mm-1060 Subject: Re: The origin of quaternions - Hamiltonıs 1844 paper big snip In hindsight one may guess that Hamiltonıs 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 donıt 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 deŜned 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 Hamiltonıs 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 ŝow 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 OıDonnell: 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 Hamiltonıs >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 - Hamiltonıs 1844 paper > big snip In hindsight one may guess that Hamiltonıs 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 donıt 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 deŜned 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 Hamiltonıs 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 ŝow 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 OıDonnell: 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 Hamiltonıs >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 itıs sufŜcient to Ŝnd 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 Ŝnitely-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 Ŝve 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] deŜned 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 Ŝxed by phi. I would have told you it was easy to compute this, except that in a thread several weeks ago I realized I didnıt 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 Ŝnd phi(Z1)=Z1 + 3u, phi(Y1) = Y1 + 3v ; so itıs not too hard to use Z1 and Y1 to cancel out any additional u, v terms. For example, phi(Z2) = Z2 + 2u Z1 + 3 u^2 and so we compute that Z2ı = Z2 - Z1^2/3 is phi-invariant. Likewise Z3ı = Z3 - Z1^2*Z2 + (2/9)*Z1^3 and similar expressions for the Yıs. Therefore our invariant subring R may be written F[ Z1, Y1, Z2ı, Y2ı, Z3ı, Y3ı ] on which it seems pretty clear that the portion preserved by phi is just F[ Z2ı, Y2ı, Z3ı, Y3ı ]. Likewise we may replace our module generators by phi-invariant ones: W1ı = W1 - Z1*Y1/3, W2ı = W2 - (Y1*Z2/3)-2*(Z1*W1)/3 + 2*(Y1*Z1^2)/9, W3ı = 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, Z2ı, Y2ı, Z3ı, Y3ı] spanned by 1, W1ı, W2ı, W3ı, W4ı = (W1ı)^2, and W5=(W2ı)(W3ı); the action of phi is easily given since all but two generators are phi-invariant. Itıs obvious (isnıt 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 Ŝxed 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 Ŝnd another Ŝve 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 donıt expect most of them to be _rotation_ - invariant.) You said something about taking cross products but Iım going to ignore that because if you REALLY meant to do that, then Iıd 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, itıs going to cost you... dave Here are the Ŝve 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 youıre 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 doesnıt 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), deŜned such that mag_n(z)^n + mag_n(z) = z? (mag means magic)? I know you can express the quinticıs solution in terms of mag_5(z), but could you express the 6th degree equationıs solution using mag_5(z) and mag_6(z)? -- We should have a town named Alderaan someday. No, seriously. Letıs 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), deŜned such that > mag_n(z)^n + mag_n(z) = z? (mag means magic)? > I know you can express the quinticıs solution in terms of > mag_5(z), but could you express the 6th degree > equationıs solution using mag_5(z) and mag_6(z)? You canıt. 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), deŜned such that > mag_n(z)^n + mag_n(z) = z? (mag means magic)? > I know you can express the quinticıs solution in terms of > mag_5(z), but could you express the 6th degree > equationıs solution using mag_5(z) and mag_6(z)? > You canıt. 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), deŜned such that >mag_n(z)^n + mag_n(z) = z? (mag means magic)? >I know you can express the quinticıs solution in terms of >mag_5(z), but could you express the 6th degree >equationıs solution using mag_5(z) and mag_6(z)? >>You canıt. 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 dıHermite, 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), deŜned such that >mag_n(z)^n + mag_n(z) = z? (mag means magic)? I know you can express the quinticıs solution in terms of >mag_5(z), but could you express the 6th degree >equationıs solution using mag_5(z) and mag_6(z)? >>You canıt. 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 > dıHermite, 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 >>dıHermite, 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 > dıHermite, 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), deŜned such that > mag_n(z)^n + mag_n(z) = z? (mag means magic)? I know you can express the quinticıs solution in terms of > mag_5(z), but could you express the 6th degree > equationıs solution using mag_5(z) and mag_6(z)? > You canıt. 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 Kingıs _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 canıt 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 canıt > 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 youıre 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 youıre 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 deŜnes the Laplace Transform as an integral from -inŜnity to inŜnity, instead of the more usual 0 to inŜnity? >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 deŜnes the Laplace Transform as an integral from -inŜnity > to inŜnity, instead of the more usual 0 to inŜnity? >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 yearsı standing? > C.L. Phillips, J.M. Parr, Signals, Systems & Transforms, 2nd ed., > Prentice > Hall, Copyright 1999, p. 46-50, 210-211 > really deŜnes the Laplace Transform as an integral from -inŜnity > to inŜnity, instead of the more usual 0 to inŜnity? === 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 yearsı standing? I donıt think I BOASTED of that, but yes, I mentioned it. I have enough experience to know that something Iıve 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 deŜnes the Laplace Transform as an integral from -inŜnity >> to inŜnity, instead of the more usual 0 to inŜnity? ************************ David C. Ullrich === Subject: Re: JSH: Simple proof X-RFC2646: Format=Flowed; Original > 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. > Itıs 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, itıs x = +1 and x = -3). > The name x is commonly called a variable, but in this case, > itıs 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 > Letıs 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 deŜnition 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 havenıt 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 Deckerıs 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 Deckerıs 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 couldnıt see how to Ŝx 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 journalıs website with no public explanation. This of course prompted howls of righteous indignation from Harris (with some reason), and he began defending the paperıs 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 deŜnition of constant term. Yesterday he started a related thread called ŒFinal examı. 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 donıt 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 havenıt 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 Deckerıs > example, but refusing to give the