > degree =< 3 and one equation: > X^6 + b_4*X^2 + b_5*X + b_6. > For details see e.g. A. Mostowski, M. Stark, Introduction to higher > algebra, Pergamon Press, 1964, pp. 366-370. > Pawel Gladki === Subject: Re: 6th degree equations |I believe that the closed form solutions of any degree polynomial can |be expressed in terms of Theta functions. See, for example, R. Bruce KingÕs _Beyond the Quintic Equation_. Keith Ramsay === Subject: Re: 6th degree equations ETAsAhQlwrPvgjmWG5EhtaAyCc7pYAfvvQIUBfkPEdY3MpJE3P9+ ximEzcRKK8o= There is not likey to be a simple combination. The transformation which, in quintic equations, leads to x^5 + ax + b = 0 gives x^6 + ax^2 + bx + c = 0 in degree 6. The magic functions you describe canÕt handle that x^2 term. Should b in the transformed 6th-degree equation happen to be zero, you get a solution involving not mag_5 or mag_6 but mag_3, where mag_3 can be rendered into ordinary radicals or trigonometric functions. --OL === Subject: Re: 6th degree equations What if you used a quintic substitution into the 6th degree equation? Yes I know it would be so damn complicated it is pretty much useless, but would it do anything? > There is not likey to be a simple combination. The transformation > which, in quintic equations, leads to x^5 + ax + b = 0 gives x^6 + ax^2 > + bx + c = 0 in degree 6. The magic functions you describe canÕt > handle that x^2 term. > Should b in the transformed 6th-degree equation happen to be zero, you > get a solution involving not mag_5 or mag_6 but mag_3, where mag_3 can > be rendered into ordinary radicals or trigonometric functions. > --OL === Subject: Re: 6th degree equations ETAtAhUAhcAIkoEbfScAdoDfO37aZthjrTICFGwHUhIFlE369E4mjQZ2o37/ 8kyH No idea, really. Yup youÕre in complicated territory. Once I get above degree 3 I go for numerical solutions. You basically need numerical methods to evaluate radicals and magic funcitons anyway. --OL === Subject: Re: 6th degree equations > No idea, really. Yup youÕre in complicated territory. Once I get above > degree 3 I go for numerical solutions. You basically need numerical > methods to evaluate radicals and magic funcitons anyway. > --OL Yep. Actually, I think you can solve any nth degree equation with radicals and magic functions, but the length of the formula increases exponentially (or maybe even superexponentially, not sure though). Numeric methods are how you check the answer anyway, and since the solutions become so horribly complicated, you might as well just ditch the exact-solve approach. Anyway, I was just curious. === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? My bad. I called it by the wrond name but it is the right function. Dwayne === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >My bad. I called it by the wrond name but it is the right function. So C.L. Phillips, J.M. Parr, Signals, Systems & Transforms, 2nd ed., Prentice Hall, Copyright 1999, p. 46-50, 210-211 really defines the Laplace Transform as an integral from -infinity to infinity, instead of the more usual 0 to infinity? >Dwayne ************************ David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? Yes. Dwayne >My bad. I called it by the wrond name but it is the right function. > So > C.L. Phillips, J.M. Parr, Signals, Systems & Transforms, 2nd ed., > Prentice > Hall, Copyright 1999, p. 46-50, 210-211 > really defines the Laplace Transform as an integral from -infinity > to infinity, instead of the more usual 0 to infinity? >Dwayne > ************************ > David C. Ullrich === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? The limits of the bilateral LT are from -oo to +oo. The limits of the unilateral LT are from 0 to +oo. Both are usual. I thought you _BOASTED_ of being a mathematician of 20 yearsÕ standing? > C.L. Phillips, J.M. Parr, Signals, Systems & Transforms, 2nd ed., > Prentice > Hall, Copyright 1999, p. 46-50, 210-211 > really defines the Laplace Transform as an integral from -infinity > to infinity, instead of the more usual 0 to infinity? === Subject: Re: Derivation Of The Spectrum Due To f(t).d(t - T)????? >The limits of the bilateral LT are from -oo to +oo. >The limits of the unilateral LT are from 0 to +oo. >Both are usual. >I thought you _BOASTED_ of being a mathematician >of 20 yearsÕ standing? I donÕt think I BOASTED of that, but yes, I mentioned it. I have enough experience to know that something IÕve never seen before is not necessarily wrong. >> C.L. Phillips, J.M. Parr, Signals, Systems & Transforms, 2nd ed., >> Prentice >> Hall, Copyright 1999, p. 46-50, 210-211 >> really defines the Laplace Transform as an integral from -infinity >> to infinity, instead of the more usual 0 to infinity? ************************ David C. Ullrich === Subject: Re: JSH: Simple proof > In sci.math, o[CapitalYAcute]in > : >> There are no variables in mathematics. >> Ummmm. What? > There are sets with more then one element, and symbols which denote > arbitrary elements of such sets. But nothing varies. Honest. Believe > me. >> I believe that is true. But that is not the same as saying there are no >> variables in mathematics. > ItÕs an unfortunate term, in some respects, but when one has > an equation such as > x^2 - 3*x + 2 = 0 > and asks what is/are the values of x to make this true?, then > one can solve this equation (in this case, itÕs x = +1 and x = -3). > The name x is commonly called a variable, but in this case, > itÕs more like a mysterious pair of constants (since the > equation has more than one root). I learned that in this case, x is an indeterminite. === Subject: Re: JSH: Simple proof > LetÕs make this as simple as possible. > You consider a function > r(x) = s(x) + c, > where you can assume s(0) = 0. > That is, c is the constant term of r(x). This is from > your own definition of constant term. > Then you divide r(x) by another function w(x). You know > that w(0) = c, but you also know that w(x) is not a constant > function. The quotient is > t(x) = s(x)/w(x) + c/w(x). > > Multiple choice test: > The constant term of t(x) is > (a) c/w(0) > (b) c/w(x) > Please specify (a) or (b). If both answers are the same, > please explain why in detail. > Nora B. > In case you havenÕt noticed, JSH never responds to questions. Your > efforts, though praiseworthy, are misplaced. Gib, Actually he was replying to posts on this topic until quite recently. He stopped replying to MY posts a few days ago and then began a familiar pattern of starting a number of somewhat-related new threads, evidently to distract people from the fact that he was not answering substantive objections. The same thing happened back in January-February of this year when he encountered an example put forward by Rick Decker, a quadratic example which showed very clearly that the method he used in Advanced Polynomial Factorization did not work. This example and some related ones clearly bothered him very deeply. He would start a new thread, very carefully documenting that Decker was a prof at Hamilton College and describing DeckerÕs example, but refusing to give the punch line, which was a factorization of a kind that he had said was impossible. He may have started 15-20 threads on that topic, describing DeckerÕs polynomial but never quite describing what it implied and thus never really explaining why he was trying to respond to it. It was bizarre. Clearly he understood on some level that his method was wrong and he couldnÕt see how to fix it. He knew it spelled trouble for his overall argument. He was desperate and beginning to panic. Finally he simply abandoned it all and went silent for several months: pretty much until sometime in June, when the Southwest Journal of Pure and Applied Mathematics (a strictly electronic journal) through what was apparently a gross editorial error, the editor and provided counterexamples. The editor instantly caved in and removed the paper from the journalÕs website with no public explanation. This of course prompted howls of righteous indignation from Harris (with some reason), and he began defending the paperÕs methodology all over again. He did answer some objections. He submitted the paper to a Hungarian journal (which rejected it). He again for a while stopped posting to sci.math, but he continued posting arguments putatively aimed at undergrad math majors in altern.math.undergrad. This did not attract much attention. He resumed posting to sci.math with variants of his constants-are- constant argument, and again got involved with me and others who pointed out errors in that argument: including, for example, that he does not follow his own definition of constant term. Yesterday he started a related thread called ŌFinal examÕ. He asserted that sci.math readers were stupid and of course implied that us regular posters are dishonest. There were several replies, at least two of which pointed out that the math he claimed in his post was wrong. Whether he read these or not I donÕt know. Today he noted that he was wrong (with no real explanation why) and said he had been in a bad mood. And, I believe, started some new threads. Nora B. > Gib === Subject: Re: JSH: Simple proof In case you havenÕt noticed, JSH never responds to questions. Your >>efforts, though praiseworthy, are misplaced. > Gib, > Actually he was replying to posts on this topic until > quite recently. He stopped replying to MY posts a few days > ago and then began a familiar pattern of starting a number > of somewhat-related new threads, evidently to distract people > from the fact that he was not answering substantive objections. > The same thing happened back in January-February of this > year when he encountered an example put forward by Rick Decker, > a quadratic example which showed very clearly that the method > he used in Advanced Polynomial Factorization did not work. > This example and some related ones clearly bothered him > very deeply. He would start a new thread, very carefully documenting > that Decker was a prof at Hamilton College and describing DeckerÕs > example, but refusing to give the punch line, which was a factorization > of a kind that he had said was impossible. Which, IÕm sure you noticed, I recently more or less reposted in JamesÕs Point of logic thread (November 11). I didnÕt expect a response, since I had to post the original (several months ago) at least twice to get him to acknowledge it. Basically, I reposted it now for the benefits of those who are new to the Harris Magnum Opus. Rick === Subject: Re: JSH: Simple proof > Why donÕt you simply and brießy state the >>problematic<< property > of the ring of algebraic integers and/or the statement that you > want to prove. The ring of algebraic integers is determined by roots of *monic* polynomials with integer coefficients. It is possible to show with basic algebra that there are numbers which are properly units but because their multiplicative inverse is not the root of some monic polynomial with integer coefficients they are not units in the ring of algebraic integers. To see how it works consider that in rationals you can have (3x + 1)(x + 1) = 3x^2 + 4x + 1 where, of course, one of the roots is a unit in the ring of algebraic integers. But now consider (3x + u_1)(x + u_2) = 3x^2 + kx + 1, where u_1 u_2 =1, and k is an integer. You find that if the uÕs are irrational, then u_1, while an algebraic integer is not a unit in the ring, while u_2 cannot then even be an algebraic integer. My research shows though that both u_1 and u_2 can be units in a ring where -1 and 1 are the only rational units, and no non-unit member of the ring is a factor of any two integers that are coprime in the ring of integers. You see, I abstracted out two key properties of rings like the ring of integers and the ring of algebraic integers. > Using standard mathematical terms and notions (for example from > commutative algebra) this should be possible in a few lines instead > of making a long story. ItÕs not complicated. Basically you canÕt just rely on whether or not some number is in the ring of algebraic integers when considering factors of roots of a polynomial. The mathematics is mostly REALLY simple. > Why do we have to discuss things like >>what is a polynomial?<< IÕm not discussing that, other posters made a big deal out of it. > here? The notion of a polynomial is defined since a long time and > can be found in every introductory book on algebra. So? > If we consistently use the common definitions of mathematical > objects like polynomials we should rather quickly be able to > clarify the situation and avoid all the frustration that frequently > seems to culminate in personal attacks. IÕve seen posters come and go, and every once in a while thereÕs a poster like you who claims to care about working things out. When it turns out that IÕm right, you go over to the other side, and either run away, or turn to bizarre behavior. Psychologists call it cognitive dissonance. Basically, deep down you believe that I must be wrong, so your post is not really in good faith. But simply *saying* certain things that indicate objectivity or willingness to be objective sets you up psychologically. That is, you feel a need to be consistent with what you said. But later, when you run into the rigid mathematics, which goes against what you wish to believe, you basically kind of break. Your mind breaks, and you run away or behave weird. IÕve seen it lots of times. Do yourself a favor, and just walk away now. James Harris === Subject: Re: JSH: Simple proof >>Why donÕt you simply and brießy state the >>problematic<< property >>of the ring of algebraic integers and/or the statement that you >>want to prove. > The ring of algebraic integers is determined by roots of *monic* > polynomials with integer coefficients. ThatÕs the definition. It is not a property of a ring. What is the problem with the ring? What prevents it from being everything thatÕs been claimed? > It is possible to show with basic algebra that there are numbers which > are properly units but because their multiplicative inverse is not the > root of some monic polynomial with integer coefficients they are not > units in the ring of algebraic integers. No, it is not possible to show this, unless the term properly a unit is defined. You have not done this. An element of a ring is or is not a unit. If it *is* a unit, then it has a multiplicative inverse in the ring. If it isnÕt a unit, then its multiplicative inverse fails to be in the ring. ThereÕs no ambiguity here. How about providing a definition for your bogus terminology? > To see how it works consider that in rationals you can have > (3x + 1)(x + 1) = 3x^2 + 4x + 1 > where, of course, one of the roots is a unit in the ring of algebraic > integers. And the polynomial is of course reducible over the integers. > But now consider > (3x + u_1)(x + u_2) = 3x^2 + kx + 1, where u_1 u_2 =1, and k is an > integer. > You find that if the uÕs are irrational, then u_1, while an algebraic > integer is not a unit in the ring, while u_2 cannot then even be an > algebraic integer. So what? HereÕs what you get from the linear term of the product: u_1 + 3 u_2 = k Letting u = u_2 (since we already know that u_1 is an algebraic integer): 1/u + 3 u = k 3 u^2 - k u + 1 = 0 We solve for u, via the quadratic formula u = (k +/- sqrt(k^2 - 12))/6 Now, letÕs look at a specific example: k = 5 25 - 12 = 13 u = (5 +/- sqrt(13))/6 IÕll let the root with + be u, and the root with - be ubar. Note that you canÕt have *both* solutions in your ring: u ubar = (25 - 13)/36 = 12/36 = 1/3 Considering your recent rant about how you canÕt get rid of the evil ambiguity of root extractions, how do you determine *which* of these roots is in your ring? Note that the fields Q(u) and Q(ubar) are isomorphic: any arithmetic statement you make in Q(u) can be translated automatically (by mechanically replacing each occurrence of u with ubar) into an equivalent arithmetic statement in Q(ubar), so that if one is true then so is the other one. That means that you cannot use algebraic means (e.g., any algebraic formula involving the integers and the desired root) to determine which root to select. If memory serves me correctly, the issue becomes even more problematic once you start trying to throw in roots of more and more non-monic polynomials. > My research shows though that both u_1 and u_2 can be units in a ring > where -1 and 1 are the only rational units, and no non-unit member of > the ring is a factor of any two integers that are coprime in the ring > of integers. Which u_2 are you going to choose? What IÕve called u, or what IÕve called ubar? You canÕt have both, and you canÕt tell them apart using algebra. You have not constructed such a ring. You have not even shown how to establish whether a given complex number is an element of the ring (such as: which of u,ubar is in the ring). Many here have agreed that such a ring can exist, but it is not unique. Note the above example. You may in fact be able to admit one of every candidate quadratic, one or two from every candidate cubic, and so forth, but you *cannot* admit all roots of *any* non-monic polynomial, without admitting a non-integral rational number. Once that happens, youÕll get invertible integers other than +/- 1. BTW, if you have two coprime elements, u and v, of a ring R, then they remain coprime in any ring RÕ that contains R. It is also not difficult to show that any common factor of coprime elements of a ring must be a unit. > You see, I abstracted out two key properties of rings like the ring of > integers and the ring of algebraic integers. Your second property: no non-unit member of the ring is a factor of any two integers that are coprime in the ring of integers. is a corollary of the remarks I made above. This has been known for far longer than I have been alive; it was most likely obvious to Dedekind. It has nothing to do with key properties like the ring of integers. It holds for arbitrary commutative rings. Hooray for you for having abstracted it. >>Using standard mathematical terms and notions (for example from >>commutative algebra) this should be possible in a few lines instead >>of making a long story. > ItÕs not complicated. Basically you canÕt just rely on whether or not > some number is in the ring of algebraic integers when considering > factors of roots of a polynomial. ThatÕs what you say. However, every specific example youÕve proposed that is supposed to highlight the ßaws of the ring of algebraic integers (such as your polynomial factorization in your ill-fated paper Advanced Polynomial Factorization) has been proven to have been incorrect. > The mathematics is mostly REALLY simple. >>Why do we have to discuss things like >>what is a polynomial?<< > IÕm not discussing that, other posters made a big deal out of it. >>here? The notion of a polynomial is defined since a long time and >>can be found in every introductory book on algebra. > So? >>If we consistently use the common definitions of mathematical >>objects like polynomials we should rather quickly be able to >>clarify the situation and avoid all the frustration that frequently >>seems to culminate in personal attacks. > IÕve seen posters come and go, and every once in a while thereÕs a > poster like you who claims to care about working things out. > When it turns out that IÕm right, you go over to the other side, and > either run away, or turn to bizarre behavior. It has not turned out youÕre right. What has invariably happened is that you have turned to some form of antisocial behavior, such as what you are doing right now. > Psychologists call it cognitive dissonance. > Basically, deep down you believe that I must be wrong, so your post is > not really in good faith. But simply *saying* certain things that > indicate objectivity or willingness to be objective sets you up > psychologically. This must be part of your morning Stuart Smalley chant. > That is, you feel a need to be consistent with what you said. > But later, when you run into the rigid mathematics, which goes against > what you wish to believe, you basically kind of break. Your mind > breaks, and you run away or behave weird. > IÕve seen it lots of times. Do yourself a favor, and just walk away > now. I think youÕre trying to censor the poster. Shame on you. > James Harris Dale. === Subject: Re: JSH: Simple proof !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> It is possible to show with basic algebra that there are numbers >> which are properly units but because their multiplicative inverse >> is not the root of some monic polynomial with integer coefficients >> they are not units in the ring of algebraic integers. > No, it is not possible to show this, unless the term > properly a unit > is defined. > You have not done this. An element of a ring is or is not > a unit. If it *is* a unit, then it has a multiplicative > inverse in the ring. If it isnÕt a unit, then its multiplicative > inverse fails to be in the ring. > ThereÕs no ambiguity here. > How about providing a definition for your bogus terminology? How about an example? When considering the ring of ordinary integers, 1/2 is properly a unit(TM), while its inverse (which is 2) is not a unit in the ordinary integers. And this is a problem with the ordinary integers. Or something. Why youÕd be considering 1/2 in the first place when talking about the ring of integers, is a different question. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Simple proof Discussion, linux) >> Why donÕt you simply and brießy state the >>problematic<< property >> of the ring of algebraic integers and/or the statement that you >> want to prove. > The ring of algebraic integers is determined by roots of *monic* > polynomials with integer coefficients. > It is possible to show with basic algebra that there are numbers which > are properly units but because their multiplicative inverse is not the > root of some monic polynomial with integer coefficients they are not > units in the ring of algebraic integers. What is the meaning of properly units? Can you give an example of a non-zero complex number which is not properly a unit? (Note: Answering the second question does not make the first question irrelevant.) -- Come on people!!! The US just blew up a lot of people in Iraq, donÕt you realize that a person with my exposure might just end up dead, by mysterious circumstances? --James Harris, on the dangers of proving FermatÕs last theorem === Subject: Re: JSH: Simple proof !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Why donÕt you simply and brießy state the >>problematic<< property >> of the ring of algebraic integers and/or the statement that you >> want to prove. > The ring of algebraic integers is determined by roots of *monic* > polynomials with integer coefficients. Well, thatÕs the definition. ThatÕs what makes them interesting in the first place. If you donÕt like the particular class of numbers defined by this property, you are free to define a different class of numbers defined by different properties. You just canÕt expect it to obey the same laws then. > It is possible to show with basic algebra that there are numbers > which are properly units This is nonsense. Units are always units in a specified ring. > but because their multiplicative inverse is not the root of some > monic polynomial with integer coefficients they are not units in the > ring of algebraic integers. You have no clue what units means. In analogy to your argument, I could say that the ring of plain integers has a problem because 1/2 is a proper unit, but its multiplicative inverse 2 is not a unit in the ring of integers. This is complete folly, because _of course_ 1/2 is not even subject to discussion when I am talking about integers. > But later, when you run into the rigid mathematics, which goes > against what you wish to believe, you basically kind of break. Your > mind breaks, and you run away or behave weird. Good self-description. > IÕve seen it lots of times. Do yourself a favor, and just walk away > now. Good self-advice. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Simple proof >>Why donÕt you simply and brießy state the >>problematic<< property >>of the ring of algebraic integers and/or the statement that you >>want to prove. > The ring of algebraic integers is determined by roots of *monic* > polynomials with integer coefficients. > It is possible to show with basic algebra that there are numbers which > are properly units but because their multiplicative inverse is not the > root of some monic polynomial with integer coefficients they are not > units in the ring of algebraic integers. Has it occurred to you that the problem might be with your undefined notion of properly units? What you just said is that something can be properly a unit, but not actually a unit. To me it sounds like non-sense. > To see how it works consider that in rationals you can have > (3x + 1)(x + 1) = 3x^2 + 4x + 1 > where, of course, one of the roots is a unit in the ring of algebraic > integers. Which is obvious from the fact that x+1 is a monic polynomial. > But now consider > (3x + u_1)(x + u_2) = 3x^2 + kx + 1, where u_1 u_2 =1, and k is an > integer. > You find that if the uÕs are irrational, then u_1, while an algebraic > integer is not a unit in the ring, while u_2 cannot then even be an > algebraic integer. Not surprising. > My research shows though that both u_1 and u_2 can be units in a ring > where -1 and 1 are the only rational units, and no non-unit member of > the ring is a factor of any two integers that are coprime in the ring > of integers. True. But that ring is likely to vary based on the values of the uÕs. You are probably talking about a set of rings, rather than a single ring. > You see, I abstracted out two key properties of rings like the ring of > integers and the ring of algebraic integers. Now, is there a maximal ring in the set? No. So it seems unlikely you have anything ground-breaking, nor do you have a ßaw in the algebraic integers. What made you think you did? >>Using standard mathematical terms and notions (for example from >>commutative algebra) this should be possible in a few lines instead >>of making a long story. > ItÕs not complicated. Basically you canÕt just rely on whether or not > some number is in the ring of algebraic integers when considering > factors of roots of a polynomial. > The mathematics is mostly REALLY simple. HereÕs the problem as I see it: You would rather force something to be a unit, thus going to a ring extension, rather than discuss the properties of a number in the algebraic integers. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Simple proof > It is possible to show with basic algebra that there are numbers which > are properly units but because their multiplicative inverse is not the > root of some monic polynomial with integer coefficients they are not > units in the ring of algebraic integers. What does properly units mean? === Subject: Re: JSH: Simple proof >>It is possible to show with basic algebra that there are numbers which >>are properly units but because their multiplicative inverse is not the >>root of some monic polynomial with integer coefficients they are not >>units in the ring of algebraic integers. > What does properly units mean? It appears to mean things I want to be units, because they make my argument work. The fact that there are such things that are not algebraic integers leads James to conclude that there is some fundamental problem with the algebraic integers. ItÕs sort of like complaining that thereÕs no solution to 3x - 1 = 0 in the ring of integers and then going on to conclude that thereÕs a fundamental problem with the integers. Rick === Subject: Re: JSH: Simple proof !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >It is possible to show with basic algebra that there are numbers which >are properly units but because their multiplicative inverse is not the >root of some monic polynomial with integer coefficients they are not >units in the ring of algebraic integers. >> What does properly units mean? > It appears to mean things I want to be units, because they make > my argument work. The fact that there are such things that are > not algebraic integers leads James to conclude that there is some > fundamental problem with the algebraic integers. > ItÕs sort of like complaining that thereÕs no solution to 3x - 1 = 0 > in the ring of integers and then going on to conclude that thereÕs a > fundamental problem with the integers. Oh, but there is. ThatÕs why we have rationals in the first place. In a similar vein, we have algebraic numbers as a superset of algebraic integers. What James does not realize is that the shortcomings and their structures are what actually makes study of those fields (uh, pun gone bad) interesting in the first place. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Simple proof > My proof that there is a problem with the accepted understanding of > the ring of algebraic integers hinges on numbers like 7 and 22 being > constant, and NOT functions of x, or any other variable, and on the > the distributive property. > ThatÕs it. > The outline of the proof is that I take a polynomial, factor it into > non-polynomial factors, get the constant terms of those factors, and > note that dividing the polynomial by a constant multiple, divides the > factors of the constant term in a very specific way. So for example: x^2 + 5x + 6 => (x+2)(x+3) YouÕre claiming x+2 and x+3 are NOT polynomials? === Subject: Re: Pseudo random 2D white noise algorithm? The polar randon algoritm was published in Byte in the 80Õs. by Alain Latour (1986 august) Pseudocode algoithm is: Repeat 1) get 2 independent random variables v1,v2, uniformly distributed between -1,1. Mathematica: v1=Random[Real,{-1,1}] v2=Random[Real,{-1,1}] True basic: v1=2*Rnd-1 v2=2*Rnd-1 2) Evaluate: S=v1^2+v2^2 Until S<1 Evaluate S=Sqrt[-2*Log[S]/S] Set x1=v1*S x2=v2*S It works pretty good. Another methosd is the random average method: x=(Sum[U(i),{i,1,n}]-n/2)/Sqrt[n/12] Where U(i)=rnd-->0<=U(i)<=1 and n is 12 or greater. >> IÕm trying to design a function I can use to generate 2D white noise. >> This will be used in a graphics application and will take a pair of >> integers (x, y) and map it to a pseudorandom real value on [0 1]. >> However, this function must be consistant and always map (x, y) to the >> same pseudo random value. It must also be able to handle very large >> values of x and y, so simply pregenerating an array of random values >> will require too much memory to be useful. >> IÕve tried to design my own function, but not had much luck. Would >> anyone be able to help me out? >> Mark McKay >> -- >> http://wwww.kitfox.com > Assuming youÕre writing in some C derivative, have you tried just: > double my_rand (int x, int y) > srand (x + y*X_MAX); > return double (rand()) / RAND_MAX; > Although it seems like most built-in rand() functions display certain > patterns in 2D arrays. > -- > Kevin -- 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: sin(x^x) Hi. Is there a formula that describes the density of the graph of sin(x^x) for real x>0 as x increases without bound? -- We should have a town named Alderaan someday. No, seriously. LetÕs put it on the table. === Subject: Re: sin(x^x) > Hi. > Is there a formula that describes the density of > the graph of sin(x^x) for real x>0 as x increases > without bound? [...] Topologically, it is nowhere dense. In terms of two-dimensional area (say, Lebesgue measure), it has measure 0. Fractal-wise, it is of Hausdorff dimension 1. (And it is a real-analytic submanifold of the plane, perhaps with some care taken about the missing left endpoint). Any other criteria? If it means how close the subsequent roots (or the increasing and decreasing segments) are, sit down and do the arithmetic, using the Lambert W-function. For your convenience, the explicit formula for y^y=x (x sufficiently large) is y = ln(x) / lambertW(ln(x)) === Subject: Re: sin(x^x) > Hi. > Is there a formula that describes the density of > the graph of sin(x^x) for real x>0 as x increases > without bound? > [...] > Topologically, it is nowhere dense. > In terms of two-dimensional area (say, > Lebesgue measure), it has measure 0. > Fractal-wise, it is of Hausdorff dimension 1. > (And it is a real-analytic submanifold of the plane, > perhaps with some care taken about the missing left endpoint). > Any other criteria? > If it means how close the subsequent roots (or the increasing and > decreasing segments) are, sit down and do the arithmetic, using the > Lambert W-function. > For your convenience, the explicit formula for y^y=x (x sufficiently > large) is > y = ln(x) / lambertW(ln(x)) By density I meant distance between the peaks of the waves as x -> inf. Obviously, the distance -> 0 (and so the density -> inf) as x -> inf but what is the equation that describes that density? I would guess itÕs something like this: sin(u) is equal to 1 at u = pi/2, so sin(x^x) has a peak whenever x^x = npi/2, with n a positive integer. Then the nth peak is at x = ln(npi/2)/W(ln(npi/2)) So the ŌdensityÕ function IÕm looking for would then be D(n) = 1/(ln((n+1)pi/2)W(ln((n+1)pi/2)) - ln(npi/2)/W(ln(npi/2))). Would that work? === Subject: Re: sin(x^x) the graph of sin(x^x) for real x>0 as x increases > without bound? > [...] > Topologically, it is nowhere dense. > In terms of two-dimensional area (say, > Lebesgue measure), it has measure 0. > Fractal-wise, it is of Hausdorff dimension 1. > (And it is a real-analytic submanifold of the plane, > perhaps with some care taken about the missing left endpoint). > Any other criteria? > If it means how close the subsequent roots (or the increasing and > decreasing segments) are, sit down and do the arithmetic, using the > Lambert W-function. > For your convenience, the explicit formula for y^y=x (x sufficiently > large) is > y = ln(x) / lambertW(ln(x)) > By density I meant distance between the peaks of the waves > as x -> inf. > Obviously, the distance -> 0 (and so the density -> inf) as > x -> inf but what is the equation that describes that density? > I would guess itÕs something like this: > sin(u) is equal to 1 at u = pi/2, so sin(x^x) has a peak > whenever x^x = npi/2, with n a positive integer. Caution: you have listed not only local maxima, but also roots and local minima. Sort them out: u = (4*k+1)*pi/2 (k integer) gives local maxima of sin(u). You can complete the list. [nothing added by me below] > Then the nth peak is at x = ln(npi/2)/W(ln(npi/2)) > So the ŌdensityÕ function IÕm looking for would then > be > D(n) = 1/(ln((n+1)pi/2)W(ln((n+1)pi/2)) - ln(npi/2)/W(ln(npi/2))). > Would that work? === Subject: Re: sin(x^x) Hi. Is there a formula that describes the density of > the graph of sin(x^x) for real x>0 as x increases > without bound? > [...] Topologically, it is nowhere dense. In terms of two-dimensional area (say, > Lebesgue measure), it has measure 0. Fractal-wise, it is of Hausdorff dimension 1. (And it is a real-analytic submanifold of the plane, > perhaps with some care taken about the missing left endpoint). Any other criteria? If it means how close the subsequent roots (or the increasing and > decreasing segments) are, sit down and do the arithmetic, using the > Lambert W-function. For your convenience, the explicit formula for y^y=x (x sufficiently > large) is y = ln(x) / lambertW(ln(x)) By density I meant distance between the peaks of the waves > as x -> inf. > Obviously, the distance -> 0 (and so the density -> inf) as > x -> inf but what is the equation that describes that density? > I would guess itÕs something like this: > sin(u) is equal to 1 at u = pi/2, so sin(x^x) has a peak > whenever x^x = npi/2, with n a positive integer. > Caution: you have listed not only local maxima, but also roots and > local minima. Sort them out: Ohh... because of n = 2, then you get sin(pi) = 0, then n = 3 -> sin(3pi/2) = -1, then n = 4 -> sin(2pi) = 0, n = 5 -> sin(5pi/2) = 1, etc. > u = (4*k+1)*pi/2 (k integer) gives local maxima of sin(u). > You can complete the list. So it should be D(n) = 1/(ln((4n+2)pi/2)W(ln((4n+2)pi/2)) - ln((4n+1)pi/2)/W(ln((4n+1)pi/2))). > [nothing added by me below] > Then the nth peak is at x = ln(npi/2)/W(ln(npi/2)) > So the ŌdensityÕ function IÕm looking for would then > be > D(n) = 1/(ln((n+1)pi/2)W(ln((n+1)pi/2)) - ln(npi/2)/W(ln(npi/2))). > Would that work? === Subject: Re: sin(x^x) Hi. Is there a formula that describes the density of > the graph of sin(x^x) for real x>0 as x increases > without bound? > [...] Topologically, it is nowhere dense. In terms of two-dimensional area (say, > Lebesgue measure), it has measure 0. Fractal-wise, it is of Hausdorff dimension 1. (And it is a real-analytic submanifold of the plane, > perhaps with some care taken about the missing left endpoint). Any other criteria? If it means how close the subsequent roots (or the increasing and > decreasing segments) are, sit down and do the arithmetic, using the > Lambert W-function. For your convenience, the explicit formula for y^y=x (x sufficiently > large) is y = ln(x) / lambertW(ln(x)) > By density I meant distance between the peaks of the waves > as x -> inf. Obviously, the distance -> 0 (and so the density -> inf) as > x -> inf but what is the equation that describes that density? I would guess itÕs something like this: sin(u) is equal to 1 at u = pi/2, so sin(x^x) has a peak > whenever x^x = npi/2, with n a positive integer. Caution: you have listed not only local maxima, but also roots and > local minima. Sort them out: > Ohh... because of n = 2, then you get sin(pi) = 0, then n = 3 -> sin(3pi/2) > = -1, > then n = 4 -> sin(2pi) = 0, n = 5 -> sin(5pi/2) = 1, etc. > u = (4*k+1)*pi/2 (k integer) gives local maxima of sin(u). > You can complete the list. > So it should be > D(n) = 1/(ln((4n+2)pi/2)W(ln((4n+2)pi/2)) - > ln((4n+1)pi/2)/W(ln((4n+1)pi/2))). Replace (4n+2) by (4n+5), because you need to use 4(n+1)+1, rather than (4n+1)+1. > [nothing added by me below] > Then the nth peak is at x = ln(npi/2)/W(ln(npi/2)) So the ŌdensityÕ function IÕm looking for would then > be D(n) = 1/(ln((n+1)pi/2)W(ln((n+1)pi/2)) - ln(npi/2)/W(ln(npi/2))). Would that work? > === Subject: Re: sin(x^x) >Is there a formula that describes the density of >the graph of sin(x^x) for real x>0 as x increases >without bound? As x increases, the period decreases to 2*pi/x^x and the amplitude stays the same. Is that what youÕre looking for? --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: sin(x^x) >Is there a formula that describes the density of >the graph of sin(x^x) for real x>0 as x increases >without bound? > As x increases, the period decreases to 2*pi/x^x and the amplitude stays > the same. Is that what youÕre looking for? > --Keith Lewis klewis {at} mitre.org > The above may not (yet) represent the opinions of my employer. === Subject: Re: logic is innate? > >>The term Paradox of Material Implication refers to the fact that if >>P is false, then P -> Q for any value of Q. This is unintuitive, >>since many natural English if... then... statements are parsed as >>biconditionals. I think in these cases itÕs usually more accurate to say that the >sentences are understood as relevant implications. TheyÕre not >really if and only if; theyÕre just not truth functional at all. >> Sure, but thatÕs on the natural language side. Suppose a person, >> during normal discourse, tells his son: If you eat your vegetables, >> IÕll give you some ice cream. Conversationally implied is the >> sentence If you donÕt eat your vegetables, I wonÕt give you ice >> cream. > No, I donÕt really think it is. I think thatÕs an oversimplification. > WhatÕs understood conversationally is that thereÕs a relationship > between the *meaning* of the hypothesis and the conclusion, not > just between their *truth*values*. That leads to a reasonable > inference that, if the father were planning to fork over (spoon over?) > the ice cream anyway, he wouldnÕt have put it as he did. But > if he caves in at the end and allows ice cream with no vegetables, > itÕs not generally considered that heÕs been inaccurate. > These intensional relationships are much more difficult to > model mathematically than truth-functional relations such > as material implication. > I think this sums it up very well. > After all, the father doesnÕt even mean that if the son eats his > vegetables, he will be guaranteed ice cream. The conditional operator > of natural language is non-monotonic. > Eat veggies ----> get ice cream > but presumably *not* > Eat veggies & kill sister ----> get ice cream > Even though eat veggies & kill sister implies eat veggies, one > shouldnÕt conclude that the kid still gets the ice cream after eating > his veggies and killing his sister. Depending, of course, on DadÕs > opinions of sis. > These arenÕt paradoxes of material implication. They are problems of > natural language, relevance, and so on. They are called Paradoxes of Material Implication for historical reasons. Note that these arenÕt problems for natural language *speakers*, just people who wish to translate a natural language into the FOL. >If theyÕre relevant at all in > the current discussion, then they cast doubt on the claim that logic > is innate. How does one claim that logic is part of the fundamental > features of the human brain, when natural language conditionals are so > obviously not truth-functional? Because there are different natural language conditionals, some of which are truth functional. I still stand by my Eat veggies, get ice cream example, since in my experience, if the father gives the kid ice cream without him having eaten his vegetables, people would say that the caved. Of course, your experience could be different. Assuming that this reaction is universal is too far out. Granting this, your example of the non-monotonicity of this conversational conditional doesnÕt really show what you think it does. It shows that the context generated (this may not be the best phrase for what I mean, but...) by the sentences Eat veggies, get ice cream and Eat veggies and kill sis, get ice cream are different -- very different. In short, youÕve shown the existence of two classes of conversational conditionals. This isnÕt new -- we should not forget that logical english, like what a mathematician would use when stating a theorem, is a fragment of the whole of english. For what itÕs worth, my view is that our natural language abilities combined with rough, though strong intuitions of truth and falsity constitute most peopleÕs logic abilities. Conversational implicatures can lead people to stray from classical logic because they cannot differentiate between a truth functional and a non-truth functional conditional. Ōcid Ōooh === Subject: Re: logic is innate? > Granting this, your example of the non-monotonicity of this > conversational conditional doesnÕt really show what you think it does. > It shows that the context generated (this may not be the best phrase > for what I mean, but...) by the sentences Eat veggies, get ice cream > and Eat veggies and kill sis, get ice cream are different -- very > different. In short, youÕve shown the existence of two classes of > conversational conditionals. Can you explain why you think this example involves two different kinds of conditional? ŌDifferent contexts generatedÕ is almost always what is going on in this kind of examples of non-monotony. -- Herman Jurjus === Subject: Re: logic is innate? > Granting this, your example of the non-monotonicity of this > conversational conditional doesnÕt really show what you think it does. > It shows that the context generated (this may not be the best phrase > for what I mean, but...) by the sentences Eat veggies, get ice cream > and Eat veggies and kill sis, get ice cream are different -- very > different. In short, youÕve shown the existence of two classes of > conversational conditionals. > Can you explain why you think this example involves two different kinds > of conditional? ŌDifferent contexts generatedÕ is almost always what is > going on in this kind of examples of non-monotony. Well, letÕs see. The Wittgensteinian in me wants to say that the meaning of an occurence of the words if and then are their use in a context. Fix a context in which the first makes intuitive sense--a father and son (and daughter) eating dinner. Now, in context, the first sentence is often parsed as a biconditional, whereas if the second is uttered, it will hopefully be parsed as a joke, and not a conditional at all. Ceteris paribus, the conditionals must be different. This may or may not be satisfactory (IÕm not too happy with it as an argument, since things arenÕt actually equal -- one *could* argue that the conditionals are in fact the same and that itÕs the phrase and kill sis that is doing the trouble. Shooting from the hip, my response would be that such a position commits one to a sort of realism about meaning that although possibly coherent, shows that one has thought about this too much.) For specificity, what I meant by context generated by a sentence is that the use of a sentence joins the contextual pool for the situation in which it occurs and governs the meanings of sentences that occur after it. (Not that you asked) Ōcid Ōooh === Subject: Re: logic is innate? >Granting this, your example of the non-monotonicity of this >conversational conditional doesnÕt really show what you think it does. > It shows that the context generated (this may not be the best phrase >for what I mean, but...) by the sentences Eat veggies, get ice cream >and Eat veggies and kill sis, get ice cream are different -- very >different. In short, youÕve shown the existence of two classes of >conversational conditionals. >>Can you explain why you think this example involves two different kinds >>of conditional? ŌDifferent contexts generatedÕ is almost always what is >>going on in this kind of examples of non-monotony. > Well, letÕs see. The Wittgensteinian in me wants to say that the > meaning of an occurence of the words if and then are their use in > a context. Fix a context in which the first makes intuitive > sense--a father and son (and daughter) eating dinner. Now, in > context, the first sentence is often parsed as a biconditional, > whereas if the second is uttered, it will hopefully be parsed as a > joke, and not a conditional at all. Ceteris paribus, the conditionals > must be different. This may or may not be satisfactory (IÕm not too > happy with it as an argument, since things arenÕt actually equal -- > one *could* argue that the conditionals are in fact the same and that > itÕs the phrase and kill sis that is doing the trouble. Shooting > from the hip, my response would be that such a position commits one to > a sort of realism about meaning that although possibly coherent, shows > that one has thought about this too much.) > For specificity, what I meant by context generated by a sentence is > that the use of a sentence joins the contextual pool for the situation > in which it occurs and governs the meanings of sentences that occur > after it. (Not that you asked) > Ōcid Ōooh Hmmm. IÕd say: we have here two meanings for the conditional connective. And the example shows that both these conditionals are non-monotonic. ThereÕs just one additional aspect of this example that may obfuscate that non-monotony, though: the first sentence is typically interpreted using one meaning, and the second sentence is typically interpreted using the other meaning. But, imo, the example clearly shows non-monotony, for both meanings of the conditional connective involved. (Not that it matters a lot for this thread.) -- Herman Jurjus === Subject: Re: logic is innate? <30hnelF2vr3i3U1@uni-berlin.de> <310mqkF35dgm5U1@uni-berlin.de> <87u0r73f4i.fsf@phiwumbda.org> <313slaF34ah79U1@uni-berlin.de> <314djvF377aqnU1@individual.net> <871xeasayt.fsf@phiwumbda.org> Discussion, linux) >>If theyÕre relevant at all in >> the current discussion, then they cast doubt on the claim that logic >> is innate. How does one claim that logic is part of the fundamental >> features of the human brain, when natural language conditionals are so >> obviously not truth-functional? > Because there are different natural language conditionals, some of > which are truth functional. Can you give a simple example of a natural language conditional that is best interpreted truth-functionally? LetÕs leave informal mathematical discussions out, if we may. > I still stand by my Eat veggies, get ice cream example, since in my > experience, if the father gives the kid ice cream without him having > eaten his vegetables, people would say that the caved. Of course, > your experience could be different. Assuming that this reaction is > universal is too far out. I didnÕt disagree with your claim that this example is better understood as a biconditional, but even then, IÕd say that each of the two conditionals is probably non-monotonic, rather than material implication. > Granting this, your example of the non-monotonicity of this > conversational conditional doesnÕt really show what you think it > does. IÕm not sure what youÕre correcting here. I didnÕt want to claim that the fact natural language conditionals arenÕt truth-functional *proves* logic is not innate. But if natural language gives us any hints on that question, IÕd guess it pushes us towards the negative at least as strongly as to the positive. To be honest, though, itÕs not the kind of philosophical question I spend much thought on. Matter of taste, I suppose. > It shows that the context generated (this may not be the best phrase > for what I mean, but...) by the sentences Eat veggies, get ice cream > and Eat veggies and kill sis, get ice cream are different -- very > different. In short, youÕve shown the existence of two classes of > conversational conditionals. This isnÕt new -- we should not forget > that logical english, like what a mathematician would use when > stating a theorem, is a fragment of the whole of english. I certainly didnÕt claim that anything I mentioned was new. -- Sure, [my Usenet presence is] like Shaq playing against you in your backyard, but that has its perks, as I find ways to have my fun *and* I can send messages to certain people in the United States Government without concern that the rest of you understand them. -- James Harris === Subject: Re: logic is innate? >>If theyÕre relevant at all in >> the current discussion, then they cast doubt on the claim that logic >> is innate. How does one claim that logic is part of the fundamental >> features of the human brain, when natural language conditionals are so >> obviously not truth-functional? > Because there are different natural language conditionals, some of > which are truth functional. > Can you give a simple example of a natural language conditional that > is best interpreted truth-functionally? LetÕs leave informal > mathematical discussions out, if we may. How about informal philosophical discussions? This might seem like cheating, since this sort of discourse shares many of the features of mathematical discussions (and includes more than a small amount of equivocation). IÕll ask my linguistics buddies if they can think of any. > I still stand by my Eat veggies, get ice cream example, since in my > experience, if the father gives the kid ice cream without him having > eaten his vegetables, people would say that the caved. Of course, > your experience could be different. Assuming that this reaction is > universal is too far out. > I didnÕt disagree with your claim that this example is better > understood as a biconditional, but even then, IÕd say that each of the > two conditionals is probably non-monotonic, rather than material > implication. > Granting this, your example of the non-monotonicity of this > conversational conditional doesnÕt really show what you think it > does. > IÕm not sure what youÕre correcting here. Well, you claimed that _The_ conditional operator of natural language is non-monotonic, whereas your example shows that it isnÕt unique. (emph mine) > I didnÕt want to claim that > the fact natural language conditionals arenÕt truth-functional > *proves* logic is not innate. But if natural language gives us any > hints on that question, IÕd guess it pushes us towards the negative at > least as strongly as to the positive. I donÕt have any strong opinions (that would settle the issue, anyway). Like IÕve said before, in my experience -- say, during freshman year analysis, I would reason by internally thinking things along the lines of So ŌXÕ is true, and ŌX->YÕ is true and using my intuitions about what truth meant. My abilities have grown in sophistication since then (not to toot my own horn, but the fact that we can have intelligent discourse on abstract topics is testament). But one can mean many things by logic. Mentally applying an algorithm to solve an NP-complete game would be equivalent to applying an algorithm to solve the satisfiability problem -- it would be difficult to say that this person wasnÕt using logic, even though all theyÕve done is raw, likely brute calculation. This sort of data crunching ability might seem to be innate, but it seems like a different skill than *finding* the algorithm. The best I think we can say about this issue is sort of a compromise -- we each start with certain mental faculties (the raw data crunching stuff) that we can develop into higher level skills with intense critical thinking. Maybe we can even improve our raw data crunching skills with lots of practice (yuck). Developmentally, we might find peopleÕs rough, intuitive notions of how logic works somewhere between the two. But this isnÕt philosophical or empirical. ItÕs downright ßakey. > To be honest, though, itÕs not the kind of philosophical question I > spend much thought on. Matter of taste, I suppose. Me neither, usually. But you and Mitch brought up some interesting points regarding my posts. Ōcid Ōooh === Subject: Re: logic is innate? <310mqkF35dgm5U1@uni-berlin.de> <87u0r73f4i.fsf@phiwumbda.org> <313slaF34ah79U1@uni-berlin.de> <314djvF377aqnU1@individual.net> <871xeasayt.fsf@phiwumbda.org> <87vfblyqx5.fsf@phiwumbda.org> Discussion, linux) >> Granting this, your example of the non-monotonicity of this >> conversational conditional doesnÕt really show what you think it >> does. >> IÕm not sure what youÕre correcting here. > Well, you claimed that _The_ conditional operator of natural language > is non-monotonic, whereas your example shows that it isnÕt unique. > (emph mine) -- Jesse F. Hughes LOL. How arrogant you are. Now when you realize that I DID prove GoldbachÕs conjecture and that I proved FermatÕs Last Theorem as well, how are you going to feel then? -- James Harris === Subject: Complex analysis power series question IÕm rather new to complex analysis, so IÕm wondering if somebody can help me with this question? Any hints are greatly appreciated, thank you! Determine all analytic functions f(z) on the complex plane that satisfy f(z^2) = (f(z))^2 I guess if f(z) is holomorphic everywhere, that means that it has its power representation valid everywhere...but the power series of these thoughts? KH === Subject: Re: Complex analysis power series question > IÕm rather new to complex analysis, so IÕm wondering if somebody can > help me with this question? Any hints are greatly appreciated, thank > you! > Determine all analytic functions f(z) on the complex plane that > satisfy > f(z^2) = (f(z))^2 > I guess if f(z) is holomorphic everywhere, that means that it has its > power representation valid everywhere...but the power series of these > thoughts? > KH I think that consideration of the the power series representation of f(z^2) is insightful and the way to crack this problem. Note that g^(n) = 0 if n is odd when g=f(z^2). === Subject: Re: Complex analysis power series question > IÕm rather new to complex analysis, so IÕm wondering if somebody can > help me with this question? Any hints are greatly appreciated, thank > you! > Determine all analytic functions f(z) on the complex plane that > satisfy > f(z^2) = (f(z))^2 > I guess if f(z) is holomorphic everywhere, that means that it has its > power representation valid everywhere...but the power series of these > thoughts? Suppose f is nonconstant. Then you can write f(z) = a + z^m*g(z), where m is a positive integer, g is entire, and g(0) is nonzero. Now equate f(z^2) and (f(z))^2 using this form for f, and see what happens. === Subject: Liouville theorem corollary? I read in a complex analysis book that one can use LiouvilleÕs theorem to classify all analytic functions f:C -> C such that |f(z)| leq K*|z|^n Does anybody know about this corollary, or how I might be able to Shin === Subject: Re: Liouville theorem corollary? > I read in a complex analysis book that one can use LiouvilleÕs theorem > to classify all analytic functions f:C -> C such that > |f(z)| leq K*|z|^n > Does anybody know about this corollary, or how I might be able to > derive it? Sure, let f(z) = sum(m=0,oo) c_m*z^m. Use CauchyÕs estimates to prove something interesting about c_m for large m. === Subject: Re: Liouville theorem corollary? >> I read in a complex analysis book that one can use LiouvilleÕs theorem >> to classify all analytic functions f:C -> C such that >> |f(z)| leq K*|z|^n >> Does anybody know about this corollary, or how I might be able to >> derive it? >Sure, let f(z) = sum(m=0,oo) c_m*z^m. Use CauchyÕs estimates to prove >something interesting about c_m for large m. Given the way the hypothesis was stated you can in fact use CauchyÕs estimates to prove something interesting about c_m for every m <> n, both large and small. ************************ David C. Ullrich === Subject: The joy of plain text > Please, no TEX when you post here; this is a plain text newsgroup. > On the contrary, this is a mathematics newsgroup, where TeX is quite > common. IÕm not sure what common is supposed to imply: Poorly stated questions are much more common on sci.math than is TeX; that is hardly an argument for them. I was under the impression that sci.math is a plain text ng. IsnÕt this forum for the worldwide mathematics community at large? This would include many who are unfamiliar with TeX: kids, high-school teachers, all sorts of amateurs and hobbyists, engineers, math Ph.D.s who got their degrees decades ago and are in other fields now, etc. I once knew TeX well enough to write a few papers in it, but I left academia years ago and today I much prefer plain old text to TeX on sci.math. === Subject: Re: The joy of plain text >> Please, no TEX when you post here; this is a plain text newsgroup. >> On the contrary, this is a mathematics newsgroup, where TeX is quite >> common. >IÕm not sure what common is supposed to imply: Poorly stated questions >are much more common on sci.math than is TeX; that is hardly an argument >for them. >I was under the impression that sci.math is a plain text ng. IsnÕt this >forum for the worldwide mathematics community at large? This would include >many who are unfamiliar with TeX: kids, high-school teachers, all sorts of >amateurs and hobbyists, engineers, math Ph.D.s who got their degrees >decades ago and are in other fields now, etc. I once knew TeX well enough >to write a few papers in it, but I left academia years ago and today I much >prefer plain old text to TeX on sci.math. That is if the plain old text can intelligently state the problem. Too often, it 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: The joy of plain text > Please, no TEX when you post here; this is a plain text newsgroup. > On the contrary, this is a mathematics newsgroup, where TeX is quite > common. > IÕm not sure what common is supposed to imply: Poorly stated questions > are much more common on sci.math than is TeX; that is hardly an argument > for them. > I was under the impression that sci.math is a plain text ng. IsnÕt this > forum for the worldwide mathematics community at large? This would include > many who are unfamiliar with TeX: kids, high-school teachers, all sorts of > amateurs and hobbyists, engineers, math Ph.D.s who got their degrees > decades ago and are in other fields now, etc. I once knew TeX well enough > to write a few papers in it, but I left academia years ago and today I much > prefer plain old text to TeX on sci.math. Plain text to me means no binaries; no .jpg files, or Word files, etc. TeX is plain text, at least the way I understand the term. Occasionally I post something here that IÕve cut ŌnÕ pasted from another source, where it was in TeX. Going through the TeX, especially if itÕs at all long, and manually editing it into a form youÕd approve is not an option; your options are, I post the TeX, or I donÕt post at all. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: The joy of plain text > Occasionally I post something here that IÕve cut ŌnÕ pasted from > another source, where it was in TeX. Going through the TeX, especially > if itÕs at all long, and manually editing it into a form youÕd approve > is not an option; your options are, I post the TeX, or I donÕt post > at all. Then donÕt post it. Unless itÕs simple Tex, likely I may never read it. Of the many many problems in the 7 groups I read, IÕve little time to translate hard to read stuff. So if youÕve not time to make your post readable, why should I take the time to make it readable and then even more time to solve the problem and yet more time to present an answer? You are asking too much of free tutoring. Those who post readable problems are those I read and (when able) answer first. === Subject: Re: The joy of plain text > Occasionally I post something here that IÕve cut ŌnÕ pasted from > another source, where it was in TeX. Going through the TeX, especially > if itÕs at all long, and manually editing it into a form youÕd approve > is not an option; your options are, I post the TeX, or I donÕt post > at all. > Then donÕt post it. Unless itÕs simple Tex, likely I may never read it. > Of the many many problems in the 7 groups I read, IÕve little time to > translate hard to read stuff. So if youÕve not time to make your post > readable, why should I take the time to make it readable and then even > more time to solve the problem and yet more time to present an answer? > You are asking too much of free tutoring. Those who post readable > problems are those I read and (when able) answer first. I see that I didnÕt make myself clear. Most of what I post here is not problems but answers to other peopleÕs problems. IÕve been providing, not asking for, the free tutoring for over a decade now (for the most part), though I guess itÕs too much to expect that even regulars contributors such as yourself would have noticed. The typical situation in which IÕll post in TeX is when someone asks a research-level question, and I happen to know that thereÕs a paper on the topic already in the literature, and IÕll go cut ŌnÕ paste the review from Math Reviews, which review will be written in TeX. If the person who asked the question canÕt decode the TeX, too bad - how much can she ask of free tutoring? You are hereby excused from reading anything I post in TeX. I think IÕll survive. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: The joy of plain text >The typical situation in which IÕll post in TeX is when someone >asks a research-level question, and I happen to know that thereÕs >a paper on the topic already in the literature, and IÕll go cut >ÕnÕ paste the review from Math Reviews, which review will be written >in TeX. For WilliamÕs sake, perhaps you could use Zentralblatt instead. Of course, then youÕd have to schneidÕnÕklebb. Lee Rudolph === Subject: Re: The joy of plain text >> Occasionally I post something here that IÕve cut ŌnÕ pasted from >> another source, where it was in TeX. Going through the TeX, especially >> if itÕs at all long, and manually editing it into a form youÕd approve >> is not an option; your options are, I post the TeX, or I donÕt post >> at all. >Then donÕt post it. [...] No, please do. Derek HoltÕs post says it all, so I wonÕt labour the point. I just wanted to add another vote for sanity (not that IÕm claiming to be sane). -- Angus Rodgers Contains mild peril === Subject: Re: The joy of plain text >> Occasionally I post something here that IÕve cut ŌnÕ pasted from >> another source, where it was in TeX. Going through the TeX, especially >> if itÕs at all long, and manually editing it into a form youÕd approve >> is not an option; your options are, I post the TeX, or I donÕt post >> at all. >Then donÕt post it. Unless itÕs simple Tex, likely I may never read it. The fact that you may never read it is a poor argument for not posting it. There may be others who do want to read it. My impression is that the most generally preferred method of writing mathematics on this newsgroup is to use a simplified version of TeX, removing all symbols (such as dollars) that are unnecessary for comprehension - for example: alpha^2 + beta^{5/2} = sum_{i=0} ^infty gamma_i ^{-3}. is not difficult to read. How would you prefer that to be written? With a longer chunk of material, cutting and pasting from a TeX document is not ideal, but as long as a handful of people are interested in reading it, it is worth posting it. There is so much total junk on this newsgroup already, that it seems strange to attmept to outlaw potentially interesting material on account of minor shortcomings. Derek Holt. >Of the many many problems in the 7 groups I read, IÕve little time to >translate hard to read stuff. So if youÕve not time to make your post >readable, why should I take the time to make it readable and then even >more time to solve the problem and yet more time to present an answer? >You are asking too much of free tutoring. Those who post readable >problems are those I read and (when able) answer first. === Subject: Re: The joy of plain text > My impression is that the most generally preferred method of writing > mathematics on this newsgroup is to use a simplified version of TeX, > removing all symbols (such as dollars) that are unnecessary for > comprehension - for example: > alpha^2 + beta^{5/2} = sum_{i=0} ^infty gamma_i ^{-3}. > is not difficult to read. How would you prefer that to be written? My sentiment more or less exactly! Everybody understands TeX (or can guess the meaning). My practice is to also leave out the backslashes (ŌÕ), if it looks like that wonÕt lead to any misunderstandings, so I might write: alpha^2 + beta^{5/2} = sum_{i=0}^infty gamma_i^{-3}. I think that this is a reasonable compromise and IMVHO slightly more readable. I also prefer not to use ŌfracÕ or ŌoverÕ (me the plainTeX-fan:) at all. I think that {daadaa}/{doobedoo} is better than the alternative ways of writing a quotient:) I do feel that cutting and pasting from TeX-source has certain other drawbacks. E.g. if I were to cut and paste the above from a TeX-file I had written, it probably wouldnÕt have that extra space surrounding ŌplusÕ and Ōequal toÕ signs. I feel that this extra space does enhance legibility a bit, so I would normally do it that way. Ok. Sometimes I relax on that rule, if IÕm in a hurry and donÕt have the time to edit or proofread my postings. But to summarize: Degustibus non est disputandum. Jyrki Lahtonen, Turku, Finland === Subject: Re: The joy of plain text is not difficult to read. How would you prefer that to be written? > I might write: > alpha^2 + beta^{5/2} = sum_{i=0}^infty gamma_i^{-3}. Much easier to read. Other hard to read stuff is ax^2+bx+c=(x-r)(x-s)=x^2-(r+s)x+rs=hardtoread === Subject: Re: JSH:Understanding constant terms > [.snip.] >> So... I addressed your comment as written. Since that comment, you >> have now changed your mind about what the g_i(x) should be; based on >> that change of heart, you complain that my correction was wrong, >> because it failed to meet your newfound and not-previously asserted >> conditions on the g_i(x). > How... interesting... >The disingenuousness of your attempted >exculpation is breathtaking. >I am unversed in exegesising the >pronouncements of the Supreme Coprolocutor. >You, however, analyse His productions with >monomaniacal intensity. Therefore, >to claim that the conditions on the g_i(x), >although new to me, were unknown to you >is sophistry of the most egregious kind. > Sigh. I addressed YOUR comment. I noted an error in YOUR comment, > based on YOUR hypothesis. You acknowledge that error, but complain > that I noted it. I did not address anything the original poster said. Typical No, I wasnÕt complaining that you noted it at all. My soul was suffused with the light that you brought unto me. I wasnÕt saying that you explicitly addressed anything the OP said, but that you must have known implicitly what the OP said about the g_i(x) due to your regular enlightening intercourse with the OP. Perhaps a ßeeting fit of insanity overcame you as you pondered these deep matters causing you to cast aside your usual intellectual caution and make that rash and impetuous statement about ANY polynomial. > Presumably, you are either joking, or dense. Joking ? Gay persißage ? Badinage of an ironic nature ? Egad Sir,I would rather ritually disembowel myself than make light of your forensic quest for Truth. > I prefer to assume the former, though it seems hard to ignore the > latter possibility. Your sharpness of mind is matched only by your generosity of spirit. === Subject: Re: JSH:Understanding constant terms Discussion, linux) > No, I wasnÕt complaining that you noted it at all. > My soul was suffused with the light that you > brought unto me. Jesus, give it a rest and put away the thesaurus. WeÕre already ing impressed. At least I know I am. -- Jesse F. Hughes To be honest, I donÕt have enough interest in math to spend the time it would take to clean up the mess that I believe has been created in the past 100 or so years. -- Curt Welch lets the world down. === Subject: Re: Powers in goup ETAuAhUAukxUcK3T5xsdweFGRmj4S9YanjECFQCNb6o0tEqy7LTghNy7chL1hw cZmQ== I would take the fifth-power equation and divide it by the third-power equation. Thus (ab)^2 = a^2b^2, meaning abab = aabb. Left multiply the last equation by a^(-1) and right multiply by b^(-1). --OL === Subject: Re: Powers in goup > I would take the fifth-power equation and divide it by the third-power > equation. What do you mean by divide it by the third-power equation. Thus (ab)^2 = a^2b^2, I donÕt think so!How did you get this? I donÕt think this is correct. meaning abab = aabb. Left multiply the > last equation by a^(-1) and right multiply by b^(-1). > --OL === Subject: Re: Is a circle just a 2-dimensional sphere? ETAsAhQednPc1mwXks/KZ8X2AHnQLh+ NUgIUe2X02DgoB4t9MkQsvzlUkHTFIWI= I think of a circle as the circumference of the area I call a disk. Likewise a sphere is the surface of a ball. ThatÕs the way I see it, FWIW. --OL === 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, isnÕt applicable, because sets are abstract mathematical ideas which canÕt 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, doesnÕt seem useful in this context. So thereÕs 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 itÕs 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 arenÕt 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 arenÕt 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 arenÕt 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 arenÕt 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 PeanoÕs 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 donÕt have an infinite amount of time to type them all: {apat, isa, lima, delawa, tatlo}. Unless you know that IÕve used Tagalog names for those five integers, and unless you actually know what those five Tagalog words mean, and unless IÕm actually using the corect Tagalog names instead of shufßing them, you canÕt 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 arenÕt 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 arenÕt 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 donÕt 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 arenÕt? No matter how you guess which are even and which arenÕt, 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 wouldnÕt 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 youÕd be definining even based on their position within some sort of structue, in this case a regular lattice. But suppose they werenÕt 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, thereÕs 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 arenÕt? No, the best you can do is make arbitrary decisions, which wouldnÕt 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 IÕve specified above? No, you canÕt. The best you can do is randomly decide for each new integerObject whether itÕs even or odd, and keep a cache of such decisions so you donÕt contradict yourself later, but thatÕs not a definition, thatÕs a random sampler, and again, just like the manual demo, youÕd get different results with each run of the program, so your function/method doesnÕt *define* a function, it merely generates a new random sample each time itÕs run. By the way, with *all* the integers, not just the positive integers, even if you know the total ordering, thatÕs still not enough to define what is even and what isnÕt 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 itÕs 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 donÕt know whether two integers are adjacent or not you canÕt in a finite number of steps determine the two alternating sets. > The set of all sets ... ThereÕs no such thing. If you understood proof-by-contradiction I could present you a very simple proof to that fact, but you donÕt 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 thereÕs a corresponding even integer, and vice versa. What you said is equivalent to saying thereÕs 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: >ThereÕs no such thing. There are set theories that include a set of all sets. They donÕt, 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 > There are set theories that include a set of all sets. > They donÕt, however, satisfy, e.g., the axiom of foundation. Do they include an axiom that allows subsets of a given set to be defined in terms of a property (boolean-valued function) defined on that set, i.e.: if S is already known to be a set, and P is a property defined on that set, do they allow a new set to be defined as: {x in S such that P(x)} Do they include an axiom that allows ordered pairs to be defined with the usual uniqueness semantics? I.e. if S and T are sets, and s,t are elements from those sets respectively, then is an ordered pair, and for all s1,s2,t1,t2 s1=s2 & t1=t2 iff = ? (With ordered pairs, predicates on single parameters can be used to define predicates with two parameters: P2(x,y) iff P1()) Given that they allow a set to contain itself as an element, do they consider subsetof to be a predicate on ordered pairs, i.e. P() is defined to be true iff S is a subset of T? If P is a predicate, then is lambda.x not P(x) likewise a predicate? If all the above are valid in such a theory of sets, how do they avoid RusselÕs paradox? === Subject: Re: Proposed definition for comparing the sizes of two sets <41b24220$12$fuzhry+tra$mr2ice@news.patriot.net> at 01:22 PM, rem642b@Yahoo.Com (tinyurl.com/uh3t) said: >Do they include an axiom that allows subsets of a given set to be >defined in terms of a property (boolean-valued function) defined on >that set, i.e.: if S is already known to be a set, and P is a >property defined on that set, do they allow a new set to be defined >as: > {x in S such that P(x)} Not for arbitrary P. Which answers your question below. >Do they include an axiom that allows ordered pairs to be defined >with the usual uniqueness semantics? Yes. >Given that they allow a set to contain itself as an element, do they >consider subsetof to be a predicate on ordered pairs, i.e. >P() is defined to be true iff S is a subset of T? Yes. >If P is a predicate, then is lambda.x not P(x) likewise a >predicate? >If all the above are valid in such a theory of sets, how do they >avoid RusselÕs paradox? By the fact that {x: P(x)} does not exist for arbitrary predicates, but only for predicates that are well formed in accordance with rules that were explicitly devised to make it impossible to shave the Spanish barber. -- 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, isnÕt applicable, because sets are > abstract mathematical ideas which canÕt 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, doesnÕt seem useful in this context. So thereÕs 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 itÕs 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. ItÕs definitely not a lie. There are obviously many ordering relations, pick one. > completely intuitive natural counting numbers. > The natural numbers arenÕt 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 arenÕt > 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 arenÕt 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 arenÕt > 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 PeanoÕs 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 donÕt have an infinite amount of time to type them > all: {apat, isa, lima, delawa, tatlo}. Unless you know that IÕve used > Tagalog names for those five integers, and unless you actually know > what those five Tagalog words mean, and unless IÕm actually using the > corect Tagalog names instead of shufßing them, you canÕt 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 arenÕt > 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 arenÕt 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 donÕt 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 arenÕt? I donÕt care. I wouldnÕt. > No matter how you guess which are even and which arenÕt, 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 wouldnÕt 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 youÕd be definining even based on their position > within some sort of structue, in this case a regular lattice. But > suppose they werenÕt 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, thereÕs > 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 arenÕt? No, the best you can do is make > arbitrary decisions, which wouldnÕt 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 IÕve specified above? > No, you canÕt. The best you can do is randomly decide for each new > integerObject whether itÕs even or odd, and keep a cache of such > decisions so you donÕt contradict yourself later, but thatÕs not a > definition, thatÕs a random sampler, and again, just like the manual > demo, youÕd get different results with each run of the program, so your > function/method doesnÕt *define* a function, it merely generates a new > random sample each time itÕs run. ThatÕs 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, thatÕs still not enough to define > what is even and what isnÕt 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 itÕs 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 donÕt know whether two integers are adjacent or not you > canÕt in a finite number of steps determine the two alternating sets. > The set of all sets ... > ThereÕs 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 donÕt 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, youÕre 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 thereÕs a > corresponding even integer, and vice versa. YouÕre wrong, it shows exactly that thing. > What you said is equivalent to saying thereÕs 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 doesnÕt. 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 setÕs 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 youÕve been about all this? Half of the integers are even, true or false? ItÕs true. If itÕs 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: Point of logic I still have a few questions... > A mathematical proof must be logical. What I aim to do here is > elaborate on the logic, which proves IÕm correct, and letÕs see what > happens. > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. If the aÕs are not polynomials, then how do you define constant term? Presumably you mean the value of each a_*(x) at the point x=0. In other words, the intercept of the functions a_*(x). Is that correct? > That corresponds with the constant term of P(x) being 1078, which is > 7(7)(22). > Now I note that dividing both sides by 49 gives me > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, > which factors as > P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 > and I note that the constant term is now 22. > Now what do I know about the constant terms? > Well they are constant, and are specific numbers, specifically 7, 7 > and 22, for the factors of P(x). But for P(x)/49, the constant term > is 22. > So, logically, two of the constant terms were divided by 7. > That is the essential point on which everything hinges. > Notice how simple it is, the constant terms are actual numbers. > Numbers like 7 and 22 are NOT variables, and they remain the same > without regard to the value of x. Well, you can certainly express each a_*(x) as a_*(x) = b_*(x) + const, such that b_*(0)=0. > Dividing P(x) by 49 changes the constant terms. > They go from being 7, 7 and 22 to being 1, 1 and 22. > Therefore, exactly two of them were divided by 7. > There is no other way to get from 7 to 1, and you can write it out > algebraically. > 7/w = 1, giving w = 7. > By the distributive property for the constant terms of two of > (5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) > to be divided by 7, the full factor has to be divided by 7, which > gives > P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) > indicating that two of the aÕs have 7 as a factor, and IÕve > arbitrarily selected a_1(x) and a_2(x). IÕm a bit confused here. When you talk about having 7 as a factor, this is usually applied to an integer value, i.e. (49 has 7 as a factor, etc). Are you saying that a_*(x) is integer valued for every x? It seems like you have only shown this for the case x=0? Maybe you can elaborate here. > What can be shown is that in the ring of algebraic integers, if > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > is irreducible over Q, with integer coefficients, then the aÕs cannot > have 7 as a factor in the ring of algebraic integers. So, are you setting x to be an integer value in order to get integer coefficients? For which x it is irreducible? Darren === Subject: Re: Point of logic > I still have a few questions... >>A mathematical proof must be logical. What I aim to do here is >>elaborate on the logic, which proves IÕm correct, and letÕs see what >>happens. >>What I have is a polynomial >>P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 >>which has 49 as a multiple, as >>P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). >>I factor the polynomial as >>P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) >>when the aÕs are the three roots of >>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). >>I determine the constant terms of the three factors by setting x=0, >>which reveals that for two of them the constant term is 7, while for >>one it is 22. > If the aÕs are not polynomials, then how do you define constant term? > Presumably you mean the value of each a_*(x) at the point x=0. In other > words, the intercept of the functions a_*(x). > Is that correct? >>That corresponds with the constant term of P(x) being 1078, which is >>7(7)(22). >>Now I note that dividing both sides by 49 gives me >>P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22, >>which factors as >>P(x)/49 = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)/49 >>and I note that the constant term is now 22. >>Now what do I know about the constant terms? >>Well they are constant, and are specific numbers, specifically 7, 7 >>and 22, for the factors of P(x). But for P(x)/49, the constant term >>is 22. >>So, logically, two of the constant terms were divided by 7. >>That is the essential point on which everything hinges. >>Notice how simple it is, the constant terms are actual numbers. >>Numbers like 7 and 22 are NOT variables, and they remain the same >>without regard to the value of x. > Well, you can certainly express each a_*(x) as a_*(x) = b_*(x) + const, such > that b_*(0)=0. >>Dividing P(x) by 49 changes the constant terms. >>They go from being 7, 7 and 22 to being 1, 1 and 22. >>Therefore, exactly two of them were divided by 7. >>There is no other way to get from 7 to 1, and you can write it out >>algebraically. >>7/w = 1, giving w = 7. >>By the distributive property for the constant terms of two of >>(5 a_1(x) + 7), (5 a_2(x)+ 7), and (5 a_3(x) + 7) >>to be divided by 7, the full factor has to be divided by 7, which >>gives >>P(x)/49 = (5 a_1(x)/7 + 1)(5 a_2(x)/7+ 1)(5 a_3(x) + 7) >>indicating that two of the aÕs have 7 as a factor, and IÕve >>arbitrarily selected a_1(x) and a_2(x). > IÕm a bit confused here. When you talk about having 7 as a factor, this is > usually applied to an integer value, i.e. (49 has 7 as a factor, etc). > Are you saying that a_*(x) is integer valued for every x? It seems like you > have only shown this for the case x=0? Maybe you can elaborate here. For any ring R, and elements b and c in R, b is a factor of c if there is an element d in R such that b*d=c. Most of the discussion that James works with occurs in the algebraic integers (roots of monic polynomials with integer coefficients). However, division by 7 is actually defined as multiplying by 1/7, which is not an algebraic integer. So he is working in some larger ring (possibly the algebraic numbers) when does the multiplication by 1/7 and claiming that the result is again in the algebraic integers. How he draws this conclusion is not entirely clear to me. Then again, IÕm not sure he is fully aware of all of the above. >>What can be shown is that in the ring of algebraic integers, if >>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) >>is irreducible over Q, with integer coefficients, then the aÕs cannot >>have 7 as a factor in the ring of algebraic integers. > So, are you setting x to be an integer value in order to get integer > coefficients? For which x it is irreducible? That is the key question he has never examined. For x=0, it clearly is reducible. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Point of logic > I was tearing my hair out trying to see how he got the cubic equation :) > Incidently, is equating terms in powers of 5 a mathematically valid thing to > do? Yes. Basically, it would be perfectly valid if one replaced the number 5 with the indeterminate y--all James has done is then formally set y equal to 5. This is why it looks so strange, but thereÕs nothing really wrong with his approach. As Arturo said above, [d]epends on what you do with it. If you were then to deduce a property that depends on what we might call the fiveness of that implicit y, all bets would be off. Rick === Subject: Re: Point of logic > can someone help me out here? I was trying to follow this at home and got > stuck out of the gate :( > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > Is there anything special about the polynomial other than it has 49 as a > factor. Does it have to be a square factor? > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > Err, where did the 5 and 7 come from? How do we know that the a_*(x) such > functions exist, and what can we infer about them - are they polynomials? > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > Where did this come from? For each x, let a_1(x), a_2(x), and a_3(x) be the three roots (counting multiplicities) of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Then we have a_1(x) + a_2(x) + a_3(x) = 3(1 - 49x), a_1(x)a_2(x) + a_1(x)a_3(x) + a_2(x)a_3(x) = 0, a_1(x)a_2(x)a_3(x)=49(2401 x^3 - 147 x^2 + 3x), and itÕs not hard to check that (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) does indeed equal JamesÕ polynomial. > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. > Err, I didnÕt follow this. Setting x=0 in the above equation yields > a^3 + 3(-1)a^2 =0 implies a=0, 0, or 3? Yes, so if you let a_1(0)=0, a_2(0)=0, a_3(0)=3, then 5a_1(0)+7=7, 5a_2(0)+7=7, 5a_3(0)+7=22. These are the constant terms of 5a_1(x)+7, 5a_2(x)+7, 5a_3(x)+7. By the constant term of a function James just means its value when x=0. > Darren === Subject: Re: Point of logic > can someone help me out here? I was trying to follow this at home and got > stuck out of the gate :( > What I have is a polynomial > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > which has 49 as a multiple, as > P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22). > Is there anything special about the polynomial other than it has 49 as a > factor. Does it have to be a square factor? No, it does not have to be a square factor. The polynomial is special in that it can be expanded out so that I can factor it into non-polynomial factors. > I factor the polynomial as > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > Err, where did the 5 and 7 come from? How do we know that the a_*(x) such > functions exist, and what can we infer about them - are they polynomials? The aÕs are not polynomials. The factorization comes from P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1+49 x )(5)(7^2) + 7^3 where if you multiply everything out and simplify you will just get P(x) = 14706125 x3 - 900375 x2 - 17640 x + 1078 which you saw above, and thatÕs part of why itÕs special. > when the aÕs are the three roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > Where did this come from? P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1+49 x )(5)(7^2) + 7^3 with respect to 7 and 5, that is, factoring it as if 7 and 5 were the polynomial variables. ItÕs a neat analysis tool which I discovered from which everything follows. IÕll stick in some variables in key places to show: P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (Y^3) - 3(-1+49 x )(Y)(Z^2) + Z^3 where IÕve used Y and Z. > I determine the constant terms of the three factors by setting x=0, > which reveals that for two of them the constant term is 7, while for > one it is 22. > Err, I didnÕt follow this. Setting x=0 in the above equation yields > a^3 + 3(-1)a^2 =0 implies a=0, 0, or 3? Yes, where each root gives one of the aÕs, so if you try that with P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) youÕll notice that it gives the same answer as P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 as they are exactly equal. > Darren YouÕre welcome. James Harris === Subject: Re: Point of logic > ... each root gives one of the aÕs, so if you try that with > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) > youÕll notice that it gives the same answer as > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078 > as they are exactly equal. Yes, and youÕll also notice that P(x) = (245x + 15.013...)(245x - 8.480... )(245x + 8.467...) where the unterminated numbers are 245 times the roots of P(x) = 0. Where are your Ō7Õs and Ō22Õ now? You seem to be unaware that there an unlimited number of ways to express the factors of a polynomial. -- 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: Abstract Algebra IÕm missing something on this seemingly trivial problem. Let G be a group with 3000 elements and H be a subgroup with 1000 elements. If x is in G, show that either x^2 or x^3 is in H. This is what I got so far. If x is in H then not only is x^2 in but x^3 is also in H. So now we assume that x is not in H. IF I can show that H, x^2H and x^3H are pairwise disjoint then G=H U x^2H U x^3H where each coset has o( H ) =1000 elements. So if x is in G then x is in H or x^2H or x^3H. But x is not in H. So x is in x^2H or x^3H. So all that remains to show is that H, x^2H and x^3H are pairwise disjoint. Since x is not in H we have H /= xH-->xH /= x^2H-->x^2H /= x^3H. So x^2 /= x^3H. I just canÕt seem to show that H /= x^2H ( or x^2 is not in H) and H /= x^3H (or x^3 is not in H). What am I not seeing? === Subject: Re: Abstract Algebra > IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 elements. > If > x is in G, show that either x^2 or x^3 is in H. > This is what I got so far. > If x is in H then not only is x^2 in but x^3 is also in H. > So now we assume that x is not in H. > IF I can show that H, x^2H and x^3H are pairwise disjoint then G=H U x^2H U > x^3H where each coset has o( H ) =1000 elements. So if x is in G then x is > in H or x^2H or x^3H. But x is not in H. So x is in x^2H or x^3H. > So all that remains to show is that H, x^2H and x^3H are pairwise disjoint. > Since x is not in H we have H /= xH-->xH /= x^2H-->x^2H /= x^3H. So x^2 /= > x^3H. I just canÕt seem to show that H /= x^2H ( or x^2 is not in H) and H > /= x^3H (or x^3 is not in H). > What am I not seeing? You are not seeing that it is H, xH, and x^2H which are mutually disjoint. Write out explicitly what it means for an element of G to be in two of these subsets of G and determine a contradiction based on the fact that H is not merely a subset of G but a subgroup of G. Achava === Subject: Re: Abstract Algebra > IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If > x is in G, show that either x^2 or x^3 is in H. Hint: If [G:H] = 3, then either H is normal in G, or H has a subgroup of index 2 thatÕs normal in G. [...] -- Jim Heckman === Subject: Re: Abstract Algebra > IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If x is in G, show that either x^2 or x^3 is in H. > Hint: If [G:H] = 3, then either H is normal in G, or H has a > subgroup of index 2 thatÕs normal in G. IÕve never seen this before - IÕm not sure I believe it - and I donÕt think anything like it is necessary for the problem at hand. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Abstract Algebra > IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If x is in G, show that either x^2 or x^3 is in H. > Hint: If [G:H] = 3, then either H is normal in G, or H has a > subgroup of index 2 thatÕs normal in G. > IÕve never seen this before - IÕm not sure I believe it - and > I donÕt think anything like it is necessary for the problem at hand. Well, of course itÕs not *necessary*. It just makes the problem easy to solve. And itÕs certainly true. The action of G on the set of cosets of H gives a surjective homomorphism from G onto a transitive degree-3 subgroup of S_3, whose kernel lies in H (since the latter is the point stabilizer of the coset H in this action). S_3 has only two transitive subgroups in its natural action, so either H is normal in G and {H, xH, x^2H} =~ C_3, or [H:K] = 2 with K normal in G and {K, xK, x^2K, HK, xHK, x^2HK} =~ S_3 (with the latter 3 cosets squaring to K). -- Jim Heckman === Subject: Re: Abstract Algebra IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If x is in G, show that either x^2 or x^3 is in H. Hint: If [G:H] = 3, then either H is normal in G, or H has a > subgroup of index 2 thatÕs normal in G. > IÕve never seen this before - IÕm not sure I believe it - and > I donÕt think anything like it is necessary for the problem at hand. > Well, of course itÕs not *necessary*. It just makes the problem > easy to solve. > And itÕs certainly true. The action of G on the set of cosets of > H gives a surjective homomorphism from G onto a transitive > degree-3 subgroup of S_3, whose kernel lies in H (since the > latter is the point stabilizer of the coset H in this action). > S_3 has only two transitive subgroups in its natural action, so > either H is normal in G and {H, xH, x^2H} =~ C_3, or [H:K] = 2 > with K normal in G and {K, xK, x^2K, HK, xHK, x^2HK} =~ S_3 > (with the latter 3 cosets squaring to K). But. I suspect that anyone having trouble with the original problem is not going to understand any of what youÕve written here. Knowing that a problem on the second ßoor is easy to solve from the perspective of the 12th ßoor doesnÕt help the guy who has been struggling to reach the mezzanine. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Abstract Algebra > IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If x is in G, show that either x^2 or x^3 is in H. [...] > And itÕs certainly true. The action of G on the set of cosets of > H gives a surjective homomorphism from G onto a transitive > degree-3 subgroup of S_3, whose kernel lies in H (since the > latter is the point stabilizer of the coset H in this action). > S_3 has only two transitive subgroups in its natural action, so > either H is normal in G and {H, xH, x^2H} =~ C_3, or [H:K] = 2 > with K normal in G and {K, xK, x^2K, HK, xHK, x^2HK} =~ S_3 > (with the latter 3 cosets squaring to K). > But. > I suspect that anyone having trouble with the original problem > is not going to understand any of what youÕve written here. > Knowing that a problem on the second ßoor is easy to solve > from the perspective of the 12th ßoor doesnÕt help the guy > who has been struggling to reach the mezzanine. Actually, IÕm having trouble thinking of a way to solve the original problem that *doesnÕt* essentially use the action of G on the cosets of H to show that G has a surjective homomorphism to C_3 or S_3 whose kernel lies in H. I notice the OP said he figured it out, so I wonder just what ßoor heÕs on. Or maybe IÕm missing some more elementary approach... -- Jim Heckman === Subject: Re: Abstract Algebra [...] IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with > 1000 elements. > If x is in G, show that either x^2 or x^3 is in H. [...] > Actually, IÕm having trouble thinking of a way to solve the > original problem that *doesnÕt* essentially use the action of G > on the cosets of H to show that G has a surjective homomorphism > to C_3 or S_3 whose kernel lies in H. I notice the OP said he > figured it out, so I wonder just what ßoor heÕs on. Or maybe > IÕm missing some more elementary approach... obvious elementary approach. DÕoh! -- Jim Heckman === Subject: Re: Abstract Algebra > Actually, IÕm having trouble thinking of a way to solve the > original problem that *doesnÕt* essentially use the action of G > on the cosets of H to show that G has a surjective homomorphism > to C_3 or S_3 whose kernel lies in H. I notice the OP said he > figured it out, so I wonder just what ßoor heÕs on. Or maybe > IÕm missing some more elementary approach... Well, I donÕt know about a method that avoids cosets entirely, but hereÕs an outline of what I came up with: Any two cosets aH and bH are either disjoint or equal. There is also a bijection between them (making them the same size if H is finite). We can see that if aH = bH, a^-1b must be in H. Then consider the 4 cosets H = x^0H, xH = x^1H, x^2H, x^3H. Since G is 3 times the size of H, the pigeonhole principle forces at least one pair of equal cosets; call them x^mH and x^nH where m < n. Since x^mH = x^nH, we have that x^(n-m) is in H. If n-m is 2 or 3, we are done. If n-m is 1, we use the fact that H is a group one last time to get the desired result. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Abstract Algebra Originator: grubb@lola > IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 > elements. > If x is in G, show that either x^2 or x^3 is in H. >Actually, IÕm having trouble thinking of a way to solve the >original problem that *doesnÕt* essentially use the action of G >on the cosets of H to show that G has a surjective homomorphism >to C_3 or S_3 whose kernel lies in H. I notice the OP said he >figured it out, so I wonder just what ßoor heÕs on. Or maybe >IÕm missing some more elementary approach... ItÕs easy to do this on a case by case basis: If x is in H, then x^2 is in H, so weÕre done. If x is not in H, then H and xH are disjoint. Now consider x^2. If x^2 is in H, weÕre done. If x^2 is in xH, then x is in H and weÕre done. If x^2 is in neither, then x^2H is disjoint from both H and xH. Furthermore, since [G:H]=3, these three cosets cover G. Now look at x^3. If x^3 is in H, weÕre done. If it is in xH, then x^2 is in H and weÕre done. If x^3 is in x^2H, then x is in H and weÕre done. --Dan Grubb === Subject: Re: Abstract Algebra > IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 elements. > If > x is in G, show that either x^2 or x^3 is in H. > This is what I got so far. > If x is in H then not only is x^2 in but x^3 is also in H. > So now we assume that x is not in H. > IF I can show that H, x^2H and x^3H are pairwise disjoint then G=H U x^2H U > x^3H where each coset has o( H ) =1000 elements. So if x is in G then x is > in H or x^2H or x^3H. But x is not in H. So x is in x^2H or x^3H. > So all that remains to show is that H, x^2H and x^3H are pairwise disjoint. > Since x is not in H we have H /= xH-->xH /= x^2H-->x^2H /= x^3H. So x^2 /= > x^3H. I just canÕt seem to show that H /= x^2H ( or x^2 is not in H) and H > /= x^3H (or x^3 is not in H). > What am I not seeing? The above is not completely correct. I see now how to do this. Steven === Subject: Re: Abstract Algebra > IÕm missing something on this seemingly trivial problem. > Let G be a group with 3000 elements and H be a subgroup with 1000 elements. > If > x is in G, show that either x^2 or x^3 is in H. > This is what I got so far. > If x is in H then not only is x^2 in but x^3 is also in H. > So now we assume that x is not in H. > IF I can show that H, x^2H and x^3H are pairwise disjoint then G=H U x^2H U > x^3H where each coset has o( H ) =1000 elements. So if x is in G then x is > in H or x^2H or x^3H. But x is not in H. So x is in x^2H or x^3H. > So all that remains to show is that H, x^2H and x^3H are pairwise disjoint. > Since x is not in H we have H /= xH-->xH /= x^2H-->x^2H /= x^3H. So x^2 /= > x^3H. I just canÕt seem to show that H /= x^2H ( or x^2 is not in H) and H > /= x^3H (or x^3 is not in H). > What am I not seeing? If two elements of G are in the same coset of H then their quotient is in H. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Putting FLT to Rest In sci.math, John Coleman So give us ONE counter example! > Easy: nonsense^handwave + absurdity^handwave = crank^handwave > what could be clearer? You forgot to divide by the incomprehensibility factor. That would have made it a *lot* clearer. ThereÕs also the fudge factor, which is the amount of chocolate one eats during a certain timeframe... -- #191, ewill3@earthlink.net ItÕs still legal to go .sigless. === Subject: Re: Putting FLT to Rest ... > So give us ONE counter example! That is easy. E. Escultura is in the clan of people that think that the set of integers also contains infinite integers, and are in fact 10-adics. Finding counter-examples in the 10-adics is simple (and I think he has posted them already long ago; at least I did). Consider the 10-adics as the direct sum of the 2-adics and the 5-adics. We find two idempotents in the 10-adics and their sum is 1. So this is a counterexample for each power you may wish. -- 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: the beautiful TOE In sci.math, Tapio : >> TOE:i. Theory of Everything:i. i is the simplest explanation there is. > No, there is more simple explanation. ;-) Well, the simplest explanation I can think of is that someone lost his medication. :-) -- #191, ewill3@earthlink.net ItÕs still legal to go .sigless. === Subject: Re: the beautiful TOE i take the medication to keep the world sane. that alone is the true explanation. > In sci.math, Tapio > : > TOE:i. Theory of Everything:i. i is the simplest explanation there is. >> No, there is more simple explanation. ;-) > Well, the simplest explanation I can think of is that someone lost > his medication. :-) > -- > #191, ewill3@earthlink.net > ItÕs still legal to go .sigless. === Subject: Re: lattice algorithm Oops, I overlooked one of your questions: > What do you mean by a lexicographic scan in this context? Lexicographic scan is nothing more than the result of nested loops where the outermost loop does the leftmost variable and the next inner loop does the next-to-leftmost variable etc. In this case thereÕs just a nesting two-deep because you have only two variables. for u = ulow to uhigh for v = vlow to vhigh previsit(u,v) (ThatÕs vaguely algolish/pseudocode. I hate C/java syntax, and if I === Subject: Re: lattice algorithm > I want to avoid any visits outside the diamond if possible. Why? Is a visit outside the diamond extremely expensive, so that even one such visit dominates all the compute time of your process? (For example, a visit may mean doing a remote access to a WebPage, where only the WebPages within the diamond are valid, and attempt to connect to anything outside the diamond sits there for several minutes waiting for the host to respond before it finally times out, whereas each visit within the diamond responds within a fraction of a second most of the time so that a hundred inside visits consumes only a small fraction of a minute.) If thatÕs the case, all you have to do to speed it up is do a pre-visit that simply checks whether the point is inside the diamond or not, if so then do a real visit, if not then omit the real visit. Even if you have to use multi-precision exact integer (bignum) arithmetic to make the inside/outside decision, thatÕs just a tiny fraction of a second for each point, much faster to do that calculation than to sit for minutes waiting for a non-existant WebSite or whatever. > this process is repeated for many thousands of lattices In your original query, I assumed you wanted to process rather large diamonds, and not too many of them, so going to extra work to set up the scanning, so that the scan of the large diamond can be efficient, was worth the effort. But if youÕre dealing with a very large number of very tiny diamonds, like where the bounding rectangle is no larger than 3-by-3 most of the time and never much larger than that, and if outside-diamond visits cost no more than inside-diamond visits in fact maybe they actually cost slightly less, the overhead of the set-up is more than the total cost of outside-diamond visits, so thereÕs no need to do anything more than lexicographic scan of the bounding rectangle and test for inside/outside diamond at each point, itÕs best to avoid the overhead of the setup. In fact if the bounding rectangles are always very tiny, it may be best not to bother computing the bounding rectangle at all. Instead, use this algorithm: (1) Find the center of the x,y rectangle, map to u,v coordinates and round x,y to nearest integer each. (2) Start with that point and start a spiral path around it, keeping track on each once-around whether you hit any inside points or not. (3) So long as your last once-around path hit at least one inside point, repeat. When a complete once-around didnÕt hit even one inside point the whole trip, you know youÕre done. That method minimizes the setup to an absolute minimum, and for tiny rectangles/diamonds the total cost of the spiral visiting, even that last once-around that had 0% success rate, isnÕt a lot. > I also wasnÕt quite clear that I want to find the (u,v) values rather > than the (x,y) values, and that IÕd prefer not to calculate the (x,y) > values at all if possible. You really need to try it both ways and see which turns out faster for typical data. One way, compute actual diamond shape in u,v coordinates, which is more work of setup but 50% saving on visiting time. Other way, compute only the u,v bounding rectangle, which is less set-up time but not the 50% saving on visiting time. For large diamonds, extra setup is better, but for small diamonds extra setup costs more than it saves, and for some middle size both methods take the same time, the tradeoff crossover point. Only by actually timing your code both ways on typical data can you know which side of the tradeoff-point your data is. > I was thinking that this problem is similar to the ßood-fill problem > in computer graphics, and that could use Bresenham to find the > boundary points. Yes, if you carefully use FOUR instances of BresenhamÕs method, starting your scan for example from top scanline, hence using top-left and top-right bounds, and carefully noticing when youÕve reached left or right corner and have to switch from top-left to bottom-left or from top-right to bottom-right, and noticing when you reach the bottommost point when the two lines come together, you get a big lot of setup cost but very efficient innermost loop and not too bad middle loop. (That middle loop needs to make the decision when youÕve reached either of the left/right corners, and when youÕve reached bottom corner.) If the corners of the diamond had integer u,v values, all youÕd have to do is find an existing ßood-fill-polygon subroutine where the paint-pixel call was parameterized, and simply replace the paint-pixel call by whatever your visit-point function is, and call the subroutine given a four-point polygon which is your diamond shape. But since u,v values at corners are not integers, you may have to get the source of such a ßood-fill-polygon subroutine and modify it to deal with non-integer u,v values, i.e. initialize BresenhamÕs method for each edge of the polygon in an unsual way, but the rest of the ßood-fill algorithm would be the same. Also, since youÕre modifying the code, given each pair of opposite sides of diamond have the same slope, you can optimize out the four slope calculations to do only two and just copy the result from one side to the opposite site. === Subject: Re: lattice algorithm >> I want to avoid any visits outside the diamond if possible. > Why? Is a visit outside the diamond extremely expensive, so that even > one such visit dominates all the compute time of your process? ItÕs just wasted processing, which IÕd like to avoid, if possible. Also, the outside points canÕt be actually visited, because thereÕs no memory allocated for them, so I need to check whether a point is inside or outside. Ideally IÕd like an algorithm that guaranteed that all points are inside. >> this process is repeated for many thousands of lattices > In your original query, I assumed you wanted to process rather large > diamonds, and not too many of them, so going to extra work to set up > the scanning, so that the scan of the large diamond can be efficient, > was worth the effort. The size of the diamond varies from quite large (several hundred by several hundred) to quite small, so maybe I need to use a different algorithm depending on the size of the diamond. >> I was thinking that this problem is similar to the ßood-fill problem >> in computer graphics, and that could use Bresenham to find the >> boundary points. > Yes IÕve tried using Bresenham to find the lattice points near the edges of the diamond, and then sorting the resulting set of points - this then gives me bounds for the double loop, but I have to adjust the set of points since some are outside the diamond. Currently it doesnÕt give quite the same results as my original algorithm, so I think thereÕs still a bug in there somewhere. Also, IÕm using a generic winding number algorithm to check whether a point is inside the diamond, but I suspect thereÕs a more efficient algorithm for this, since a diamond is such a simple shape. Any pointers? Chris === Subject: Help soving this problem? Trying to get back into school after many years away. I have to take an admitance exam which I am currently studying for. Here is a problem on a sample exam that is stumping me . I belive it is a y intercept problem Any ideas on how to solve? I am thinking you solve for y but that didnÕt seem to work for me. M and N are the X and Y coordinates, respectively, of a point in a coordinate plane. If the points (M,N) and (M + P, N +4) both lie on the line defined by the equation X = y/2 - 2/5, what is the value of P? === Subject: Re: Help soving this problem? So each point must solve the line equation (since they both lie on it) I.e. For point (M,N), we have: M=N/2 - 2/5 And for (M + P, N+4), we have: M+P = (N+4)/2 -2/5 We substitute the value for M from our first equation into our second equation: (N/2 - 2/5) + P = (N+4)/2 -2/5 implies N/2 - 2/5 + P = N/2 + 2 - 2/5 implies P = 2 -Darren > Trying to get back into school after many years away. I have to take > an admitance exam which I am currently studying for. Here is a problem > on a sample exam that is stumping me . I belive it is a y intercept > problem Any ideas on how to solve? I am thinking you solve for y but > that didnÕt seem to work for me. > M and N are the X and Y coordinates, respectively, of a point in a > coordinate plane. If the points (M,N) and (M + P, N +4) both lie on > the line defined by the equation X = y/2 - 2/5, what is the value of > P? === Subject: Re: Help soving this problem? still think the problem is way too hard for liberal arts admitance test :) Fletch > So each point must solve the line equation (since they both lie on it) > I.e. > For point (M,N), we have: > M=N/2 - 2/5 > And for (M + P, N+4), we have: > M+P = (N+4)/2 -2/5 > We substitute the value for M from our first equation into our second > equation: > (N/2 - 2/5) + P = (N+4)/2 -2/5 > implies > N/2 - 2/5 + P = N/2 + 2 - 2/5 > implies > P = 2 > -Darren > Trying to get back into school after many years away. I have to take > an admitance exam which I am currently studying for. Here is a problem > on a sample exam that is stumping me . I belive it is a y intercept > problem Any ideas on how to solve? I am thinking you solve for y but > that didnÕt seem to work for me. > M and N are the X and Y coordinates, respectively, of a point in a > coordinate plane. If the points (M,N) and (M + P, N +4) both lie on > the line defined by the equation X = y/2 - 2/5, what is the value of > P? === Subject: Re: Help soving this problem? >Trying to get back into school after many years away. I have to take >an admitance exam which I am currently studying for. Here is a problem >on a sample exam that is stumping me . I belive it is a y intercept >problem Any ideas on how to solve? I am thinking you solve for y but >that didnÕt seem to work for me. >M and N are the X and Y coordinates, respectively, of a point in a >coordinate plane. If the points (M,N) and (M + P, N +4) both lie on >the line defined by the equation X = y/2 - 2/5, what is the value of 1. Solve your equation for y to get the form y = mx + b and read off the slope m. 2. Compute the slope between your two points and set it equal to your value of m. 3. Solve for P --Lynn === Subject: Re: Help soving this problem? >Trying to get back into school after many years away. I have to take >an admitance exam which I am currently studying for. Here is a problem >on a sample exam that is stumping me . I belive it is a y intercept >problem Any ideas on how to solve? I am thinking you solve for y but >that didnÕt seem to work for me. >M and N are the X and Y coordinates, respectively, of a point in a >coordinate plane. If the points (M,N) and (M + P, N +4) both lie on >the line defined by the equation X = y/2 - 2/5, what is the value of >P? > 1. Solve your equation for y to get the form y = mx + b and read off > the slope m. > 2. Compute the slope between your two points and set it equal to your > value of m. > 3. Solve for P > --Lynn === Subject: Re: Help soving this problem? >Trying to get back into school after many years away. I have to take >an admitance exam which I am currently studying for. Here is a problem >on a sample exam that is stumping me . I belive it is a y intercept >problem Any ideas on how to solve? I am thinking you solve for y but >that didnÕt seem to work for me. >M and N are the X and Y coordinates, respectively, of a point in a >coordinate plane. If the points (M,N) and (M + P, N +4) both lie on >the line defined by the equation X = y/2 - 2/5, what is the value of >P? M = N/2 - 2/5 M+P = (N+4)/2 - 2/5 Subtract the first from the second. === Subject: Re: Help soving this problem? I donÕt think I follow you... Fletch Trying to get back into school after many years away. I have to take >an admitance exam which I am currently studying for. Here is a problem >on a sample exam that is stumping me . I belive it is a y intercept >problem Any ideas on how to solve? I am thinking you solve for y but >that didnÕt seem to work for me. M and N are the X and Y coordinates, respectively, of a point in a >coordinate plane. If the points (M,N) and (M + P, N +4) both lie on >the line defined by the equation X = y/2 - 2/5, what is the value of >P? > M = N/2 - 2/5 > M+P = (N+4)/2 - 2/5 > Subtract the first from the second. === 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 > 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 isnÕt immovable or the force isnÕt unstoppable. > Any finite mass on which a net force is exerted will accelerate in the > direction of the net force. NewtonÕs Second Law. > Bob Kolker Yes, I know. In that case, itÕs 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, itÕs 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, itÕs 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? WhatÕs wrong with the argument? === Subject: Re: Unstoppable Force vs Immovable Object > Why? WhatÕs 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? WhatÕs 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 donÕt think youÕve identified a problem with my reasoning. === Subject: Re: Unstoppable Force vs Immovable Object > I donÕt think youÕve 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 donÕt think youÕve 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 WhatÕs that supposed to mean, I address a physical concept in a non-physical way? I conclude that itÕs logically impossible for an unstoppable force to meet an immovable object. Do you disagree? Do you think itÕs logically possible? === Subject: Re: Unstoppable Force vs Immovable Object > WhatÕs that supposed to mean, I address a physical concept in a > non-physical way? > I conclude that itÕs logically impossible for an unstoppable force to > meet an immovable object. Do you disagree? Do you think itÕs 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 donÕt exist, but canÕt exist. Bob Kolker === Subject: Re: Unstoppable Force vs Immovable Object > WhatÕs that supposed to mean, I address a physical concept in a > non-physical way? > I conclude that itÕs logically impossible for an unstoppable force to > meet an immovable object. Do you disagree? Do you think itÕs 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 donÕt exist, > but canÕt exist. > Bob Kolker So you conclude that itÕs logically impossible for an unstoppable force or an immovable object to exist. IÕm not sure thatÕs correct, but in any event itÕs consistent with my conclusion. So whatÕs to argue about? === Subject: Re: Deviration Of The Spectrom Deu To f(t).d(t - T)????? Neener, and Neener that... > Nothing wrong with top-posting. ThatÕs what clueless newbies always say. > It is the preferred method, because you donÕt have to > page down through much repeated and already-seen > history. > skipped over unread. > In a previous post. > Top-posting makes the whole thing impractical. -- Checkmate all rights reserved === Subject: Graphing polynomial equations Hi. Houw would one go about plotting a 3D surface graph of a quintic or higher equation in three variables? Like this: xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0 Obviously you would loop through x and y and calculate z, and weÕd use a numerical approximation technique to do that. IÕm trying to write a program that would plot the graph. The problem boils down to applying a numerical algorithm to a quintic like this: z^5 + a4 z^4 + a3 z^3 + a2 z^2 + a1 z + a0 = 0. But the problem IÕm having is this: Most numerical approximation algorithms require an initial guess. So, what would be a good algorithm to find a good initial guess? -- We should have a town named Alderaan someday. No, seriously. LetÕs put it on the table. === Subject: Re: Graphing polynomial equations >xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0 >Obviously you would loop through x and y and calculate >z, and weÕd use a numerical approximation technique to >do that. IÕm trying to write a program that would plot >the graph. The problem boils down to applying a >numerical algorithm to a quintic like this: >z^5 + a4 z^4 + a3 z^3 + a2 z^2 + a1 z + a0 = 0. >But the problem IÕm having is this: Most numerical >approximation algorithms require an initial guess. So, what >would be a good algorithm to find a good initial >guess? You probably realize that a degree-5 polynomial can have up to 5 zeroes, so you would want 5 (or maybe more) initial guesses. The high and low guesses should be fairly easy, because thereÕs no such thing as too high or too low for the tangential method (sorry I forgot who discovered that or IÕd give him credit!). Even if youÕre using some other numerical method you can still use the tangential results to seed it. You can get the other possibilities by looking at the zeroes of the derivative, which are potentially relative mimima and maxima. Any zeroes other than the first and last will be between a relative maximum and a relative minimum. You may have to apply this process recursively until you get down to degree 2 where you can use the quadratic equation. Once youÕre started, you can use results from neighboring (x,y) as guesses. They wonÕt always work out, but when they do it will save time. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Graphing polynomial equations >Houw would one go about plotting a 3D surface graph of >a quintic or higher equation in three variables? >Like this: >xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0 Try surf.sourceforge.net . === Subject: Re: Graphing polynomial equations >Houw would one go about plotting a 3D surface graph of >a quintic or higher equation in three variables? >Like this: >xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0 > Try surf.sourceforge.net . === Subject: Re: Graphing polynomial equations > Houw would one go about plotting a 3D surface graph of > a quintic or higher equation in three variables? > Like this: > xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0 > Obviously you would loop through x and y and calculate > z, and weÕd use a numerical approximation technique to > do that. The particular equation youÕve given is linear in y, so if I were to follow your approach IÕd loop through x and z and calculate y by baby algebra. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Graphing polynomial equations > Houw would one go about plotting a 3D surface graph of > a quintic or higher equation in three variables? > Like this: > xyz^3 + 3xz + 3zy + xyz + zyx^2 + x - y + 3 = 0 > Obviously you would loop through x and y and calculate > z, and weÕd use a numerical approximation technique to > do that. > The particular equation youÕve given is linear in y, > so if I were to follow your approach IÕd loop through > x and z and calculate y by baby algebra. But what about a GENERAL algorithm for a GENERAL polynomial? > -- > Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Embedding theorem and uniqueness? >> For example, for an arbitrary 3+1-dimension Lorentzian >> manifold to be isometrically embedded into an N+M-dimension >> ßat manifold, the lowest known dimension is N=88,M=2 (!). > That is terrible. Is this lower bound considered > conservative, or are there particular examples that are known to need > this? I donÕt know. There are definitely 3+1-d manifolds that cannot be embedded isometrically in a 4+1-d ßat manifold. But of course there are also 2-d manifolds that cannot be embedded in a 3-d ßat Euclidean space (Klein bottle). As this last example shows, you need to understand issues of orientability.... > Part of the problem is that I am still trying to determine the > more rigerous definition. IÕm not knowledgeable enough to help. And not interested enough to pursue it. >Basically I am trying to build an intuitive > view of GR, by picturing metrics as surfaces in higher dimensional > space. I suspect that is either useless or doomed. AFAICT the important thing about GR is that you really want to discuss and understand it intrinsically. After all, we clearly have no way to go outside the entire universe to examine any such embedding. In my experience, modern tensor analysis goes a long way here, and the old component notation is obscure at best, and often misleading. > For instance I know that the robertson walker metric at a > particular time slice has constant curvature everywhere. I can > picture this as a sphere in a n+1 dimensional space. Hmmm. That completely misses the time evolution, which is the interesting part. In fact, for all but trivial manifolds the time dependence is the intriguing part. And to handle that you must inherently use semi-Riemannian techniques. Tom Roberts tjroberts@lucent.com === Subject: Re: Embedding theorem and uniqueness? <9oxrd.37038$Qv5.34139@newssvr33.news.prodigy.com> posting-account=skOyXQwAAADl3JUBhIA36tRgey9Rva9d >Basically I am trying to build an intuitive > view of GR, by picturing metrics as surfaces in higher dimensional > space. > I suspect that is either useless or doomed. AFAICT the important thing > about GR is that you really want to discuss and understand it > intrinsically. After all, we clearly have no way to go outside the > entire universe to examine any such embedding. In my experience, modern > tensor analysis goes a long way here, and the old component notation is > obscure at best, and often misleading. IÕve been doing GR with modern tensor analysis for years now. The problem with tensor analysis is the guage invariance. I find it frusterating that you can completely obscure the manifold in question. For instance, the seemingly interesting metric ds^2=dv^2-v^2du^2 is merely the ßat minkowski space with a change of coordinates. If someone gave me that metric I could see myself wasting time trying to figure out properties of it before realizing what I was dealing with. And the metric could be made much more complicated than that (choose nonorthogonal coordinates). As a shape in a higher dimensional space this ambiguity would be immediately leapfrogged. Or take the Schwarzschild solution. For years it was believed that the solution was divergent near the event horizon, because the metric blew up. Then it was shown that this is not at all the case- the only reason for the divergance was a bad choice of coordinates. The manifold is completely well behaved at the event horizon (not so at the singularity or course). The process of visualizing this presented in Missner, Thorn, Wheeler involves something akin to an embedding into higher space (although they donÕt explicitely call it this). > For instance I know that the robertson walker metric at a > particular time slice has constant curvature everywhere. I can > picture this as a sphere in a n+1 dimensional space. > Hmmm. That completely misses the time evolution, which is the > interesting part. In fact, for all but trivial manifolds the time > dependence is the intriguing part. And to handle that you must > inherently use semi-Riemannian techniques. Correct, I was making a point. Even the most trivial solutions (a time slice in the Robertson Walker metric) can not be dealt with in a straightforward manner. A simple sphere can be completely obscured by choosing strange coordinates, and necessarily always will have a zero or singularity in the metric (hairy ball theorem). Any by remaining in the lower dimensional space I donÕt even know if there is a choice of coordinates that can exploit the symetries of the solution (the solution has perfect translation symmetry, but that is not evident from the usual metric). In a higher dimensional space better coordinates could be chosen. There has to be a better way than dealing with the usual coordinates. -I === Subject: Re: Embedding theorem and uniqueness? <9oxrd.37038$Qv5.34139@newssvr33.news.prodigy.com> posting-account=skOyXQwAAADl3JUBhIA36tRgey9Rva9d >Basically I am trying to build an intuitive > view of GR, by picturing metrics as surfaces in higher dimensional > space. > I suspect that is either useless or doomed. AFAICT the important thing > about GR is that you really want to discuss and understand it > intrinsically. After all, we clearly have no way to go outside the > entire universe to examine any such embedding. In my experience, modern > tensor analysis goes a long way here, and the old component notation is > obscure at best, and often misleading. IÕve been doing GR with modern tensor analysis for years now. The problem with tensor analysis is the guage invariance. I find it frusterating that you can completely obscure the manifold in question. For instance, the seemingly interesting metric ds^2=dv^2-v^2du^2 is merely the ßat minkowski space with a change of coordinates. If someone gave me that metric I could see myself wasting time trying to figure out properties of it before realizing what I was dealing with. And the metric could be made much more complicated than that (choose nonorthogonal coordinates). As a shape in a higher dimensional space this ambiguity would be immediately leapfrogged. Or take the Schwarzschild solution. For years it was believed that the solution was divergent near the event horizon, because the metric blew up. Then it was shown that this is not at all the case- the only reason for the divergance was a bad choice of coordinates. The manifold is completely well behaved at the event horizon (not so at the singularity or course). The process of visualizing this presented in Missner, Thorn, Wheeler involves something akin to an embedding into higher space (although they donÕt explicitely call it this). > For instance I know that the robertson walker metric at a > particular time slice has constant curvature everywhere. I can > picture this as a sphere in a n+1 dimensional space. > Hmmm. That completely misses the time evolution, which is the > interesting part. In fact, for all but trivial manifolds the time > dependence is the intriguing part. And to handle that you must > inherently use semi-Riemannian techniques. Correct, I was making a point. Even the most trivial solutions (a time slice in the Robertson Walker metric) can not be dealt with in a straightforward manner. A simple sphere can be completely obscured by choosing strange coordinates, and necessarily always will have a zero or singularity in the metric (hairy ball theorem). Any by remaining in the lower dimensional space I donÕt even know if there is a choice of coordinates that can exploit the symetries of the solution (the solution has perfect translation symmetry, but that is not evident from the usual metric). In a higher dimensional space better coordinates could be chosen. There has to be a better way than dealing with the usual coordinates. -I === Subject: Re: Embedding theorem and uniqueness? <9oxrd.37038$Qv5.34139@newssvr33.news.prodigy.com> posting-account=skOyXQwAAADl3JUBhIA36tRgey9Rva9d >Basically I am trying to build an intuitive > view of GR, by picturing metrics as surfaces in higher dimensional > space. > I suspect that is either useless or doomed. AFAICT the important thing > about GR is that you really want to discuss and understand it > intrinsically. After all, we clearly have no way to go outside the > entire universe to examine any such embedding. In my experience, modern > tensor analysis goes a long way here, and the old component notation is > obscure at best, and often misleading. IÕve been doing GR with modern tensor analysis for years now. The problem with tensor analysis is the guage invariance. I find it frusterating that you can completely obscure the manifold in question. For instance, the seemingly interesting metric ds^2=dv^2-v^2du^2 is merely the ßat minkowski space with a change of coordinates. If someone gave me that metric I could see myself wasting time trying to figure out properties of it before realizing what I was dealing with. And the metric could be made much more complicated than that (choose nonorthogonal coordinates). As a shape in a higher dimensional space this ambiguity would be immediately leapfrogged. Or take the Schwarzschild solution. For years it was believed that the solution was divergent near the event horizon, because the metric blew up. Then it was shown that this is not at all the case- the only reason for the divergance was a bad choice of coordinates. The manifold is completely well behaved at the event horizon (not so at the singularity or course). The process of visualizing this presented in Missner, Thorn, Wheeler involves something akin to an embedding into higher space (although they donÕt explicitely call it this). > For instance I know that the robertson walker metric at a > particular time slice has constant curvature everywhere. I can > picture this as a sphere in a n+1 dimensional space. > Hmmm. That completely misses the time evolution, which is the > interesting part. In fact, for all but trivial manifolds the time > dependence is the intriguing part. And to handle that you must > inherently use semi-Riemannian techniques. Correct, I was making a point. Even the most trivial solutions (a time slice in the Robertson Walker metric) can not be dealt with in a straightforward manner. A simple sphere can be completely obscured by choosing strange coordinates, and necessarily always will have a zero or singularity in the metric (hairy ball theorem). Any by remaining in the lower dimensional space I donÕt even know if there is a choice of coordinates that can exploit the symetries of the solution (the solution has perfect translation symmetry, but that is not evident from the usual metric). In a higher dimensional space better coordinates could be chosen. There has to be a better way than dealing with the usual coordinates. -I === Subject: Re: Embedding theorem and uniqueness? > > [... interesting stuff] > I just want to confirm it is interesting stuff indeed. I am trying to > follow through to the best of my ability. My background in math is not > topology - it was functional analysis, distribution theory and non standard > analysis although I did take standard courses such as introduction to > algebraic topology (largely forgotten). > Do you think anyone really gives a , you arrogant ass? When people go to the trouble of posting detailed references those that have courtesy thank them and give them feedback. Of course such is lost on a troll like you whose only jollies in life is posting drivel on what should be serious newsgroups and watching what ensures. > You must think this whole NG is about you. Of course not - nor is it about your semantically motivated idiotic spew concerning time and motion that is a pretext for your troll antics. Bill === Subject: Re: Embedding theorem and uniqueness? And the obsession continues........ === Subject: Re: Embedding theorem and uniqueness? > Hi- > IÕm not sure what you mean by the embedding theorem. There are several > embedding theorems for manifolds, from WhitneyÕs theorem embedding n- > dimensional manifolds in R^(2n), to NashÕs isometric embedding theorem, > to a number of refinements. > Sorry, I was referring to isometric embedding. > You probably meant the sphereÕs radius to be 1/a: a larger sphere has > smaller curvature. > Sorry again, I was being stupid. I definitely meant 1/a. > The Gauss-Bonnet Theorem expresses the Euler characteristic of a > manifold in terms of the integral of the Gaussian curvature. This > is sufficient to guarantee that the surface has Euler characteristic > equal to 2, since it must be positive, and all compact connected > surfaces have Euler characteristic given by 2 - 2g, where g is the > genus (or number of handles, for a sphere with n handles). > For higher-dimensional manifolds, a manifold with sectional curvature > strictly between 1 and 4 is homeomorphic to a sphere. > If I am correct this is only applicable to topological > embedding (am I misreading?). What about isometric? Is it clear for > example that any 2 manifold with constant curvature a^2 embedded in 3 > space is exactly a sphere of radius 1/a? I suspect that there might > be other surfaces with constant curvature a^2 at local patches but not > necessarily ones that globally so (I suspect so because curvature of a > 2 dimensional manifold in 3 space = product of curvature of a line > going through the point in two orthogonal directions, so a point can > have curvature a^2 not only by having curvature a for two lines going > through the point, but also b and c such that b*c=a). > I am interested in this because I am trying to build a more > intuitive view of GR, and in many cases I can better visualize a > manifold as a surface in euclidean space than in the original lower > dimensional curved space (my brain evolved to like euclidean space). > Of course if the euclidean space is 40 dimensions I might be better > off not doing this but there are many cases where the euclidean space > is of low dimension (Robertson-Walker, even the Schwarzschild can be > embedded in low dimensional space). I am even considering for my own > purposes looking at how the Einstein equation looks from this > paradigm, but if there is no unique surface in a higher dimensional > space corresponding to a metric, the whole process might be doomed. You may be interested in what WessonÕs theory STM has to say on the issue of embeddings. See http://arxiv.org/abs/gr-qc/0302015. It is interesting to note that in such 5D embeddings FRW models are computationally found to be a hypersuface of a ßat 5D space - http://arxiv.org/abs/gr-qc/0202010. Bill > One of the reasons I like this approach is because of the ugly gauge > invariance of the manifold metric (ie- the same manifold can be > described by many different metrics all differing by simple change of > coordinates). If the surface of the embedded manifold is unique I can > visualize a manifold in a very specific way. > -I Supersedes: === Subject: Invariant Galilean Transformations (FAQ) On All Laws Summary: All laws/equations are Galilean invariant when expressed in the generalized cartesian coordinates demanded by basic analytic geometry, vector algebra, and measurement theory. Originator: faqserv@penguin-lust.mit.edu Disclaimer: approval for *.answers is based on form, not content. Opponents of the content should first actually find out what it is, then think, then request/submit-to arbitration by the appropriate neutral mathematics authorities. Flaming the hard- working, selßess, *.answers moderators evidences ignorance and despicable netiquette. Archive-Name: physics-faq/criticism/galilean-invariance Version: 0.04.03 Posting-frequency: 15 days Invariant Galilean Transformations (FAQ) On All Laws (c) Eleaticus/Oren C. Webster Thnktank@concentric.net An obvious typo or two corrected. The Brittanica section revised to less Ōpussy-footingÕ and to more directly anticipate the elementary measurement theory and basic analytic geometry that is applied to the transformation concept. ------------------------------ === Subject: 1. Purpose The purpose of this document is to provide the student of Physics, especially Relativity and Electromagnetism, the most basic princ- iples and logic with which to evaluate the historic justification of Relativity Theory as a necessary alternative to the classical physics of Newton and Galileo. We will prove that all laws are invariant under the Galilean transformation, rather than some being non-invariant, after we show you what that means. We shall also show that another primal requirement that SR exist is nonsense: Michelson-Morley and Kennedy-Thorndike do indeed fit Galilean (c+v) physics. ------------------------------ === Subject: 2. Table of Contents 1. Foreword and Intent 2. Table of Contents 3. The Principle of Relativity 4. The Encyclopedia Brittanica Incompetency. 5. Transformations on Generalized Coordinate Laws 6. The data scale degradation absurdity. 7. The CrackpotsÕ Version of the Transforms. 8. What does sci.math have to say about x0Õ=x0-vt? 9. But DoesnÕt x.cÕ=x.c? 10. But IsnÕt (xÕ-x.cÕ)=(x-x.c) Actually Two Transformations? 11. But DoesnÕt (xÕ-x.c+vt) Prove The Transformation Time Dependent? 12. But IsnÕt (xÕ-x.cÕ)=(x-x.c) a Tautology? 13. But IsnÕt (xÕ-x.cÕ)=(x-x.c) Almost the Definition of a Linear Transform? 14. But The Transform WonÕt Work On Time Dependent Equations? 15. But The Transform WonÕt Work On Wave Equations? 16. But MaxwellÕs Equations ArenÕt Galilean Invariant? 17. First and Second Derivative differential equations. ------------------------------ === Subject: 3. The Principle of Relativity and Transformation If a law is different over there than it is here, it is not one law, but at least two, and leaves us in doubt about any third location. This is the Principle of Relativity: a natural law must be the same relative to any location at which a given event may be perceived or measured, and whether or not the observer is moving. The idea of location translates to a coordinate system, largely because any object in motion could be considered as having a coordinate system origin moving with it. If you perceive me moving relative to you - who have your own coordinate system - will your measurements of my position and velocity fit the same laws my own, different measurements fit? If a law has the same form in both cases it is called covariant. If it is identical in form, var- ables, and output values, it is called invariant. What weÕre asking is that if the x-coordinate, x, on one coordinate axis works in an equation, does the coordinate, xÕ, on some other, parallel axis work? Speaking in terms of the axis on which x is the coordinate, xÕ is the ŌtransformedÕ coordinate. The situation is complicated because weÕre talking about coordinates - locations - but in most mean- ingful laws/equations, it is lengths/distances (and time intervals) the equations are about, and x coord- inates that represent good, ratio scale measures of distances are only interval scale measures on the xÕ axis. [See Table of Contents for discussion of scales.] So, if we have an x-coordinate in one system, then we can call the xÕ value that corresponds to the same point/location the transform of x. In particular, the Principle of Relativity is embodied in the form of the Galilean transformation, which relates the original x, y, z, t to xÕ, yÕ, zÕ, tÕ by the transform equations xÕ=x-vt, yÕ=y, zÕ=z, tÕ=t in the simplified case where attention is focused only on transforming the x-axis, and not y and z. In the case of Special Relativity, the xÕ transform is the same except that xÕ is then divided by sqrt(1-(v/c)^2), and tÕ=(t-xv/cc)/sqrt(1-(v/c)^2). In either case, v is the relative velocity of the coordinate systems; if there is already a v in the equations being trans- formed use u or some other variable name. ------------------------------ === Subject: 4. The Encyclopedia Brittanica Incompetency. One example of the traditional fallacious idea that an equation is not invariant under the galilean transformation comes from the Encyclopedia Brittanica: Before EinsteinÕs special theory of relativity was published in 1905, it was usually assumed that the time coordinates measured in all inertial frames were identical and equal to an Ōabsolute timeÕ. Thus, t = tÕ. (97) The position coordinates x and xÕ were then assumed to be related by xÕ = x - vt. (98) The two formulas (97) and (98) are called a Galilean transformation. The laws of nonrelativ- istic mechanics take the same form in all frames related by Galilean transformations. This is the restricted, or Galilean, principle of relativity. The position of a light wave front speeding from the origin at time zero should satisfy x^2 - (ct)^2 = 0 (99) in the frame (t,x) and (xÕ)^2 - (ctÕ)^2 = 0 (100) in the frame (tÕ,xÕ). Formula (100) does not transform into formula (99) using the transform- ations (97) and (98), however. ................................................. Besides the trivially correct statement of what the Galilean ŌtransformÕ equations are, there is exactly one thing they got right. I. Eq-100 is indeed the correct basis for discussing the question of invariance, given that eq-99 is the correct ŌstationaryÕ (observer S) equation. [Let observer M be the ŌmovingÕsystem observer.] In particular, eq-100 is of exactly the same form [the square of argument one minus the square of argument two equals zero (argument three).] II. It is nonsense to say eq-99 should be derivable from eq-100; for one thing, the transforms are TO xÕ and tÕ from x and t, not the other way around, and the idea that either observerÕs equation should contain within itself the terms to simplify or rearrange to get to the other is ridiculous. As the transform equations say, the relationship of tÕ, xÕ to t, x is based on the relative velocity between the two systems, but neither the original (eq-99) equation nor the M observer equation is about a relationship between coordinate systems or observers. One might as well expect the two equations to contain banana export/import data; there is no relevancy. The ŌtransformÕ equations are the relationships between xÕ and x, tÕ and t and have nothing to do with what one equation or the other ought to ŌsayÕ. The equationsÕ content is the rate at which light emitted along the x-axes moves. III. Most remarkable, the True Believer SR crackpots who most despise the consequences of measurement theory (demonstrable fact) contained in this document are those who want to argue against our saying the Britt- anica got eq-100 right; They insist that the correct equation is derived directly from xÕ=x-vt and tÕ=t. Solve for x=xÕ+vt and replace t with tÕ, then substitute the result in eq-99: (xÕ+vtÕ)^2 - (ctÕ)^2 = 0. Besides the fact that this results in an equation with arguments exactly equal to eq-99, they will insist the transform is not invariant. IV. A major justification they have for their idea of the correct M system equation on which to base the the discussion of invariance, is that the variables are M system variables, never mind the fact that the arguments are S system values. That argument of theirs is arrant nonsense. The velocity v that S sees for the M system relative to herself is the negative of what the M system sees for the S system relative to himself. In other words, xÕ+vtÕ is a mixed frame expression and it is xÕ+(-v)tÕ that would be strictly M frame notation, and that equation is far off base. [Work it out for yourself, but make sure you try out an S frame negative v so as not to mislead yourself.] V. In I. we said: given that eq-99 is the correct ŌstationaryÕ equation. LetÕs look at it closely: x^2 - (ct)^2 = 0 (99) This whole matter is supposed to be about coordinate transforms. Is that what t is, just a coordinate? No. It isnÕt, in general. Suppose you and I are both modelling the same light event and you are using EST and IÕm using PST. ŌJust a time coordinateÕ is just a clock reading amd your t clock reading says the light has been moving three hours longer than my clock reading says. Well, thatÕs what the idea that t is a coordinate means. Eq-99 works if and only if t is a time interval, and in particular the elapsed time since the light was emitted. Thus, that equation works only if we understand just what t is, an elapsed time, with emissioon at t=0. However, we donÕt have to ŌunderstandÕ anything if we use a more intelligent and insightful form of the equation: (x)^2 - [ c(t-t.e) ]^2 = 0, where t.e is anyoneÕs clock reading at the time of light emission, and t is any subsequent time on the same clock. Similarly, x is not just a coordinate, but a distance since emission. (x-x.e)^2 - [ c(t-t.e) ]^2 = 0 (99a) VI. In the spirit of Ōthere is exactly one thing they got rightÕ, the correct M system version of eq-99a is eq-100a: (xÕ-x.eÕ)^2 - [ c(tÕ-t.eÕ) ]^2 = 0 (100a) Every observer in the universe can derive their eq-100a from eq-99a and vice versa, not to mention to and from every other observerÕs eq-99a. Now, THATÕs invariance. [You do realize that every eq-100a reduces to eq-99a, when you back substitute from the transforms, right? t.eÕ=t.e, x.eÕ=x.e-vt.] ------------------------------ === Subject: 5. Transformations on Generalized Coordinate Laws The traditional Gallilean transform is correct: tÕ = t xÕ = x - vt. But remember this: a transform of x doesnÕt effect just some values of x, but all of them, whether they are in the formula or not. This is important if you want to do things right. The crackpot position is strongly against this sci.math verified position, and the apparently standard coordinate pseudo-transformation they suggest is perhaps the result. {See Table of Contents.] LetÕs use a simple equation: x^2 + y^2 = r^2, which is the formula for a circle with radius r, centered at a location where x=0. But what if the circle center isnÕt at x=0? Well, weÕd want to use the form analytic geometry, vector algebra, and elementary measurement theory tells us to use, a form where we make explicit just where the circle center is, even if it is at x=x0=0: (x-x0)^2 + (y-y0)^2 = r^2. The circle center coordinate, x0, is an x-axis coordinate, just like all the x-values of points on the circle. So, in proper generalized cartesian coordinate forms of laws/equations we want to transform every occurence of x and x0 - by whatever name we call it: x.c, x_e, whatever. So, what is the transformed version of (x-x0)? Why, (xÕ-x0Õ); both x and x0 are x-coordinates, and every So, what is the value of (xÕ-x0Õ) in terms of the original x data? is also true for x0Õ=x0-vt: (xÕ-x0Õ)=[ (x-vt)-(x0-vt) ]=(x-x0). In other words, when we use the generalized coordinate form specified by analytic geometry, we find that the value of (xÕ-x0Õ) does not depend on either time or velocity in any way, shape, form, or fashion. Similarly for (y-y0). We can treat time the same way if necessary: (t-t0). The above is a proof that any equation in x,y,z,t is invariant under the galilean transforms. Just use the generalized coordinate form, with (x-x0)/etc, in the transformation process, not the incompetently selected privileged form, with just x/etc. [The form is privileged because it assumes the circle center, point of emission, whatever, is at the origin of the axes instead at some less convenient point. After transform the coordinate(s) of the circle center/origin are also changed but the privileged form doesnÕt make this explicit and screws up the calculations, which should be based on (xÕ-x0Õ) but are calculated as (xÕ-0).] The value of (xÕ-x0Õ) is the same as (x-x0). That makes sense. Draw a circle on a piece of paper, maybe to the right side of the paper. On a transparent sheet, draw x and y coordinate axes, plus x to the right, plus y at the top. Place this axis sheet so the y-axis is at the left side of the circle sheet. Now answer two questions after noting the x-coordinate of the circle center and then moving the axis sheet to the right: (a) did the circle change in any way because you moved the axis sheet (ie because you transformed the coordin- nate axis)? (b) did the coordinate of the circle center change? The circle didnÕt change [although SR will say it did]; that means that (xÕ-x0Õ) does indeed equal (x-x0). The coordinate of the circle center did change, and it changed at the same rate (-vt) as did every point on the circle. That means that x0Õ<>x0, and the fact the circle center didnÕt change wrt the circle, means that the relationship of x0Õ with x0 is the same as that of any xÕ on the circle with the corresponding x: xÕ=x-vt; x0Õ=x0-vt. This is to prepare you for the True Believer crackpots that say ŌconstantÕ coordinates canÕt be transformed; some even say they arenÕt coordinates. These crackpots include some that brag about how they were childhood geniuses, btw. QED: The galilean transformation for any law on generalized Cartesian coordinates is invariant under the Galilean transform. The use of the privileged form explains HOW the transformed equation can be messed up, the next Subject explains what the screwed up effect of the transform is, and how use of the generalized form corrects the screwup. ------------------------------ === Subject: 6. The data scale degradation absurdity. The SR transforms and the Galilean transforms both convert good, ratio scale data to inferior interval scale data. The effect is corrected, allowed for, when the transforms are conducted on the generalized coordinate forms specified by analytic geometry and vector algebra. Both sets of transforms are ŌtranslationsÕ - lateral movements of an axis, increasing over time in these cases - but with the SR transform also involving a rescaling. It is the translation term, -vt in the x transform to xÕ, and -xv/cc in the t transform to tÕ, that degrades the ratio scale data to interval scale data. In general, rescaling does not effect scale quality in the size-of-units sense we have here. SR likes to consider its transforms just rotations, however - in spite of the fact Einstein correctly said they were ŌtranslationsÕ (movements) - and in the case of ŌgoodÕ rotations, ratio scale data quality is indeed preserved, but SR violates the conditions of good ro- tations; they are not rigid rotations and they donÕt appropriately rescale all the axes that must be rescaled to preserve compatibility. The proof is in the pudding, and the pudding is the combination of simple tests of the transformations. We can tell if the transformed data are ratio scale or interval. Ratio scale data are like absolute Kelvin. A measure- ment of zero means there is zero quantity of the stuff being measured. Ratio scale data support add- ition, subtraction, multiplication, and division. The test of a ratio scale is that if one measure looks like twice as much as another, the stuff being measured is actually twice as much. With absolute Kelvin, 100 degrees really is twice the heat as 50 degrees. 200 degrees really is twice as much as 100. Interval scale data are like relative Celsius, which is why your science teacher wouldnÕt let you use it in gas law problems. There is only one mathematical operation interval scales support, and that has to be between two measures on the same scale: subtraction. 100 degrees relative (household) Celsius is not twice as much as 50; we have to convert the data to absolute Kelvin to tell us what the real ratio of temperatures is. However, whether we use absolute Kelvin or relative Celsius, the difference in the two temperature readings is the same: 50 degrees. Thus, if we know the real quantities of the ŌstuffÕ being measured, we can tell if two measures are on a ratio scale by seeing if the ratio of the two measures is the same as the ratio of the known quant- ities. If a scale passes the ratio test, the interval scale test is automatically a pass. If the scale fails the ratio test, the interval scale test becomes the next in line. It isnÕt just the bare differences on an interval scale that provides the test, however. Differences in two interval scale measures are ratio scale, so it is ratios of two differences that tell the tale. LetÕs do some testing, and remember as we do that our concern is for whether or not the data are messed up, not with ŌreasonsÕ, excuses, or avoidance. ------------------------------------------------------ Are we going to take a transformed length (difference) and see whether that length fits ratio or interval scale definitions? Of course, not. Interval scale data are ratio after one measure is subtracted from another. That is the major reason the SR transforms can be used in science. Let there be three rods, A, B, C, of length 10, 20, 40, respectively. These lengths are on a known ratio scale, our original x-axis, with one end of each rod at the origin, where x=0, and the other end at the coordinate that tells us the correct lengths. Note that these x-values are ratio scale only because one end of each rod is at x=0. That may remind you of the correct way to use a ruler or yard/meter-stick: put the zero end at one end of the thing you are measuring. Put the 1.00 mark there instead of the zero, and you have interval scale measures. Let A,B,C, be 10, 20, 40. Let a,b,c be xÕ at v=.5, t=10. xÕ=x-vt. A B C a b c ---------------- -------------------- 10 20 40 5 15 35 ---------------- -------------------- B/A = 2 b/a = 3 C/A = 4 c/a = 7 C/B = 2 c/b = 2.333 Obviously, the transformed values are no longer ratio scale. The effect is less on the greater values. C-A = 10 b-a = 10 C-A = 30 c-a = 30 C-B = 20 c-b = 20 Obviously, the transformed values are now interval scale. This will hold true for any value of time or velocity. (C-A)/(B-A) = 3 (c-a)/(b-a) = 3 (C-B)/(B-A) = 2 (c-b)/(b-a) = 2 Obviously, the ratios of the differences are ratio scale, being identical to the ratios of the corresponding original - ratio scale - differences. The main difference between these results and the SR results is that the differences do not correspond so neatly to the original, ratio scale, differences. This is due only to the rescaling by 1/sqrt(1-(v/c)^2). The ratios of the differences on the transformed values do correspond neatly and exactly to the ratio scale results. Using the generalized coordinate form, such as (x-x0), the transform produces an interval scale xÕ and an interval scale x0Õ. That gives us a ratio scale (xÕ-x0Õ), just like - and equal to - (x-x0). ------------------------------ === Subject: 7. The CrackpotsÕ Version of the Transforms. It has become apparent - whether misleading or not - that the crackpot responses to the obvious derive from a common source, whether it be bandwagoning or their SR instructors. Below, in the sci.math subject, we see that all sci.math respondents agree with the basic controversial position of this faq: every coordinate is transformed, whether a supposed constant or not. Think about it, the generalized coordinate of a circle center, x0, applies to infinities upon infinities of circle locations (given y and z, too); it is a constant only for a given circle, and even then only on a given coordinate axis. And even variables are often held ŌconstantÕ during either integration or differentiation. The utility of a variable is that you can discuss all possible particular values without having to single out just one. That utility does not make particular - singled out - values on the variableÕs axis not values of the variable just because they have become named values. In any case, all that is preamble to the incompetent idea they have proposed for a transform of coordinates. It is based on the idea that the circle center, point of emission, whatever, has coordinates that cannot be transformed. Let there be an equation, say (x)^2 - (ict)^2 = 0. What is the transformed version of that equation? Answer: (xÕ)^2 - (ictÕ)^2 = 0. ThatÕs the one thing the Brittanica got right. Note that the leading crackpot just criticized this faq for presuming to correct the Britt- anica, but it then and before poses the incompetent pseudo- transform we analyze here in this section. x to xÕ and t to tÕ are obviously coordinate transforms; the x and t coordinates have been replaced by the coord- inates in the primed system. A tranform of an equation from one coordinate system to another is NOT a substitution of the/a definition of x for itself; that is not a coordinate transformation. The most that can said for such a substitution is that it is a change of variable. But the crackpots are calling this a coordinate trans- form of the original equation: (xÕ+vt)^2 - (ictÕ)^2 = 0. It is not a coordinate transform, of course, except accidentally. (xÕ+vt) is not the primed system coordinate, it is another form/expression of x. They get that substitution by solving xÕ=x-vt for x; x=xÕ+vt. So, by incompetent misnomer, they accomplish what they have been railing against all along. It has been the generalized coordinate form in question all this time: (x-x0)^2 - (ict)^2 = 0. Here they substitute for x instead of transforming to the primed frame: (xÕ+vt-x0)^2 - (ictÕ)^2. ----- ^ | ^ | It is still x ^ but see what they have accomplished by their mis/malfeasance: [xÕ+vt-x0]=[xÕ+(vt-x0)]=[xÕ-(x0 -vt)]. =[xÕ-x0Õ] The crackpots have been bragging about how you donÕt have to transform the circle centerÕs coordinate to transform the circle centerÕs coordinate. Bragging that what they were doing was not what they said they were doing. This does give us insight as to some of the crackpot variations on their x0Õ<>x0-vt theme, which in all the variations will be discussed in later sections.. They are used to seeing the mixed coordinate form, (xÕ+vt-x0) without realizing what it respresented, so - accompanied with a lack of understanding of the term ŌdependentÕ - they are used to seeing just the one vt term, and not the one hidden in the defi- nition of xÕ and are used to imagining it makes the whole expression time dependent and thus not invariant. About which, let x=10, let, x0=20, v=10, and t variously 10 and 23: (x-x0)=-10. Using their (xÕ+vt-x0): For t=10, we have (xÕ+vt-x0) = [ (10-10*10) + (10*10) - (20) ] = -90 + 100 - 20 = -10 = (x-x0) For t=23, we have (xÕ+vt-x0) = [ (10-10*23) + (10*23) - (20) ] = -220 + 230 - 20 = -10 = (x-x0) The result depends in no way on the value of time; we showed the obvious for a couple of instances of t just so you can see that the crackpots not only do not understand the obvious logic of the algebra { (xÕ-x0Õ)=[ (-vt)-(x0-vt) ]=(x-x0) } - which shows that the transform has no possible time term effect - but they donÕt understand even a simple arithmetic demonstration of the facts. Oh. Their (xÕ+vt-x0) or (xÕ+vtÕ-x0) reduces the same way since tÕ=t: (x-vt+vt-x0)=(x-x0). Their process, which says (xÕ+vtÕ) is the transform of x, says that (xÕ+vtÕ) is the moving system location of x, but it canÕt be because x is moving further in the negative direction from the moving viewpoint. That formula will only work out with v<0 which is indeed the velocity the primed system sees the other moving at. However, that formula cannot be derived from xÕ=x-vt, the formula for transformation of the coordinates from the unprimed to the primed, ------------------------------ === Subject: 8. What does sci.math have to say about x0Õ=x0-vt? The crackpotsÕ positions/arguments were put to sci.math in such a way that at least two or three who posted re- sponses thought it was your faq-er who was on the idiotÕs side of the questions. Their responses: ---------------------------------------------------------- I. x0Õ = x0. In other words: x0Õ <> x0-vt, or constant values on the x-axis are not subject to the transform. No. x0Õ = x0 - vt. Well, if you want, you could define constant values on the but in the context of the question that is not relevant. The relevant fact is that if the unprimed observer holds an object at point x0, then the primed observer assigns to that object a coordinate x0Õ which is numerically related to x0 by x0Õ= x0 -vt. What does this mean? The line x=x0 will give xÕ=x-v*t=x0-vtÕ, so if x0Õ is to give the coordinate in the (xÕ,tÕ,)-system, it will be given by x0Õ=x0-v*tÕ: ie., it is not given by a constant. Thus, being at rest (constant x-coordinate) is a coordinate-dependent concept. Sounds very false. We can say that the representation of the point X0 is the number x0 in the unprimed system, and x0Õ in the primed system. Clearly x0 and x0Õ are different, if vt is not zero. However one may say that (though it sounds/is stupid) the point X0 itself is the same throughout the transformation. However that expression sounds meaningless, since a transform (ok, maybe we should call it a change of basis) is only a function that takes the pointÕs representation in one system into the same pointÕs representation in another system. It is preferrable to use three notations: X0 for the point itself and x0 and x0Õ for the pointsÕ representations in some coordinate systems. ------------------------------ === Subject: 9. But DoesnÕt x.cÕ=x.c? That idea is one of the most idiotic to come up, and it does so frequently. And in a number of guises. The idea being that x.cÕ <> x.c-vt, with x.c being what we have called x0 above; the notation makes no difference. Some crackpots have managed to maintain that position even after graphs have illustrated that such an idea means that after a while a circle center represented by x.cÕ could be outside the circle. The leading crackpot just make that explicit, as far as one can tell from his befuddled post in response to a line about active transforms, which are actually moving body situations, not coordinate transformations: ------------------------------------------------------------- ------- e>An active transform is not a coordinate transform, ... Right, it is a transform of the center (in the opposite direction) done to effect the change of coordinates without a coordinate transform. ... E: Transform of the center? Center of a circle? He really is saying a circle center moves in the opposite direction of the circle! Right? ------------------------------------------------------------- ------- If r=10 and x.c was at x.c=0, then the points on the circle (10,0), (-10,0), (0,10) and (0,-10) could at some time become (-10,0), (-30,0), (-20,10), and (-20,-10), but with x.cÕ=x.c, the circle center would be at (0,0) still! The circle is here but its center is way, way over there! Indeed, although a change of coordinate systems is not movement of any object described in the coordinates, the x.cÕ=x.c crackpottery is tantamount to the circle staying put but the center moving away. Or vice versa. ------------------------------ === Subject: 10. But IsnÕt (xÕ-x.cÕ)=(x-x.c) Actually Two Transformations? One crackpot puts the (xÕ-x.cÕ)=(x-vt - x.c+vt) relationship like this: (x-vt+vt - x.c). See, he says, that is transforming x (with x-vt - x.c) and then reversing the transform (x-vt+vt - x.c). ThatÕs just another crackpot form of the idiocy that x.cÕ <> x.c-vt. YouÕll have noticed the implication is that there is no transform vt term relating to x.c. ------------------------------ === Subject: 11. But DoesnÕt (xÕ-x.c+vt) Prove The Transformation Time Dependent? That particular crackpottery is perhaps more corrupt than moronic, since it includes deliberately hiding a vt term from view, and pretending it isnÕt there. [However, we have seen above that it is a familiar incompetency, and not likely an original.] Look, the crackpots say, there is a time term in the transformed (xÕ - x.c+vt). The transform isnÕt invariant! ItÕs time dependent! Just put xÕ in its original axis form, also, which reveals the other time term, the one they hide: (xÕ-x.c+vt) = (x-vt - x.c+vt) = (x-x.c). So, at any and all times, the transform reduces to the original expression, with no time term on which to be dependent. Then there is the fact that if you leave the equation in any of the various notation forms - with or without reducing them algebraicly - the arithmetic always comes down to the same as (x-x.c). That means nothing to crack- pots, but may mean something to you. ------------------------------ === Subject: 12. But IsnÕt (xÕ-x.cÕ)=(x-x.c) a Tautology? My dictionary relates ŌtautologyÕ to needless repetition. ThatÕs another form of the x.cÕ <> x.c-vt idiocy. The repetition involved is the vt transformation term. Apply the -vt term to the x term, and it is needless repetition to apply it anywhere again? The ŌagainÕ is to the x.c term. The x.cÕ = x.c crackpot idiocy. The repetition of the vt terms is required by the presence of two x values to be transformed. Be sure to note the next section. ------------------------------ === Subject: 13. But IsnÕt (xÕ-x.cÕ)=(x-x.c) Almost the Definition of a Linear Transform? Now, how on earth can we relate a tautology to a basic definition in math? we get this definition: -------------------------------------------------------------- A linear transformation, A, on the space is a method of corr- esponding to each vector of the space another vector of the space such that for any vectors U and V, and any scalars a and b, A(aU+bV) = aAU + bAV. ------------------------------------------------------------- Let points on the sphere satisfy the vector X={x,y,z,1}, and the circle center satisfy C={x.c,y.c,z.c,1}. Let a=1, and b=-1. Let A= ( 1 0 0 -ut ) ( 0 1 0 -vt ) ( 0 0 1 -wt ) ( 0 0 0 1 ) A(aX+bC) = aAX + bAC. aX+bC = (x-x.c, y-y.c, z-z.c, 0 ). The left hand side: A( x - x.c , y - y.c, z - z.c, 0 ) = ( x-x.c , y-y.c, z-z.c, 0 ). The right hand side: aAX= ( x-ut, y-vt, z-wt, 1 ). bAC= (-x.c+ut, -y.c+vt, -z.c+wt, -1 ). and aAX+bAC = ( x-x.c, y-y.c, z-z.c, 0 ). Need it be said? Sure: QED. On the galilean transform the definition of a linear transform, A(aU+bV)=aAU + bAV, is completely satisfied. The generalized form transforms exactly and non-redundantly - with ONE TRANSFORM, not a transform and reverse transform - and non- tautologically, just as the very definition of a linear transform says it should. And does so with absolute invariance, with this galilean transformation. ------------------------------ === Subject: 14. But The Transform WonÕt Work On Time Dependent Equations? The main crackpot that has asserted such a thing was referring to equations such as in Subject 4, above. The Light Sphere equation; for which we have shown repeatedly elsewhere that the numerical calculations are identical for any primed values as for the unprimed values. The presence - before transformation - of a velocity term seems to confuse the crackpots. It turns out there is ex- treme historical reason for this, as you will see in the subject on MaxwellÕs equations. ------------------------------ === Subject: 15. But The Transform WonÕt Work On Wave Equations? See Subject 17, below, for a discussion of Second Derivative forms and the galilean transforms. ------------------------------ === Subject: 16. But MaxwellÕs Equations ArenÕt Galilean Invariant? Oh? Just what is the magical term in them that prevents (xÕ-x.cÕ)=(x-vt - x.c+vt)=(x-x.c) from holding true? It turns out not to be magic, but reality, that interferes with the application of the galilean transforms to the gen- eralized coordinate form(s) of Maxwell: there are no coordi- nates to transform! When True Believer crackpots are shown the simple demonstration that the galilean transform on generalized cartesian coordinates is invariant, their first defense is usually an incredibly stupid x0Õ=x0, because the coordinate of a circle center, or point of emission, etc, is a constant and canÕt be transformed. The last defense is but MaxwellÕs equations are not invariant under that coordinate transform. When asked just what magic occurs in Maxwell that would prevent the simple algebra (xÕ-x0Õ)=[ (x-vt)-(x0-vt) ]=(x-x0) from working, and when asked them for a demonstration, they will never do so, however many hundreds of times their defense is asserted. The reason may help you understand part of EinsteinÕs 1905 paper in which he gave us his absurd Special Relativity derivation: THERE ARE NO COORDINATES IN THE EQUATIONS TO BE TRANSFORMED. Einstein gave the electric force vector as E=(X,Y,Z) and the magnetic force vector as B=(L,M,N), where the force components in the direction of the x axis are X and L, Y and M are in the y direction, Z and N in the z direction. Those values are not, however, coordinates, but values very much like acceleration values. BTW, the current fad is that E and B are ŌfieldsÕ, having been Ōforce fieldsÕ for a while, after being ŌforcesÕ. So, when Einstein says he is applying his coordinate transforms to the Maxwell form he presented, he is either delusive or lying. (a) there are no coordinates in the transform equations he gives us for the Maxwell transforms, where B=beta=1/sqrt(1-(v/c)^2): XÕ=X. LÕ=L. YÕ=B(Y-(v/c)N). MÕ=B(M+(v/c)Z). ZÕ=B(Z+(v/c)M). NÕ=B(N-(v/c)Y). X is in the same direction as x, but is not a coordinate. Ditto for L. They are not locations, coordinates on the Similarly for Y and M and y, Z and N and z. (b) the v of the coordinate transforms is in Maxwell before any transform is imposed; EinsteinÕs transform v is the velocity of a coordinate axis, not the velocity he touched it. (c) if they were honest Einsteinian transforms, theyÕd be x, which means it is X and L that are supposed to be transformed, not Y and M, and Z and N. And when SR does transform more than one axis, each axis has its own velocity term; using the v along the x-axis as the v for a y-axis and z-axis transform is thus trebly absurd: the axes perpendicular to the motion are not changed according to SR, the v used is not their v, and the v is not a transform velocity anyway. (d) as everyone knows, the effect of E and B are on the direction. Both the speed and direction are changed by E and B, but v - the speed - is a constant in SR. As absurd as are the previously demonstrated Einsteinian blunders, this one transcends error and is an incredible example of True Believer delusion propagating over decades. The components of E and B do differ from point to point, and in the variations that are not coordinate free, they are subject to the usual invariant galilean trans- formation when put in the generalized coordinate form. ------------------------------------------------------------- The SR crackpots donÕt know what coordinates are. The various things they call coordinates include coordin- nates, but also include a variety of other quantities. ------------------------------------------------------ 1. One may express coordinates in a one-axis-at-a-time manner [like x^2+y^2=r^2] but it is the use of vector notation that shows us what is going on. In vector notation the triplet x,y,z [or x1,x2,x3, whatever] represents the three spatial coordinates, but there are so-called basis vectors that underlie them. Those may be called i,j,k. Thus, what we normally treat as x,y,z is a set of three numbers TIMES a basis vector each. 2. These e*i, f*j, g*k products can have a lot of meanings. If e, f, j are distances from the origin of i,j,k then e*i, f*j, g*k are coordinates: distances in the directions of i,j,k respectively, from their origin. That makes the triplet a coordinate vector that we describe as being an x,y,z triplet; perhaps X=(x,y,z). The e*i, f*j, g*k products could be directions; take any of the other vectors described above or below and divide the e,f,g numbers by the length of the vector [sqrt(e^2+f^2+g^2)]. That gives us a vector of length=1.0, the e,f,g values of which show us the direction of the original vector. That makes the triplet a direction vector that we describe as being an x,y,z triplet; perhaps D=(x,y,z). The e*i, f*j, g*k products could be velocities; take any of the unit direction vectors described above and multiply by a given speed, perhaps v. That gives a vector of length v in the direction specified. That makes the triplet a velocity vector that we describe as being an x,y,z triplet; perhaps V=(x,y,z). Each of the three values, e,f,g, is the velocity in the direction of i,j,k respectively. The e*i, f*j, g*k products could be accelerations; take any of the unit direction vectors described above and multiply by a given acceleration, perhaps a. That gives a vector of length a in the direction specified. That makes the triplet an acceleration vector that we describe as being an x,y,z triplet; perhaps A=(x,y,z). Each of the three values, e,f,g, is the acceleration in the direction of i,j,k respectively. The e*i, f*j, g*k products could be forces (much like accel- erations); take any of the unit direction vectors described above and multiply by a given force, perhaps E or B. That gives a vector of length E or B in the direction specified. That makes the triplet a force vector that we describe as being an x,y,z triplet; perhaps E=(x,y,z) or B=(x,y,z). Each of the three values, e,f,g, is the force in the direction of i,j,k respectively. EinsteinÕs - and MaxwellÕs - E and B are not coordinate vectors. There is another variety of intellectual befuddlement that misinforms the idea that Maxwell isnÕt invariant under the galilean transform: confusions about velocities. Velocities With Respect to Coordinate Systems. ----------------------------------------------- Aaron Bergman supplied the background in a post to a sci.physics.* newsgroup: Imagine two wires next to each other with a current I in each. Now, according to simple E&M, each current generates a magnetic field and this causes either a repulsion or attraction between the wires due to the interaction of the magnetic field and the current. LetÕs just use the case where the currents are parallel. Now, suppose you are running at the speed of the current between the wires. If you simply use a galilean transform, each wire, having an equal number of protons and electrons is neutral. So, in this frame, there is no force between the wires. But this is a contradiction. First of all, the invariance of the galilean transform (xÕ-x.cÕ) =(x-x.c), insures that it is an error to imagine there is any difference between the data and law in one frame and in another; the usual, convenient rest frame is the best frame and only frame required for universal analysis. [Well, (xÕ<>x, x,cÕ<>x.c, but (xÕ-x.cÕ)=(x-x.c).] Second, given that you decide unnecessarily to adapt a law to a moving frame, donÕt confuse coordinate systems with meaningful physical objects, like the velocity relative to a coordinate system instead of relative to a physical body or field. In other words, what does current velocity with respect to a coordinate system have to do with physics? Nothing. Certainly not anything in the example Bergman gave. What is relevant is not current velocity with respect to a coordinate system, but current velocity with respect to wires and/or a medium. The velocity of an imaginary coordinate sys- tem has absolutely nothing to do with meaningful physical vel- ocity. You can - if you are insightful enough and donÕt violate item (e) - identify a coordinate system and a relevant physical object, but where some v term in the pre-transformed law is in use, donÕt confuse it with the velocity of the coordinate transform. Velocities With Respect to ... What? ----------------------------------------------- Albert Einstein opened his 1905 paper on Special Relativity with this ancient incompetency: The equations of the day had a velocity term that was taken as meaning that moving a magnet near a conductor would create a current in the conductor, but moving a conductor near a wire would not. This was belied by fact, of course. The important velocity quantity is the velocity of the magnet and conductor with respect to each other, not to some absolute coordinate frame (as far as we know) and not to an arbitrary coordinate system. One possible cause was the idea: but the equation says the magnet must be moving wrt the coordinate system or ... the absolute rest frame. There not being anything in the equation(s) to say either of those, it is amazing that folk will still insist the velocity term has nothing to do with velocity of the two bodies wrt each other. ----------------------------------------------------------- ------------------------------ === Subject: 17. First and Second Derivative differential equations. One of the intellectually corrupt ways of denying the very simple demonstration of galilean invariance of all laws expressed in the generalized coordinate form demanded by analytic geometry, vector analysis, and measurement theory [ (xÕ-x.cÕ)=[ (x-vt)-(x.c-vt) ]=(x-x.c) ] is the assertion that those equations Ōover thereÕ (usually Maxwell or wave) are somehow immune to the elementary laws of algebra used to demon- strate the invariance. [Unfortunately, the assertions are never accompanied by reference to the magical math that makes elementary al- gebra invalid. Wonder why that is?] Part of the time it is based on the old lore based on the incompetent transformation of the privileged form of an equation instead of the correct form. [Evidence of this is any reference to an effect due to the velocity of the transform; it falls out algebraicly metically - as you can see above.] But usually it is just whistling in the dark, waving the cross (zwastika, IÕd say) at the mean old vampire. The most general equation that could be conjured up is a differential with either First or Second Derivatives. LetÕs examine the plausibility of such magical magical, non-invariance assertions. (a) to get a Second Derivative you must have a First Derivative. (b) to get a First Derivative you must have a function to differentiate. (c) to get a Second Derivative you must have a function in the second degree. So, let us examine the question as to whether any such common Maxwell/wave equation will differ for (a) the common, privileged form, represented as ax^2, with a being an unknown constant function. (b) the generalized cartesian form, represented as a(x-x.c)^2 = ax^2 -2ax(x.c) + ax.c^2, with a being an unknown constant function. (c) the transformed generalized cartesian form, represented as a(x-vt -x.c+vt)^2, same as for (b), = ax^2 -2ax(x.c) + ax.c^2, of course, with a being an unknown constant function. I. for (a), remembering that x.c is a constant, and that this version is only correct because x.c=0, otherwise (b) is the correct form: d/dx ax^2 = 2ax (d/dx)^2 ax^2 = 2a II. for (b), remembering that x.c is a constant. d/dx (ax^2 -2ax(x.c) + ax.c^2) = 2ax - 2ax.c (d/dx)^2 (ax^2 -2ax(x.c) + ax.c^2) = 2a III. for (c); same as for (b). So, what we have seen so far is (1) differential equations in the second degree - the wave equations - must clearly be the same for all forms: the privileged form in x, the generalized cartesian form in x and the centroid, x.c, or the transformed generalized cartesian form. That is, anyone who imagines that correct usage gives different results for galilean transformed frames is at first showing his ignorance, and in the end showing his intellectual corruption. (2) As far as the First Derivatives are concerned, the only cases in which there really is a difference between the two forms is where x.c <> 0, and in that case, the use of the privileged form is obviously incompetent. So, how do you correctly use the differential equations? If you are using rest frame data with the centroid at x=0, etc, you canÕt go wrong without trying to go wrong. If you are using rest frame data with the centroid not at x=0, you must use (x-x.c) anyplace x appears in the equation. If you are using moving frame data, you must use the moving frame centroid as well as the light front (or whatever) moving frame data itself, perhaps first calculating (xÕ-x.cÕ), which equals (x-x.c) which is obviously correct, and which is obviously the plain old correct x of the privileged form. Unless, of course, there really is some magical term or expression that invalidates the obvious and elemen- tary algebra of the invariance demonstration. Or maybe you just whistle when you donÕt want basic algebra to hold true. Eleaticus !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ---!---? ! Eleaticus Oren C. Webster ThnkTank@concentric.net ? ! Anything and everything that requires or encourages systematic ? ! examination of premises, logic, and conclusions ? !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ---!---? Supersedes: === Subject: (SR) Lorentz tÕ, xÕ = Intervals Summary: The Lorentz transforms themselves are proof tÕ and xÕ cannot possibly be just coordinates. Examination of their derivation verifies their identity as intervals. Originator: faqserv@penguin-lust.mit.edu Disclaimer: approval for *.answers is based on form, not content. Opponents should first actually find out what the content is, then think, then request/submit-to arbitration by the appropriate neutral mathematics authorities. Flaming the hard- working, selßess, *.answers moderators evidences ignorance and atrocious netiquette. Version: 0.02.1 Archive-name: physics-faq/criticism/lorentz-intervals Posting-frequency: 15 days (SR) Lorentz tÕ, xÕ = Intervals (c) Eleaticus/Oren C. Webster Thnktank@concentric.net ------------------------------ === Subject: 1. Introduction with the obvious debunking of the use of Ōjust coordinatesÕ in any scientific formula. Defenders of the Special Relativity faith are especially fond of telling opponents of their space-time fairy tales that they do not know the difference between coordinates and magnitudes. That may often be so, but the fault lies ultimately with SR dogma. The Lorentz-Einstein transformations cannot possibly be Ōjust coordinatesÕ, which is the interpre- tation required to support the many sideshow carnival acts with which the SR faithful bedazzle the public, and establish their moral and intellectual superiority. If I get in my car and drive steadily for a few hours at 50 kilometers per hour, is 50t the distance I travel? Of course not. The last time my hours-counting Ōjust coord- inatesÕ clock was set to zero was when Zeno first reported one of his paradoxes to Parmenides. That was a long time ago, so my t is not useful for such purposes unless you also use my clock to established the starting time, perhaps t0, and use the formula 50(t-t0) to calculate the distance. In any case, my t is even then not Ōjust a coordinateÕ because it always represents particular elapsed times that can be used in the (t-t0) form to calculate perfectly good time intervals (elapsed times). Alternatively, I could (re)set my clock to zero at the start of some meaningful time interval, in which case my t shows a scientifically perfect current and/or end time. In which case, the Lorentz-Einstein tÕ=(t-vx/cc)/g is a function of an elapsed time interval (not Ōjust a coordinateÕ) and a time interval (-vx/cc; the interval amount the tÕ clock is being screwed up at time t) and thus cannot be Ōjust a coordinateÕ since neither of the independent variables is such a ŌjustÕ thing. {Their meaning is shown below, step-by-step.] If it takes me 50 minutes to cross the Interstate highway, was x/50 my velocity crossing it? Of course not. The origin of all my axes is at the very spot where Zeno first presented his first paradox to Parmenides. That makes my x equal a couple of thousands of miles, plus, and is not useful for such purposes unless you establish the starting x value, perhaps x0, and use the formula (x-x0)/50 to calculate my velocity. In any case, even then my x is not Ōjust a coordinateÕ because it always repesents particular distance intervals that can always be used in the (x-x0) form for any and every scientific purose. Alternatively, I could move my x-axis origin to the starting (zero) point of some meaningful distance, in which case my x shows a scientifically perfect current and/or end distance. In which case, the Lorentz-Einstein xÕ=(x-vt)/g is a function of a current/ending distance interval (not Ōjust a coordinateÕ) and a distance interval (-vt; the interval amount the xÕ axis is being screwed up at time t) and thus cannot be Ōjust a coordinateÕ since neither of the independent variables is such a ŌjustÕ thing. {Their meaning is shown below, step-by-step.] ------------------------------ === Subject: 2. Table of Contents 1. Introduction with the obvious debunking of the use of Ōjust coordinatesÕ in any scientific formula. 2. Table of Contents. 3. The Lorentz-Einstein transforms. 4. The Ōjust coordinatesÕ argument. 5. Single-system, little-purpose ambiguity. 6. Relating two coordinate measures/systems. 7. Distances and moving coordinate axes. 8. Time intervals. 9. EinsteinÕs (1905) derivations. 10. A word about intervals. 11. Intervals versus the Twins Paradox. 12. Summary ------------------------------ === Subject: 3. The Lorentz-Einstein transforms Special RelativityÕs space-time circus is based on the ŌtransformationÕ equations by which it is believed one can relate a nominally ŌstationaryÕ systemÕs space and time coordinates to those of an inertially (not accelerating) moving other observer. That moving observerÕs own physical body and coordinate system might have been identical in size to those of the stationary observer before the traveller began moving, but are ŌseenÕ as very different by the stationary observer when the relative velocity of the two is great enough, a high percentage of the velocity of light. Concerning ourselves - as is customary - with just the spatial coordinate axis that lies parallel to the direction of motion, and with time, Einstein arrived at these formulas that relate the moving system measures or coordinates (xÕ and tÕ) to the stationary system coordinates (x and t): xÕ = (x - vt)/sqrt(1-vv/cc) (Eq 1x) tÕ = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) The v is for the two systemsÕ relative velocity as seen by the stationary observer, and is positive if the dir- ection is toward higher values of x. By concensus, the moving system xÕ-axis higher values also lie in that direction, and all axes parallel the other systemÕs corresponding axis. We used vv to mean the square of v but might use v^2 for that purpose below. Similarly for c. Because it is believed that no physical object can reach or exceed c, the square-root term in both denominators is presumed always less than one, which means that the formulas say both xÕ and tÕ will tend to be greater than x and t, respectively. However, SRians call the xÕ result ŌcontractionÕ - which means shortening - and the tÕ result ŌdilationÕ - which means increasing. ------------------------------ === Subject: 4. The Ōjust coordinatesÕ argument The Ōjust coordinatesÕ argument is so patently ridiculous that even opponents have a hard time accepting just how simple and obvious its debunking can be, as shown in this section. However, further sections take a more arithmet- ical approach that youÕll maybe find more professorial. The Ōjust coordinatesÕ argument is that t is mot a duration, not a time interval; itÕs just an arbitrary clock reading. But what if the moving system observer comes speeding by while you make your annual Ōspring forwardÕ or Ōfall backÕ change? The formula says that the moving system clockÕs Ōjust coordinateÕ reading can be calculated from yours: tÕ = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Imagine the moving system oberverÕs confusion if his clock changes its reading while heÕs looking at it! If his clock doesnÕt change when yours does, the formula is wrong; if it is truly a Ōjust coordinatesÕ formula. And then what happens if you realize you were a day early and put your clock back to what it had said previously? And suppose you are in NYC and your twin in LA and both are watching the moving observer. YouÕll both be using the same v because you are at rest wrt (with respect to) each other. YouÕre on Eastern Standard Time and your twin is on Pacific Standard Time maybe. You have three hours more on your clock than does your twin. On which Ōjust coordinateÕ clock will the Lorentz transforms base the Ōjust coordinateÕ time the moving system clock says? The formula applies to both of your t-times: tÕ = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Sure, the idea that you can change someone elseÕs clock with no connection of any kind is really ridiculous, but Eqs 1x and 1t arenÕt MY equations. Are they yours? And we arenÕt the ones to say x, t, xÕ, and tÕ are just coordinates. If the tÕ formula is actually either an elapsed time formula, or the basis of a tÕ/t ratio, then there is no implication that one clockÕs reading has anything to do with the otherÕs. It can only be rates of clock ticking, or how one time INTERVAL compares to the other that the formula is about. ------------------------------ === Subject: 5. Single-system, little-purpose ambiguity. Since weÕre going to be comparing measurements on two coordinate systems in the next section, letÕs go to our supply cabinet and get our yard-stick (which we use to measure things in inches) and our meter-stick (which we use to measure things in centimeters). Here, IÕm getting mine. Oh! Oh! ThereÕs an ant on mine, and he ... she ... sure is hanging on, right at the 3.5 inch mark of the yard- stick. LetÕs see if I can wave the stick around enough that sheÕll let go. Nope. However, before I gave up I waved the stick and the ant Ōall over the place. Always, however, the ant was at the 3.5 mark on the yard-stick, and always 3.5 away from the end of the stick, however far and wide I have transported her. Neither of those 3.5 facts means very much. Of the two, the distance aspect meant almost nothing. So the distance was 3.5 from the end. So what? That length, distance, was not in use. And only maybe the ant might have been concerned with just what location, Ōjust coordinateÕ, on the stick she was at. Just so with x and t. So, is the 3.5 reading just a coordinate? Or a distance/length? ItÕs ambiguous in and of itself, and really makes no difference what you say until you try to make use of the number. Hey, my address is 5047 Newton Street. If you are looking for me and youÕre at 4120 Newton, it is helpful information, because it tells you which direction to go. Is that Ōjust coordinateÕ? Where it really becomes useful, perhaps, is in telling you how far away I am. ThatÕs not just a coordinate value, thatÕs a distance, length, interval. However, it is subtracting 4120 from 5047 that tells you which direction and how far. It is only because both 5047 and 4120 are distances from the same point - ANY same point - that the result means anything. My x - my yardstick reading - is always a distance or length; it is impossible to be otherwise with an honest, competently designed yardstick. Whether or not its reading is of good use in some particular scientific formula depends on whether I put the zero end of the yardstick at some useful place. As in the introduction, we should either put it at the starting location/end, or use two readings from it: (x-x0). ------------------------------ === Subject: 6. Relating two coordinate measures/systems. Taking care to not damage our brave little ant, I place my yard-stick onto the table, zero end to the left, 36 end to the right. Now I place the Ōjust coordinateÕ meter-stick on the table in the same orientation, in a random location, and find that the antÕs coordinate on the meter-stick is 51. The formula relating centimeters to inches is cm=i*2.54 but we want a formula similar to xÕ=(x-vt)/sqrt(1-vv/cc). That would be c=i/.03937 approximately, but letÕs use xÕ for the meter-stick reading, and x for the inch reading: xÕ=x/.3937. 3.5/.3937 = 8.89 Wait a minute. ItÕs not just science but definition that says c=i/.3937=8.89, so something is wrong. 8.89 is not 51. We already knew that 51 cm was just an arbitrary coordinate. Arbitrary not because that point isnÕt 51 cm from the zero end of the meter-stick, but because the zero point was in an arbitrary position. LetÕs put the meter-stick in a position where itÕs zero point is at the yard-stick zero point. What is the centimeter coordinate now? Hey. 8.89, just like the formula says. The only way for a ŌtransformÕ like xÕ=x/g to work, whatever g might be, is for both coordinate systems to have their zero points aligned, in which case saying the two measures are not intervals is pure idiocy. Noe that with both zero points at the same position both xÕ and x are great measures for scientific purposes, in any and every case where we were smart enough to put those zero points at a useful location. There is one extension of xÕ=x/g that will let us use the meter-stick in arbitrary position. When the cm reading was 51, the zero point of the yard-stick read (51-8.89=) 42.11 cm. If we call that point x.zÕ we get xÕ = x.zÕ + x/.3937. = 42.11 + 3.5/.3937 = 42.11 + 8.89 = 51. Obviously, in this formula x/.3937 is the distance from the xÕ coordinate of the location where x=0. An interval. Just as obviously, the fact that we now have the correct formula for relating an x interval to an arbitrary xÕ coordinate, does not mean that xÕ is anything more than nonsense for use in any scientific formula. Unless we were smart enough to put the x zero point in a useful location, and use (xÕ-x.zÕ) in the scientific formula. (xÕ-x.zÕ) equals the useful, Ratio Scale value x/.3937. So, we have discovered a basic fact: a transformation formula like xÕ=x/g works only if the two zero points of the coordinate systems coincide. That makes it non- sense to say the two coodinates are only coordinates and not intervals. Both must be values that represent distances from their respective zero points unless you take the proper steps to adjust for the discrepancy. Make sure you understand that although the inclusion of x.zÕ made it possible to correctly calculate xÕ, the result is nonsense when it comes to use of xÕ for general length/distance purposes; it is xÕ-x.zÕ that is a useful number in such cases. It could be that weÕre measuring a sheet of paper with one end at x=0 and the other at x=3.5; xÕ=51 is nonsense as a centimeter measure of the paper. But, you say, the Lorentz transform contain a -vt term. ------------------------------ === Subject: 7. Distances and moving coordinate axes. We discovered xÕ=x.zÕ + x/g as the correct formula for relating one coordinate to another systemÕs. But the Lorentz transform contains another term, -vt/sqrt(1-vv/cc). What is it? LetÕs start with our xÕ=51 cm, x=3.5, x.zÕ=42.11 example. Every minute, letÕs move the meter-stick one inch to our right. At minute 0, the cm reading was 51 cm. At minute 1, the cm reading is now 50 cm. At minute 2, the cm reading is now 49 cm. In this instance, v=1 inch/minute. And t was 0, 1, 2. What has happened is that we have made our x.zÕ a lie, and increasingly so. -vt/.3937 is the change in x.zÕ. xÕ = (x.z - vt/.3937) + x/.3937. Obviously, vt/.3937 is not a coordinate; even most SRians wouldnÕt imagine it was. It is an interval, the distance over which the moving system has moved since t=0. And, of course, x/.3937 is the distance of our brave little ant from the point where x=0 and the centimeter reading is x.zÕ-vt/.3937. Yes, every minute the meter- stick moves to the right and the meter-stick coordinate of the spot where x=0 gets less and less - and eventually negative. Make sure you understand that every minute the xÕ coordinate, because of -vt/g, becomes a better measure of, say, the 3.5 paper we might be measuring with the yard-stick, given that 51 was too big a number and -vt is negative. That is, until the two origins coincide at xÕ=x=0, and then it gets worse and worse. With -vt positive (because v<0) the situation is different. With 51 and -vt positive, xÕ just gets worse and worse over time. Quite obviously, the fact that we now have the correct formula for relating an x interval to an arbitrary xÕ coordinate even when the xÕ axis is moving, does not mean that xÕis anything more than nonsense for use in any scientific formula. Unless we were smart enough to put the x zero point in a useful location, and use (xÕ-x.zÕ+vt/.3937) in the scientific formula. (xÕ-x.zÕ+vt/.3937) equals the useful, Ratio Scale value x/.3937. ------------------------------ === Subject: 8. Time intervals. Instead of using our sticks, letÕs get out two clocks. Mind you, weÕre not going to deal with different clock rates here, just establish the same basics as for distance. Your clock says 9:00 Eastern Standard Time (EST) and we note that t=540 minutes when we put down the clock. Blindly, letÕs turn the setting knob of your twinÕs Pacific Standard Time clock and put it down before us. According to what we see, ESTÕs 540 minutes (9:00) corre- sponds to PSTÕs 14:30; tÕ=870. We know the formula relating PST to EST is tÕ (pacific) = t (eastern) - 180 (minutes). Thus, it is not correct that the second clock can have an arbitrary setting, because 870 <> 540-180. We know that the two clocks are related by tÕ = t/1 since both are using the same second, hour, etc units. But 870 (14:30 in minutes) is not 540/1-180, so once again we know something is wrong. However, tÕ=t.zÕ + t/1 works. EST midnight equals PST 0.0 (midnite) - 180, so t.zÕ = -180, and tÕ = -180 + 540/1 = 360. Since EST-180=PST, 9:00 EST is 6:00 PST = 360 minutes. We see thus that like distance measures/coordinates, time axis origins (zero points) must either be Ōlined upÕ or adjusted for. So, the Lorentz/Einstein tÕ=t/sqrt(1-vv/cc) must be the moving system elapsed time interval since the time axes were both at a common zero. There is no t.zÕ adjustment: tÕ = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t) Make sure you understand that in the clock case, if the EST is showing a good number for elapsed time since the travelling observer passed NYC, then the PST clock is silliness. t.zÕ must be zero or must be taken out of time lapse calculations for the PST clock to be used intelligently, just as was true for x.zÕ. What is lacking as yet for Lorentz tÕ is the -vx/cc term that corresponds to the xÕ formula -vt term. Break it up into two parts: v/c and x/c. v/c is a scaling factor that changes velocity from whatever kind of unit you are using over to fractions of c. x/c is distance divided by velocity, which is time. x/c is thus the time interval since the two time axes had a common zero point - which they have to have in the Lorentz transforms which do not have the t.zÕ term we learned to use above. Thus, (-vx/cc)/sqrt(1-vv/cc) is the interval amount the moving system clock has been changed - since the common/ adjusted time - over and beyond the elapsed time interval represented by x/sqrt(1-vv/cc). We have discovered that the only way for tÕ to be t/g is for tÕ and t to have a common zero point, just as for xÕ and x. It would be otherwise if the tÕ formula contained an adjustment t.zÕ under some name or other, but the necessity to include such a term correlates 100% with tÕ numbers that arenÕt directly usable. As for x and xÕ, our knowledge of how to setup a proper formula relating t and tÕ is of no use unless we use the knowledge in scientific formulas; (tÕ-t.zÕ+xv/gcc) gives us the only directly useful value: t/g. ------------------------------ === Subject: 9. EinsteinÕs (1905) derivations. When we return to EinsteinÕs derivations of the transform formulas with a well-focused eye, we find he was a wee bit confused - or at least self-contradictory. When he set up his (at first unknown) tau=moving system time formulas, he created three particular instances of tau. Tau.0 is the time at which light is emitted at the moving origin toward a mirror to the right that is moving at rest wrt that moving origin and at a constant distance from that origin. He lets the stationary time slot have the value t, a constant, the stationary system starting time. Tau.1 is the time at which the light is reßected. He lets the stationary time be t+xÕ/(c-v); t is still a constant and xÕ/(c-v) is the time interval since t. Tau.2 is the time at which the light gets (back) to the moving origin. The stationary time value is put as t + xÕ/(c-v) + xÕ/(c+v); t is still a constant and xÕ/(c-v) + xÕ/(c+v) is the time interval since t. On the thesis that the moving observer sees the time to the mirror as the same as the time back to the origin, he sets .5[ tau.0 + tau.2 ] = tau.1. Tau.0 completely drops out of the analysis and leaves no trace, and has no effect. Further, the t you see in tau.0, tau.1, and tau.2 also completely drops out with no trace and no effect, leaving us with exactly what youÕd get if you had explicilty said tÕ is an interval and so is t. What doesnÕt drop out in the stationary time values is xÕ/(c-v) and xÕ/(c+v), the time interval it takes for light to get to the ßeeing mirror, and the time interval it takes for light to get back to the approaching origin. Thus, his resultant tÕ formula is strictly based on time intervals in the stationary system. Time intervals since some starting time, yes, but time intervals. There is absolutely nothing in the derived formulas that depends on arbitrary coordinates like the constant t in the stationary time arguments. LetÕs look at the x dimension; it is xÕ=x-vt [as x increases by vt, the effect over time is xÕ=(x+vt)-vt)], which Einstein explicitly sets up as a constant stationary distance. He uses that xÕ not just in the time interval parts of the stationary time arguments, but also in the x (distance) stationary system argument for the tau at the time light is reßected. xÕ canÕt be the stationary system coordinate of the mirror at that time. That value is xÕ+vt. xÕ is explicitly an interval, distance. Thus, the whole tau derivation of the tÕ formula is fully and explicitly based on xÕ - a spatial length/distance/interval - and the two time interals xÕ/(c-v) and xÕ/(c+v). While weÕre at it, if the starting t is not zero, his xÕ=x-vt formula is complete nonsense also. Given that there was some L that was the mirror x-location and length when the light is emitted, if t was already, say, 500, then xÕ=L-vt could have been a very negative length. ------------------------------ === Subject: 10. A word about intervals. There are intervals, and there are intervals. If we put our yard stick zero point at one end of a piece of paper and read off the coordinate at the other end of the paper, we have a good measure of the paperÕs length, a Ratio Scale measure. [Absolute temperature scales are ratio scale.] If instead we put the one end of the paper at the one inch mark (or the zero end of the stick one inch ŌintoÕ the length of the paper) we get measures that are one inch off the true, ratio scale length. The two messed up measures are still intervals, but they are Interval Scale measures. [Household temperature scales are interval scale, which is why your physics and chemistry professors wonÕt let you use them without first converting to the ratio scale absolute temperatures.) tÕ=t/g and xÕ=x/g represent ratio scale measures, given that t and x were ratio scalae to start with. tÕ=t.zÕ+t/g and tÕ=t/g-vx/gcc are both interval scale measures, even given a good ratio scale t and a good ratio scale x. xÕ=x.zÕ+x/g and xÕ=x/g-vt/g are both interval scale measures, even given a good ratio scale x and a good ratio scale t. Look for the (SR) Lorentz tÕ, xÕ = degraded measures document soon at a newsgroup near you. ------------------------------ === Subject: 11. Intervals versus the Twins Paradox. tÕ=(t-vx/cc)/g shows tÕ being greater than t. The reason Special Relativity will not allow the use of its basic time equation in determining what SR has to say about the twinsÕ ages, is that tÕ and xÕ are supposedly just coordinates, and they say you have to take the coordinate pairs (tÕ,xÕ) and (x,t) into consideration in both the time and place the twinsÕ separation started and the time and place the twins reunited. Since tÕ and xÕ are actually both intervals, not just coordinates, the ŌexcuseÕ is spurious, and is so even without use of the obvious (x_b-x_a) and (t_b-t_a) usages. However, SR is right to be embarrassed by their transformation formulas. Look for the (SR) Lorentz tÕ, xÕ = degraded measures document at a newsgroup near you. ------------------------------ === Subject: 12. Summary A. tÕ=t/g and xÕ=x/g can be almost Ōjust coordinatesÕ in the sense that the values obtained may not be of much use except in the most primal and useless way: how long and how far since/from the time/ place they were zero. Even here, however, the zero points within each of the two scale pairs (tÕ,t) and (xÕ.x) must have been lined up. If the zero points have been intelligently selected (such as at the starting point and time of a trip) they can be rationally used Ōas isÕ in any valid sci- entific equation. B. Even the interval scale tÕ=t.zÕ - xv/gcc + t/g and xÕ=x.zÕ - vt/g + x/g are not Ōjust coordinatesÕ. They can be used to good effect by establishing the relevant starting times/points and using (tÕ-t.zÕ+xv/gcc) and (xÕ-x.zÕ+vt/g), as the situation may require. C. When you see vx/gcc or vt/g in use in any guise with non-zero values, you know the resultant tÕ or xÕ is a degraded, interval scale value. E-X: Anytime you do not see what amounts to t.zÕ and xv/gcc in the time case, or x.zÕ and vt/g in the distance case, you know that the tÕ and/or xÕ in use are intervals. Period. Y: Either set your clock to zero at the start of the relevant time interval, or use (t-t0), with both being readings on the same clock. Either move your x-axis origin to the starting end or point, or use (x-x0), with both being readings on the same axis. Z: In _(SR) Lorentz tÕ, xÕ = Degraded (Interval) Scales_ we see that tÕ and xÕ satisfy the mathematical tests for/of interval scales when -vt and -vx/cc are not zero; thus, they must be intervals. When -vt and -vx/cc are zero, tÕ and xÕ satisfy the much better mathematical definition of ratio scales, and are thus not just mere intervals, but (rescaled) good ones. Eleaticus !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ---!---? ! Eleaticus Oren C. Webster ThnkTank@concentric.net ? ! Anything and everything that requires or encourages systematic ? ! examination of premises, logic, and conclusions ? !---?---!---?---!---?---!---?---!---?---!---?---!---?---!---? ---!---? === Subject: Partial Products Hello Liu Pingkai, In the following UBASIC example I used different names for the variables than before. The code based on your suggestion needs 3M-1 multiplications (mod N) and 1 inversion (mod N) to calculate M inverses (mod N): 110 Ō Inversion (mod N) of M integers X(0),...X(M-1). 120 Ō Needed 3*(M-1) modular multiplications and 1 modular inversion. 130 Ō A(), B() are auxiliary arrays. V() array holds inverses (mod N). 140 Ō 150 word -20 160 M=2^10:Õ this example uses M=1024 samples 170 dim X(M),A(M),B(M),V(M) 180 Ō Now fill up array X() with 1024 samples and define N. 190 A(0)=X(0) 200 for I=1 to M-2 210 A(I)=A(I-1)*X(I)@N 220 next I 230 U=A(M-2)*X(M-1)@N 240 V=modinv(U,N) 250 A(M-2)=V*A(M-2)@N 260 Ō 270 B(0)=X(M-1)*V@N 280 for I=1 to M-2 290 B(I)=B(I-1)*X(M-I-1)@N 300 next I 310 Ō 320 V(0)=B(M-2) 330 for I=1 to M-2 340 V(I)=A(I-1)*B(M-I-2)@N 350 next I 360 V(M-1)=A(M-2) 370 end This method may be faster than direct method based on M modular inversions. The improvement depends on the size of M, choice of N, (also on CPU architecture and language/compiler). This method has the time cost of 1 inversion equivalent to 3 multiplications. A.D. === Subject: A New Way To Solve Cubics Using A Linear Fractional Transformation Hello all, HereÕs a new take at a well-studied mathematical object: A New Way To Solve Cubics Using a Linear Fractional Transformation ABSTRACT: We provide a new method to solve the general cubic equation by using a linear fractional transformation. This transformation, sometimes referred to as a Mobius transformation, can transform the general cubic into the binomial form, though in a manner different from the traditional Tschirnhausen transformation that can also transform the cubic into the binomial form. The resulting binomial is then simply solved by the extraction of a cube root. Mathematics Subject Classification. Primary: 12E12; Secondary: 15A04 http://www.geocities.com/titus_piezas/cubics.html Just click at the link to the pdf file. P.S. ThereÕs also a new paper on sextics. Just go to the index. --Titus === Subject: Re: A New Way To Solve Cubics Using A Linear Fractional Transformation > Hello all, > HereÕs a new take at a well-studied mathematical object: > A New Way To Solve Cubics Using a Linear Fractional Transformation new, eh? How much of the 19th century theory of equations did you consult before deciding your method had not been done before? === Subject: Re: A New Way To Solve Cubics Using A Linear Fractional Transformation > A New Way To Solve Cubics Using a Linear Fractional Transformation > ABSTRACT: We provide a new method to solve the general cubic equation by > using a linear fractional transformation. This transformation, sometimes > referred to as a Mobius transformation, can transform the general cubic > into the binomial form, though in a manner different from the traditional > Tschirnhausen transformation that can also transform the cubic into the > binomial form. The resulting binomial is then simply solved by the extraction > of a cube root. > Mathematics Subject Classification. Primary: 12E12; Secondary: 15A04 http://www.fc.up.pt/mp/jcsantos/PDF/artigos/Mobius.pdf Well, OK, itÕs written in portuguese, but I think that youÕll be able to get the general picture from the equations. Jose Carlos Santos === Subject: Re: A New Way To Solve Cubics Using A Linear Fractional Transformation > A New Way To Solve Cubics Using a Linear Fractional Transformation > ABSTRACT: We provide a new method to solve the general cubic equation by > using a linear fractional transformation. This transformation, sometimes > referred to as a Mobius transformation, can transform the general cubic > into the binomial form, though in a manner different from the traditional > Tschirnhausen transformation that can also transform the cubic into the > binomial form. The resulting binomial is then simply solved by the extraction > of a cube root. > Mathematics Subject Classification. Primary: 12E12; Secondary: 15A04 > http://www.fc.up.pt/mp/jcsantos/PDF/artigos/Mobius.pdf > Well, OK, itÕs written in portuguese, but I think that youÕll be able > to get the general picture from the equations. > Jose Carlos Santos I guess it was bound to happen that the same idea would occur to different people independently. I canÕt read Portuguese, but, yes, mathematics is the universal language and I can understand some of your ideas from the equations. I can see you also brought up the case when the discriminant is zero. I hope you translate your paper. That way, itÕll be accessible to more people. --Titus === Subject: Re: A New Way To Solve Cubics Using A Linear Fractional Transformation It wasnÕt published but it has already been accepted for publication. It will be published in May 2005. Jose Carlos Santos === Subject: Jesus christ WhatÕs going on? posting-account=bPt6DQwAAAAhtlVqEGFRIrnEMpUABDpg I just got on Gooleg where I use to post mesasges and theyreÕ thing is all fuceked up now! and i canÕt nkow how to vie waticles properly For god sake are they trying tko Kikc us off or what!! Andrew Usher === Subject: Re: Jesus christ WhatÕs going on? posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs > I just got on Gooleg where I use to post mesasges and theyreÕ thing is > all fuceked up now! and i canÕt nkow how to vie waticles properly > For god sake are they trying tko Kikc us off or what!! > Andrew Usher Some of us have been seeing this for awhile, ever since we discovered that Google had a beta version of their server. Apparently theyÕve taken the beta version live. It seems to be having a hard time under the increased load. IÕm getting a lot of server errors today. - Randy === Subject: Re: Jesus christ WhatÕs going on? > I just got on Gooleg where I use to post mesasges and theyreÕ thing > is > all fuceked up now! and i canÕt nkow how to vie waticles properly > For god sake are they trying tko Kikc us off or what!! > Andrew Usher > Some of us have been seeing this for awhile, a while > ever since > we discovered that Google had a beta version of their > server. Apparently theyÕve taken the beta version live. Not here. Or maybe, as someone suggested, it went live for a few hours and then they changed their minds. IÕve tried the new version now and it does my head in. I think a few more horizontal lines might help. I can see that something like this is the way to go though (Wow! Look! My nameÕs highlighted** in a colour box!), so IÕll have to get used to it. In my Favourites I have a GG search for my name and handle. I notice that the new GG lists the hits in a completely different order, and if I click on search by date the number of hits more than halves. Odd. Given that the two forms of GG have such completely different algorithms, it would worry me if the original version did not continue to be available. **highlit, anyone? Adrian (UK) === Subject: Re: Jesus christ WhatÕs going on? >I just got on Gooleg where I use to post mesasges and theyreÕ thing is >all fuceked up now! and i canÕt nkow how to vie waticles properly >For god sake are they trying tko Kikc us off or what!! >Andrew Usher I still canÕt view them properly - I have to use my other news reader and itÕs not posting my name at the subject header of the messages. Why in the hell are they doing this? Andrew Usher === Subject: Re: Jesus christ WhatÕs going on? >>I just got on Gooleg where I use to post mesasges and theyreÕ thing is >>all fuceked up now! and i canÕt nkow how to vie waticles properly >>For god sake are they trying tko Kikc us off or what!! >>Andrew Usher >and itÕs not posting my name at the subject header of the messages. >Why in the hell are they doing this? Google is not a news reader. Once you have used a real news reader, you will not news :) ). Bruce ------------------------------ Health nuts are going to feel stupid someday, lying in hospitals dying of nothing. -Redd Foxx (if there were any) === Subject: Re: Jesus christ WhatÕs going on? >>I just got on Gooleg where I use to post mesasges and theyreÕ thing is >>all fuceked up now! and i canÕt nkow how to vie waticles properly >>For god sake are they trying tko Kikc us off or what!! >>Andrew Usher >and itÕs not posting my name at the subject header of the messages. >Why in the hell are they doing this? >Andrew Usher I hope everyone has been sending feedback to Google instead of just to sci groups and a hapless english group. -- Is that plutonium on your gums? Shut up and kiss me! -- Marge and Homer Simpson === Subject: Re: Jesus christ WhatÕs going on? > I hope everyone has been sending feedback to Google instead of just to sci > groups and a hapless english group. I have written to them in the past; I think IÕll try to compose a useful critiqe of the new version, when they put it back up. Andrew Usher === Subject: Re: Jesus christ WhatÕs going on? > I hope everyone has been sending feedback to Google instead of just to sci > groups and a hapless english group. > I have written to them in the past; I think IÕll try to compose a > useful critiqe of the new version, when they put it back up. > Andrew Usher I simply emailed them a link to this thread. They probably donÕt browse everywhere, but plenty of stuff that was said here should be seen by GoogleÕs folks. === Subject: Re: Jesus christ WhatÕs going on? Gregory L. Hansen typed thus: > I hope everyone has been sending feedback to Google instead of just to sci > groups and a hapless english group. IÕll have you know that we here at AUE are well provided with haps. In fact weÕve got enough haps between us to keep ourselves hapy for some considerable time to come. -- David replace the first component of address === Subject: Re: Jesus christ WhatÕs going on? >>I just got on Gooleg where I use to post mesasges and theyreÕ thing is >>all fuceked up now! and i canÕt nkow how to vie waticles properly >>For god sake are they trying tko Kikc us off or what!! >>Andrew Usher > and itÕs not posting my name at the subject header of the messages. > Why in the hell are they doing this? date (somewhere near the top right), it should return a list similar to the old format, but this format looks somewhat better somehow, though I donÕt like the way text and links run into other blocks of text. be nice. Dyl. ------------------------------- Une rhapsodie pour grapheus. http://tinylink.com/?J0TDOmaIWK === Subject: Re: Jesus christ WhatÕs going on? > be nice. IÕve been using Gougle Groups 2 Beta as much as the original. I like the new one better. === Subject: Re: Jesus christ WhatÕs going on? too, >> thatÕd be nice. > IÕve been using Gougle Groups 2 Beta as much as the original. I like > the new one better. What happens when you use it to post a message? Mike. === Subject: Re: Jesus christ WhatÕs going on? > too, >> thatÕd be nice. > IÕve been using Gougle Groups 2 Beta as much as the original. I > like > the new one better. > What happens when you use it to post a message? Hmmm, I see. Something has changed in the last day or so. Any press release from Google on it? === Subject: Re: Jesus christ WhatÕs going on? >> too, >> thatÕd be nice. > IÕve been using Gougle Groups 2 Beta as much as the original. I >> like > the new one better. >> What happens when you use it to post a message? > Hmmm, I see. Something has changed in the last day or so. Any press > release from Google on it? Not that I know of. When I was taking an interest a few months ago I emailed them with some comments, and they replied courteously but didnÕt appear to do anything. I think, as I said the other day, theyÕre into something the project managers didnÕt quite understand; or, which is highly likely, the project team understand but the commercials made them go public too early (hand up anybody who hasnÕt seen this happen elsewhere). Mike. === Subject: Re: Jesus christ WhatÕs going on? <3ddtq0lmsgld92i29pkrinf7otk6i7qrmt@4ax.com> <31825aF37p6pkU1@individual.net> posting-account=YlfmdwwAAACC77JOQxzd52GbZnwIMh2h thatÕd > be nice. I am able to use both the old Google groups and the beta groups with no problem. On the other hand, IÕm not sure that I appreciate the current inconsistency in how one navigates. On one visit I may get a handy list, off to the left, of recently visited groups. Next time around, no such thing. It does, in fact, change in several different ways, as if the software is trying to guess where I wish to go before I, myself, know. It also doesnÕt seem to allow you to click on a posterÕs name and get an instant archive of everything theyÕve posted. within a few minutes, whereas the old groups take several hours. -Mark Martin === Subject: Re: Jesus christ WhatÕs going on? >>I just got on Gooleg where I use to post mesasges and theyreÕ thing is >>all fuceked up now! and i canÕt nkow how to vie waticles properly >>For god sake are they trying tko Kikc us off or what!! >>Andrew Usher > and itÕs not posting my name at the subject header of the messages. > Why in the hell are they doing this? > date (somewhere near the top right), it should return a list similar to the > old format, but this format looks somewhat better somehow, though I donÕt > like the way text and links run into other blocks of text. > be nice. I use newreader software, in my case MT-Newswatcher on Mac OS. IÕm evidently missing all the fun. Do most people get their Usenet from Google these days? === Subject: Re: Jesus christ WhatÕs going on? > I use newreader software, in my case MT-Newswatcher on Mac OS. IÕm > evidently missing all the fun. Do most people get their Usenet from > Google these days? There are a couple of benefits, the best one is the easy click on the authorÕs name, so you can see all their other achived posts by title and newsgroup. It makes obvious trolls easy to spot. And having a web interface to a usenet archive stretching back to the begining of the Usenet helps when you need old info and donÕt want to get ßamed for the crime of asking the incredibly obvious. Oh, well, thereÕs still Groupsrv, http://www.groupsrv.com/science/ , I guess. === Subject: Re: Jesus christ WhatÕs going on? <3ddtq0lmsgld92i29pkrinf7otk6i7qrmt@4ax.com> <31825aF37p6pkU1@individual.net> posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs Only temporarily. My preferred newsreader is Agent (free version is FreeAgent), but my windoze box is down right now. HereÕs a bug IÕm surprised they didnÕt fix in the beta: The only way to get quoted text in your reply and to edit it, is to hit preview before entering any text of your own. - Randy === Subject: Re: Jesus christ WhatÕs going on? <3ddtq0lmsgld92i29pkrinf7otk6i7qrmt@4ax.com> <31825aF37p6pkU1@individual.net> posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs Only temporarily. My preferred newsreader is Agent (free version is FreeAgent), but my windoze box is down right now. HereÕs a bug IÕm surprised they didnÕt fix in the beta: The only way to get quoted text in your reply and to edit it, is to hit preview before entering any text of your own. - Randy === Subject: Re: Jesus christ WhatÕs going on? >I just got on Gooleg where I use to post mesasges and theyreÕ thing is >all fuceked up now! and i canÕt nkow how to vie waticles properly >For god sake are they trying tko Kikc us off or what!! Andrew Usher >> I still canÕt view them properly - I have to use my other news reader >> and itÕs not posting my name at the subject header of the messages. >> Why in the hell are they doing this? >> by >> date (somewhere near the top right), it should return a list similar to >> the >> old format, but this format looks somewhat better somehow, though I donÕt >> like the way text and links run into other blocks of text. >> thatÕd >> be nice. > I use newreader software, in my case MT-Newswatcher on Mac OS. IÕm > evidently missing all the fun. Do most people get their Usenet from > Google these days? Dyl. ------------------------------- Une rhapsodie pour grapheus. http://tinylink.com/?J0TDOmaIWK === Subject: Re: Jesus christ WhatÕs going on? >I just got on Gooleg where I use to post mesasges and theyreÕ thing is >all fuceked up now! and i canÕt nkow how to vie waticles properly >For god sake are they trying tko Kikc us off or what!! Andrew Usher >>I still canÕt view them properly - I have to use my other news reader >>and itÕs not posting my name at the subject header of the messages. >>Why in the hell are they doing this? >by >date (somewhere near the top right), it should return a list similar to >the >old format, but this format looks somewhat better somehow, though I donÕt >like the way text and links run into other blocks of text. >thatÕd >be nice. >>I use newreader software, in my case MT-Newswatcher on Mac OS. IÕm >>evidently missing all the fun. Do most people get their Usenet from >>Google these days? -- Dirk The Consensus:- The political party for the new millenium http://www.theconsensus.org === Subject: Re: Jesus christ WhatÕs going on? <31825aF37p6pkU1@individual.net> <318k2hF37h1s9U1@individual.net> <318onkF38ntouU4@individual.net> Discussion, linux) > groups? The easiest thing is to have a newsreader that can send HTML requests to Google and parse the results. Gnus does this. For now. Changing the interface is bound to temporarily break this feature, I reckon. -- Jesse F. Hughes Knowing about logic is not the same as being in touch with reality. -- David Kastrup === Subject: Re: Jesus christ WhatÕs going on? Well, for now, at least. Dyl. === Subject: Re: Jesus christ WhatÕs going on? <3ddtq0lmsgld92i29pkrinf7otk6i7qrmt@4ax.com> <31825aF37p6pkU1@individual.net> posting-account=0QrkrwwAAABDyQGPKX7NtkkaKfngvovA ItÕs free. David Ames === Subject: Re: Jesus christ WhatÕs going on? <3ddtq0lmsgld92i29pkrinf7otk6i7qrmt@4ax.com> <31825aF37p6pkU1@individual.net> posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L hoping its back to the usual 8 hour delay. message goes up instantly, different view options, once you find the [subject only], [sort by most recent], [show thread view], its just the same as old groups but instant! is a pain but can make for quick reading of long threads. if they could ßag the new messages instead of just listing your subscribed threads it would be better than outlook, bit hard to find new threads in 200+ posts but still usable. Hope your reading this Google I know you follow my posts! now groups2-beta is replaced with groups-beta which seems like read only at the moment, see if this goes up. pity, been writing my cantor rebuttal all week, absolutely classic make you change sides! Herc === Subject: Re: Jesus christ WhatÕs going on? > message goes up instantly, different view options, once you find the > [subject only], [sort by most recent], [show thread view], its just the > same as old groups but instant! Goes up instantly? To all Usenet or only to Google? Yes, I got those viewing options too. > is a pain but can make for quick reading of long threads. > if they could ßag the new messages instead of just listing your > subscribed threads it would be better than outlook, bit hard to find > new threads in 200+ posts but still usable. Hope your reading this > Google I know you follow my posts! I think you should be able to disable hide quote. > now groups2-beta is replaced with groups-beta which seems like read > only at the moment, see if this goes up. > pity, been writing my cantor rebuttal all week, absolutely classic make > you change sides! Back in the real world, pigs donÕt ßy, and C(R) > C(N). Andrew Usher === Subject: Re: Jesus christ WhatÕs going on? kinds of numbnuts running around there... But there is a way to use the old system. Remove the beta- from the URL, eliminate all from the last slash, hit enter, and youÕre back in.-Jitney === Subject: Re: Jesus christ WhatÕs going on? > ItÕs free. > David Ames ItÕs back, AFAIK, at this time, Noon, NYC time :-) Mark (And as the popularly elected president, I would raze the White House with itÕs oval orifice and replace it with a mud hut :-) === Subject: Re: Jesus christ WhatÕs going on? > ItÕs back, AFAIK, at this time, Noon, NYC time :-) Yes, itÕs back! But for how long ... Actually the new version isnÕt quite as bad as I thought at first. I got to a view pretty similar to the old one, and the new one now works properly for threads longer than 250 messages. I didnÕt try to post using it, though, can anyone comment on that? Andrew Usher === Subject: Re: Jesus christ WhatÕs going on? k_over_hbarc@yahoo.com exposited: >I just got on Gooleg where I use to post mesasges and theyreÕ thing is >all fuceked up now! and i canÕt nkow how to vie waticles properly >For god sake are they trying tko Kikc us off or what!! >Andrew Usher > and itÕs not posting my name at the subject header of the messages. > Why in the hell are they doing this? IsnÕt this where you turn to the camera and say, Why am I asking you? -- dg (domain=ccwebster) === Subject: Re: Jesus christ WhatÕs going on? <3ddtq0lmsgld92i29pkrinf7otk6i7qrmt@4ax.com> posting-account=3vN21wwAAABdx4qzvSrOjmV8X2Szf3e3 Agreed. ItÕs a total disaster. The easy navigation is gone, there are a lot of aggravating requests, Must activate scripts, etc. A good reasons for giving up usenet. === Subject: Re: Real Analysis >> If f is a continuous function and e >0, can you find a linear combination >> of >> 1, x, x^2, x^3, ... which approximates f on [-1,1]. >> WouldnÕt the Taylor series expansion for f about x=0 fit the bill? >No. Not every continuous function has a Taylor series. Even if the function has a Taylor series, that series doesnÕt necessarily converge anywhere. And youÕd want the series to converge uniformly to f on the whole interval [-1,1]: thatÕs not true even for functions that are real-analytic. >What is needed >is a proof of WeierstrassÕs approximation theorem. Indeed. But notice that the question was can you find, not prove that there is... Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Real Analysis >> If f is a continuous function and e >0, can you find a linear combination >> of >> 1, x, x^2, x^3, ... which approximates f on [-1,1]. >> WouldnÕt the Taylor series expansion for f about x=0 fit the bill? >No. Not every continuous function has a Taylor series. > Even if the function has a Taylor series, that series doesnÕt > necessarily converge anywhere. And youÕd want the series to converge > uniformly to f on the whole interval [-1,1]: thatÕs not true even > for functions that are real-analytic. >What is needed >is a proof of WeierstrassÕs approximation theorem. > Indeed. But notice that the question was can you find, not > prove that there is... Oh! I gave a proof of WAT that shows how to construct an appropriate poly. === Subject: inverse modulos If two numbers x and m are relatively prime, so that gcd(x, m) = 1 is true, then x has a unique multiplicative inverse modulo m, a, so that ax = 1 (mod m). Knowing only the multiplicative inverse, a modulo m, and m, is it possible to find x? Is it true that a*x - 1 = k*m, for some k, possibly negative? Or multiple kÕs? How to find the multiple kÕs so that x could be found? Is that possible at all, so that the candidate kÕs can be found knowing only the inverse a modulo m and m? If there is no single way to find it, is there a reasonable trial and error process to go through to find x given a and m? What I suppose I am saying is, what is the relationship between x and its inverse a modulo m? Well, if you caught all that, hope you can help me out :) Johnathan === Subject: Re: inverse modulos posting-account= y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g The Euclidean algorithm constructively produces coefficients a,b such that ax + bm = gcd(x,m). So given x,m with gcd(x,m) = 1, one easily finds the multiplicative inverse a of x modulo m. You are asking if a were known, could you find x (at least, up to modulo m), and the answer is yes. This is just turning the question around; x is again the multiplicative inverse of a modulo m. If you do a Google search for Euclidean algorithm, youÕll see that it is a pretty iteration of the division algorithm. === Subject: Re: inverse modulos posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK > If two numbers x and m are relatively prime, so that gcd(x, m) = 1 is > true, then x has a unique multiplicative inverse modulo m, a, so that ax > = 1 (mod m). > Knowing only the multiplicative inverse, a modulo m, and m, is it > possible to find x? X is the modulo-inverse of a and m. > Is it true that a*x - 1 = k*m, for some k, possibly negative? Or > multiple kÕs? How to find the multiple kÕs so that x could be found? > Is that possible at all, so that the candidate kÕs can be found knowing > only the inverse a modulo m and m? > If there is no single way to find it, is there a reasonable trial and > error process to go through to find x given a and m? > What I suppose I am saying is, what is the relationship between x and > its inverse a modulo m? > Well, if you caught all that, hope you can help me out :) > Johnathan HereÕs a Python demo: Ō > import gmpy Ō > for i in range(1,20): Ō twoo = 2**i Ō twee = 3**i Ō a = gmpy.invert(twoo,twee) # modulo-inverse function Ō x = gmpy.invert(a,twee) # run it again to find x Ō k = divmod(a*x-1,twee) # returns quotient, remainder Ō print twoo,twee,a,x,k Ō Ō Ō 2 3 2 2 (mpz(1), mpz(0)) Ō 4 9 7 4 (mpz(3), mpz(0)) Ō 8 27 17 8 (mpz(5), mpz(0)) Ō 16 81 76 16 (mpz(15), mpz(0)) Ō 32 243 38 32 (mpz(5), mpz(0)) Ō 64 729 262 64 (mpz(23), mpz(0)) Ō 128 2187 1589 128 (mpz(93), mpz(0)) Ō 256 6561 4075 256 (mpz(159), mpz(0)) Ō 512 19683 11879 512 (mpz(309), mpz(0)) Ō 1024 59049 35464 1024 (mpz(615), mpz(0)) Ō 2048 177147 17732 2048 (mpz(205), mpz(0)) Ō 4096 531441 363160 4096 (mpz(2799), mpz(0)) Ō 8192 1594323 181580 8192 (mpz(933), mpz(0)) Ō 16384 4782969 90790 16384 (mpz(311), mpz(0)) Ō 32768 14348907 9611333 32768 (mpz(21949), mpz(0)) Ō 65536 43046721 11980120 65536 (mpz(18239), mpz(0)) Ō 131072 129140163 92083502 131072 (mpz(93461), mpz(0)) Ō 262144 387420489 175181914 262144 (mpz(118535), mpz(0)) Ō 524288 1162261467 862431935 524288 (mpz(389037), mpz(0)) The gcd of a power of 2 and a power of 3 is always 1, so there is a valid ŌaÕ for every pair twoo and twee. Once I determined the modulo-inverse (a) of twoo, twee; I plugged ŌaÕ into the gmpy.invert function to get ŌxÕ (which is the same as twoo, but weÕre pretending we donÕt know that). of the calculation is always 0, so the quotient is the legit ŌkÕ. CC: mensanator@aol.com === Subject: Re: inverse modulos > X is the modulo-inverse of a and m. Ah-ha, I thought there must be some kind of relationship (not bright enough to find it myself ;-) > HereÕs a Python demo: Lovely, thankyou!!! But why doesnÕt it work with: x = 2385150540210225032 m = 3517 x and m are relatively prime. a = 3500 (multiplicative inverse of x and m) ? Or am I misunderstanding something? (Do I need to get k to find the original x?) Running with this sample of x, m and a gives me 1655 as the inverse of a and m, which should have been equal to x. > Ō > import gmpy > Ō > for i in range(1,20): > Ō twoo = 2**i > Ō twee = 3**i > Ō a = gmpy.invert(twoo,twee) # modulo-inverse function > Ō x = gmpy.invert(a,twee) # run it again to find x > Ō k = divmod(a*x-1,twee) # returns quotient, remainder > Ō print twoo,twee,a,x,k > Ō 2 3 2 2 (mpz(1), mpz(0)) > Ō 4 9 7 4 (mpz(3), mpz(0)) > Ō 8 27 17 8 (mpz(5), mpz(0)) > Ō 16 81 76 16 (mpz(15), mpz(0)) > Ō 32 243 38 32 (mpz(5), mpz(0)) > Ō 64 729 262 64 (mpz(23), mpz(0)) > Ō 128 2187 1589 128 (mpz(93), mpz(0)) > Ō 256 6561 4075 256 (mpz(159), mpz(0)) > Ō 512 19683 11879 512 (mpz(309), mpz(0)) > Ō 1024 59049 35464 1024 (mpz(615), mpz(0)) > Ō 2048 177147 17732 2048 (mpz(205), mpz(0)) > Ō 4096 531441 363160 4096 (mpz(2799), mpz(0)) > Ō 8192 1594323 181580 8192 (mpz(933), mpz(0)) > Ō 16384 4782969 90790 16384 (mpz(311), mpz(0)) > Ō 32768 14348907 9611333 32768 (mpz(21949), mpz(0)) > Ō 65536 43046721 11980120 65536 (mpz(18239), mpz(0)) > Ō 131072 129140163 92083502 131072 (mpz(93461), mpz(0)) > Ō 262144 387420489 175181914 262144 (mpz(118535), mpz(0)) > Ō 524288 1162261467 862431935 524288 (mpz(389037), mpz(0)) > The gcd of a power of 2 and a power of 3 is always 1, so there > is a valid ŌaÕ for every pair twoo and twee. Once I determined > the modulo-inverse (a) of twoo, twee; I plugged ŌaÕ into the > gmpy.invert function to get ŌxÕ (which is the same as twoo, but > weÕre pretending we donÕt know that). > of the calculation is always 0, so the quotient is the legit ŌkÕ. I wish I could get these results with my numbers... :) only a and m. === Subject: Re: inverse modulos === >Subject: Re: inverse modulos >Message-id: <41af7774$0$25776$5a62ac22@per-qv1-newsreader-01.iinet.net.au > X is the modulo-inverse of a and m. >Ah-ha, I thought there must be some kind of relationship (not bright >enough to find it myself ;-) >> HereÕs a Python demo: >Lovely, thankyou!!! >But why doesnÕt it work with: >x = 2385150540210225032 >m = 3517 Yeah, I guess that only works when xx and m are relatively prime. >a = 3500 (multiplicative inverse of x and m) >? Or am I misunderstanding something? (Do I need to get k to find the >original x?) Running with this sample of x, m and a gives me 1655 as >the inverse of a and m, which should have been equal to x. 1655 is the first of an infinite series of xÕs where the inverse modulo is 3500. When x> Ō > import gmpy >> Ō > for i in range(1,20): >> Ō twoo = 2**i >> Ō twee = 3**i >> Ō a = gmpy.invert(twoo,twee) # modulo-inverse function >> Ō x = gmpy.invert(a,twee) # run it again to find x >> Ō k = divmod(a*x-1,twee) # returns quotient, remainder >> Ō print twoo,twee,a,x,k >> Ō >> Ō >> Ō 2 3 2 2 (mpz(1), mpz(0)) >> Ō 4 9 7 4 (mpz(3), mpz(0)) >> Ō 8 27 17 8 (mpz(5), mpz(0)) >> Ō 16 81 76 16 (mpz(15), mpz(0)) >> Ō 32 243 38 32 (mpz(5), mpz(0)) >> Ō 64 729 262 64 (mpz(23), mpz(0)) >> Ō 128 2187 1589 128 (mpz(93), mpz(0)) >> Ō 256 6561 4075 256 (mpz(159), mpz(0)) >> Ō 512 19683 11879 512 (mpz(309), mpz(0)) >> Ō 1024 59049 35464 1024 (mpz(615), mpz(0)) >> Ō 2048 177147 17732 2048 (mpz(205), mpz(0)) >> Ō 4096 531441 363160 4096 (mpz(2799), mpz(0)) >> Ō 8192 1594323 181580 8192 (mpz(933), mpz(0)) >> Ō 16384 4782969 90790 16384 (mpz(311), mpz(0)) >> Ō 32768 14348907 9611333 32768 (mpz(21949), mpz(0)) >> Ō 65536 43046721 11980120 65536 (mpz(18239), mpz(0)) >> Ō 131072 129140163 92083502 131072 (mpz(93461), mpz(0)) >> Ō 262144 387420489 175181914 262144 (mpz(118535), mpz(0)) >> Ō 524288 1162261467 862431935 524288 (mpz(389037), mpz(0)) >> The gcd of a power of 2 and a power of 3 is always 1, so there >> is a valid ŌaÕ for every pair twoo and twee. Once I determined >> the modulo-inverse (a) of twoo, twee; I plugged ŌaÕ into the >> gmpy.invert function to get ŌxÕ (which is the same as twoo, but >> weÕre pretending we donÕt know that). >> of the calculation is always 0, so the quotient is the legit ŌkÕ. >I wish I could get these results with my numbers... :) >only a and m. -- Mensanator Ace of Clubs === Subject: Re: inverse modulos > If two numbers x and m are relatively prime, so that gcd(x, m) = 1 is > true, then x has a unique multiplicative inverse modulo m, a, so that ax > = 1 (mod m). > Knowing only the multiplicative inverse, a modulo m, and m, is it > possible to find x? IÕm not sure if I understand what you mean... We are given two numbers a and m and we want to find two numbers x and k such that: a*x - 1 = k*m or, equivalently: a*x - k*m = 1. But this is a very simple linear diophantine equation - so, where is the problem...? Pawel Gladki CC: gladki@math.usask.ca === Subject: Re: inverse modulos > IÕm not sure if I understand what you mean... We are given two numbers a > and m and we want to find two numbers x and k such that: > a*x - 1 = k*m > or, equivalently: > a*x - k*m = 1. > But this is a very simple linear diophantine equation - so, where is the > problem...? Ok... hang on! I understand what *you* are asking now. Ok, just did a mathworld.wolfram.com. This is exactly what I am looking for. Solving for two unknowns. Is there some way I can solve a diophantine equation computationally, i.e. I am using the GNU GMP library and writing programs in C. Now I need to do it in code. My papers show: px + sn = 1, where x and s are unknown, and p = modular inverse of x and n, and n is relatively prime to x of course for p to exist. So rearranging this I get px = 1 + (-s)n, is that the same thing? (I have done this one to verify when I do know x, but now I want to find x and s for px + sn = 1 knowing only p and n.) Johnathan === Subject: Re: inverse modulos >> IÕm not sure if I understand what you mean... We are given two numbers >> a and m and we want to find two numbers x and k such that: >> a*x - 1 = k*m >> or, equivalently: >> a*x - k*m = 1. >> But this is a very simple linear diophantine equation - so, where is >> the problem...? > Ok... hang on! I understand what *you* are asking now. Ok, just did a > mathworld.wolfram.com. This is exactly what I am looking for. Solving > for two unknowns. > Is there some way I can solve a diophantine equation computationally, > i.e. I am using the GNU GMP library and writing programs in C. > Now I need to do it in code. My papers show: > px + sn = 1, where x and s are unknown Solving diophantine equations is easy... it uses the extended Euclidean algorithm, for details see e.g.: http://www.win.tue.nl/~ida/demo/c1s3f2.html Pawel Gladki === Subject: Re: inverse modulos > Solving diophantine equations is easy... it uses the extended Euclidean > algorithm, for details see e.g.: > http://www.win.tue.nl/~ida/demo/c1s3f2.html use of the extended Euclidean algorithm, but itÕs not giving the result I want. I thought that the inverse modulo was supposed to be unique where it existed, but I must have misunderstood that. Apparently there are many solutions sometimes (or always?) I havenÕt yet found a way to grab all the solutions and test them for the particular value of x that I want. And even then, if the relationship between x and n and its inverse modulo is not unique (same for every possible x solution) then perhaps I canÕt know what the original x was. Johnathan === Subject: Re: inverse modulos :> Solving diophantine equations is easy... it uses the extended Euclidean :> algorithm, for details see e.g.: :> http://www.win.tue.nl/~ida/demo/c1s3f2.html : use of the extended Euclidean algorithm, but itÕs not giving the result : I want. I thought that the inverse modulo was supposed to be unique : where it existed, but I must have misunderstood that. The inverse of a modulo m is unique *modulo m*, if it exists. For example, a = 17, m = 5. The inverse of 17 modulo 5 is 3, since 17*3 = 51 = 1 + 5*10. (k=10 in the notation of your earlier posts). I could add any multiple of m (in this case 5) to 3, and it would still be an inverse: for example 8,13,18,... but these are all congruent modulo 5 (i.e. they differ from each other by a multiple of 5). The inverse is unique modulo 5. In general, if a-b is divisible by m, then we say that a and b are congruent modulo m, and write a = b (mod m). Ted === Subject: Re: inverse modulos >> Solving diophantine equations is easy... it uses the extended >> Euclidean algorithm, for details see e.g.: >> http://www.win.tue.nl/~ida/demo/c1s3f2.html > use of the extended Euclidean algorithm, but itÕs not giving the result > I want. I thought that the inverse modulo was supposed to be unique > where it existed, but I must have misunderstood that. Apparently there > are many solutions sometimes (or always?) There are _always_ infinitely many solutions. However, since: a*x - k*m = 1 it follows that gcd(x, m) = 1, so there exists _exactly_ _one_ x which satisfies the above equation and is less than m (and bigger than 0, too). Indeed, suppose that we have two such x-es, x and xÕ(and two respective k and kÕ). Then: a*x - k*m = 1 a*xÕ - kÕ*m = 1 Subtracting the second equation from the first gives: a*(x - xÕ) - m*(k - kÕ) = 0 that is a*(x-xÕ) = m*(kÕ-k), which yields x-xÕ = m*(kÕ-k)/a. But also gcd(a,m) = 1, which implies that a divides kÕ-k, so that (kÕ-k)/a is an integer. That means x and xÕ differ by multiplicity of m, which is possible only if one of them is bigger than m. Pawel Gladki === Subject: Re: inverse modulos > Solving diophantine equations is easy... it uses the extended > Euclidean algorithm, for details see e.g.: > http://www.win.tue.nl/~ida/demo/c1s3f2.html >> makes use of the extended Euclidean algorithm, but itÕs not giving >> the result I want. I thought that the inverse modulo was supposed >> to be unique where it existed, but I must have misunderstood that. >> Apparently there are many solutions sometimes (or always?) > There are always infinitely many solutions. However, since: > ax - km = 1 > it follows that gcd(x, m) = 1, so there exists exactly one x which > satisfies the above equation and is less than m (and bigger than 0, > too). Indeed, suppose that we have two such x-es, x and xÕ(and two > respective k and kÕ). Then: > ax - k m = 1 > axÕ - kÕm = 1 > Subtracting the second equation from the first gives: > a(x - xÕ) - m(k - kÕ) = 0 > that is a(x-xÕ) = m(kÕ-k), which yields x-xÕ = m(kÕ-k)/a. But also > gcd(a,m) = 1, which implies that a divides kÕ-k, so that (kÕ-k)/a is > an integer. That means x and xÕ differ by multiplicity of m, which is > possible only if one of them is bigger than m. Simpler: Inverses are always unique in any abelian group G. For if B has inverses A,C then A = A(BC) = (AB)C = C Here G = U(Z/m) = unit group of Z/m, the ring of integers (mod m), i.e. the units (invertibles) of the multiplicative monoid of Z/m. By the extended Euclidean/Bezout algorithm the units are simply the cosets n (mod m) := n + mZ such that n is coprime to m since then and only then does there exist integers j,k such that j n + k m = 1 <=> j n = 1 (mod m) <=> n in U(Z/m) Uniqueness theorems are powerful tools for proving equalities. For many further less trivial examples see my prior posts --Bill Dubuque === Subject: Re: inverse modulos >I thought that the inverse modulo was supposed >to be unique where it existed, but I must have misunderstood that. >Apparently there are many solutions sometimes (or always?) >>There are always infinitely many solutions. However, since: >> ax - km = 1 >>it follows that gcd(x, m) = 1, so there exists exactly one x which >>satisfies the above equation and is less than m (and bigger than 0, >>too). Indeed, suppose that we have two such x-es, x and xÕ(and two >>respective k and kÕ). Then: >> ax - k m = 1 >> axÕ - kÕm = 1 >>Subtracting the second equation from the first gives: >> a(x - xÕ) - m(k - kÕ) = 0 >>that is a(x-xÕ) = m(kÕ-k), which yields x-xÕ = m(kÕ-k)/a. But also >>gcd(a,m) = 1, which implies that a divides kÕ-k, so that (kÕ-k)/a is >>an integer. That means x and xÕ differ by multiplicity of m, which is >>possible only if one of them is bigger than m. > Simpler: Inverses are always unique in any abelian group G. (...) Well, yeah, I know that much :-) But it seems that the person who asked that question didnÕt know much about groups, rings etc., so I was trying to explain the uniqueness of an inverse element in Z_m in the simplest possible terms... hope I havenÕt confused him too much. Sometimes one has to say difficult things, but one ought to say them as simply as one knows how - G. H. Hardy Pawel Gladki === Subject: Re: inverse modulos > IÕm not sure if I understand what you mean... Let me see if I can explain it a bit better. Two numbers, IÕll call them x and n, and the gcd is 1, so they are relatively prime. I compute the modulo inverse of x and n, to get p. I know only n and p: how do I find x? See: xp = 1 (mod n), so: xp - 1 is exactly divisible by n. HereÕs a concrete example: x = 2385150540210225032 n = 3517 x and n are relatively prime. p = 3500, the inverse modulo of x and n. x*p = 1 (mod n), and x*p - 1 is divisible by n exactly. If I know only n and p here, how do I get back the x? > But this is a very simple linear diophantine equation - so, where is the > problem...? IÕm after a numerical solution. Is it possible to find one? That is, given an unknown x and a known n with a known inverse modulo of those two p, can I find x? (Is there only one x?) So I can find a/the numerical solution by solving a linear diophantine equation? I hope that I explained it correctly on the first attempt! Johnathan === Subject: Re: inverse modulos > IÕm not sure if I understand what you mean... > Let me see if I can explain it a bit better. Two numbers, IÕll call > them x and n, and the gcd is 1, so they are relatively prime. I compute > the modulo inverse of x and n, to get p. > I know only n and p: how do I find x? > See: xp = 1 (mod n), so: > xp - 1 is exactly divisible by n. > HereÕs a concrete example: > x = 2385150540210225032 > n = 3517 Drat! My previous post was not quite correct. There are an infite number of xÕs whose modulo inverse 3517 result in 3500. Doing the gmpy.invert on (3500,3517) yields 1655, the first of that infinite list of xÕs. Call the first one x_0. To find x_k: x_k = k*n + x_0 HereÕs how the list starts out. Note the modulo inverse is 3500 for each x_k: > for k in range(20): print k, k*n+r,gmpy.invert(k*n+r,n) 0 1655 3500 1 5172 3500 2 8689 3500 3 12206 3500 4 15723 3500 5 19240 3500 6 22757 3500 7 26274 3500 8 29791 3500 9 33308 3500 10 36825 3500 11 40342 3500 12 43859 3500 13 47376 3500 14 50893 3500 15 54410 3500 16 57927 3500 17 61444 3500 18 64961 3500 19 68478 3500 Your example happens to occur at k = 678177577540581. > x and n are relatively prime. > p = 3500, the inverse modulo of x and n. > x*p = 1 (mod n), > and x*p - 1 is divisible by n exactly. > If I know only n and p here, how do I get back the x? Now that I think IÕm doing it correctly, I donÕt see how. If you saw a clock with a missing hour hand, you would know how many minutes past the hour it was, but not what hour. > But this is a very simple linear diophantine equation - so, where is the > problem...? > IÕm after a numerical solution. Is it possible to find one? That is, > given an unknown x and a known n with a known inverse modulo of those > two p, can I find x? (Is there only one x?) > So I can find a/the numerical solution by solving a linear diophantine > equation? ThatÕs beyond me. > I hope that I explained it correctly on the first attempt! > Johnathan === Subject: Re: Anybody else get VirusW32.Beagle.AV@mm from Muslim-hating Robert Kolker? posting-account=cNo16gwAAABRPsTqLAPMwzEEq_aUXC-- > You think a virus would use a real return address? > Viruses steal legitimate e-mail addresses to use in > their spoof headers. Get a clue. > This appears to have come via segreteria.org (62.98.222.246) > but that may be spoofed also. > - Randy If you receive a virus today, you can be almost certain that the of the mail header is not infected with a virus. Current viruses scan the hard disk of the infected machine. They look through the email address book, the webbrowser cache and just about any other file on the disk. Basically any string with an Ō@Õ in it becomes fair game. The virus mails itself to such addresses, using another address from the same pool as the fake sender. That is the reason why the sender address of a virus is often related to the recipient - they may share a common contact, the person whose computer got infected. YouÕre right: the e-mail address is most likely spurious. Sorry for the misattribution. And those murderous, anti-Moslem diatribes--mightnÕt they be the jism of some cryptoid ethnic cleanser? --John === Subject: equilateral triangles and their mirror image posting-account=P2P6jA0AAABbIkifFRIAe-RxXLVkbmLq My book says word for word: Suppose a given triangle is directly congruent to its mirror image. We can be absoilutely certain that this triangle is: Equilateral Isosceles acute obtuse The answer, according to my book, is acute. Why? Why not an Equilateral triangle? Guess what happens when I construct an Equilateral triangle and stick it in front of a mirror? ItÕs mirror image is directly congruent to the actual triangle. So, how in the world does my book figure that we canÕt be absolutely certain that a mirror image of an Equilateral triangle will always be directly congruent with its mirror image? Is my book wrong? Or am I missing something? === Subject: Re: equilateral triangles and their mirror image I am afraid that your book is wrong. Any isosceles triangle with a vertical angle >= 90 degrees is a counterexample. The correct alternative is isosceles, not the more restrictive equilateral. Proof: (1) Figures that are directly congruent to their mirror images possess a symmetry axis, which acts as a mirror by definition. (2) In a symmetric triangle one of the sides is necessarily perpendicular to the symmetry axis; the other two are each otherÕs mirror images and are therefore equal. Done! To answer your question about what you missed: just nothing; but the relevant issue here is that equilateral triangles are not the only symmetric ones. Johan Ernest Mebius >My book says word for word: >Suppose a given triangle is directly congruent to its mirror >image. We can be absoilutely certain that this triangle is: >Equilateral >Isosceles >acute >obtuse >The answer, according to my book, is acute. Why? >Why not an Equilateral triangle? Guess what happens >when I construct an Equilateral triangle and stick >it in front of a mirror? ItÕs mirror image is directly >congruent to the actual triangle. So, how in the world >does my book figure that we canÕt be absolutely certain >that a mirror image of an Equilateral triangle will >always be directly congruent with its mirror image? >Is my book wrong? Or am I missing something? === Subject: Re: equilateral triangles and their mirror image > My book says word for word: > Suppose a given triangle is directly congruent to its mirror > image. We can be absoilutely certain that this triangle is: > Equilateral > Isosceles > acute > obtuse > The answer, according to my book, is acute. Why? > Why not an Equilateral triangle? Guess what happens > when I construct an Equilateral triangle and stick > it in front of a mirror? ItÕs mirror image is directly > congruent to the actual triangle. So, how in the world > does my book figure that we canÕt be absolutely certain > that a mirror image of an Equilateral triangle will > always be directly congruent with its mirror image? > Is my book wrong? Or am I missing something? Could br picked from a tree A) fruit B) apple C) green D) red answer of your book is red, your answer is apple conclude... -- philippe (chephip at free dot fr) === Subject: Re: A rather complicated pseudorandom number generator >In my explorations of FTL I have discovered that associated with each solution Looks like a homework problem to me... >a^n + b^n = c^n there is a sum >(a^n + b^n) mod c + >(c^n - a^n) mod b + >(c^n - b^n) mod a = 0 Of course. Consider the following: a^n + b^n = c^n therefore (a^n +c^n) mod c = c^n mod c = 0 The others should follow logically. I hope this has made it easier for you to figure out where you are going with this... === Subject: Re: Limits for functions defined on arbitrary sets (not intervals) > Let g be a real valued function on a set S of real numbers. Let a and l > be real numbers . The statement > lim g(x) = l > x->a > x in S > means that when S subset R, g:S -> R. for all e > 0, some d with for all x, (0 < |x - a| < d ==> |g(x) - l| < e) In topology lim(x->a) f(x) = l when for all open U nhood l, some open V nhood a with f(Va) subset U > (b) For each posative number epsilon there is a posative number delta > such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon > However there is no requirement in (b) that x not equal a. Is this a > typo or there is a reason for it? Another sloppy, quick profit text book? It should be there except when g is continuous at a, then the 0 < is redundant. > let f(x) = x for all x in Natural numbers. This function is continuous > under a metric topology. And the above definition can also be used to > prove the same. However if we add the requirement that x not equal > a then this function does not have a limit at any point. Pick any e. Let d = e. Then for all x 0 < |x - a| < d ==> |f(x) - a| < e lim(x->a) f(x) = a. > On the other hand consider the function g(x)=x for all x in natural On the other hand for readablity, consider the function g(x) = x. > numbers except 1, and g(1)=2. We would like a definition that yeilds > that the limit of g(x) as x tends to 1 is 1. However the above > definition implies that this limit is 2. It does not. Let e = 1/2, whereÕs the d such that for all x 0 < |x - 1| < d ==> |g(x) - 2| < 1/2 ? === Subject: Re: Limits for functions defined on arbitrary sets (not intervals) I suspect you are correct about the typo, though Kennan T SmithÕs book should not be that bad; it has been around for a while and springer-verlag released the second ed. >> On the other hand consider the function g(x)=x for all x in natural >> numbers except 1, and g(1)=2. We would like a definition that yeilds >> that the limit of g(x) as x tends to 1 is 1. However the above >> definition implies that this limit is 2. > It does not. Let e = 1/2, whereÕs the d such that for all x > 0 < |x - 1| < d ==> |g(x) - 2| < 1/2 ? If as in the definition there was no requirement that 0 < |x - 1| then let d = 1 and we could always let x = 1 so |g(x) - 2| = 0 === Subject: Limits for functions defined on arbitrary sets (not intervals) Correction made to the last part of my previous post. > Let g be a real valued function on a set S of real numbers. Let a and l > be real numbers . The statement > lim g(x) = l > x->a > x in S > means that > when S subset R, g:S -> R. > for all e > 0, some d with for all x, (0 < |x - a| < d ==> |g(x) - l| < e) > In topology lim(x->a) f(x) = l when > for all open U nhood l, some open V nhood a with > f(Va) subset U It is becoming apparent from the discussion below, why the above isnÕt a standard definition, but instead f continuous at a, ie lim(x->a) f(x) = f(a) when for all open U nhood f(a), some open V nhood a with f(V) subset U. > (b) For each posative number epsilon there is a posative number delta > such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon > However there is no requirement in (b) that x not equal a. ThatÕs for when g is continuous at a. BTW, youÕve a typo at g(x) - a. > let f(x) = x for all x in Natural numbers. This function is continuous > under a metric topology. And the above definition can also be used to > prove the same. However if we add the requirement that x not equal > a then this function does not have a limit at any point. > Pick any e. Let d = e. Then for all x > 0 < |x - a| < d ==> |f(x) - a| < e > lim(x->a) f(x) = a. > On the other hand consider the function g(x)=x for all x in natural > On the other hand for readability, consider the function g(x) = x. > numbers except 1, and g(1)=2. We would like a definition that yeilds > that the limit of g(x) as x tends to 1 is 1. However the above > definition implies that this limit is 2. Let g(1) = b. Then for any e > 0, let d = 1/2. Thus for all integers x, (0 < |x - 1| < 1/2 ==> |g(x) - c| < e) because there is no integer x with 0 < |x - 1| < 1/2. Hence lim(x->1) g(x) = c for any and all c. This is a rather perplexing result showing that this calculus definition of limit of a function at a point has to be limited to exclude discrete domains. You will notice that for the domain D = (-oo,-1] / { 0 } / [1,oo) the same problem will occur but not for domains like (-oo,0] / (1,2) / [5,oo) which donÕt have any isolated points like 0 in the first example. Thus the 0 < requirement needs be included with a requirement that for x->a, that a is not an isolated point of the domain of the function. When a is an isolated point, then there is no sequence of points /= a that will converge to a. When there is a sequence of points /= a, a1, a2, ... that converge to a, then the definition of limit will show f(a1), f(a2), ... will converge to the limit point. But thatÕs the intuition in lim(x->a) f(x), that f(x) will approach a number as x approaches a. Now you be an imp and ask teacher whatÕs the limit of a function at 0 with the domain D as described above and report to us his answer. The function could be x^2, x, 1 or even 0 for spectacular results. On domain D, lim(x->0) 0 = 1,000,000,000 or any number a what ever. If e > 0, let d = 1/2 for all x in D, (0 < |x - 0| < d ==> |0 - a| < e) again by vacuous implication. === Subject: Re: Limits for functions defined on arbitrary sets (not intervals) The vacuous implication argument is indeed a convincing one. bt > Correction made to the last part of my previous post. >> Let g be a real valued function on a set S of real numbers. Let a and l >> be real numbers . The statement >> lim g(x) = l >> x->a >> x in S >> means that >> when S subset R, g:S -> R. >> for all e > 0, some d with for all x, (0 < |x - a| < d ==> |g(x) - l| < e) >> In topology lim(x->a) f(x) = l when >> for all open U nhood l, some open V nhood a with >> f(Va) subset U > It is becoming apparent from the discussion below, why the above isnÕt a > standard definition, but instead > f continuous at a, ie lim(x->a) f(x) = f(a) when > for all open U nhood f(a), some open V nhood a with f(V) subset U. >> (b) For each posative number epsilon there is a posative number delta >> such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon >> However there is no requirement in (b) that x not equal a. > ThatÕs for when g is continuous at a. BTW, youÕve a typo at g(x) - a. >> let f(x) = x for all x in Natural numbers. This function is continuous >> under a metric topology. And the above definition can also be used to >> prove the same. However if we add the requirement that x not equal >> a then this function does not have a limit at any point. >> Pick any e. Let d = e. Then for all x >> 0 < |x - a| < d ==> |f(x) - a| < e >> lim(x->a) f(x) = a. >> On the other hand consider the function g(x)=x for all x in natural >> On the other hand for readability, consider the function g(x) = x. >> numbers except 1, and g(1)=2. We would like a definition that yeilds >> that the limit of g(x) as x tends to 1 is 1. However the above >> definition implies that this limit is 2. > Let g(1) = b. Then for any e > 0, let d = 1/2. > Thus for all integers x, (0 < |x - 1| < 1/2 ==> |g(x) - c| < e) > because there is no integer x with 0 < |x - 1| < 1/2. > Hence lim(x->1) g(x) = c for any and all c. > This is a rather perplexing result showing that this calculus > definition of limit of a function at a point has to be limited > to exclude discrete domains. You will notice that for the domain > D = (-oo,-1] / { 0 } / [1,oo) the same problem will occur but not > for domains like (-oo,0] / (1,2) / [5,oo) which donÕt have any > isolated points like 0 in the first example. Thus the 0 < requirement > needs be included with a requirement that for x->a, that a is not an > isolated point of the domain of the function. > When a is an isolated point, then there is no sequence of points /= a > that will converge to a. When there is a sequence of points /= a, > a1, a2, ... that converge to a, then the definition of limit will show > f(a1), f(a2), ... will converge to the limit point. But thatÕs the > intuition in lim(x->a) f(x), that f(x) will approach a number as x > approaches a. > Now you be an imp and ask teacher whatÕs the limit of a function at > 0 with the domain D as described above and report to us his answer. > The function could be x^2, x, 1 or even 0 for spectacular results. > On domain D, lim(x->0) 0 = 1,000,000,000 or any number a what ever. > If e > 0, let d = 1/2 > for all x in D, (0 < |x - 0| < d ==> |0 - a| < e) > again by vacuous implication. === Subject: Re: Limits for functions defined on arbitrary sets (not intervals) > For functions defined on arbitrary sets (such as integers) one usually > talks of continuity in terms of topological concepts. However I was > reading Kennan T Smiths Primer of Modern analysis where he defines > limit as follows : > Let g be a real valued function on a set S of real numbers. Let a and l > be real numbers . The statement > lim g(x) = l > x->a > x in S > means that > (a) For each posative number delta there is at least one point x in S > with |x-a| < delta > (b) For each posative number epsilon there is a posative number delta > such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon > For functions defined on intervals (b) is ALMOST the definition of > limits since (a) is trivially satisfied. However there is no requirement > in (b) that x not equal a. Is this a typo or there is a reason for it? > Notices : > (1) > let f(x) = x for all x in Natural numbers. This function is continuous > under a metric topology. And the above definition can also be used to > prove the same. However if we add the requirement that x not equal > a then this function does not have a limit at any point. > (2) > On the other hand consider the function g(x)=x for all x in natural > numbers except 1, and g(1)=2. We would like a definition that yeilds > that the limit of g(x) as x tends to 1 is 1. However the above > definition implies that this limit is 2. > Prof Smith then goes on define limits on deleted neighbourhoods > (intervals) in the usual way. Could some one kindly shed some > light on the apparently unusual way for defining limits and the > motivation for it. > sincerely > B Thomas === Subject: Re: Limits for functions defined on arbitrary sets (not intervals) > (a) For each posative number delta there is at least one point x in S > with |x-a| < delta > (b) For each posative number epsilon there is a posative number delta > such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon >(2) >On the other hand consider the function g(x)=x for all x in natural >numbers except 1, and g(1)=2. We would like a definition that yeilds >that the limit of g(x) as x tends to 1 is 1. However the above >definition implies that this limit is 2. That doesnÕt make sense to me. You can certainly interpolate (g(0)+g(2))/2 but that isnÕt the same as a limit. The one case where a limit on a function which is in a discrete domain would make sense is a=oo. Then you can test successively closer deltas for validation. For example: lim(x->oo) n!/(n-1)!/(n+1) The epsilons are then infinite, but other than that it works. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Limits for functions defined on arbitrary sets (not intervals) >> (a) For each posative number delta there is at least one point x in S >> with |x-a| < delta >> (b) For each posative number epsilon there is a posative number delta >> such that if |x-a|< delta and x in S, then |g(x)-a|< epsilon >>(2) >>On the other hand consider the function g(x)=x for all x in natural >>numbers except 1, and g(1)=2. We would like a definition that yeilds >>that the limit of g(x) as x tends to 1 is 1. However the above >>definition implies that this limit is 2. > That doesnÕt make sense to me. You can certainly interpolate (g(0)+g(2))/2 > but that isnÕt the same as a limit. No I am not interpolating. Just using the definition as in a epsilon-delta proof of a limit to show the limit is indeed 2. However if the domain was real numbers then the limit would be 1, for the same function. Which is why I said we would like a definition that yeilds 1 as the limit. > The one case where a limit on a function which is in a discrete domain would > make sense is a=oo. Then you can test successively closer deltas for > validation. For example: > lim(x->oo) n!/(n-1)!/(n+1) > The epsilons are then infinite, but other than that it works. One can talk of continuity in arbitrary topological spaces. And if the topology is induced by a metric then we can certainly talk about limits even if we are dealing with point sets and not intervals. sincerely b thomas === Subject: Re: censorship screen in sci.math posting-account=0QrkrwwAAABDyQGPKX7NtkkaKfngvovA > It appears to me that posts with titles containing some words as > Optimal, Strategy, StockMarket, VonNeumann, OS, free are being > censored and not registering in sci.math or sci.econ. > Some person or group of persons is deliberately censoring the posts > of author Archimedes Plutonium to these newsgroups. Some of your problems are perhaps addressed at http://www.geocities.com/ResearchTriangle/Lab/1131/ua.txt and see Q.11 and A.11. David Ames === Subject: Re: censorship screen in sci.math ItÕs true some messages ( a few) do seem to get lost. Not only my experience , but a friendÕs as well. I suggest reposting ones that donÕt appear until they do. >>It appears to me that posts with titles containing some words as >>Optimal, Strategy, StockMarket, VonNeumann, OS, free are being >>censored and not registering in sci.math or sci.econ. >>Some person or group of persons is deliberately censoring the posts >>of author Archimedes Plutonium to these newsgroups. >> >Some of your problems are perhaps addressed at >http://www.geocities.com/ResearchTriangle/Lab/1131/ua.txt >and see Q.11 and A.11. >David Ames -- 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: Mathematical calendar (long) posting-account=qemkjQwAAACJJO3EwwT9L4LMAQP02OUw Having too much time in my hands, I decided to compile a mathematical calendar, where each day is celebrated with a mathematical concept somehow related to the numbers of the date. This is the result of my waste of time. Note that the dates are written in the European format (day/month). January 1/1 - Fibonnaci series: The first two terms of the Fibonacci series are 1,1... 2/1 - Continued fraction for KhinchinÕs constant: The continued fraction for KhinchinÕs constant is [2,1...] 3/1 - Trefoil knot: The symbol for the trefoil knot is 3sub1 4/1 - Square pyramid: A square pyramid is formed of 4 triangles and 1 square 5/1 - Pentagonal pyramid: A pentagonal pyramid is formed of 5 triangles and 1 square 6/1 - StevedoreÕs knot: The symbol for StevedoreÕs knot is 6sub1 7/1 - Cube line picking: The minimum straight line distance between any two of 7 points placed as far apart as possible in a unit cube is 1 8/1 - Kaprekar numbers: The first Kaprekar number is 9 because 9^2=81 and 8+1=9 9/1 - TarskiÕs plank problem: Given a circular table of diameter 9 feet, which is the minimal number of planks, 1 foot wide and length greater that 9 feet, needed in order to completely cover the tabletop? 10/1 - Biaugmented triangular prism: A biaugmented triangular prism has 10 triangles and 1 square 11/1 - Herschel graph: There is only 1 non-Hamiltonian polyhedral graph in 11 nodes (the least possible), and this is the Herschel graph 12/1 - Gyroelongated square pyramid: A gyroelongated square pyramid has 12 triangles and 1 square 13/1 - Clock solitaire: The probability of winning a clock solitaire is 1/13 14/1 - Stella octangula numbers: The first two stella octangula numbers are 1,14... 15/1 - Rhombic dodecahedral numbers: The first two rhombic dodecahedral numbers are 1,15... 16/1 - Biquadratic numbers: The first two biquadratic numbers are 1,16... 17/1 - Magic tesseract constants: The first two magic tesseract constants are 1,17... 18/1 - Smallest Salem number: The smallest Salem number is 1.18... 19/1 - Foias constant: The value of Foias constant is 1.19... 20/1 - AperyÕs constant (zeta(3)): The value of AperyÕs constant is 1.20... 21/1 - Smallest simple perfect square: There is only 1 simple perfect square of order 21 (the least possible) 22/1 - Inradius of the small rhombicuboctahedron: The inradius of the small rhombicuboctahedron is 1.22... 23/1 - Ways of coloring an octahedron: There is 1 way of coloring an octahedron with 1 color, and 23 with 2 colors 24/1 - Midradius of the snub cube: The midradius of the snub cube is 1.24... 25/1 - Bifurcation diagram: The onset of the second bifurcation in the bifurcation diagram is 1.25 26/1 - Delian constant (cubic root of 2): The value of the Delian constant is 1.26... 27/1 - In-shufße: The ordering of cards after an in-shufße is 27,1... 28/1 - Glaisher-Kinkelin constant: The value of the Glaisher-Kinkelin constant is 1.28... 29/1 - C29: There is only 1 finite group of order 29, and it is C29 30/1 - Midradius of the dodecahedron: The midradius of the dodecahedron is 1.30... 31/1 - MillÕs constant: The value of MillÕs constant is 1.31... February 1/2 - Natural numbers: The first two natural numbers are 1,2... 2/2 - Continued fraction of the square root of 2: The continued fraction of the square root of 2 is periodical with 2: [1,2,2,2,2...] 3/2 - CatalanÕs conjecture: 2^3 and 3^2 are the only consecutive powers 4/2 - Subsets of {1,2}: There are 4 subsets of {1,2} 5/2 - Piet HeinÕs superellipse: Piet HeinÕs superellipse formula is ax^(5/2)+by^(5^2)=1 6/2 - Hexagram: The symbol of the hexagram is {6/2} 7/2 - 2nd Mersenne prime: The 2nd Mersenne prime is 7 8/2 - Star of Lakshmi: The symbol of the star of Lakshmi is {8/2} 9/2 - Tetrahedron circumscribing: The smallest volume of a tetrahedron circumscribing a parallelepiped of volume 1 is 9/2 10/2 - Binary: 2 is written 10 in binary 11/2 - Decimal expansion of 11: The decimal expansion of 11 is the first unit fraction with period 2 12/2 - Sphenocorona: A sphenocorona has 12 triangles and 2 squares 13/2 - Satellite knots with 13 crossings: There are 2 satellite knots with 13 crossings 14/2 - Satellite knots with 14 crossings: There are 2 satellite knots with 14 crossings 15/2 - 3 circle packing: The minimum diameter of a circle containing 3 unit circles is 2.15... 16/2 - Sphenomegacorona: A sphenomegacorona has 16 triangles and 2 squares 17/2 - Quadratic surfaces: There are 17 types of quadratic surfaces (defined by polynomials of grade 2) 18/2 - Small universal Turing machine: The smallest known universal Turing machine with 2 states has 18 colors 19/2 - Monotonic matrices of order 2: There are 19 monotonic matrices of order 2 20/2 - Inradius of the great rhombicuboctahedron: The inradius of the great rhombicuboctahedron is 2.20... 21/2 - Mean number of edges for graphs with 7 vertices: The mean number of edges for graphs with 7 vertices is 21/2 22/2 - Possible last 2 digits of a square: There are 22 possible combinations for the las 2 digits of a square number 23/2 - Birthday problem: The minimum number of people in a meeting so that there is more than 50% chance that 2 have their birthdays on the same day is 23 24/2 - Snub square antiprism: A snub square antiprism has 24 triangles and 2 squares 25/2 - Smallest hypothenuse with 2 integer right-angle triangles: The smallest hypothenuse length that is shared by 2 integer right-angle triangles is 25 26/2 - Midradius of the great rhombicuboctahedron: The midradius of the great rhombicuboctahedron is 2.26... 27/2 - Sattelite knots with wrapping number bigger than 2: The minimum crossings of a sattelite knot with wrapping number bigger than 2 is 27 28/2 - 2nd perfect number: The 2nd perfect number is 28 29/2 - Prime generating polynomial: The polynomial 2x^2+29 generates primes for 0,...,28 March 1/3 - Odd numbers: The first two odd numbers are 1,3... 2/3 - Prime numbers: The first two prime numbers are 2,3... 3/3 - Tetrahedron: The symbol of the tetrahedron is {3,3} 4/3 - Cube: The symbol of the cube is {4,3} 5/3 - Dodecahedron: The symbol of the dodecahedron is {5,3} 6/3 - Hexagonal tesselation: The symbol of the hexagonal tesselation is {6,3} 7/3 - Plane division by 3 lines: 3 lines can divide the plane in 7 regions 8/3 - Octagram: The symbol of the octagram is {8/3} 9/3 - Sum of cubes: Every integer can be expressed as a sum of at most 9 3rd powers 10/3 - Ternary: 3 is 10 in ternary 11/3 - Mirror images in space groups: There are 11 space groups on 3 dimensions that are mirror images 12/3 - Tricylinder: The intersection of 3 cylinders has 12 curved faces 13/3 - Delannoy numbers: The first two Delannoy numbers are 3,13... 14/3 - Pi: The value of pi is 3.14... 15/3 - Tritriangular numbers: The first two tritriangular numbers are 3,15... 16/3 - Simple directed graphs with 3 nodes: There are 16 simple directed graphs with 3 nodes 17/3 - Tri-bes: There are 17 tri-bes 18/3 - Least common rolls of 3 dice: The least common rolls of 3 dice are 3 and 18 19/3 - Braced polygon problem: The minimum number of rods needed to make a triangle rigid is 3, and for a square is 19 20/3 - Steiner points: The 20 points of the 3 by 3 intersections of the Pascal lines 21/3 - Binary trees of 3 nodes height: There are 21 binary trees of 3 nodes height 22/3 - Isohedral tilings: There are 3 isohedral tilings with heptominoes and 22 with octominoes 23/3 - First cluster primes: The 23 first primes starting with 3 are cluster primes 24/3 - 24-cell: The 24-cell is a polychoron composed of 24 octahedra, with 3 to an edge 25/3 - Silver constant: The silver constantÕs value is 3.25... 26/3 - RubikÕs cube: A RubikÕs cube is a 3x3x3 cube in which the 26 subcubes on the outside can be rotated on any plane of cubes 27/3 - Power tower of 3: 3^3=27 28/3 - Khinchin-Levy constant: The value of Khinchin-Levy constant is 3.28... 29/3 - Confidence interval for a normal distribution: The confidence interval for a normal distribution with p=0.999 is 3.29 30/3 - Smallest Giuga number: The smallest Giuga number is 30 and has 3 factors 31/3 - 3rd Mersenne prime: The 3rd Mersenne prime is 31 April 1/4 - Square numbers: The first two square numbers are 1,4... 2/4 - Even numbers: The first two even numbers are 2,4... 3/4 - Octahedron: The symbol of the octahedron is {3,4} 4/4 - Square tesselation: The symbol of the square tesselation is {4,4} 5/4 - First Brown pair: The first Brown pair is 4,5 6/4 - Bichromatic graphs with 4 nodes: There are 6 bichromatic graphs with 4 nodes 7/4 - Archimedean solids: There are 13 Archimedean solids, 7 obtained by truncation and 4 by expansion 8/4 - Eight-point circle theorem: If a quadrilateral has perpendicular diagonals, the 4 midpoints of the sides and the 4 feet of the perpendiculars from the midpoints to the opposite sides are in a circle that includes the 8 points. 9/4 - Tangent circles of the triangle: There are 4 tangent circles to a given triangle: the incircle and the 3 excircles, and they are all touched by the 9-point circle. 10/4 - Quaternary: 4 is 10 in quaternary 11/4 - Bracketings of 4 letters: 4 letters have 11 possible bracketings 12/4 - Construction of an icosahedron from octahedron: If the edges of an octahedron are divided in the golden ratio proportion such that the points of division for any face form an equilateral triangle, the 12 points form an icosahedron. This can be done in 4 ways, resulting in 4 possible icosahedra. 13/4 - Busy beaver of 4 states: A busy beaver of 4 states will write 13 1s before halting 14/4 - Tetrabolos: There are 14 tetrabolos, or figures formed by 4 isosceles right triangles joined at the sides 15/4 - 15 puzzle: The 15 puzzle consists of 15 squares numbered from 1 to 15 in a 4x4 square 16/4 - Even square numbers: The first two even square numbers are 4,16... 17/4 - Area of the heptadecagon: The area of the regular heptadecagon with unit side is 17/4cot(pi/17) 18/4 - Party problem with 4 people: The minimum number of guests that must be invited to a party so that at least 4 will know each other or at least 4 wonÕt know each other is 18 19/4 - Sum of biquadratics: Every integer can be expressed as a sum of 19 4th powers 20/4 - Disphenocingulum: A disphenocingulum has 20 triangles and 4 squares 21/4 - Necklaces of 4 white, black or red beads: There are 21 possible necklaces with 4 white, black or red beads 22/4 - Tetraplets: There are 22 tetraplets, or figures formed by 4 squares joined at the sides or corners 23/4 - Triangle dissections in 4 triangles: There are 23 topologically different ways of dissecting a triangle into 4 triangles 24/4 - Kissing hyperspheres: 24 hyperspheres can touch another one in 4 dimensions 25/4 - 4th automorphic number day: The 4th automorphic number is 25 26/4 - Plane division by 4 ellipses: 4 ellipses can divide the plane in 26 regions 27/4 - Nine mice problem walk length: If nine mice start at the corners of a regular nonagon, each heading towards it closest neighboring mouse at constant speed, they will trace a logarithmic spiral of length 4.27... 28/4 - EuclidÕs postulates: Euclid proved the first 28 propositions of the Elements using only the first 4 geometry postulates 29/4 - Universal cellular automaton of von Neumann: von Neumann proved that a cellular automaton consisting of cells with 4 orthogonal neighbours and 29 states could simulate a Turing machine 30/4 - Sangaku problem: One sangaku problem asks: how can you distribute 30 identical spheres such that they are tangent to a single central sphere and to 4 other small spheres? May 1/5 - Euler numbers: The first two Euler numbers are 1,5... 2/5 - Sums of two squares: The first two numbers that are sum of two squares are 2,5... 3/5 - Icosahedron: The symbol of the icosahedron is {3,5} 4/5 - Pentiamonds: There are 4 pentiamonds, or figures formed by 5 equilateral triangles joined at the sides 5/5 - Wong graph: The Wong graph is one of the (5,5) cage graphs 6/5 - WhippleÕs identity: WhippleÕs identity is given by the generalized hypergeometric function 6F5 7/5 - Triakis tetrahedron construction: The triakis tetrahedron can be constructed by cumulation of a tetrahedron by pyramids of height 7/5 8/5 - Queens problem: The maximum number of queens that can be placed on a chessboard without any two attacking each other is 8, and the minimum needed to occupy or attack every square is 5 9/5 - Last digit of odd Catalan numbers: The last digit of the odd Catalan numbers is 9 or 5 from the second one up to at least the 30th number 10/5 - Pentatope: A pentatope has 10 triangular faces and 5 tetrahedral cells 11/5 - Second Brown pair: The second Brown pair is (11,5) 12/5 - Pentominoes: There are 12 pentominoes, or figures formed by 5 squares joined at the sides 13/5 - Wilson primes: The first two Wilson primes are 5,13... 14/5 - Pentagon tilings: There are 14 known tilings with pentagons (5-sided polygons) 15/5 - Go-moku: Go-moku is a game similar to tic-tac-toe played on a 15x15 board, trying to place 5 pieces in a row 16/5 - Plane division by 5 lines: 5 lines can divide the plane in 16 regions 17/5 - Simple connected tricolorable graphs on 5 nodes: There are 17 simple connected tricolorable graphs on 5 nodes 18/5 - One-sided pentominoes: There are 18 one-sided pentominoes, or figures formed by 5 squares joined at the sides, when mirror images are considered different 19/5 - Continued fraction for ChampernowneÕs constant: The continued fraction for ChampernowneÕs constant has sporadic extremely large terms, the first two being at positions 5 and 19 20/5 - Tetrahedron 5-compound: The tetrahedron 5-compound is composed of 5 tetrahedra occupying the 20 vertices of a dodecahedron 21/5 - Connected graphs with 5 nodes: There are 21 connected graphs with 5 nodes 22/5 - Pentahexes: There are 22 pentahexes, or figures formed by 5 regular hexagons joined at the sides 23/5 - EuclidÕs Elements: EuclidÕs Elements started with 23 definitions, 5 postulates and 5 common notions 24/5 - Ruth-Aaron pairs: The first two numbers giving Ruth-Aaron pairs are 5 and 24 25/5 - Isohedra: The isohedra are 25 finite solids and 5 classes of infinite solids 26/5 - Space division by 5 planes: 5 planes can divide space in 26 parts 27/5 - Pentakites: There are 27 pentakites, or figures formed by 5 kite shapes joined at the sides 28/5 - Orders of sociable numbers: The smallest sets of sociable numbers are two sets with 5 and 28 elements 29/5 - Pentacubes: There are 29 pentacubes, or solids formed by 5 cubes joined by the faces 30/5 - Octahedron 5-compound: The octahedron 5-compound is composed of 5 octahedra occupying the 30 vertices of an icosidodecahedron 31/5 - Dervish double points: The dervish is a quintic surface (defined by a polynomial of degree 5) that has the maximum possible number of double points for a quintic, that is 31 June 1/6 - Golden ratio: The value of the golden ratio is 1.6... 2/6 - Pronic numbers: The first two pronic numbers are 2,6... 3/6 - Triangular tesselation: The symbol of the triangular tesselation is {3,6} 4/6 - Interprimes: The first two interprimes are 4,6... 5/6 - Dissections of a regular hexagon into a golden rectangle: There are 5 ways of dissecting a regular hexagon (6-sided polygon) into a golden rectangle 6/6 - Dissections of a regular hexagon into a Latin cross: There are 6 ways of dissecting a regular hexagon (6-sided polygon) into a Latin cross 7/6 - Dice: Ordinary dice have the numbers from 1 to 6, in such a way that opposite faces add up to 7 8/6 - Cuboctahedron: A cuboctahedron has 8 triangles and 6 squares 9/6 - 6-multiperfect numbers: 6-multiperfect numbers have at least 9 prime factors 10/6 - Dodecahedron 6-compound: A dodecahedron 6-compound is composed of 6 dodecahedra, each rotated by 1/10 of a turn about the line joining the centroids of opposite faces 11/6 - Nets for the cube: There are 11 nets for the cube, which has 6 faces 12/6 - Hexiamonds: There are 12 hexiamonds, or figures formed by 6 equilateral triangles joined by the sides 13/6 - Sums of 6 cubes: The first two numbers that are a sum of six cubes are 6,13... 14/6 - Necklaces of 6 black or white beads: There are 14 possible necklaces of 6 black or white beads, counting mirror images 15/6 - Odd elements in PascalÕs triangle: There are 15 odd elements in the first 6 rows of PascalÕs triangle 16/6 - Random walk of 4 steps: The probability of being back to the origin after a random walk of 4 steps is 6/16 17/6 - Golomb ruler with 6 marks: The length of the optimal Golomb ruler with 6 marks is 17 18/6 - Even pentagonal pyramidal numbers: The first two even pentagonal pyramidal numbers are 6,18... 19/6 - Graphs with two centres on 6 nodes: There are 19 graphs with two centres in 6 nodes 20/6 - 6 equilateral triangles in a convex figure: The maximum area possible for 6 equilateral triangles forming a convex figure is 20 times the size of the smallest one 21/6 - Hyperperfect numbers: The first two hyperperfect numbers are 6,21... 22/6 - Plane division by 6 lines: 6 lines can divide the plane in 22 regions 23/6 - Home prime for 6: The home prime for 6 is 23 24/6 - Tetrahedron cutting: A tetrahedron cut by 6 planes, each passing through an edge and bisecting the opposite edge, is sliced in 24 pieces 25/6 - Smallest 6th power that is a sum of 6th powers day: The smallest 6th power that is a sum of 6th powers is 25^6 26/6 - 6-snake: The maximum length of a 6-snake is 26 27/6 - SolomonÕs seal lines: SolomonÕs seal lines are 27 real or imaginary lines which lie on the general cubic surface, that can put into a one-to-one correspondence with the vertices of a polytope in 6-dimensional space 28/6 - Perfect numbers: The first two perfect numbers are 6,28... 29/6 - Graphs with one centre on 6 nodes: There are 29 graphs with one centre on 6 nodes 30/6 - Longimeter: A longimeter is a transparent sheet of plastic with a regular grid of lines at an angle of 30 degrees. By counting the number of squares occupied by a linear feature on a map for 6 different rotations, the length of the feature can be determined July 1/7 - Hex numbers: The first two hex numbers are 1,7... 2/7 - e: The value of e is 2.7... 3/7 - Cousin primes: The first pair of cousin primes is 3 and 7 4/7 - Feigenbaum constant: The value of the Feigenbaum constant is 4.7... 5/7 - Ass and mule problem: A classic by Euclid. The mule says to the ass: If you gave me one of your sacks, I would have as many as you. The ass replies, If you gave me one of your sacks, I would have twice as many as you. The solution is 5 sacks for the mule and 7 for the ass. 6/7 - 7 tree plantation problem: The maximum number of rows of three trees you can have with 7 trees is 6 rows 7/7 - Sum of prime factors of 7: The sum of prime factors of 7 is 7 8/7 - Compositions of 7 of length 2: There are 8 compositions of 7 of length 2 9/7 - Angles that are solved using a bicubic equation: The trigonometric functions of the angles pi/7 and pi/9 are resolved using a bicubic equation 10/7 - Sum of random numbers exceeding 5: The expected number of picks in (0,1) so that the sum exceeds 5 is 10.7... 11/7 - Inequality of RobinÕs theorem: The first two numbers that hold the inequality of RobinÕs theorem are 7,11... 12/7 - Harmonic mean of the divisors of 4: The harmonic mean of the divisors of 4 is 12/7 13/7 - EulerÕs 6n+1 theorem: Every prime of the form 6n+1 (the first two of them are 7,13...) can be written in the form x^2+3y^2 14/7 - Heawood graph: The Heawood graph has 14 nodes and represents the 7-color torus map 15/7 - Sums of 4 squares: The first two numbers that are sums of 4 squares are 7,15... 16/7 - Collatz sequence for 7: The Collatz sequence for 7 has 16 steps 17/7 - Full reptend primes: The first two full reptend primes are 7,17... 18/7 - Multiplication of 2x2 matrices: The minimum number of operations needed to multiply two 2x2 matrices is 7 multiplications and 18 additions 19/7 - e approximation: A good fractional approximation of e is 19/7 20/7 - Steiner points: The Pascal lines intersect three at a time in 20 Steiner points. In the case of a regular hexagon inscribed in a cirle, they degenerate into 7 points (the vertices and center of a regular hexagon) 21/7 - Kobon triangles constructed with 7 lines: The maximum number of Kobon triangles that can be constructed with 7 lines is 21 22/7 - Pi approximation: A good fractional approximation of pi is 22/7 23/7 - Pi Bible reference: There is a reference of pi in the Bible that gives it a value of 3, which is Kings 7:23 24/7 - Heptiamonds: There are 24 heptiamonds, or figures formed by 7 equilateral triangles joined at the sides 25/7 - Smallest pseudoprime in base 7: The smallest pseudoprime in base 7 is 25 26/7 - Langford sequences with 7 digits: There are 26 Langford sequences with 7 digits 27/7 - Giuga sequences of length 7: There are 27 Giuga sequences of length 7 28/7 - Bitangents of the quartic curve: There are 28 bitangents on the quartic curve, that can be put into a one-to-one correspondence with the vertices of a polytope in 7-dimensional space 29/7 - Plane division by 7 lines: 7 lines can divide the plane in 29 regions 30/7 - 7th 3-almost prime: The 7th 3-almost prime is 30 31/7 - Graph cycles in a wheel graph with 7 nodes: There are 31 graph cycles in a wheel graph with 7 nodes August 1/8 - Cubic numbers: The first two cubic numbers are 1,8... 2/8 - Fransen-Robinson constant: The value of the Fransen-Robinson constant is 2.8... 3/8 - Levi graph: The Levi graph is the unique (3,8) cage graph 4/8 - Squareful numbers: The first two squareful numbers are 4,8... 5/8 - Sums of two primes: The first two sums of two primes are 5,8... 6/8 - Truncated octahedron: The truncated octahedron has 6 squares and 8 hexagons 7/8 - Incomparable rectangles tiling: At least 7 and at most 8 mutually incomparable rectangles are needed to tile a given rectangle 8/8 - Rooks problem: 8 rooks can be placed on a chessboard without any two attacking, and 8 is also the minimum number of pieces needed to occupy or attack every square 9/8 - BaileyÕs transformation: BaileyÕs transformation is given by the generalized hypergeometric function 9F8 10/8 - Octal: 8 is written 10 in octal 11/8 - Nets for the octahedron: There are 11 nets for the octahedron, which has 8 faces 12/8 - Matchstick graph of degree 3: The matchstick graph of degree 3 has 12 edges and 8 vertices 13/8 - Octohexes with holes: There are 13 octohexes with holes, or figures formed by 8 regular hexagons joined at the sides 14/8 - Bishops problem: 14 bishops can be placed on a chessboard without any two attacking, and 8 bishops are the the minimum number of pieces needed to occupy or attack every square 15/8 - Pawn positions: There are 15^8 possible pawn possitions in chess without captures 16/8 - Cubeful numbers: The first two cubeful numbers are 8,16... 17/8 - Wallpaper groups: There are 17 wallpaper groups, 8 of them pure translation 18/8 - Small rhombicuboctahedron: The small rhombicuboctahedron has 8 triangles and 18 squares 19/8 - 8th prime: The 8th prime is 19 20/8 - Stella octangula construction from the dodecahedron: The stella octangula can be constructed using 8 of the 20 vertices of the dodecahedron 21/8 - Prime knots with 8 crossings: There are 21 prime knots with 8 crossings 22/8 - Partitions of 8: There are 22 partitions of 8 23/8 - Sums of 8 biquadratics: The first two numbers that are sums of 8 biquadratics are 8,23... 24/8 - Tesseract: A tesseract has 24 square faces and 8 cubic cells 25/8 - 8th lucky number: The 8th lucky number is 25 26/8 - Even heptagonal pyramidal numbers: The first two even heptagonal pyramidal numbers are 8,26... 27/8 - Prime third powers: The first two prime third powers are 8,27... 28/8 - 8th positive fundamental discriminant: The 8th positive fundamental discriminant is 28 29/8 - 8th Hilbert number: The 8th Hilbert number is 29 30/8 - Rule 30: Rule 30 is one of the elementary linear cellular automaton rules, where each of the 8 possible combinations of colors of a cell and its neighbours are encoded in the binary representation 00011110=30 31/8 - GomoryÕs theorem: Regardless of where one white and one black square are deleted from an 8x8 chessboard, the reduced board can always be covered exactly with 31 dominoes September 1/9 - Centered cube numbers: The first two centered cube numbers are 1,9... 2/9 - Sums of two cubes: The first two sums of two cubes are 2,9... 3/9 - Cullen numbers: The first two Cullen numbers are 3,9... 4/9 - Prime second powers: The first two prime second powers are 4,9... 5/9 - Triangular dipyramid: A triangular dipyramid has 5 vertices and 9 edges 6/9 - Sum of prime factors of 9: The sum of prime factors of 9 is 6 7/9 - Percentage of numbers starting with 5: 7.9% of numbers in listings will start with 5 8/9 - Isolated singularities on a cubic surface: There are 9 possible types of isolated singularities on a cubic surface, 8 of them rational double points 9/9 - Anisohedral nonominoes: There are 9 anisohedral nonominoes, or figures formed by 9 squares joined at the sides 10/9 - 9 tree plantation problem: The maximum number of rows of three threes you can have with 9 trees is 10 rows 11/9 - Dissections of the eneagon into a Greek cross: There are 11 ways of dissecting a regular eneagon (9-sided polygon) into a Greek cross 12/9 - Partitions of 9! of length 9: There are 12 partitions of 9! of length 9 13/9 - Chromatic number of a surface of genus 9: The chromatic number of a surface of genus 9 is 13 14/9 - Coin tossing paradox: The expected wait until one sees the combination THTH is 20, and to see HTHH is 18, but the probability that THTH occurs before HTHH is 9/14 15/9 - Triaugmented dodecahedron: The triaugmented dodecahedron has 15 triangles and 9 pentagons 16/9 - Kings problem: 16 kings can be placed in a chessboard without any two attacking each other, and 9 kings is the minimum number needed to occupy or attack every square 17/9 - 9th number with an odd number of prime factors: 17 is the 9th number with an odd number of prime factors 18/9 - Mean number of edges for a graph with 9 vertices: The mean number of edges for a graph with 9 vertices is 18 19/9 - Collatz sequence for 9: The Collatz sequence for 9 has 19 steps 20/9 - Harmonic mean of the divisors of 10: The harmonic mean of the divisors of 10 is 20/9 21/9 - Generalized Moore graphs with 9 nodes: There are 21 generalized Moore graphs with 9 nodes 22/9 - Kempner series for 9: The value of the Kempner series for 9 is 22.9... 23/9 - Sum of 9 positive cubes: 23 is the smallest number that is sum of 9 positive cubes 24/9 - Sum of 9 biquadratics: The first two numbers that are sum of 9 biquadratics are 9,24... 25/9 - Nonattacking kings in a 9x9 chessboard: 25 nonattacking kings can be placed in a 9x9 chessboard 26/9 - KatonaÕs problem on the letters of the alphabet: The solution of KatonaÕs problem for the 26 letters of the alphabet is that they can be separated by a family of 9 27/9 - van der Waerden numbers: The smallest nontrivial van der Waerden numbers are 9 and 27 28/9 - Smallest pseudoprime in base 9: The smallest pseudoprime in base 9 is 28 29/9 - VargaÕs constant: The value of VargaÕs constant is 9.29... 30/9 - Partitions of 9: There are 30 partitions of 9 October 1/10 - CatalanÕs constant: The value of CatalanÕs constant is 1.10... 2/10 - Unordered factorizations of 10: There are 2 unordered factorizations of 10 3/10 - Sums of 3 cubes: The first two sums of 3 cubes are 3,10... 4/10 - Even tetrahedral numbers: The first two even tetrahedral numbers are 4,10... 5/10 - Hypotenuses for a right integer triangle: The smallest two possible hypothenuses for a right integer triangle are 5 and 10 6/10 - Regular polychora: There are 6 convex polychora and 10 stellated polychora 7/10 - Sum of prime factors of 10: The sum of prime factors of 10 is 7 8/10 - Square bicupolas: The square bicupolas have 8 triangles and 10 squares 9/10 - Star polychora: 9 of the 10 star polychora have the same vertices as the hexacosichoron 10/10 - Biaugmented dodecahedrons: The biaugmented dodecahedrons have 10 triangles and 10 pentagons 11/10 - Totient function of 11: The totient function of 11 is 10 12/10 - 10 tree plantation problem: The maximum number of rows of three trees you can have with 10 trees is 12 rows 13/10 - Smallest sum of 10th powers equal to another 10th power: 13 is the smallest known number of 10th powers whose sum equals another 10th power 14/10 - Possible solitary numbers: The two smallest numbers that are suspected to be solitary, but not known for certain, are 10 and 14 15/10 - 10th non-Egyptian number: The 10th non-Egyptian number is 15 16/10 - Hexadecimal: 16 in hexadecimal is 10 17/10 - Cumulative digit sum of binary numbers up to 10: The cumulative digit sum of binary numbers up to 10 gives 17 18/10 - 10th composite number: The 10th composite number is 18 19/10 - Reverse-then-add algorithm: The smallest number to require one iteration to reach a palindrome in the reverse-then-add algorithm is 10, and the smallest to require two iterations is 19 20/10 - Decreasing decimal numbers: The first two decreasing decimal numbers are 10,20... 21/10 - Cubic graphs on 10 nodes: There are 21 cubic graphs on 10 nodes 22/10 - Square dissection into acute isosceles triangles: There are two known dissections of the square into acute isosceles triangles, one requiring 10 triangles and the other 22 23/10 - HilbertÕs problems: HilbertÕs problems are a set of originally unsolved problems in mathematics proposed by Hilbert. 10 of them were presented at the Second International Congress in Paris on 1900. 24/10 - Gyroelongated square bicupola day: The gyroelongated square bicupola has 24 triangles and 10 squares 25/10 - Sums of 10 biquadratics: The first sums of 10 biquadratics are 10,25... 26/10 - Noncototients: The first two noncototients are 10,26... 27/10 - Eckardt points day: The Eckardt points are the 10 points where three of the 27 SolomonÕs seal lines meet 28/10 - Non-hamiltonian symmetric cubic graphs: There two smallest non-hamiltonian symmetric cubic graphs are the Petersen graph, with 10 nodes, and the Coxeter graph, with 28 nodes 29/10 - 10th prime: The 10th prime is 29 30/10 - Percentage of numbers starting with 1: 30.10% of all numbers in listings will start with 1 31/10 - Partitions of 10! of length 10: There are 31 partitions of 10! of length 10 November 1/11 - Repunits: The first two repunits are 1,11... 2/11 - Dihedral primes: The first two dihedral primes are 2,11... 3/11 - Unique primes day: The first two unique primes are 3,11... 4/11 - Sums of 4 cubes: The first two sums of 4 cubes are 4,11... 5/11 - Sexy primes: The first pair of sexy primes is 5 and 11 6/11 - Russian roulette problem: If a bullet is put in one of the six chambers of a revolver, and two duelist alternately spin the chamber and fire at themselves until one is killed, what is the probability that the first duelist is killed? The answer is 6/11 7/11 - Josephus problem: 11 men are arranged in a circle. Every second man will be executed going around the circle, until only one remains. Where should you stand to be the survirvor? The answer is in the 7th place 8/11 - Perfect shufßes: After 11 perfect shufßes, the first card will be in position 8 9/11 - Remainder when the 11th composite number is divided by 11: When the 11th composite number (21) is divided by 11, the remainder is 9 10/11 - Most common rolls of 3 dice: The most common rolls of 3 dice are 10 and 11 11/11 - Sum of prime factors for 11: The sum of prime factors for 11 is 11 12/11 - Partitions of 11 into distinct parts: There are 12 partitions of 11 into distinct parts 13/11 - Amphichiral Archimedean solids: 11 of the 13 Archimedean solids are amphichiral 14/11 - Collatz sequence for 11 day: The Collatz sequence for 11 has 14 steps 15/11 - Smallest pseudoprime in base 11: The smallest pseudoprime in base 11 is 15 16/11 - 11 tree plantation problem: The maximum number of rows of three trees you can have with 11 trees is 16 17/11 - 11th prime power: The 11th prime power is 17 18/11 - Herschel graph day: The Herschel graph has 11 nodes and 18 edges 19/11 - Euler-Mascheroni constant approximation: A good fractional approximation of the Euler-Mascheroni constant is 11/19 20/11 - Uniform tessellations: There are 11 1-uniform tesselations and 20 2-uniform tesselations 21/11 - 11th odd number: The 11th odd number is 21 22/11 - Perfect rectangles of order 11: There are 22 perfect rectangles of order 11 23/11 - Smallest prime dividing the 11th Mersenne number: The smallest prime that divides the 11th Mersenne number is 23 24/11 - Strictly Egyptian numbers: The first two strictly Egyptian numbera are 11,24... 25/11 - Hendecaiamonds with holes: There are 25 hendecaiamonds with holes, or figures formed by 11 equilateral triangles joined at the sides 26/11 - Odd perfect numbers: If an odd perfect number exists, and it isnÕt divisible by 3, it must have at least 11 distinct prime factors. If it isnÕt divisible by 3, 5 or 7, it has at least 26 distinct prime factors 27/11 - 11th plaindrome: The 11th plaindrome is 27 28/11 - 11th squareful number: The 11th squareful number is 28 29/11 - Least primes such that ßoor((n^n)/p) is prime: The two first terms greater than two in this sequence are 11,29... 30/11 - Unattacked squares that 11 queens can leave in a 11x11 chessboard: The maximum number of unattacked squares that 11 queens can leave in an 11x11 chessboard is 30 December 1/12 - Triangle triangle picking problem: The mean area of a triangle with vertices picked inside a triangle with unit area is 1/12 2/12 - Least common rolls of 2 dice: The least common rolls of two dice are 2 and 12 3/12 - Kissing spheres: 12 spheres can touch another one of the same size in 3 dimensions 4/12 - Rational values of sin(2pi/n): The value of sin(2pi/n) is irrational for all n except 4 and 12 5/12 - Sums of 5 cubes day: The first two sums of 5 cubes are 5,12... 6/12 - Octahedral graph: The octahedral graph has 6 nodes and 12 edges 7/12 - Dissections of a dodecagon into a golden rectangle: There are 7 dissections of a dodecagon (12-sided polygon) into a golden rectangle 8/12 - Cubical graph: The cubical graph has 8 nodes and 12 edges 9/12 - Collatz sequence for 12: The Collatz sequence for 12 has 9 steps 10/12 - Duodecimal: 12 is 10 in duodecimal 11/12 - Three book stacking problem: A stack of three books can protrude 11/12 book lengths over the edge of a table without falling over 12/12 - Dodecadodecahedron day: A dodecadodecahedron is formed of 12 pentagrams and 12 pentagons 13/12 - Sparse polynomial square: The minimum degree for a polynomial having a sparse square is 13, with a square of degree 12 14/12 - Stomachion: The stomachion is a 14-piece dissection puzzle made from a 12x12 square 15/12 - Partitions of 12 into distinct parts: There are 15 partitions of 12 into distinct parts 16/12 - 12th Stormer number: The 12th Stormer number is 16 17/12 - Home plate error: The shape of the home plate in baseball is an irregular pentagon. The specification of the shape requires the existence of an isosceles right triangle of legs 12 and hypothenuse 17, which isnÕt correct. 18/12 - Abundant numbers: The first two abundant numbers are 12,18... 19/12 - 12 tree plantation problem: The maximum number of rows of three trees you can have with 12 trees is 19 20/12 - Icosidodecahedron: The icosidodecahedron has 20 triangles and 12 pentagons 21/12 - 12th non-Egyptian number: The 12th non-Egyptian number is 21 22/12 - Uniquely three-colorable graph: The smallest uniquely three-colorable graph has 12 vertices and 22 edges 23/12 - 12th odd number: The 12th odd number is 23 24/12 - Cuboctahedral graph: The cuboctahedral graph has 12 nodes and 24 edges 25/12 - 4th harmonic number: The 4th harmonic number is 25/12 26/12 - Totient function of 26: The totient function of 26 is 12 27/12 - Sums of 12 biquadratics: The first two sums of 12 biquadratics are 12,27... 28/12 - 12th negabinary number: The 12th negabinary number is 28 29/12 - 12th metadrome: The 12th metadrome is 29 30/12 - Icosahedral graph: The icosahedral graph has 12 vertices and 30 edges 31/12 - 12th product of an odd number of distinct prime factors: The 12th product of an odd number of distinct prime factors is 31 === Subject: Re: Mathematical calendar (long) A very nice idea! Richard Schorn === Subject: Re: Mathematical calendar (long) <31eoqhF3bffsqU1@individual.net> posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L hmmm modern day numerology introduced by Doly and received with SIN X and SCORN. Despite my personal disposition on the idea, it seems the heavens may consider this false idol Herc === Subject: Re: Mathematical calendar (long) > Having too much time in my hands, I decided to compile a mathematical > calendar, where each day is celebrated with a mathematical concept > somehow related to the numbers of the date. > This is the result of my waste of time. Note that the dates are written > in the European format (day/month). ThatÕs really great! If there were additional pictures and maybe a little more info, I would definitelly buy one for myself (and 6 more for gifts for my teachers :-) sirix. P.S. There could be also some info about current events, but of course only as an addition to the main day motto. === Subject: Re: PROOF that 0.99999... = 1 >Hi. >Theorem: 0.9999... = 1 >Proof: >A number x = 0.aaaaaa..., where 0 <= a <= 9 is >x = sum_{n=1...inf} a/(10^n) >If a = 9, then >x = sum_{n=1...inf} 9/(10^n) >Then, we have the sequence of partial sums >S_n = sum_{i=1...n} 9/(10^n). >We can then derive the _general formula_ as follows: >S_1 = 9/10 >S_2 = 9/10 + 9/100 = 99/100 >S_3 = 9/10 + 9/100 + 9/1000 = 999/1000 >... >S_n = ((10^n)-1)/(10^n) >... >Then we take the limit as n -> inf: >lim_{n->inf} ((10^n)-1)/(10^n). >Direct susbtitution produces the indeterminate form inf/inf, so we rewrite >the limit: >lim_{n->inf} ((10^n)-1)/(10^n) = lim_{n->inf} 1 - (1/(10^n)) > = lim_{n->inf} 1 - lim_{n->inf} (1/(10^n)) > = 1 - 0 > = 1 >Thus, 0.9999... = 1. QED. Hey thank you very much, this proves .999... diverges. According to the Basic Convergence Test according to Advanced Engineering Mathematics, you just proved that, Since, n-->oo Lim Z_n =/= 0 So it diverges, thank you. SmartÕs 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: PROOF that 0.99999... = 1 posting-account=XHGPzAwAAACBzDwz8l_9KYOmofDRUylj IÕm afraid not. lim Z_n =/= 0 implies that sum_i (Z_i) diverges. However the S_n in the original post were already the partial sums. === Subject: Re: PROOF that 0.99999... = 1 > IÕm afraid not. > lim Z_n =/= 0 implies that > sum_i (Z_i) diverges. > However the S_n in the original post were already the partial sums. Yes, so 0.9999... = 1. === Subject: Re: Most Distorted Triangle === Subject: Re: Most Distorted Triangle > Is there a most scalene/ asymmetric/distorted triangle, most remote > from the equilateral triangle? If so, what may be a proper criterion > or definition of asymmetry? > Petitjean gives a result,but I could not follow > what the criterion is from links to his work he provided. The general criterion applies for figures more complicated that triangles (distributions). It is why it cannot be summarized quickly (done on my web site). In the case of a set of n points, we look for the sum of the sqaured distances between the n points and its mirrored image, and we minimize for all rotations and translations. If we find this sum is zero, it means that the set is achiral. For a triangle in the plane, it means that it is isocele. The sum is normalized to the inertia (sum of squred distances to the center of gravity). There is a family of triangles (all have the same sides ratios, i.e. the same shape), for which the normalized sum is maximized: this is the most chiral triangle. Strictly speaking, this triangle (or any scaled or/and rotated or/and translated copy) is the farthest from the isocele triangle. There is an other one with a similar criterion for direct symmetry rather than indirect symmetry, which, in your case could be viewed as the farthest from the equilateral triangle. The shape is simpler, and buildable with ruler and compass: angles are pi/4, pi/8, 5pi/8. Alas, for reasons a bit long to write here, the theory does not work for continuous sets (nevertheless work for n points in the euclidean d-space). Details on: http://petitjeanmichel.free.fr/itoweb.petitjean.symmetry.html Final remark: the most distorted triangle could mean: - farthest from isocele: compare it to its mirror image - farthest from equilateral: compare it to itself, rotating - fartest from half rectangle (any criterion you like) My question is: which relative weights would you set for these three criterions, in order to define an overall criterion ? Michel Petitjean, Email: petitjean@itodys.jussieu.fr ITODYS (CNRS, UMR 7086) ptitjean@ccr.jussieu.fr http://petitjeanmichel.free.fr/itoweb.petitjean.html === Subject: Re: .99999... still=/= 1 >>.999... =/= 1 >If so, then there must be a real number X such that 0.999... < X < 1. >What >is the decimal expansion of X? >> X = ..............999... = asymptote toward 1 never reaching it. >In what digit position does that differ from 0.999...? >>It shows somewhere away from the starting point. >Could you be more specific? .999... < X < 1 X = .999... SmartÕs 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: .99999... still=/= 1 > .999... < X < 1 > X = .999... === Subject: Re: .99999... still=/= 1 >> .999... < X < 1 >> X = .999... .999... may converge to 1, but it never equals 1 and thatÕs what the whole debate has been about. In fact, I did state Partial Sums did show convergence to 1, but that was with only so many significant digits and showed and approximate convergence value. The total value itself never equals 1. If you take the total SUM lim of z_m it doesntÕ converge to 1 or equal 1. If you take the mth term lim then it converges to 1. for .999... m-->oo lim SUM z_m = divergence m-->oo lim z_m = convergence The point is, .999... NEVER reaches something perfectly equal to 1. .999... still=/= 1 Why donÕt you admit you made a mistake? SmartÕs 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: .99999... still=/= 1 After a little more thought about this convergence test that I used, I donÕt think I really made a mistake. Because the way you people have been stating that .999... = 1, means the the entire SUM at the lim of infinity must converge to 1 and it doesnÕt. I really used this form of the test based on what you people described. z_m = SUM z_m = .999... For the TOTAL value of .999..., because you people said, .999... EQUALS 1. I used, m-->oo lim ( 1 - 1/10^m) = 1 divergence But if you say a basic convergence test with the mth term, then m-->oo lim 9/10^m = 0 equals convergence. So in both cases, the original debate was does, .999... equal to 1 and it DOESNÕT. It converges to 1, which means an approximate nearest value to, is reached. So I donÕt really think I made a mistake after all. You people inßuenced my way of thinking, by taking the convergence test to the area that the TOTAL value .999... = 1 and it doesnÕt. To SUM .999..., the TOTAL value would be used and it would have to converge to 1, for it to equal 1. So I was right after all. SmartÕs 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: .99999... still=/= 1 > So in both cases, the original debate was does, > .999... equal to 1 and it DOESNÕT. It converges to 1, which means an > approximate nearest value to, is reached. How that approximate nearest value is reached? > To SUM .999..., the TOTAL value would be used and it would have to converge to > 1, for it to equal 1. What means the TOTAL value? How many values have one number? === Subject: Re: .99999... still=/= 1 posting-account=AE-QyQ0AAAC84T96q9_yI_Fj9ThoZQPi > So in both cases, the original debate was does, > .999... equal to 1 and it DOESNÕT. It converges to 1, which means an > approximate nearest value to, is reached. An approximate nearest value to what? > To SUM .999..., the TOTAL value would be used and it would have to converge to > 1, for it to equal 1. What means the TOTAL value? === Subject: Re: .99999... still=/= 1 >> So in both cases, the original debate was does, >> .999... equal to 1 and it DOESNÕT. It converges to 1, which means an >> approximate nearest value to, is reached. >An approximate nearest value to what? nearest to 1 >> To SUM .999..., the TOTAL value would be used and it would have to >converge to >> 1, for it to equal 1. >What means the TOTAL value? m-->oo .9 + .09 + .009 + .... = lim SUM 9/10^m If you add up the total .999... series as it approaches infinity does it equal 1? If this particular limit is tested for convergence, which is, m-->oo lim ( 1 - 1/10^m ) = ? does it converge to an equality of 1? According to the basic convergence test, it converges to an equality state if it equals zero otherwise it diverges. If this test is used for the mth term, then we can see that it does converge. But this doesnÕt mean equal to 1. It means it approaches 1 near the limits of infinity. .999... can never equal 1, but it can converge to 1. Does 1/3 really equal .333... ? no .333... converges to 1/3. So the closest approximate value of 1/3 is .333... . SmartÕs 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: .99999... still=/= 1 Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >.999... equal to 1 and it DOESNÕT. It converges to 1, which means an >approximate nearest value to, is reached. As youÕve been told over and over again, .999... *means* the value that .9, .99, .999, ... converges to. And that value is 1. -- Richard === Subject: Re: .99999... still=/= 1 >>.999... equal to 1 and it DOESNÕT. It converges to 1, which means an >>approximate nearest value to, is reached. > As youÕve been told over and over again, .999... *means* the value > that .9, .99, .999, ... converges to. And that value is 1. People who have a head full of do not hear well at all and they understand even less. Bob Kolker === Subject: Re: .99999... still=/= 1 > After a little more thought about this convergence test that I used, I donÕt > think I really made a mistake. Because the way you people have been stating > that .999... = 1, means the the entire SUM at the lim of infinity must converge > to 1 and it doesnÕt. One more time. 1 is the -limit- of the sequence of partial sums. The seqeunce of partial sums -converges- to 1. Got it? The n-th partial sum is (10^n -1)/10^n which equals 1 - 1/10^n. The limit of the sequence a_n = (1 - 1/10^n) is 1. That is why we say the infinite series equals 1. How many times must this be explained to you, you ing dunce? Bob Kolker === Subject: Re: .99999... still=/= 1 >>.999... < X < 1 >>X = .999... Archimedean Axiom. Bob Kolker === Subject: Re: .99999... still=/= 1 >>.999... < X < 1 >>X = .999... > Archimedean Axiom. Please be more specific. Note that S. Enterprize is claiming that: 0.999... < 0.999... === Subject: Re: .99999... still=/= 1 > Please be more specific. Note that S. Enterprize is claiming that: > 0.999... < 0.999... Archimedean Axiom roughly states that there are no nonzero numbers which are infinitely small. To stick a non zero X in the interval .999... < X < 1 you would require a non-zero quantity 1 - X such that 1 - X < 1/10^n for all n. For the standard real number system this cannot be done. If 0 <= x < a_n and a_n converges to 0 then x must be 0. Suppopse not. then x is small but still > 0. But a_n converges to 0 means there must exist N such that if n > N then 0 <= a_n < x which is a contradiction. Another way of seeing this is take any very samll postive number eps and any very large positive number Big. There exists an integer n such that n*eps > Big which is the same as eps > Big/n, so that eps cannot be an infinitely small number greater than 0. Ordered fields which have this property are call Archimedean fields. Bob Kolker === Subject: Re: .99999... still=/= 1 In sci.math, S. Enterprize Company >If so, then there must be a real number X such that 0.999... < X < 1. >>What >>is the decimal expansion of X? > X = ..............999... = asymptote toward 1 never reaching it. >>In what digit position does that differ from 0.999...? >It shows somewhere away from the starting point. >>Could you be more specific? > .999... < X < 1 > X = .999... So .999... < .999... < 1. Now whatÕs Y = .999... - .999... ? [.sigsnip] -- #191, ewill3@earthlink.net ItÕs still legal to go .sigless. === Subject: Re: .99999... still=/= 1 >In sci.math, S. Enterprize Company >.999... =/= 1 >If so, then there must be a real number X such that 0.999... < X < 1. >What >is the decimal expansion of X? >> X = ..............999... = asymptote toward 1 never reaching it. >In what digit position does that differ from 0.999...? >>It shows somewhere away from the starting point. >Could you be more specific? >> .999... < X < 1 >> X = .999... >So .999... < .999... < 1. >Now whatÕs Y = .999... - .999... ? >[.sigsnip] The point I have been trying to make was, .999... =/= 1 The convergence value never equals that value unless you want to use an approximate value. >-- >#191, ewill3@earthlink.net >ItÕs still legal to go .sigless. SmartÕs 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: .99999... still=/= 1 >> It diverges, I didnÕt need the calculator. >> That series is of the form, >> 1 + 1/2 + 1/4 + 1/8+ ... >> This diverges. >You lame brain . The series adds up to 2. >You are truly one of the Incompetents of Great Note. You and JSH should >sing duet a capella. >Bob Kolker A Harmonic Series is an exception to the rule. The above shown series appears to be a harmonic series similar to, 1 + 1/2 + 1/3 + 1/4 + .... I stated one view of it. According to the Advanced Engineering Math book I have, it satisfies the condition of convergence but it diverges according to, Thomas, G.B. and R.L. Finney, Calculus and Analytic Geometry, 6th edition So I guess you could say both. or, 1 + ( series) m-->oo lim (series) --> 0 So 1 + 1 = 2 Which shows it converges, but with respect to the total series and the basic convergence test, m--> lim Z_m = 0 for convergence m-->oo lim ( 1 + (series) ) = 2 This shows that it is divergent. Because if Z_m =/= 0 then it is divergent. SmartÕs 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: .99999... still=/= 1 > A Harmonic Series is an exception to the rule. The above shown > series > appears to be a harmonic series similar to, > 1 + 1/2 + 1/3 + 1/4 + .... Yes, this is harmonic and diverges. However, the series 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... is not harmonic and converges. It is geometric with r = 1/2. Look in your books for a geometric series and find out when it converges and when it diverges. Just because something (1 + 1/2 + 1/4 + 1/8 + ...) might somehow resemble something else (1 + 1/2 + 1/3 + 1/4 + ...), does not at all mean they are similar in any way. Surely an apple and a wax apple look similar. However, I would rather bite into the former than the latter. - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: .99999... still=/= 1 >> A Harmonic Series is an exception to the rule. The above shown >> series >> appears to be a harmonic series similar to, >> 1 + 1/2 + 1/3 + 1/4 + .... >Yes, this is harmonic and diverges. However, the series > 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... >is not harmonic and converges. It is geometric with r = 1/2. Look in your >books for a geometric series and find out when it converges and when it >diverges. Root Test ____ lim n / Z_n = L L < 1 = converge L > 1 = diverges L = 1, no conclusion possible Z_n = 1/2^n n-->oo lim 1/2^n = At n = 0, L = 1 n > 0 converges YouÕr right. I thought it was harmonic and it isnÕt, but it doesnÕt appear to diverge at all. It just seems to have L = 1 at n=0 and the rest converges. >Just because something (1 + 1/2 + 1/4 + 1/8 + ...) might somehow resemble >something else (1 + 1/2 + 1/3 + 1/4 + ...), does not at all mean they are >similar in any way. Surely an apple and a wax apple look similar. >However, I would rather bite into the former than the latter. > - Tim >-- >Timothy M. Brauch >NSF Fellow >Department of Mathematics >University of Louisville SmartÕs 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: .99999... still=/= 1 >> A Harmonic Series is an exception to the rule. The above shown >> series >> appears to be a harmonic series similar to, >> 1 + 1/2 + 1/3 + 1/4 + .... >Yes, this is harmonic and diverges. However, the series > 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... >is not harmonic and converges. It is geometric with r = 1/2. Look in your >books for a geometric series and find out when it converges and when it >diverges. > Root Test > ____ > lim n / Z_n = L > L < 1 = converge > L > 1 = diverges > L = 1, no conclusion possible > Z_n = 1/2^n > n-->oo > lim 1/2^n = > At n = 0, L = 1 > n > 0 converges > YouÕr right. I thought it was harmonic and it isnÕt, but it doesnÕt appear to > diverge at all. It just seems to have L = 1 at n=0 and the rest converges. Math triumphs again! Glad thatÕs over. === Subject: Re: .99999... still=/= 1 > > A Harmonic Series is an exception to the rule. The above shown > series > appears to be a harmonic series similar to, 1 + 1/2 + 1/3 + 1/4 + .... >>Yes, this is harmonic and diverges. However, the series >> 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... >>is not harmonic and converges. It is geometric with r = 1/2. Look in >your >>books for a geometric series and find out when it converges and when it >>diverges. >> Root Test >> ____ >> lim n / Z_n = L >> L < 1 = converge >> L > 1 = diverges >> L = 1, no conclusion possible >> Z_n = 1/2^n >> n-->oo >> lim 1/2^n = >> At n = 0, L = 1 >> n > 0 converges >> YouÕr right. I thought it was harmonic and it isnÕt, but it doesnÕt >appear to >> diverge at all. It just seems to have L = 1 at n=0 and the rest >converges. >Math triumphs again! >Glad thatÕs over. Yes math does triumph. .999... =/= 1 z_1 + z_2 + .... = Z_m m-->oo lim ( 1 -1/10^m) = 1 according to math, if this equals zero then it converges otherwise it diverges. Math works good. SmartÕs 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: .99999... still=/= 1 A Harmonic Series is an exception to the rule. The above shown > series > appears to be a harmonic series similar to, 1 + 1/2 + 1/3 + 1/4 + .... >>Yes, this is harmonic and diverges. However, the series >> 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... >>is not harmonic and converges. It is geometric with r = 1/2. Look in your >>books for a geometric series and find out when it converges and when it >>diverges. >Root Test Correction: Use The Ratio Test The answer is still the same. Not this vvvvvvv > ____ >lim n / Z_n = L >L < 1 = converge >L > 1 = diverges >L = 1, no conclusion possible >Z_n = 1/2^n >n-->oo >lim 1/2^n = >At n = 0, L = 1 > n > 0 converges > YouÕr right. I thought it was harmonic and it isnÕt, but it doesnÕt appear >diverge at all. It just seems to have L = 1 at n=0 and the rest converges. >>Just because something (1 + 1/2 + 1/4 + 1/8 + ...) might somehow resemble >>something else (1 + 1/2 + 1/3 + 1/4 + ...), does not at all mean they are >>similar in any way. Surely an apple and a wax apple look similar. >>However, I would rather bite into the former than the latter. >> - Tim >>-- >>Timothy M. Brauch >>NSF Fellow >>Department of Mathematics >>University of Louisville SmartÕs 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: .99999... still=/= 1 >In sci.math, Richard Henry >: >mister Company is going to love this. > I just found an application of 0.9999..., >over and above OVERFLOW CONDITION or TENS COMPLIMENT. > ThatÕs because, > m--> oo > Lim (1 - 1/10^m) = 1 > divergent >> Convergent. >Both. Be very very careful with your notation, guys! :-) >lim (1 - 10^m) = infinity >m-> +oo >lim (1 - 10^m) = 1 >m-> -oo >-- >#191, ewill3@earthlink.net >ItÕs still legal to go .sigless. Wrong function. But that particular function you show may be oo and divergent, but you also have to use the common logic of what a divergent series means too. First use the correct function, m-->oo lim ( 1 - 1/10^m ) = ? If 0 then convergent otherwise divergent. On my calculator, 1/(a large number) = something approaching zero, so 1 - 0 = 1. .999... is divergent and not equal to 1. SmartÕs 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: .99999... still=/= 1 In sci.math, S. Enterprize Company In sci.math, Richard Henry >>: >>mister Company is going to love this. >> I just found an application of 0.9999..., >>over and above OVERFLOW CONDITION or TENS COMPLIMENT. >> ThatÕs because, >> m--> oo >> Lim (1 - 1/10^m) = 1 >> divergent > Convergent. >>Both. Be very very careful with your notation, guys! :-) >>lim (1 - 10^m) = infinity >>m-> +oo >>lim (1 - 10^m) = 1 >>m-> -oo >>-- >>#191, ewill3@earthlink.net >>ItÕs still legal to go .sigless. > Wrong function. But that particular function you show may be oo and > divergent, but you also have to use the common logic of what a divergent series > means too. First use the correct function, > m-->oo > lim ( 1 - 1/10^m ) = ? > If 0 then convergent otherwise divergent. Divergent because of a specification error. oo is both positive and negative. lim ( 1 - 1/10^m ) = -oo m->+oo lim ( 1 - 1/10^m ) = 1 m->-oo > On my calculator, 1/(a large number) = something approaching zero, > so 1 - 0 = 1. > .999... is divergent and not equal to 1. OK, so whatÕs an X such that .999... < X < 1 or 1 < X < .999... ? [.sigsnip] -- #191, ewill3@earthlink.net ItÕs still legal to go .sigless. === Subject: Re: .99999... still=/= 1 >>mister Company is going to love this. >> I just found an application of 0.9999..., >>over and above OVERFLOW CONDITION or TENS COMPLIMENT. >> ThatÕs because, >> m--> oo >> Lim (1 - 1/10^m) = 1 >> divergent >Convergent. Ok donÕt my word for it, but at least believe the actual Advanced Engineering Math Book I have. Quoting from: 6th Edition Advanced Engineering Mathematics by Erwin Kreyszig Page 805 14.2 Convergence Tests for Series Theorem 1 ( Divergence ) If a series z_1 + z_ 2 + .... converges, then lim Z_m = 0 m-->oo Hence if the series does not satisfy this condition, it diverges. SmartÕs 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: .99999... still=/= 1 posting-account=HPGXOQ0AAAB8Vqh61Q9SABrM_YhXJe__ >>mister Company is going to love this. >> I just found an application of 0.9999..., >>over and above OVERFLOW CONDITION or TENS COMPLIMENT. >> ThatÕs because, >> m--> oo >> Lim (1 - 1/10^m) = 1 >> divergent >Convergent. > Ok donÕt my word for it, but at least believe the actual Advanced Engineering > Math Book I have. > Quoting from: > 6th Edition > Advanced Engineering Mathematics > by Erwin Kreyszig > Page 805 > 14.2 Convergence Tests for Series > Theorem 1 ( Divergence ) > If a series z_1 + z_ 2 + .... converges, then > lim Z_m = 0 > m-->oo > Hence if the series does not satisfy this condition, it diverges. So, applied to our case: 0.999... = lim_{m->oo} SUM_{i=1->m} 9*(1/10)^m, that is z_m = 9*(1/10)^m and clearly lim_{m->oo} z_m = 0 , so you cannot conclude that the series diverges with the divergence test in Theorem 1. Actually it converges, since it is a geometric series with coefficient 9*(1/10)^m < 1 Your confusing seqences with series; the partial sums form the sequence S_m = { SUM_{i=1->m} 9*(1/10)^m } = { 1 - (1/10)^m } lim_{m->oo} S_m = lim_{m->oo} { 1 - (1/10)^m } = 1, and so you have actually concluded yourself that 0.999... = lim_{m->oo} { 1 - (1/10)^m } = 1 Just admit it. > SmartÕs 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: .99999... still=/= 1 >mister Company is going to love this. > I just found an application of 0.9999..., >over and above OVERFLOW CONDITION or TENS COMPLIMENT. ThatÕs because, m--> oo > Lim (1 - 1/10^m) = 1 divergent >>Convergent. >> Ok donÕt my word for it, but at least believe the actual Advanced >Engineering >> Math Book I have. >> Quoting from: >> 6th Edition >> Advanced Engineering Mathematics >> by Erwin Kreyszig >> Page 805 >> 14.2 Convergence Tests for Series >> Theorem 1 ( Divergence ) >> If a series z_1 + z_ 2 + .... converges, then >> lim Z_m = 0 >> m-->oo >> Hence if the series does not satisfy this condition, it diverges. >So, applied to our case: >0.999... = lim_{m->oo} SUM_{i=1->m} 9*(1/10)^m, >that is z_m = 9*(1/10)^m and clearly >lim_{m->oo} z_m = 0 , so you cannot conclude that the series diverges >with the divergence test in Theorem 1. >Actually it converges, since it is a geometric series with >coefficient 9*(1/10)^m < 1 >Your confusing seqences with series; the partial sums form the sequence >S_m = { SUM_{i=1->m} 9*(1/10)^m } = { 1 - (1/10)^m } >lim_{m->oo} S_m = lim_{m->oo} { 1 - (1/10)^m } = 1, >and so you have actually concluded yourself that >0.999... = lim_{m->oo} { 1 - (1/10)^m } = 1 >Just admit it. YouÕre right. I did make a mistake. I used the lim SUM z_m instead of the mth term z_m. But the whole debate was whether or not .999... = 1 and it doesnÕt. It might converge to 1 but it doesnÕt equal 1. So why donÕt you admit your mistake too? SmartÕs 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: .99999... still=/= 1 posting-account=HPGXOQ0AAAB8Vqh61Q9SABrM_YhXJe__ > YouÕre right. I did make a mistake. I used the lim SUM z_m instead of the > mth term z_m. But the whole debate was whether or not .999... = 1 and it > doesnÕt. It might converge to 1 but it doesnÕt equal 1. So why donÕt you admit > your mistake too? If there were any mistakes I would admit them. Your conception of convergence and equalities are different from mine. I donÕt know where youÕve learned this, or if you have any references to mathematical texts promoting this view. Adapted for the reals from Rudin:Principles of mathematical analysis, (this is what IÕve learned) Definition: A sequence {p_n} in the set of real numbers R is said to CONVERGE if there is a point p in R with the following property: For every e>0 there is an integer N such that n >= N => | p_n - p | < e In this case we SAY: {p_n} converges to p, or that p is the limit of {p_n} , and we WRITE: p_n -> p or lim_{n->oo} p_n = p That is if {p_n} converges to p then lim_{n->oo} p_n = p | , and the equality is BY DEFINITION. (1) Do you agree with this definition of convergence? If not would you please present your own definition of convergence. Our sequence is {p_n} = { 0.9, 0.99, 0.999, ... } = { 1 - (1/10)^n } for n=1,2,3, ... So by the symbol 0.999... the following is understood, (this is how I understand the symbol): 0.999... = lim_{n->oo} { 1 - (1/10)^n } | Again equality BY DEFINITION (2) Do you agree with this definition of the symbol? If not would you please present your own definition of the symbol 0.999... ? Otherwise the rest is logic: (a) By definition we have 0.999... = lim_{n->oo} { 1 - (1/10)^n } (b) It is easy to show that 1 is the limit of {1 - (1/10)^n}: For every e > 0 ( assume e is also < 0.1 ), choose N = ceil( log10( 1/e ) ), then | 1 - (1/10)^n - 1 | < e for every n>=N. QED Since 1 is the limit of { 1 - (1/10)^n }, then by definition of notation it follows that lim_{n->oo} { 1 - (1/10)^n } = 1. Using (a) and (b) we get 0.999... = lim_{n->oo} { 1 - (1/10)^n } = 1 (3) Do you agree with the logic? If not, please point out where it fails. === Subject: Re: .99999... still=/= 1 > YouÕre right. I did make a mistake. I used the lim SUM z_m instead of the > mth term z_m. But the whole debate was whether or not .999... = 1 and it > doesnÕt. It might converge to 1 but it doesnÕt equal 1. So why donÕt you admit > your mistake too? That is what equals means in the context of a series or sequence. It means being the limit of the series or sequence. It does not mean equality with a particular element of the sequence or a partial sum. To say SUM (a_n) [n >= 1= = S MEANS Lim [n >= 1] SUM (a_k) [k >=1 and k <= n] = S Idiot! Bob Kolker === Subject: Re: .99999... still=/= 1 does the use of z in the textbook have to do with complex series? > lim_{m->oo} z_m = 0 , so you cannot conclude that the series diverges > with the divergence test in Theorem 1. > Actually it converges, since it is a geometric series with > coefficient 9*(1/10)^m < 1 --Advice 0.50; free, if wrong! http://tarpley.net/bush22.htm === Subject: Re: .99999... still=/= 1 >does the use of z in the textbook have >to do with complex series? >> lim_{m->oo} z_m = 0 , so you cannot conclude that the series diverges >> with the divergence test in Theorem 1. >> Actually it converges, since it is a geometric series with >> coefficient 9*(1/10)^m < 1 >--Advice 0.50; free, if wrong! >http://tarpley.net/bush22.htm Yes but, .999... still=/= 1 The convergence of .999... to 1 doesnÕt mean it reaches 1. And that was the original debate. SmartÕs 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: .99999... still=/= 1 > Yes but, > .999... still=/= 1 > The convergence of .999... to 1 doesnÕt mean it reaches 1. And that was the > original debate. No. That was the strawman you erected. Equality does not mean the sequence or series -reaches- the limit. It only requires -convergence- to the limit. Bob Kolker === Subject: Re: .99999... still=/= 1 >> The 9Õs never stop repeating. ItÕs like this. If you say, cat repeately >> infinitely, catcatcat..., it never converges to dog. >> It just stays as cat. >When you (or your computer/calculator) successfully said cat (or >9) infinitely many times can you tell me the last decimal digit of >pi (since pi have infinitely many digits)? >0, 1, 2, 3, 4, 5, 6, 7, 8 or 9? (Hint: try with your favourite digit) I think that pi may be a circular convergence series. In other words, there may be a circle of convergence test using the Radius Of Convergence R Test. So given a value of R (the radius), I think pi converges. And like I mentioned before, a series that does converge doesnÕt mean it really equals that convergence value. So that particular last decimal digit canÕt be determined. SmartÕs 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: .99999... still=/= 1 posting-account=AE-QyQ0AAAC84T96q9_yI_Fj9ThoZQPi >> The 9Õs never stop repeating. ItÕs like this. If you say, cat repeately >> infinitely, catcatcat..., it never converges to dog. >> It just stays as cat. >When you (or your computer/calculator) successfully said cat (or >9) infinitely many times can you tell me the last decimal digit of >pi (since pi have infinitely many digits)? >0, 1, 2, 3, 4, 5, 6, 7, 8 or 9? (Hint: try with your favourite digit) > I think that pi may be a circular convergence series. In other words, there > may be a circle of convergence test using the Radius Of Convergence R Test. So > given a value of R (the radius), I think pi converges. > And like I mentioned before, a series that does converge doesnÕt mean it > really equals that convergence value. So that particular last decimal digit > canÕt be determined. So you understand that starting with 0. and adding digit after digit infinitely it is impossible to reach/construct 0.999...? === Subject: Re: .99999... still=/= 1 > The 9Õs never stop repeating. ItÕs like this. If you say, cat >repeately > infinitely, catcatcat..., it never converges to dog. > It just stays as cat. >>When you (or your computer/calculator) successfully said cat (or >>9) infinitely many times can you tell me the last decimal digit of >>pi (since pi have infinitely many digits)? >>0, 1, 2, 3, 4, 5, 6, 7, 8 or 9? (Hint: try with your favourite >digit) >> I think that pi may be a circular convergence series. In other >words, there >> may be a circle of convergence test using the Radius Of Convergence R >Test. So >> given a value of R (the radius), I think pi converges. >> And like I mentioned before, a series that does converge doesnÕt >mean it >> really equals that convergence value. So that particular last decimal >digit >> canÕt be determined. >So you understand that starting with 0. and adding digit after digit >infinitely it is impossible to reach/construct 0.999...? Nope. SmartÕs 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: .99999... still=/= 1 >So you understand that starting with 0. and adding digit after digit >infinitely it is impossible to reach/construct 0.999...? > Nope. The following program in pseudo-C code illustrates the construction of 0.999... as a string and number using the rule adding digit after digit infinitely: ------------------------------------------------------------- --------------- ---------- result = n = digit = 0 string = 0. w = 0.00079 r = 3.9 //or any combination of values for w and r within the intervals (0, 1) //and (3, 4) resp. which leads to chaotic dynamics the logistic map while (n < oo) { n++ result += 9/10^n string += 9 //concatenation of strings digit = pi_ddigit(n) //pi_ddigit(n) returns the n-th decimal digit of pi w *= r * (1 - w) } print(Infinity successfully reached:) print( the number 0.999... reached.) print( the string 0.999... constructed.) print( the last digit of pi is: , digit) print( state of chaotic ddsystem at the infinity: , round_off_to_18_ddigits(w)) ------------------------------------------------------------- --------------- ---------- If you accept that the above program halts then you must tell me the last decimal digit of pi and the state of the dynamical system at the infinity (i.e. the round off value of w). This program will never halt. Reasons: For every integers a, b => a + b is integer. (1) => if b = 1 => a + b = a + 1 is integer for every integer a. (2) In other words: integer + 1 is again integer Since every integer < oo => n < oo is always true. (3) halts. Conclusion: 0.999... can not be reached/constructed using the rule adding digit after digit infinitely. === Subject: Re: .99999... still=/= 1 >>So you understand that starting with 0. and adding digit after digit >>infinitely it is impossible to reach/construct 0.999...? >> Nope. You can use the convergence tests at the lim of oo without having to type in every oo digit. For example, m-->oo lim Z_m = 0 then it converges otherwise diverges. .999... = function ( 1 - 1/10^m) = Z_m m-->oo lim ( 1 - 1/10^m) = 1 So .999... doesnÕt converge it diverges. So .999... =/= 1. >The following program in pseudo-C code illustrates the construction >of 0.999... as a string and number using the rule adding digit >after digit infinitely: >------------------------------------------------------------ ------------- ------------- >result = n = digit = 0 >string = 0. >w = 0.00079 >r = 3.9 >//or any combination of values for w and r within the intervals (0, 1) >//and (3, 4) resp. which leads to chaotic dynamics the logistic map >while (n < oo) > n++ > result += 9/10^n > string += 9 //concatenation of strings > digit = pi_ddigit(n) > //pi_ddigit(n) returns the n-th decimal digit of pi > w *= r * (1 - w) >print(Infinity successfully reached:) >print( the number 0.999... reached.) >print( the string 0.999... constructed.) >print( the last digit of pi is: , digit) >print( state of chaotic ddsystem at the infinity: , >round_off_to_18_ddigits(w)) >------------------------------------------------------------ ------------- ------------- >If you accept that the above program halts then you must tell me the >last decimal digit of pi and the state of the dynamical system at the >infinity (i.e. the round off value of w). >This program will never halt. Reasons: >For every integers a, b => a + b is integer. (1) >=> if b = 1 => a + b = a + 1 is integer for every integer a. (2) >In other words: integer + 1 is again integer >Since every integer < oo => n < oo is always true. (3) >halts. >Conclusion: 0.999... can not be reached/constructed using the rule >adding digit after digit infinitely. I already said this. SmartÕs 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: .99999... still=/= 1 posting-account=EH2x8QsAAABu84CuyjstkC4nRyQ1ZHKW wow, IÕm tripping on this new format -- it seems to enforce *selective* quoting, instead of dumping the whole thread into the editor. anyway, I had said taht I found an *application* of your alleged dilemma. if you look at EsculturaÕs write-up for his counter-examples, he seems to use 0.9999... as a coefficient, namely to get somehting like 10-adics (as someone was saying to him .-) --le ducs dÕEnron! http://tarpley.net/bush7.htm === Subject: Re: .99999... still=/= 1 whew; back to the old format, which is good, since it was impossible to see the tree in v2. the main point is that the folks who use p-adics blithely assume that the string of b-1s (in base-b) do continue, out to infity to the left of the decimal, although that point is superßuous, except as a kind of reminder. just like one assumes about the underßow condition on the calculator, subtracting a larger numger from a smaller; one always has to make sure that it wasnÕt just the tail-end of a very long string of 9s. anyway, if that was an apt characterization of EsculturaÕs method, then it seems to be akin to APÕs. if that is so, then it seems to be outside of the bounds of common-sense, since Fermat clearly meant the last theorem in terms of Diophantus, which does not include the subject of p-adic valuations -- not that IÕve read every thing about it. but theyÕre reputedly good for other stuff -- that is not *only* known to AP, I guess. > I had said taht I found an *application* > of your alleged dilemma. if you look > at EsculturaÕs write-up for his counter-examples, > he seems to use 0.9999... as a coefficient, > namely to get somehting like 10-adics --le ducs dÕEnron! http://tarpley.net/bush22.htm === Subject: Re: .99999... still=/= 1 >> .999...* 10 - 9 = .999... => x * 10 - 9 = x where x = .999... >=> 10x - x = 9 = 9x <==> x = 1 So why you claim that .999... != 1? ThatÕs right. > x = .999... > and after his calculations are done he shows it to be equal to something > else. > And another way to look at it letÕs set x = 1 to start with instead of > .999... > x = 1 > 10x = 10 > x = 1 > Here we can see x = 1 before and after. > And, > x = .999... > 10 *x = 9.999... > x = .999... > Here we can see x = .999... before and after. > This is often given as a laymanÕs justification (not proof) that .999... > =1. However mathematically, itÕs not quite right. It is sufficient for a proof. > You canÕt justify the > step 10x = 9.999... without appealing to a theorem that says you can > multiply each term of a convergent series by a constant, and the > resulting series will converge to the constant times the sum of the > original series. Such a theorem would be a tautology of little value, since the statement follows from the directly from distributive axiom a(b+c)= ab + ac. If x = 0.99999... [repeating n times] Then, x = 9/10^1 + 9/10^2 + .... 9/10^n ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n Let a = 10 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n = 9 + x > If you could prove that, then youÕd already have enough understanding of > limits to know that .999... = 1. Limits are unnecessary. The proof is simply, x = 0.9_ 10x = 9.9_ (follows directly from the distributive axiom) = 9 + x 9x = 9 x = 1 === Subject: Re: .99999... still=/= 1 Discussion, linux) >> You canÕt justify the >> step 10x = 9.999... without appealing to a theorem that says you can >> multiply each term of a convergent series by a constant, and the >> resulting series will converge to the constant times the sum of the >> original series. > Such a theorem would be a tautology of little value, since the > statement follows from the directly from distributive axiom a(b+c)= ab > + ac. It does not follow that a(b_1 + b_2 + b_3 + b_4 + ...) = a b_1 + a b_2 + a b_3 + a b_4+ ... The distributive axiom generalizes to arbitrary finite sums, but not to arbitrary countable sums. The above theorem can be proved for convergent sums, but it does *require* proof. Hey, why donÕt you prove it and call it SchoenfeldÕs theorem? -- This confused and outraged many Matrix fans, whoÕd already spent hours on the web explaining that man and computers could never really live in such a state of harmony and mutual benefit. -- http://www.pointlesswasteoftime.com === Subject: Re: .99999... still=/= 1 >> You canÕt justify the >> step 10x = 9.999... without appealing to a theorem that says you can >> multiply each term of a convergent series by a constant, and the >> resulting series will converge to the constant times the sum of the >> original series. > Such a theorem would be a tautology of little value, since the > statement follows from the directly from distributive axiom a(b+c)= ab > + ac. > It does not follow that > a(b_1 + b_2 + b_3 + b_4 + ...) = a b_1 + a b_2 + a b_3 + a b_4+ ... > The distributive axiom generalizes to arbitrary finite sums, but not > to arbitrary countable sums. The above theorem can be proved for > convergent sums, but it does *require* proof. Not really. a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} = a b_1 + a(b_2 + b_3 + ...) {distributive} > Hey, why donÕt you prove it and call it SchoenfeldÕs theorem? Since it is simple tautology. === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> posting-account=HPGXOQ0AAAB8Vqh61Q9SABrM_YhXJe__ Theorem 3.47 in Rudin Principles of mathematical analysis: If Sum a_n = A and Sum b_n = B then Sum (a_n+b_n) = A+B and Sum( c*a_n ) = c*A for any fixed c. If the series is convergent, the distributive property holds. To apply the distributive property to a series, we therefore first need to know if it converges. === Subject: Re: .99999... still=/= 1 In sci.math, John Schoenfeld step 10x = 9.999... without appealing to a theorem that says you can > multiply each term of a convergent series by a constant, and the > resulting series will converge to the constant times the sum of the > original series. >> Such a theorem would be a tautology of little value, since the >> statement follows from the directly from distributive axiom a(b+c)= ab >> + ac. >> It does not follow that >> a(b_1 + b_2 + b_3 + b_4 + ...) = a b_1 + a b_2 + a b_3 + a b_4+ ... >> The distributive axiom generalizes to arbitrary finite sums, but not >> to arbitrary countable sums. The above theorem can be proved for >> convergent sums, but it does *require* proof. > Not really. > a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} > = a b_1 + a(b_2 + b_3 + ...) {distributive} Then 1 + 2 + 4 + 8 + ... = -1. Proof. Let a = 1 + 2 + 4 + 8 + ... Then 2*a = 2 + 4 + 8 + .... = a - 1. Since 2*a = a-1, a = -1. QED. The ßaw in this proof is admittedly rather subtle, though the biggest clue is that 1 + 2 + 4 + 8 + ... is in fact divergent, tending towards no real number, though one might make a case that it tends towards positive infinity. Even with conditionally convergent series one can play some games, though IÕd have to look up the details now. >> Hey, why donÕt you prove it and call it SchoenfeldÕs theorem? > Since it is simple tautology. Not quite that simple, methinks. -- #191, ewill3@earthlink.net ItÕs still legal to go .sigless. === Subject: Re: .99999... still=/= 1 > 1 + 2 + 4 + 8 + ... is in fact divergent, tending towards no real > number, That statement is mostly true in the sense that itÕs divergent in the real metric, so tends toward no real number, but itÕs convergent in the 2-adic metric, so it *does* tend toward a 2-adic number. > Even with conditionally convergent series one can play > some games, though IÕd have to look up the details now. Yeah. For example 1 - 1/2 + 1/3 - 1/4 + 1/5 -+... If you rearrange the terms, still hitting every term once each, you can get a different sequence of partial sums that diverges to plus or minus infinity, or which oscillates wildly, or which converges to some (finite real) number other than what the original series converges to. ItÕs actually rather easy to construct an algorithm that will converge to any desired value: Separate the positive and negative terms into independent input streams, then whenever the partial sum is below your goal you take in the next positive term, and whenever the partial sum is above your goal you take in the next negative term. Every term in each of the two input streams is guaranteed to be used exactly once, and the partial sums are guaranteed to oscillate back and forth across your chosen goal, and satisfy the condition for convergence to the goal you chose. Or if you want it to oscillate wildly, or diverge to positive infinity, etc., you merely set up a sequence of goals that force it in a particular direction or through a particular oscillating pattern, and make sure you not only pass (cross) each goal but also use at least one term from each input stream before you change to the next goal. === Subject: Re: .99999... still=/= 1 > In sci.math, John Schoenfeld > You canÕt justify the > step 10x = 9.999... without appealing to a theorem that says you can > multiply each term of a convergent series by a constant, and the > resulting series will converge to the constant times the sum of the > original series. >> Such a theorem would be a tautology of little value, since the >> statement follows from the directly from distributive axiom a(b+c)= ab >> + ac. > It does not follow that > a(b_1 + b_2 + b_3 + b_4 + ...) = a b_1 + a b_2 + a b_3 + a b_4+ ... > The distributive axiom generalizes to arbitrary finite sums, but not >> to arbitrary countable sums. The above theorem can be proved for >> convergent sums, but it does *require* proof. > Not really. > a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} > = a b_1 + a(b_2 + b_3 + ...) {distributive} > Then 1 + 2 + 4 + 8 + ... = -1. > Proof. Let a = 1 + 2 + 4 + 8 + ... > Then 2*a = 2 + 4 + 8 + .... = a - 1. > Since 2*a = a-1, a = -1. QED. 2a = a. > The ßaw in this proof is admittedly rather subtle, > though the biggest clue is that 1 + 2 + 4 + 8 + ... > is in fact divergent, tending towards no real number, > though one might make a case that it tends towards > positive infinity. > Even with conditionally convergent series one can play > some games, though IÕd have to look up the details now. >> Hey, why donÕt you prove it and call it SchoenfeldÕs theorem? > Since it is simple tautology. > Not quite that simple, methinks. The variable ŌaÕ is transfinite and not an element of an ordered ring. http://mathworld.wolfram.com/Ring.html http://mathworld.wolfram.com/TransfiniteNumber.html === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> Discussion, linux) > Not really. > a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} > = a b_1 + a(b_2 + b_3 + ...) {distributive} And after a finite number of iterations, you still have a term of the form a (b_n + b_{n+1} + ...) ThereÕs no such thing as a derivation that involves infinitely many steps, so you still havenÕt proved a damn thing. >> Hey, why donÕt you prove it and call it SchoenfeldÕs theorem? > Since it is simple tautology. Wrong. -- When you go to class today, if your professor talks about algebraic number theory, or misuses Galois Theory[,] I want you to carefully notice how you feel. Hold on to that feeling so that you never forget it. --James S. Harris, on channeling rage via Galois theory. === Subject: Re: .99999... still=/= 1 > Not really. > a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} > = a b_1 + a(b_2 + b_3 + ...) {distributive} > And after a finite number of iterations, you still have a term of the > form > > a (b_n + b_{n+1} + ...) > ThereÕs no such thing as a derivation that involves infinitely many > steps, so you still havenÕt proved a damn thing. Trivially false. >> Hey, why donÕt you prove it and call it SchoenfeldÕs theorem? > Since it is simple tautology. > Wrong. === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> Discussion, linux) >> Not really. >> a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} >> = a b_1 + a(b_2 + b_3 + ...) {distributive} >> And after a finite number of iterations, you still have a term of the >> form >> >> a (b_n + b_{n+1} + ...) >> ThereÕs no such thing as a derivation that involves infinitely many >> steps, so you still havenÕt proved a damn thing. > Trivially false. You must be talking about Schoenfeld derivations. TheyÕre not in the books yet. Maybe tomorrow. -- Jesse F. Hughes History will hate you and love me. IÕm the misunderstood and persecuted genius. YouÕre the assholes. -- James Harris === Subject: Re: .99999... still=/= 1 > Not really. >> a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} >> = a b_1 + a(b_2 + b_3 + ...) {distributive} > And after a finite number of iterations, you still have a term of the >> form >> >> a (b_n + b_{n+1} + ...) > ThereÕs no such thing as a derivation that involves infinitely many >> steps, so you still havenÕt proved a damn thing. > Trivially false. > You must be talking about Schoenfeld derivations. TheyÕre not in > the books yet. Maybe tomorrow. Recursion. === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> <87fz2mnzj0.fsf@phiwumbda.org> Discussion, linux) > > Not really. a(b_1 + b_2 + b3 + ...) = a(b_1 + (b_2 + b3 + ...)) {associative} > = a b_1 + a(b_2 + b_3 + ...) {distributive} > > And after a finite number of iterations, you still have a term of the > form > > a (b_n + b_{n+1} + ...) > > ThereÕs no such thing as a derivation that involves infinitely many > steps, so you still havenÕt proved a damn thing. >> Trivially false. >> You must be talking about Schoenfeld derivations. TheyÕre not in >> the books yet. Maybe tomorrow. > Recursion[...] ainÕt got a damn thing to do with anything here. Hint: your tautology is plainly false without some assumptions (like the sum converges). It doesnÕt follow directly from distributivity. You are confused on (at least) two points. (1) The infinite summation notation is just a shorthand for a particular limit. What you have is: a (lim_{n -> oo} Sum_{i=1}^n b_n) Distributivity doesnÕt tell you a damn thing about how multiplication (2) Even if you werenÕt confused about the meaning of the summation notation, you cannot prove what you want because derivations are finite. Your proof would require an infinite number of steps, which is just hogwash. -- Jesse F. Hughes Yesterday was Judgment Day. HowÕd you do? -- The Flatlanders === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> <87fz2mnzj0.fsf@phiwumbda.org> <87hdn1dkv2.fsf@phiwumbda.org> posting-account=I8cafwwAAACoOYfL9BiKocZY4Rsgl4L7 It has to do with your objection to the sentence 10*0.9_ = 9.9_ despite it following directly from the axioms of the implied ordered ring. a(b+c) = ab + ac {Distributive} a(b+c+d) = a(b+(c+d)) {Associative} It follows that a(b_1+ ... + b_n) = a b_1 + a (b_2 + ... + b_n). === Subject: Re: .99999... still=/= 1 Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >a(b+c) = ab + ac {Distributive} >a(b+c+d) = a(b+(c+d)) {Associative} >It follows that a(b_1+ ... + b_n) = a b_1 + a (b_2 + ... + b_n). Very good. Now prove the limit case. -- Richard === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> <87fz2mnzj0.fsf@phiwumbda.org> <87hdn1dkv2.fsf@phiwumbda.org> posting-account=I8cafwwAAACoOYfL9BiKocZY4Rsgl4L7 It has to do with your objection to the sentence 10*0.9_ = 9.9_ despite it following directly from the axioms of the implied ordered ring. a(b+c) = ab + ac {Distributive} a(b+c+d) = a(b+(c+d) {Associative} Thus, a(b_1+ ... + b_n) = a b_1 + a (b_2 + ... + b_n). === Subject: Re: .99999... still=/= 1 <87acsxynol.fsf@phiwumbda.org> <87k6rz9rb5.fsf@phiwumbda.org> <87fz2mnzj0.fsf@phiwumbda.org> <87hdn1dkv2.fsf@phiwumbda.org> Discussion, linux) > It has to do with your objection to the sentence 10*0.9_ = 9.9_ despite > it following directly from the axioms of the implied ordered ring. Moron. IÕve never objected to that equation. IÕve objected to your spurious claim that it follows from the distributive property. a Sum_{i = 1}^{oo} b_i = Sum_{i=1}^{oo} a b_i whenever Sum_{i = 1}^{oo} b_i converges. This is true, but it is not true by virtue of the distributive property alone. It requires proof. -- If you have a really big idea, you can get a measure of how big it is REALLY, REALLY, *REALLY*, BIG DISCOVERY!!! --James Harris, on being ignored === Subject: Re: .99999... still=/= 1 >> It has to do with your objection to the sentence 10*0.9_ = 9.9_ despite >> it following directly from the axioms of the implied ordered ring. >Moron. IÕve never objected to that equation. IÕve objected to your >spurious claim that it follows from the distributive property. >a Sum_{i = 1}^{oo} b_i = Sum_{i=1}^{oo} a b_i >whenever Sum_{i = 1}^{oo} b_i converges. This is true, but it is not >true by virtue of the distributive property alone. It requires proof. Or an appropriate notion of ring. Lee Rudolph (I suggest secret decoder ring as most appropriate in this context) === Subject: Re: .99999... still=/= 1 > It has to do with your objection to the sentence 10*0.9_ = 9.9_ despite > it following directly from the axioms of the implied ordered ring. > a(b+c) = ab + ac {Distributive} > a(b+c+d) = a(b+(c+d) {Associative} > Thus, a(b_1+ ... + b_n) = a b_1 + a (b_2 + ... + b_n). Notice that this holds for finite sums. You have to use a limit argument to extend that to an infinite series. Bob Kolker === Subject: Re: .99999... still=/= 1 > .999...* 10 - 9 = .999... >>=> x * 10 - 9 = x where x = .999... >>=> 10x - x = 9 = 9x <==> x = 1 >>So why you claim that .999... != 1? >> >> ThatÕs right. >> x = .999... >> >> and after his calculations are done he shows it to be equal to something >> else. >> >> And another way to look at it letÕs set x = 1 to start with instead of >> .999... >> >> >> x = 1 >> 10x = 10 >> x = 1 >> >> Here we can see x = 1 before and after. >> >> And, >> >> x = .999... >> >> 10 *x = 9.999... >> >> x = .999... >> >> Here we can see x = .999... before and after. >> This is often given as a laymanÕs justification (not proof) that .999... >> =1. However mathematically, itÕs not quite right. >It is sufficient for a proof. >> You canÕt justify the >> step 10x = 9.999... without appealing to a theorem that says you can >> multiply each term of a convergent series by a constant, and the >> resulting series will converge to the constant times the sum of the >> original series. >Such a theorem would be a tautology of little value, since the >statement follows from the directly from distributive axiom a(b+c)= ab >+ ac. > x = 0.99999... [repeating n times] >Then, > x = 9/10^1 + 9/10^2 + .... 9/10^n > ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n >Let a = 10 > 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n > = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n > = 9 + x >> If you could prove that, then youÕd already have enough understanding of >> limits to know that .999... = 1. >Limits are unnecessary. The proof is simply, >x = 0.9_ >10x = 9.9_ (follows directly from the distributive axiom) > = 9 + x > 9x = 9 > x = 1 Nope, this is incorrect. The question was does .999... converge to 1, not 9.999... . equation. Equations arenÕt used in convergence tests. The basic convergence test is like this. If, m-->oo Lim Z_m = 0 then convergence otherwise divergence. .999... is represented as (1 - 1/10^m) Z_m = ( 1 - 1/10^m) m-->oo Lim Z_m = 1 This means divergence. So not only does it not converge to 1 but it doesnÕt equal 1. And even if it did converge it wouldnÕt EQUAL 1. There isnÕt even one series in all of math that does converge that really equals that convergence value. .999... --> diverges and =/= 1 SmartÕs 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: .99999... still=/= 1 posting-account=I8cafwwAAACoOYfL9BiKocZY4Rsgl4L7 Show the error: x = 0.99999 [repeating n times] = 9/10^1 + 9/10^2 + ..... + 9/10^n 10x = 90/10^1 + 90/10^2 + ..... + 90/10^n = 9/10^0 + 9/10^1 + 9/10^2 + ..... + 9/10^(n - 1) = 9 + x - (9/10^n) For the case that n = +inf, then 10x = lim (n -> +inf) 9 + x - (9/10^n) = 9 + x - [ lim (n -> +inf) 9/10^n ] = 9 + x - 0 Since x = 0.9_, 10 * 0.9_ = 9 + 0.9_ 9 * 0.9_ = 9 0.9_ = 9/9 = 1 === Subject: Re: .99999... still=/= 1 > Show the error: > x = 0.99999 [repeating n times] > = 9/10^1 + 9/10^2 + ..... + 9/10^n > 10x = 90/10^1 + 90/10^2 + ..... + 90/10^n > = 9/10^0 + 9/10^1 + 9/10^2 + ..... + 9/10^(n - 1) > = 9 + x - (9/10^n) > For the case that n = +inf, then > 10x = lim (n -> +inf) 9 + x - (9/10^n) > = 9 + x - [ lim (n -> +inf) 9/10^n ] > = 9 + x - 0 You left out some steps. Notice that x is really a function of n, the way you defined it so: 10*lim x [n-> inf] = 9 + lim x [n -> inf] - lim 9/10^n [n -> inf] You have to show first that lim x [n -> inf] exists. (We know it does, but you have to show it). Then let L = lim x [n -> inf] and get: 10*L = 9 + L; hence 9*L = 9 and L = 1. Bob Kolker === Subject: Re: .99999... still=/= 1 > Show the error: > x = 0.99999 [repeating n times] > = 9/10^1 + 9/10^2 + ..... + 9/10^n > 10x = 90/10^1 + 90/10^2 + ..... + 90/10^n > = 9/10^0 + 9/10^1 + 9/10^2 + ..... + 9/10^(n - 1) > = 9 + x - (9/10^n) > For the case that n = +inf, then > 10x = lim (n -> +inf) 9 + x - (9/10^n) > = 9 + x - [ lim (n -> +inf) 9/10^n ] > = 9 + x - 0 > You left out some steps. Notice that x is really a function of n, the > way you defined it so: > 10*lim x [n-> inf] = 9 + lim x [n -> inf] - lim 9/10^n [n -> inf] > You have to show first that lim x [n -> inf] exists. (We know it does, > but you have to show it). Then let L = lim x [n -> inf] and get: > 10*L = 9 + L; hence 9*L = 9 and L = 1. lim(n->+inf) x = x - [ lim(n->+inf) 0 ] = x - 0 = x Most proofs are quasi-proofs as it would take thousands of statements to derive the simplest of propositions from the axioms of the relevant mathematical structure and first order logic. http://en.wikipedia.org/wiki/Mathematical_proof http://en.wikipedia.org/wiki/Proof_theory > Bob Kolker === Subject: Re: .99999... still=/= 1 > Let a_n = .999..9 [repeated n times]. > Lim a_n (n ->oo) = 1. > Proof: Lim (n->oo) 1/10^n = 0, so 1.0 - a_n = 1/10^n goes to 0 as n ->oo > Go back into your box. > Bob Kolker > You can do it simpler, > x = 0.9_ > 10x = 9.9_ > = 9 + x > 9x = 9 > x = 9/9 > = 1 > What youÕre showing with this argument is that, IF the limit exists, > then it MUST be 1. What you are not showing is that the limit exists > (which is not very difficult). It only requires the distributive axiom a(b+c)= ab + ac, which is a given. If x = 0.99999... [repeating n times] Then, x = 9/10^1 + 9/10^2 + .... 9/10^n ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n Let a = 10 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n = 9 + x > However, let me remark that it is possible to use this idea to define > the (rational) number represented by any eventually repeating decimal > expansion in this way. > That is, > 0. a_1 a_2 a_3 ... a_n { b_1 b_2 b_3 ... b_m } > (where {...} means periodic continuation) is defined to be the > rational number > A / 10^n + B / ( 10^(n+m) - 10^n ), > where A and B are the integers with decimal representation a_1 a_2 > ... a_n and > b_1 b_2 ... b_m, respectively. > Thus, > 0.99999... = 0.{9} = 9/(10 - 1 ) =1. > Similarly, > 0.33333... = 0.{3} = 3/(10 - 1 ) = 1/3, > 0.5 = 0.5{0} = 5/10 = 1/2, > 0.166666... = 0.1{6} = 1/10 + 6/(90) = 1/6, and > 0.142857142857142857... = 0.{142857} = 142857/999999 = 1/7. > The downside is that this definition may seem somewhat artificial. Of > course it is easy to show that, when defined like this, decimal > expansions have all the usual properties --- but that brings us back > to having to discuss convergence, which we were trying to avoid in the > first place. No. === Subject: Re: .99999... still=/= 1 > You can do it simpler, > x = 0.9_ > 10x = 9.9_ > = 9 + x > 9x = 9 > x = 9/9 > = 1 > What youÕre showing with this argument is that, IF the limit exists, > then it MUST be 1. What you are not showing is that the limit exists > (which is not very difficult). > It only requires the distributive axiom a(b+c)= ab + ac, which is a given. No, you are wrong. What I was referring to is the fact that the original limit x = lim_{n->infty} SUM_{k=1}^{n} 9/(10)^k exists. The distributive property has nothing to do with this --- rather, we use the statement that every bounded monotone sequence of real numbers has a limit. However, this fact uses deep properties of the real numbers. It is actually EASIER to show directly that the limit is 1 --- this can be done strictly in the realm of rational numbers. > The downside is that this definition may seem somewhat artificial. Of > course it is easy to show that, when defined like this, decimal > expansions have all the usual properties --- but that brings us back > to having to discuss convergence, which we were trying to avoid in the > first place. > No. No to having to discuss convergence, or no to trying to avoid it? In any case, you are wrong: the usual definition involves convergence, which thus has to be discussed, and the original argument was clearly intended to avoid doing this. If you disagree, please elaborate. Lasse --- === Subject: Re: .99999... still=/= 1 > x = 0.9_ > 10x = 9.9_ > = 9 + x > 9x = 9 > x = 9/9 > = 1 > If Sum (n >= 0) [a_n] = a one must prove that > k*Sum (n >= n) [a_n] = k*a. It requires a limit argument to justify the > above manipulatio. So it really isnÕt simpler. > Bob Kolker It only requires the distributive axiom a(b+c)= ab + ac, which is a given. If x = 0.99999... [repeating n times] Then, x = 9/10^1 + 9/10^2 + .... 9/10^n ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n Let a = 10 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n = 9 + x === Subject: Re: .99999... still=/= 1 >You can do it simpler, >x = 0.9_ >10x = 9.9_ > = 9 + x > That requires you to have shown that infinite sums can be multiplied > and subtracted in that way, which is harder than the result youÕre > proving. It only requires the distributive axiom a(b+c)= ab + ac, which is a given. If x = 0.99999... [repeating n times] Then, x = 9/10^1 + 9/10^2 + .... 9/10^n ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n Let a = 10 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n = 9 + x > -- Richard === Subject: Re: .99999... still=/= 1 Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >It only requires the distributive axiom a(b+c)= ab + ac, which is a given. It requires a version of it for infinite sums. > x = 0.99999... [repeating n times] >Then, > x = 9/10^1 + 9/10^2 + .... 9/10^n > ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n >Let a = 10 > 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n ^^^^^^^^^^^^^ You mean 90/10^n > = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n ^^^^^^^^^^^^^ You mean 9/10^(n-1) > = 9 + x You mean 9 + x - 9/10^n. -- Richard === Subject: Re: .99999... still=/= 1 posting-account=I8cafwwAAACoOYfL9BiKocZY4Rsgl4L7 >It only requires the distributive axiom a(b+c)= ab + ac, which is a given. > It requires a version of it for infinite sums. >If > x = 0.99999... [repeating n times] >Then, > x = 9/10^1 + 9/10^2 + .... 9/10^n > ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n >Let a = 10 > 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n > ^^^^^^^^^^^^^ > You mean 90/10^n > = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n > ^^^^^^^^^^^^^ > You mean 9/10^(n-1) > = 9 + x > You mean 9 + x - 9/10^n. > -- Richard I made a few typos, yes. The point is what I originally provided was a proof, and that you were wrong. === Subject: Re: .99999... still=/= 1 Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >I made a few typos, yes. The point is what I originally provided was a >proof, and that you were wrong. ??? What are you talking about? -- Richard === Subject: Re: .99999... still=/= 1 >I made a few typos, yes. The point is what I originally provided was a >proof, and that you were wrong. > ??? What are you talking about? Things of a trivial nature. === Subject: Re: .99999... still=/= 1 >>It only requires the distributive axiom a(b+c)= ab + ac, which is a given. >It requires a version of it for infinite sums. >>If >> x = 0.99999... [repeating n times] >>Then, >> x = 9/10^1 + 9/10^2 + .... 9/10^n >> ax = (9a)/10^1 + (9a)/10^2 + .... (9a)/10^n >>Let a = 10 >> 10x = 90/10^1 + 90/10^2 + .... + (9*10^n)/10^n > ^^^^^^^^^^^^^ >You mean 90/10^n >> = 9/10^0 + 9/10^1 + 9/10^2 + .... + (9*10^n)/10^n > ^^^^^^^^^^^^^ >You mean 9/10^(n-1) >> = 9 + x >You mean 9 + x - 9/10^n. >-- Richard So this shows that, 10x -.999... = 9 + x - .999... This doesnÕt prove .999... = 1. It shows that, 10x = 9.999... 9.999... - .999... = 9 + .999... - .999... 9 + .999... - .999.... = 9 + .999... - .999... 9 = 9 So what, 9 = 9, 1 = 1, .999... = .999..., dummy = dummy and so on. SmartÕs 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: .99999... still=/= 1 Originator: richard@cogsci.ed.ac.uk (Richard Tobin) > So this shows that, >10x -.999... = 9 + x - .999... Add .999... to both sides. Then subtract x. What do you get? -- Richard === Subject: Re: .99999... still=/= 1 >> So this shows that, >>10x -.999... = 9 + x - .999... >Add .999... to both sides. Then subtract x. What do you get? >-- Richard The point is, using this type of equation isnÕt a proof of convergence or divergence. lim ( equation below) 10x = 9 + x At x = 0, you get 0 = 9 At x = 1, you get 10 = 10 at x = .999... , you get 9.999... = 9.999... At x = 10, 100 = 19 At x = oo, you get 10(oo) = 9 + oo ???? <-- indeterminate This test fails and is inconsistent. SmartÕs 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: .99999... still=/= 1 > Just multiply 9*.999... on your calculator. ThatÕs not possible because .999... canÕt be entered on a calculator. So youÕre ordering us to do something thatÕs impossible. === Subject: Re: .99999... still=/= 1 >> Just multiply 9*.999... on your calculator. >ThatÕs not possible because .999... canÕt be entered on a calculator. >So youÕre ordering us to do something thatÕs impossible. But some people thing they can add and substract values of .999... , so what is the difference? I just used their logic on my calculator, and I still show, .999... still=/= 1 SmartÕs 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: .99999... still=/= 1 > .999... canÕt be entered on a calculator. > But some people thing [sic] they can add and substract values of > .999... , so what is the difference? The difference is that some people here actually understand the usual mathematical meaning of that notation. ItÕs too bad you donÕt. > I just used their logic on my calculator No, you didnÕt use any logic at all on your calculator, you merely punched in some buttons which caused your calculator to perform some approximate entries and calculations. > I still show, .999... still=/= 1 Nope, you havenÕt shown any such thing. All youÕve shown is that when you punch a particular sequence of keys on a particular brand of calculator it gives such-and-such approximate value as the result. You still havenÕt posted what you mean by the notatin you used in the Subject field of this thread. Since you posted to sci.math, not to alt.commercial-brand-of-calculator.games, we assumed you meant the usual mathematical meaning of that notation, but if all you meant by that notation is what you get if you punch in a period followed by a long sequence of 9Õs then you were wrong to post to this newsgroup? === Subject: Re: .99999... still=/= 1 >> .999... canÕt be entered on a calculator. >> But some people thing [sic] they can add and substract values of >> .999... , so what is the difference? >The difference is that some people here actually understand the usual >mathematical meaning of that notation. ItÕs too bad you donÕt. >> I just used their logic on my calculator >No, you didnÕt use any logic at all on your calculator, you merely >punched in some buttons which caused your calculator to perform some >approximate entries and calculations. >> I still show, .999... still=/= 1 >Nope, you havenÕt shown any such thing. All youÕve shown is that when >you punch a particular sequence of keys on a particular brand of >calculator it gives such-and-such approximate value as the result. >You still havenÕt posted what you mean by the notatin you used in the >Subject field of this thread. Since you posted to sci.math, not to >alt.commercial-brand-of-calculator.games, we assumed you meant the >usual mathematical meaning of that notation, but if all you meant by >that notation is what you get if you punch in a period followed by a >long sequence of 9Õs then you were wrong to post to this newsgroup? Bottom line. Does the convergence value perfectly equal the actual value of a series? .999... --> converges to 1, okay but, .999... =/= 1 .999... never reaches 1. It is just assumed that it is the closest approximate value it could be. Main Entry: as.87ymp.87tote Pronunciation: Ōa-s&m(p)-tOt Function: noun Etymology: probably from (assumed) New Latin asymptotus, from Greek asymptOtos not meeting, from a- + sympiptein to meet .84more at SYMPTOM : a straight line associated with a curve such that as a point moves along an infinite branch of the curve the distance from the point to the line approaches zero and the slope of the curve at the point approaches the slope of the line In other words the curve and the straight line never actually meet. This is similare to convergence. ItÕs nearest value the slope approaches. .999... still =/= 1 and never will. I won this debate a long time ago. Why donÕt you people admit you were wrong? SmartÕs 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: .99999... still=/= 1 > Bottom line. Does the convergence value perfectly equal the actual value of > a series? The series -equals- the limit of the sequence of partial sums. This is a definition of the series. Pay attention when people tell you correct stuff. Bob Kolker === Subject: Re: .99999... still=/= 1 >> Bottom line. Does the convergence value perfectly equal the actual value >> a series? >The series -equals- the limit of the sequence of partial sums. This is a >definition of the series. Pay attention when people tell you correct stuff. >Bob Kolker And this is incorrect as found in the foundations of math itself, not just you. The Partial Sums never precisely equals the total value of the series. It is only a rounded off approximate value that can be used. I simply corrected their mistake. Again, The series .3 + .03 + ...+ 3/10^n = .333... Partial Sums ( using the Online Line Partial Sums calculator not me) shows it as, Partial Sums Convergence at value, .33333333333333333333326 this doesnÕt precisely equal, .333... or 1/3 SmartÕs 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: .99999... still=/= 1 > And this is incorrect as found in the foundations of math itself, not just > you. The Partial Sums never precisely equals the total value of the series. It > is only a rounded off approximate value that can be used. I simply corrected > their mistake. Putzarooni. Do you know what the phrase converge to a limit means? Bob Kolker === Subject: Re: .99999... still=/= 1 >> And this is incorrect as found in the foundations of math itself, not >just >> you. The Partial Sums never precisely equals the total value of the series. >> is only a rounded off approximate value that can be used. I simply >corrected >> their mistake. >Putzarooni. Do you know what the phrase converge to a limit means? >Bob Kolker Looks like everyone except you is left in this debate about .999... = 1. Seems like they finally realized the convergence value of a series can never precisely equal tha actual value of the series as it approaches infinity. Even the convergence test is based on an approximation of what occurs near infinity. notice the meaning of this m--->oo. You never reach infinity, you are approaching it. So m-->oo lim 9/10^m = really canÕt be determined without assuming you can say it reached infinity. Example, 9/10^999999999999999999999... You assume this reaches, 9/10^oo, then you assume you can say, anything divided by oo equals 0. When all along it really doesnÕt. It only approaches 0. The convergence test says If m-->oo lim Z_m = 0 then convergence otherwise divergence. It never really reaches 0. Then after this you assume that a convergence of a series proves that it actually reaches that convergence value and it doesnÕt. .999... remains this way. It approaches 1. It never reaches 1. It never equals 1. To say to converge to a lim means the nearest ((approximate)) value can be reached, but it doesnÕt really reach it, like an asymptote. You might as well admit it, you made a mistake. .999... =/= 1 SmartÕs 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: .99999... still=/= 1 >> And this is incorrect as found in the foundations of math itself, not >just >> you. The Partial Sums never precisely equals the total value of the series. >> is only a rounded off approximate value that can be used. I simply >corrected >> their mistake. >Putzarooni. Do you know what the phrase converge to a limit means? >Bob Kolker Do you know what precisely equals to means with respect to a series as it approaches infinity? SmartÕs 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: Set inclusion and membership > Put differently, a consequence of this brilliant suggestion that > x = {x} is > The element relation is the same as equality. No, that is only valid for a singleton. Han de Bruijn === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> <87llciow8j.fsf@phiwumbda.org> <87fz2qov5z.fsf@phiwumbda.org> Discussion, linux) >> Put differently, a consequence of this brilliant suggestion that >> x = {x} is >> The element relation is the same as equality. > No, that is only valid for a singleton. For every set x, {x} is a singleton. Therefore, for every set x, x = {x}. If you have something else in mind, youÕll have to express it more ,---- | Now suppose that I am kind of stubborn, insisting that the 2 concepts | should denote the same thing. With other words: I find that a = { a } | everywhere. `---- What did that *mean*? Everywhere sounds fairly sweeping to me. -- I am a force of Nature. Time is a friend of mine, and We talk about things, here and there. And sometimes We muse a bit [...] and then We watch them go... in the meantime, Time and I, We play with some of them, at least for a little while. --- JSH and His pal, Time. === Subject: Re: Set inclusion and membership > If you have something else in mind, youÕll have to express it more > clearly. [ ... ] There are _a lot_ of things I would have to express more clearly. But, as far as I am concerned, the rest of this thread will be nitpicking. I have found my Precious One. ItÕs called Mereology. Which is equivalent to a Boolean Algebra without the empty set. I find this a satisfactory end-result. IÕve also learned that ZFC without the axiom of Foundation is a possibility. Han de Bruijn Note: Precious One as quoted from Lord of the Rings by Tolkien. No argument. Just a pun. Since I have to explain everything to some. === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> <87llciow8j.fsf@phiwumbda.org> <87fz2qov5z.fsf@phiwumbda.org> <874qj5ylom.fsf@phiwumbda.org> Discussion, linux) >> If you have something else in mind, youÕll have to express it more >> clearly. [ ... ] > There are _a lot_ of things I would have to express more clearly. > But, as far as I am concerned, the rest of this thread will be > nitpicking. I have found my Precious One. ItÕs called Mereology. > Which is equivalent to a Boolean Algebra without the empty set. > I find this a satisfactory end-result. Good, but it does not satisfy your desideratum that x = {x}. Neither does anti-well-founded set theory. In both cases, there are certain xÕs for which that is true: discrete xÕs for mereology and the unique x satisfying x = {x} for AczelÕs anti-well-founded set theory. But it is certainly not satisfied everywhere, which is what started this thread. -- I am one of the more important discoverers in mathematical history, but future students will have the luxury of knowing that, and may be puzzled by your behavior now. -- James Harris (At least I have the foresight to quote his pearls of wisdom.) === Subject: Re: Set inclusion and membership > So your only sets satisfy the following two conditions. > (1) x is in x > (2) x is the only element of x. I never said this. My only _singletons_ satisfy these two conditions. But there exist many sets which are not singletons, of course, even in my theory :-) Han de Bruijn === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> <87llciow8j.fsf@phiwumbda.org> Discussion, linux) >> So your only sets satisfy the following two conditions. >> (1) x is in x >> (2) x is the only element of x. > I never said this. My only _singletons_ satisfy these two conditions. > But there exist many sets which are not singletons, of course, even in > my theory :-) For any set x, {x} is a singleton. You said before that a={a} is true everywhere. Whatever did you mean? ,---- | It doesnÕt. It cannot be that (S not in S) because S = {S} implies that | (S is in S), always, unconditionally. With other words, JesseÕs would-be | (E y)(y in S and NOT y = S) is a set that simply does not exist. `---- This says that there are no sets aside from sets satisfying x={x}. Thus (1) every set is a singleton and (2) every set is an element of itself. Therefore, the is-element relation is equivalent to equality. If this isnÕt what you mean, then what *do* you mean? IÕm sure I canÕt guess. -- Jesse F. Hughes I talk with bigger fish who are playing different games. -- James S Harris has a way with the metaphor. === Subject: Re: Set inclusion and membership [ .. nitpicking as usual .. ] Two points, in case you didnÕt notice: 1. IÕm done with you 2. This thread has ended Han de Bruijn === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> <87llciow8j.fsf@phiwumbda.org> <871xe9ylka.fsf@phiwumbda.org> Discussion, linux) > [ .. nitpicking as usual .. ] > Two points, in case you didnÕt notice: > 1. IÕm done with you > 2. This thread has ended Yes, pointing out that the solution to your question has very little to do with your question is nitpicking. You are a tedious and stupid person. -- Jesse F. Hughes Well, I donÕt claim to be an expert, in fact I am a fry cook with a national burger chain, but I have solved many differential and partial differential equations numerically. --C. Bond === Subject: Re: Set inclusion and membership > You are a tedious and stupid person. Logged. Are you proud of yourself? Han de Bruijn === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> <87llciow8j.fsf@phiwumbda.org> <871xe9ylka.fsf@phiwumbda.org> <87acsvbqu0.fsf@phiwumbda.org> Discussion, linux) >> You are a tedious and stupid person. > Logged. Are you proud of yourself? I thought that the earlier vulgar language prompted you to ignore my posts. Feeling suddenly less sensitive, are we? -- Jesse F. Hughes You shouldnÕt hate Mother Mathematics. -- James S. Harris === Subject: Re: Set inclusion and membership > This says that there are no sets aside from sets satisfying x={x}. > Thus (1) every set is a singleton and (2) every set is an element of > itself. Therefore, the is-element relation is equivalent to equality. > If this isnÕt what you mean, then what *do* you mean? IÕm sure I > canÕt guess. What he means is that he is clueless :-( -- 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: Set inclusion and membership > > Look who is saying this! Just go to Google and search for the phrase > > David C. Ullrich and OFF. Then youÕll know _his_ meaning of > > being pretty convincing. (And being civilized as well.) > David C. Ullrich containing that text, in 36 it was in quotes from > at least one of them is a falsification. And that in 4 1/2 years. So? So what? Substitute idiot, fool, stupid, dishonest, ignorant, youÕre crazy for OFF, to name only a few of his favorite words. Not to mention all of his sneering comments which cannot be tracked by keyword search at all. But what can you expect from an inhabitant of a country, where calling someone an asshole is done by a candidate for presidency. > Back to the present. I take it then that > ŌBTW. Gozer: Are you a god? Dr. Raymond Stantz: No. Gozer: Then die. > (as quotet from Ghostbusters 1)Õ > is in fact the best justification you can come up with for > the curious notion that a = {a} in physics? Ullrich doesnÕt even notice the difference between a would-be argument and a would-be signature. Han de Bruijn Reporter: What do you think of Western Civilization, sir? Ghandi: Well, I think that would be a mighty good idea! === Subject: Re: Set inclusion and membership Discussion, linux) > But what can you expect from an inhabitant of a country, where > calling someone an asshole is done by a candidate for presidency. Speaking of assholes and idiots... off. -- Naomi Klein reports that Microsoft is helping develop e-government in Iraq, Ōwhich Dees admits is a little ahead of the curve, since there is no g-government in Iraq--not to mention functioning phones lines.Õ -- as reported in The Register, http://www.theregister.co.uk === Subject: Re: Set inclusion and membership > Speaking of assholes and idiots... > off. Then you are the next one I will not answer anymore. And thatÕs final. Civilization comes before the math. Han de Bruijn === Subject: Re: Set inclusion and membership > Pointfullessness > Are you appealing, to some invocation > That is revealing, a small revocation > For my repealing, your incantation? > Now changing the point, to make my point which is the point I deserve for > pointing out the most pointedly pointed point ever pointed, may I point > out to you, the point I wish you to appoint me without disappointment at > our next appointment? Hmm, your English seems to be as good as my Dutch! :-) Indeed. LetÕs finish, please, this part of the thread: > P(A) = { aÕ.bÕ , aÕ.b , a.bÕ , a.b } where (.) denotes intersection. I apologize for launching this add-on, which has caused nothing else but confusion. I shouldnÕt have opened this can of worms, at all. OK? Han de Bruijn === Subject: Re: Set inclusion and membership Are you appealing, to some invocation > That is revealing, a small revocation > For my repealing, your incantation? > Now changing the point, to make my point which is the point I deserve for > pointing out the most pointedly pointed point ever pointed, may I point > out to you, the point I wish you to appoint me without disappointment at > our next appointment? > P(A) = { aÕ.bÕ , aÕ.b , a.bÕ , a.b } where (.) denotes intersection. > I apologize for launching this add-on, which has caused nothing else > but confusion. I shouldnÕt have opened this can of worms, at all. OK? Depends. If you like the fish you catch with those worms, then eat Ōem and if you donÕt then toss them back, dump the worms into soft soil and pack up your fishing rod. === Subject: ADV: Staff Announcement boundary=1..A4_6A.FA6.D_EAB5.E38. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id iB2BIqI13956; by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, $Revision: 1.6 secondary) with SMTP id iB2BImX2012059; ------------------------------------------------------------- -------- Attention All School Staff: Teachers, Students and Faculty Members: Through a special arrangement, Avtech Direct is offering a limited allotment of BRAND NEW, top of-the-line, name-brand desktop computers at more than 50% off MSRP to all Teachers, Students,Faculty and Staff, All desktop computers are brand-new packed in their original boxes, and come with a full manufacturerÕs warranty plus a 100% satisfaction guarantee. 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All desktop computers will be available on a first come first serve basis. and we will hold the desktops you request on will call. 4. You are not obligated in any way. 5. 100% Satisfaction Guaranteed. 6. Ask for Department C. Call Avtech Direct Visit our website at http://www.avtechcomputers.com If you wish to unsubscribe from this list, please go to http://www.avtechcomputers.com/announcements.asp Avtech Direct 22647 Ventura Blvd. Suite 374 Woodland Hills, CA 91364 === Subject: Re: Staff Announcement Actech advertised: > * IBM Processor for amazing speed and performance Which IBM processor? === Subject: Re: ADV: Staff Announcement Administrator scribbled the following: > Attention All School Staff: Teachers, Students and Faculty Members: Why? What will happen if they donÕt? -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -------------------------------------------------------- rules! --------/ You have moved your mouse, for these changes to take effect you must shut down and restart your computer. Do you want to restart your computer now? - Karri Kalpio === Subject: need proof!!! posting-account=D763ug0AAAADgMar-YrVD_exvbaACoLd i resently asked the following question IÕm looking for a continuous function f:R->R with discontinuity on irrational domain and continuous on Q. the good people who responded, told me that the answer is NO. now, does anyone knows a formal proof for this... but if you dont know just tell me why === Subject: Re: need proof!!! > i resently asked the following question > IÕm looking for a continuous function f:R->R with discontinuity on > irrational domain and continuous on Q. > the good people who responded, told me that the answer is NO. > now, does anyone knows a formal proof for this... > but if you dont know just tell me why Suppose f:R->R is an arbitrary real function. For n in N, n>=1 and y in R let A_n(y) = f^-1[(y-1/n, y+1/n)] U_n(y) = int A_n(y) where int = topological interior U_n = UNION {U_n(y) | y in R} C = INTERSECTION {U_n | n in N, n>=1} If f is continuous at x: x in int(f^-1[O]) whenever O is open and f(x) in O so x in U_n(f(x)) for all n so x in U_n for all n so x in C If x in C: For all n, x in U_n So for each n we can find y_n in R with x in U_n(y_n) So for all n, x in int f^-1[(y_n-1/n, y_n+1/n)] Thus for all n, |f(x)-y_n| < 1/n and also for all n there is d_n>0 such that |x-z| < d_n => |f(z)-y_n| < 1/n Suppose e>0 and pick N such that 1/N < e/2 |x-z| < d_N => |f(z)-f(x)| <= |f(z)-y_N| + |y_N-f(x)| < 1/N + 1/N < e Hence f is continuous at x. So C is precisely the set of points at which f is continuous. Each U_n is open since it is a union of open sets and so C is a G_delta set as required. If Q were a G_delta set, R-Q would be an F_sigma set, so there would be a sequence of closed sets F_0, F_1, ... with R-Q = UNION {F_n | n in N} Each F_i contains no rational and so is nowhere dense. Let (q_n) be an enumeration of Q. Each {q_n} is also a nowhere dense set. Then R = F_0 U F_1 U ... U q_0 U q_1 U ... is a countable union of nowhere dense sets, which contradicts BaireÕs category theorem. Hence Q is not a G_delta set. === Subject: Re: need proof!!! > i resently asked the following question > IÕm looking for a continuous function f:R->R with discontinuity on > irrational domain and continuous on Q. > the good people who responded, told me that the answer is NO. > now, does anyone knows a formal proof for this... > but if you dont know just tell me why Why do you ask? When youÕve asked your previous question, Dennis May discontinuous must be an F_sigma set. The irrationals are not an F_sigma set. If thatÕs not a formal proof, then I have no clue about what a formal proof is. Of course, if you had written that you do not understand it, that would have been different. But itÕs a proof. And, please, stop putting those exclamation marks at the subjects of your posts. Jose Carlos Santos === Subject: Re: need proof!!! <318crhF35divaU1@individual.net IÕm looking for a continuous function f:R->R with discontinuity on > irrational domain and continuous on Q. > the good people who responded, told me that the answer is NO. > now, does anyone knows a formal proof for this... > There is no such function. The set of points at which f is > discontinuous must be an F_sigma set. The irrationals are not an F_sigma > set. If thatÕs not a formal proof, then I have no clue about what a > formal proof is. Of course, if you had written that you do not > understand it, that would have been different. But itÕs a proof. The maverick may want a proof that RQ isnÕt F_sigma, and that the points of continuity of a function into a metric space are G_delta. That RQ isnÕt F_sigma can be proved by the fact that if it were F_sigma, then R would be sparse or meager which it isnÕt by BaireÕs theorem for example. === Subject: Re: need proof!!! > IÕm looking for a continuous function f:R->R with discontinuity on > irrational domain and continuous on Q. > the good people who responded, told me that the answer is NO. > now, does anyone knows a formal proof for this... > There is no such function. The set of points at which f is > discontinuous must be an F_sigma set. The irrationals are not an F_sigma > set. If thatÕs not a formal proof, then I have no clue about what a > formal proof is. Of course, if you had written that you do not > understand it, that would have been different. But itÕs a proof. > The maverick may want a proof that RQ isnÕt F_sigma, and that the points > of continuity of a function into a metric space are G_delta. > That RQ isnÕt F_sigma can be proved by the fact that if it were > F_sigma, then R would be sparse or meager which it isnÕt by BaireÕs > theorem for example. can you explain me what is F_sigma set and why Q isnt a F_sigma set? === Subject: Re: need proof!!! IÕm looking for a continuous function f:R->R with discontinuity on > irrational domain and continuous on Q. > the good people who responded, told me that the answer is NO. > now, does anyone knows a formal proof for this... There is no such function. The set of points at which f is > discontinuous must be an F_sigma set. The irrationals are not an F_sigma > set. The maverick may want a proof that RQ isnÕt F_sigma, and that the points > of continuity of a function into a metric space are G_delta. > That RQ isnÕt F_sigma can be proved by the fact that if it were > F_sigma, then R would be sparse or meager which it isnÕt by BaireÕs > theorem for example. > can you explain me what is F_sigma set and why Q isnt a F_sigma set? A set is F_sigma when itÕs a countable union of closed sets. Exercise: Q is F_sigma. === Subject: Re: need proof!!! > i resently asked the following question > IÕm looking for a continuous function f:R->R with discontinuity on > irrational domain and continuous on Q. > the good people who responded, told me that the answer is NO. > now, does anyone knows a formal proof for this... > but if you dont know just tell me why Q is a subset of R, and so is the irrational numbers R - Q. If you have discontinuity on R - Q, then you have discontinuity on a subset of R, and hence on R as a whole. Therefore, no such function exists. === Subject: Re: need proof!!! > i resently asked the following question > IÕm looking for a continuous function f:R->R with discontinuity on > irrational domain and continuous on Q. > the good people who responded, told me that the answer is NO. > now, does anyone knows a formal proof for this... > but if you dont know just tell me why > Q is a subset of R, and so is the irrational numbers R - Q. If you > have discontinuity on R - Q, then you have discontinuity on a > subset of R, and hence on R as a whole. Therefore, no such > function exists. ok, so why is the oppside function exist??? if the discontinuity is on Q which is also a subset of R??? === Subject: Re: need proof!!! > ok, so why is the oppside function exist??? A partial answer to some of your questions: The set R-Q is a G_delta, i.e. a countable intersection of open subsets of R. This is because R-Q= cap_{qin Q} (R-{q}). Here the complement of a singleton is obviously open, and as Q is countable, so is this intersection. Obviously this generalizes to the statement that the complement of a countable set is a G_delta. Q is not a G_delta as explained by William Elliott (the complement of a G_delta is an F_sigma, i.e. a countable union of closed sets). I donÕt know, whether being a G_delta is a sufficient condition for a set to be equal to the points of contuinity of some function. It is relatively easy to see that it is a necessary condition, though. This is an exercise (together with some hints) in RoydenÕs book Real Analysis. IÕm sure many other first year graduate texts in real analysis have related material as well. Jyrki Lahtonen, Turku, Finland === Subject: Re: need proof!!! > ok, so why is the oppside function exist??? > A partial answer to some of your questions: > The set R-Q is a G_delta, i.e. a countable intersection > of open subsets of R. This is because > R-Q= cap_{qin Q} (R-{q}). > Here the complement of a singleton is obviously open, > and as Q is countable, so is this intersection. Obviously > this generalizes to the statement that the complement > of a countable set is a G_delta. > Q is not a G_delta as explained by William Elliott > (the complement of a G_delta is an F_sigma, i.e. a countable > union of closed sets). > I donÕt know, whether being a G_delta is a sufficient > condition for a set to be equal to the points of contuinity > of some function. It is relatively easy to see that it is > a necessary condition, though. This is an exercise (together > with some hints) in RoydenÕs book Real Analysis. IÕm sure > many other first year graduate texts in real analysis have > related material as well. > Jyrki Lahtonen, Turku, Finland but i have one more problem you are based on the fact that The set of points at which f is discontinuous must be an F_sigma set and thats o.k but i only know that: a function f(x) in a single variable x is said to be continuous at point y if lim x-->y f(x) = f(y) can you proof basing on the lim of a function????? === Subject: Re: need proof!!! > but i have one more problem > you are based on the fact that The set of points at which f is > discontinuous must be an F_sigma set and thats o.k > but i only know that: > a function f(x) in a single variable x is said to be continuous at point y if > lim x-->y f(x) = f(y) > can you proof basing on the lim of a function????? I did this as an exercise as a first year graduate student. It was good for me to work it out myself. IÕm sure itÕs good for you as well. If you get stuck, look at Dennis MayÕs reply:) Jyrki === Subject: Re: need proof!!! > I donÕt know, whether being a G_delta is a sufficient > condition for a set to be equal to the points of contuinity > of some function. It is relatively easy to see that it is > a necessary condition, though. This is an exercise (together > with some hints) in RoydenÕs book Real Analysis. IÕm sure > many other first year graduate texts in real analysis have > related material as well. It is also sufficient. The book Counterexamples in Analysis shows a construction of a function which is discontinuous at precisely the points of an arbitrary F_sigma set. === Subject: Re: need proof!!! >i resently asked the following question >IÕm looking for a continuous function f:R->R with discontinuity on >irrational domain and continuous on Q. >the good people who responded, told me that the answer is NO. >now, does anyone knows a formal proof for this... >but if you dont know just tell me why >>Q is a subset of R, and so is the irrational numbers R - Q. If you >>have discontinuity on R - Q, then you have discontinuity on a >>subset of R, and hence on R as a whole. Therefore, no such >>function exists. > ok, so why is the oppside function exist??? > if the discontinuity is on Q which is also a subset of R??? instead of posting an answer to what you meant to write. Your problem is function which was *discountinuous* at some points (namely, at the irrational numbers). Obviously, no such function exists, by the reasons stated by mike3. What you meant to write was: IÕm looking for a function f:R->R which is discontinuous on every irrational point and continuous on every point of Q. Jose Carlos Santos === Subject: Re: need proof!!! >i resently asked the following question >IÕm looking for a continuous function f:R->R with discontinuity on >irrational domain and continuous on Q. >the good people who responded, told me that the answer is NO. >now, does anyone knows a formal proof for this... >but if you dont know just tell me why >Q is a subset of R, and so is the irrational numbers R - Q. If you >>have discontinuity on R - Q, then you have discontinuity on a >>subset of R, and hence on R as a whole. Therefore, no such >>function exists. > ok, so why is the oppside function exist??? > if the discontinuity is on Q which is also a subset of R??? > instead of posting an answer to what you meant to write. Your problem is > function which was *discountinuous* at some points (namely, at the > irrational numbers). Obviously, no such function exists, by the reasons > stated by mike3. What you meant to write was: > IÕm looking for a function f:R->R which is discontinuous on every > irrational point and continuous on every point of Q. > Jose Carlos Santos Though the reverse is quite possible, continuous at irrational points and discontinuous at rational ones. === Subject: Re: Turing machines posting-account=sASPfg0AAAB32ck2Ys0CgbD_dk7GKPYH There is no such thing as an ultimate model of symbol manipulation. There are many equivalent models, and TM is but one of them. The only special thing is that it is generally thought to be the first complete model written in a mathematical paper. A TM is nothing but a computer specification that runs with a very very low level machine code. ItÕs not any more fundamental than lambda calculus or cellular automata. (to a computer scientist!) What it does capture is merely the notion of a causal graph, that univerally characterizes *any* discrete mechanism in the world, e.g. a machine that works on blocks of things. ItÕs up to you to call these symbols. Is the state of a transistor, or a wire in your VLSI chip, or a single These are philosophical questions of course, but the proper approach, in my opinion, is to see these as mere marks, that may or may not be conceived as symbols (which possibly refer to other things, etc.) They are just states, chosen from a finite alphabet like the alphabet of a TM. -- Eray Ozkural === Subject: Re: Turing machines > There is no such thing as an ultimate model of symbol manipulation. > There are many equivalent models, and TM is but one of them. The only > special thing is that it is generally thought to be the first complete > model written in a mathematical paper. > A TM is nothing but a computer specification that runs with a very very > low level machine code. ItÕs not any more fundamental than lambda > calculus or cellular automata. (to a computer scientist!) > What it does capture is merely the notion of a causal graph, that > univerally characterizes *any* discrete mechanism in the world, e.g. a > machine that works on blocks of things. ItÕs up to you to call these > symbols. > Is the state of a transistor, or a wire in your VLSI chip, or a single > These are philosophical questions of course, but the proper approach, > in my opinion, is to see these as mere marks, that may or may not be > conceived as symbols (which possibly refer to other things, etc.) They > are just states, chosen from a finite alphabet like the alphabet of a > TM. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Turing machines > There is no such thing as an ultimate model of symbol manipulation. > There are many equivalent models, and TM is but one of them. The only > special thing is that it is generally thought to be the first complete > model written in a mathematical paper. > A TM is nothing but a computer specification that runs with a very very > low level machine code. ItÕs not any more fundamental than lambda > calculus or cellular automata. (to a computer scientist!) > What it does capture is merely the notion of a causal graph, that > univerally characterizes *any* discrete mechanism in the world, e.g. a > machine that works on blocks of things. ItÕs up to you to call these > symbols. > Is the state of a transistor, or a wire in your VLSI chip, or a single > These are philosophical questions of course, but the proper approach, > in my opinion, is to see these as mere marks, that may or may not be > conceived as symbols (which possibly refer to other things, etc.) They > are just states, chosen from a finite alphabet like the alphabet of a > TM. which is the case for physical computers. ItÕs trivial to see. We have something called network. We can attach computers, and we can indefinitely, we can do it. The silly considerations of OS address space etc. are irrelevant) -- Eray Ozkural === Subject: SmullyanÕs Quiz Problem Raymond Smullyan presents the following (paraphrased) riddle in his book, Forever Undecided: On Monday a professor says to his class, I will give you a surprise examination some day this week. You will not know that there is an examination when the class begins. A student then reasoned, I canÕt get the quiz on Friday because if I havenÕt gotten it by Friday I will know the quiz must be that day. Similarly he reasoned for Thursday all the way to Monday. Where is the error in his logic? My question is this: Why is there any inconsistency in the professor giving the quiz to the class on Monday? Now if the professor had stated to his class on Friday, I will give you a surprise examiniation some day next week, then there would be inconsistency problems for all days of the week, but thatÕs not the way the problem is stated. Comments anyone? -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124 === Subject: Re: SmullyanÕs Quiz Problem > Raymond Smullyan presents the following (paraphrased) riddle in his > book, Forever Undecided: > On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > canÕt get the quiz on Friday because if I havenÕt gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? > My question is this: Why is there any inconsistency in the professor > giving the quiz to the class on Monday? > Now if the professor had stated to his class on Friday, I will give > you a surprise examiniation some day next week, then there would be > inconsistency problems for all days of the week, but thatÕs not the > way the problem is stated. > Comments anyone? Quine has quite a good discussion of this paradox in On A Supposed Antinomy in The Ways of Paradox. The student needs to allow for the logical possibility that the exam will be on Friday but he will not know this fact on Thursday. === Subject: Re: SmullyanÕs Quiz Problem days. My association with the Department is that of an alumnus. >Raymond Smullyan presents the following (paraphrased) riddle in his >book, Forever Undecided: > On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > canÕt get the quiz on Friday because if I havenÕt gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? >My question is this: Why is there any inconsistency in the professor >giving the quiz to the class on Monday? One of the hypothesis is that an examination will be given during the week. The argument shows that the hypothesis lead to a contradiction: that no examination will be given during the week. That is, H->not(H). But (H -> not(H)) -> not(H) is a tautology. That means that the only thing we can deduce from the professorÕs statements are that his conditions will not be met. Since we can only conclude that his conditions will not be met, none of the original argument showing that no test can occur is valid. The test may occur at any time. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: SmullyanÕs Quiz Problem >Raymond Smullyan presents the following (paraphrased) riddle in his >book, Forever Undecided: > On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > canÕt get the quiz on Friday because if I havenÕt gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? >My question is this: Why is there any inconsistency in the professor >giving the quiz to the class on Monday? > One of the hypothesis is that an examination will be given during the > week. The argument shows that the hypothesis lead to a contradiction: > that no examination will be given during the week. That is, > H->not(H). But > (H -> not(H)) -> not(H) > is a tautology. That means that the only thing we can deduce from the > professorÕs statements are that his conditions will not be met. > Since we can only conclude that his conditions will not be met, none > of the original argument showing that no test can occur is valid. The > test may occur at any time. I agree with your analysis of the purely logical version of the paradox. The professor can be said to be asserting a contradiction. Thus, either the professor must be considered fallible, or we have a truth teller asserting a contradiction. In this case we can prove anything, (including the facts that a test on Wednesday is both expected and unexpected). However, I think that the paradox is much deeper than this. Consider four statements the professor could make: i) there will be an unexpected test tomorrow ii) there will be an unxepected test in the next three days iii) there will be an unexpected test next week iv) there will be an unexpected test this semester All four have the same logical structure. However, i) seems absurd, ii) seems questionable, iii) seems fine, iv) is completely unremarkable. A full resoution of the paradox must explain the difference. Other versions of the paradox which do not involve prediction (e.g. the unexpected egg) seem very difficult to resolve. As I have noted elsethread, I think the problem is the deep seated conviction that a person must either have a rational basis for belief or not have a rational basis for belief. However, if you can arrange a situation where a person has a rational basis for belief, iff she does not have a rational basis for belief, this deep seated conviction is seem to be wrong. My analysis is that the unexpected egg paradox produces exactly such a situation. -William Hughes === Subject: Re: SmullyanÕs Quiz Problem days. My association with the Department is that of an alumnus. [.snip.] >> One of the hypothesis is that an examination will be given during the >> week. The argument shows that the hypothesis lead to a contradiction: >> that no examination will be given during the week. That is, >> H->not(H). But >> (H -> not(H)) -> not(H) >> is a tautology. That means that the only thing we can deduce from the >> professorÕs statements are that his conditions will not be met. >> Since we can only conclude that his conditions will not be met, none >> of the original argument showing that no test can occur is valid. The >> test may occur at any time. >I agree with your analysis of the purely logical version of >the paradox. The professor can be said to be asserting a contradiction. >Thus, either the professor must be considered fallible, or >we have a truth teller asserting a contradiction. In this case >we can prove anything, (including the facts that a test on Wednesday >is both expected and unexpected). >However, I think that the paradox is much deeper than this. Consider >four statements the professor could make: > i) there will be an unexpected test tomorrow > ii) there will be an unxepected test in the next three days > iii) there will be an unexpected test next week > iv) there will be an unexpected test this semester >All four have the same logical structure. However, i) seems absurd, >ii) seems questionable, iii) seems fine, iv) is completely unremarkable. >A full resoution of the paradox must explain the difference. The paradox here is literally that they contradict our intuition. But the only explanation needed is that ->our intuition is wrong<-. Just because some of them seem absurd and others do not does not mean that the statements actually are or are not wrong. These statements have a whole bunch of unstated assumptions. For example, I disagree with you that (i) seems absurd prima facie. It only becomes absurd if you assume that you know what time and what context such a test would take place in. The reason (iv) do not seem so absurd is that the latitude for those unknowns seems so much wider that you cannot lull yourself into a false sense of knowledge about the statement. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: SmullyanÕs Quiz Problem > [.snip.] >> One of the hypothesis is that an examination will be given during the >> week. The argument shows that the hypothesis lead to a contradiction: >> that no examination will be given during the week. That is, >> H->not(H). But > (H -> not(H)) -> not(H) > is a tautology. That means that the only thing we can deduce from the >> professorÕs statements are that his conditions will not be met. > Since we can only conclude that his conditions will not be met, none >> of the original argument showing that no test can occur is valid. The >> test may occur at any time. >>I agree with your analysis of the purely logical version of >the paradox. The professor can be said to be asserting a contradiction. >Thus, either the professor must be considered fallible, or >we have a truth teller asserting a contradiction. In this case >we can prove anything, (including the facts that a test on Wednesday >is both expected and unexpected). >However, I think that the paradox is much deeper than this. Consider >four statements the professor could make: > i) there will be an unexpected test tomorrow > ii) there will be an unxepected test in the next three days > iii) there will be an unexpected test next week > iv) there will be an unexpected test this semester >All four have the same logical structure. However, i) seems absurd, >ii) seems questionable, iii) seems fine, iv) is completely unremarkable. >A full resoution of the paradox must explain the difference. > The paradox here is literally that they contradict our intuition. Agreed, the arguments are now informal, so paradox has exactly this meaning. > But the only explanation needed is that ->our intuition is wrong<-. True, but the bald statement our intuintion is wrong is not very helpful. A full resolution of the paradox requires an explanation as to why our intuition is wrong >Just > because some of them seem absurd and others do not does not mean > that the statements actually are or are not wrong. > These statements have a whole bunch of unstated assumptions. For > example, I disagree with you that (i) seems absurd prima facie. It > only becomes absurd if you assume that you know what time and what > context such a test would take place in. Yes, there are ways to make (i) sensible (e.g. An unxepected exam is one on a different colour of paper), however, these do not seem sensible to me. If you take (i) to mean (and this seems to me the most obvious meaning) You will have an exam tomorrow but you donÕt know this, (i) is BertramÕs paradox. > The reason (iv) do not seem > so absurd is that the latitude for those unknowns seems so much wider > that you cannot lull yourself into a false sense of knowledge about > the statement. Indeed. However, both the professor and students agree that information has been communicated, and appear to agree on what has been communicated. So the questionÕs are: Do the two parties actually agree on what has been communicated? and If the professorÕs statement is self contradictory, what should he have said? -William Hughes === Subject: Re: SmullyanÕs Quiz Problem days. My association with the Department is that of an alumnus. [.snip.] >>I agree with your analysis of the purely logical version of >>the paradox. The professor can be said to be asserting a contradiction. >>Thus, either the professor must be considered fallible, or >>we have a truth teller asserting a contradiction. In this case >>we can prove anything, (including the facts that a test on Wednesday >>is both expected and unexpected). >>However, I think that the paradox is much deeper than this. Consider >>four statements the professor could make: >> i) there will be an unexpected test tomorrow >> ii) there will be an unxepected test in the next three days >> iii) there will be an unexpected test next week >> iv) there will be an unexpected test this semester >>All four have the same logical structure. However, i) seems absurd, >>ii) seems questionable, iii) seems fine, iv) is completely unremarkable. >>A full resoution of the paradox must explain the difference. >> The paradox here is literally that they contradict our intuition. >Agreed, the arguments are now informal, so paradox has >exactly this meaning. It should also be added that there is a fair amount of the Wilde-like paradox in this; you know, the The only thing worse than being talked about is not being talked about kind of statements, since we argue that if we expect the test then we donÕt expect the test (the old I got you by not getting you argument and variants thereof). >> But the only explanation needed is that ->our intuition is wrong<-. >True, but the bald statement our intuintion is wrong is not >very helpful. A full resolution of the paradox requires an >explanation as to why our intuition is wrong I think out intuition is wrong because we are making a lot of unstated assumptions about these statements. As I noted: there will be an unexpected test tomorrow seems absurd when we assume that ŌunexpectedÕ means something like ŌI wonÕt know exactly whenÕ, but that knowing the ->day<- of the exam narrows the window sufficiently for us to figure out exactly when. The other three statements donÕt lull us into that assumption because, prima facie, we have more than one option and no information about which option we should take. But that (i) if we know the day then we know when; and (ii) if we donÕt know which day in advance then we donÕt know exactly when; are common enough conclusions which are both unwarranted in this situation. >>Just >> because some of them seem absurd and others do not does not mean >> that the statements actually are or are not wrong. >> These statements have a whole bunch of unstated assumptions. For >> example, I disagree with you that (i) seems absurd prima facie. It >> only becomes absurd if you assume that you know what time and what >> context such a test would take place in. >Yes, there are ways to make (i) sensible (e.g. An unxepected exam >is one on a different colour of paper), however, these do not >seem sensible to me. IÕm not even saying that. You can even make it sensible if unexpected refers to you wonÕt know exactly when; I teach at three different hours; which hour will contain the exam? > If you take (i) to mean (and this seems >to me the most obvious meaning) You will have an exam tomorrow but >you donÕt know this, (i) is BertramÕs paradox. And here we are making another assumption about what unexpected means. Does it mean, you do not expect an exam at all? Or does it mean, you wonÕt know exactly when? It is in the latter sense that the Paradox of the Unexpected Hanging uses it, for example. >> The reason (iv) do not seem >> so absurd is that the latitude for those unknowns seems so much wider >> that you cannot lull yourself into a false sense of knowledge about >> the statement. >Indeed. However, both the professor and students agree >that information has been communicated, and appear to agree on >what has been communicated. Which, as we know, in the real world is a virtual guarantee that it is not true that what the professor meant to communicate is the communication that the students have received... (-: > So the questionÕs are: Do the two >parties actually agree on what has been communicated? and >If the professorÕs statement is self contradictory, what should >he have said? Do you mean, how can he make a statement which communicates the same information but is not self-contradictory? You may be begging the question there, as in fact the point is that you cannot communicate that information without also communicating a contradiction. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: SmullyanÕs Quiz Problem > [.snip.] >>I agree with your analysis of the purely logical version of >>the paradox. The professor can be said to be asserting a contradiction. >>Thus, either the professor must be considered fallible, or >>we have a truth teller asserting a contradiction. In this case >>we can prove anything, (including the facts that a test on Wednesday >>is both expected and unexpected). >>However, I think that the paradox is much deeper than this. Consider >>four statements the professor could make: >> i) there will be an unexpected test tomorrow >> ii) there will be an unxepected test in the next three days >> iii) there will be an unexpected test next week >> iv) there will be an unexpected test this semester >>All four have the same logical structure. However, i) seems absurd, >>ii) seems questionable, iii) seems fine, iv) is completely unremarkable. >>A full resoution of the paradox must explain the difference. > The paradox here is literally that they contradict our intuition. >Agreed, the arguments are now informal, so paradox has >exactly this meaning. > It should also be added that there is a fair amount of the > Wilde-like paradox in this; you know, the The only thing worse than > being talked about is not being talked about kind of statements, > since we argue that if we expect the test then we donÕt expect the > test (the old I got you by not getting you argument and variants > thereof). >> But the only explanation needed is that ->our intuition is wrong<-. >True, but the bald statement our intuintion is wrong is not >very helpful. A full resolution of the paradox requires an >explanation as to why our intuition is wrong > I think out intuition is wrong because we are making a lot of > unstated assumptions about these statements. As I noted: there will > be an unexpected test tomorrow seems absurd when we assume that > ŌunexpectedÕ means something like ŌI wonÕt know exactly whenÕ, but > that knowing the ->day<- of the exam narrows the window sufficiently > for us to figure out exactly when. The other three statements donÕt > lull us into that assumption because, prima facie, we have more than > one option and no information about which option we should take. But > that (i) if we know the day then we know when; and (ii) if we donÕt > know which day in advance then we donÕt know exactly when; are common > enough conclusions which are both unwarranted in this situation. You appear to be arguing that all four statements are equally contradictory, but that the latter statements donÕt seem to be as absurd because of the longer time window. While this is a reasonable explanation of why (iv) is intuitively less problematic than (i), it leaves out an important fact. A professor telling his students (i) imparts no information. A professor telling his students (iv) imparts information (or at least appears to). So: -what information does the professor impart? -is there a more accurate statement that the professor could use, or is (iv) the best he can do? >>Just >> because some of them seem absurd and others do not does not mean >> that the statements actually are or are not wrong. > These statements have a whole bunch of unstated assumptions. For >> example, I disagree with you that (i) seems absurd prima facie. It >> only becomes absurd if you assume that you know what time and what >> context such a test would take place in. >Yes, there are ways to make (i) sensible (e.g. An unxepected exam >is one on a different colour of paper), however, these do not >seem sensible to me. > IÕm not even saying that. You can even make it sensible if > unexpected refers to you wonÕt know exactly when; I teach at three > different hours; which hour will contain the exam? > If you take (i) to mean (and this seems >to me the most obvious meaning) You will have an exam tomorrow but >you donÕt know this, (i) is BertramÕs paradox. > And here we are making another assumption about what unexpected > means. Does it mean, you do not expect an exam at all? Or does it > mean, you wonÕt know exactly when? It is in the latter sense that the > Paradox of the Unexpected Hanging uses it, for example. The OP quoted Smullyan definiton of a surprize examination as You will not know that there is an examination when the class begins. I am using unexpected to mean the same as SmullyanÕs surprize. >> The reason (iv) do not seem >> so absurd is that the latitude for those unknowns seems so much wider >> that you cannot lull yourself into a false sense of knowledge about >> the statement. >>Indeed. However, both the professor and students agree >that information has been communicated, and appear to agree on >what has been communicated. > Which, as we know, in the real world is a virtual guarantee that it > is not true that what the professor meant to communicate is the > communication that the students have received... (-: > So the questionÕs are: Do the two >parties actually agree on what has been communicated? and >If the professorÕs statement is self contradictory, what should >he have said? > Do you mean, how can he make a statement which communicates the same > information but is not self-contradictory? You may be begging the > question there, as in fact the point is that you cannot communicate > that information without also communicating a contradiction. If we decide that the professor is trying to communicate the fact that there will with very high (but less than 100%) probability be an unexpected examination, then this information can be communicated without also communicating a contradiction. If we decide that the professor is trying to communicate the fact that there is a 100% probability of an unexpected examination (i.e. that he has scheduled an examination in such a way that it will be unexpected) then he cannot communicate this fact without communicating a contradiction. But, it may not be possible to communicate a contradiction. The students may conclude from the professorÕs attempt to communicate a contradiction that there will with very high (but less than 100%) probability be an unexpected examination. So it may not be possible for the professor to communicate what he wants to communicate, but be possible for him to communicate the information the students obtain without communicating a contradiction. -William Hughes === Subject: Re: SmullyanÕs Quiz Problem >> [.snip.] >I agree with your analysis of the purely logical version of >the paradox. The professor can be said to be asserting a contradiction. >Thus, either the professor must be considered fallible, or >we have a truth teller asserting a contradiction. In this case >we can prove anything, (including the facts that a test on Wednesday >is both expected and unexpected). However, I think that the paradox is much deeper than this. Consider >four statements the professor could make: i) there will be an unexpected test tomorrow > ii) there will be an unxepected test in the next three days > iii) there will be an unexpected test next week > iv) there will be an unexpected test this semester All four have the same logical structure. However, i) seems absurd, >ii) seems questionable, iii) seems fine, iv) is completely unremarkable. >A full resoution of the paradox must explain the difference. > > The paradox here is literally that they contradict our intuition. >>Agreed, the arguments are now informal, so paradox has >>exactly this meaning. >> It should also be added that there is a fair amount of the >> Wilde-like paradox in this; you know, the The only thing worse than >> being talked about is not being talked about kind of statements, >> since we argue that if we expect the test then we donÕt expect the >> test (the old I got you by not getting you argument and variants >> thereof). > But the only explanation needed is that ->our intuition is wrong<-. >>True, but the bald statement our intuintion is wrong is not >>very helpful. A full resolution of the paradox requires an >>explanation as to why our intuition is wrong >> I think out intuition is wrong because we are making a lot of >> unstated assumptions about these statements. As I noted: there will >> be an unexpected test tomorrow seems absurd when we assume that >> ŌunexpectedÕ means something like ŌI wonÕt know exactly whenÕ, but >> that knowing the ->day<- of the exam narrows the window sufficiently >> for us to figure out exactly when. The other three statements donÕt >> lull us into that assumption because, prima facie, we have more than >> one option and no information about which option we should take. But >> that (i) if we know the day then we know when; and (ii) if we donÕt >> know which day in advance then we donÕt know exactly when; are common >> enough conclusions which are both unwarranted in this situation. >You appear to be arguing that all four statements are equally >contradictory, but that the latter statements donÕt seem to >be as absurd because of the longer time window. While this >is a reasonable explanation of why (iv) is intuitively less >problematic than (i), it leaves out an important fact. A professor >telling his students (i) imparts no information. A professor telling >his students (iv) imparts information (or at least appears to). And IÕm saying: no. All four statements result, if taken to their logical analysis, in the same amount of (correct) information being given to the students: that the professorÕs statement cannot be taken at face value, and therefore, that no information ->about the test<- has in fact been given. You are simply in error in thinking that (iv) is different from (i), because (iv) imparts as little information as (i) does, ->IF WE ANALYZE THEM AS LOGICAL STATEMENTS<-. If we think of them as informal, colloquial, statements, then of course all bets are off. > So: > -what information does the professor impart? The professor has only succeeded in imparting the information that we cannot assume his statemetn true. > -is there a more accurate statement that the professor > could use, or is (iv) the best he can do? Accurate? What is he trying to say in the first place? You seem to be having the unwarranted and UNSPOKEN assumption that the professor has some actual piece of information that he is attempting to convey. I have made no such assumption, and I do not see why you are making such an assumption. What is it that you think the professor is actually trying to convey, exactly? [.snip.] >> So the questionÕs are: Do the two >>parties actually agree on what has been communicated? and >>If the professorÕs statement is self contradictory, what should >>he have said? >> Do you mean, how can he make a statement which communicates the same >> information but is not self-contradictory? You may be begging the >> question there, as in fact the point is that you cannot communicate >> that information without also communicating a contradiction. >If we decide that the professor is trying >to communicate the fact that there will with very high >(but less than 100%) probability be an unexpected examination, >then this information can be communicated without also communicating a >contradiction. I donÕt see how. What makes you so sure that this information can accurately be communicated? There are statements that simply cannot be communicated without also setting up a contradiction. Surprise, since it hinges on lack of information, is rife with them, since communicating involves an exchange of information. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu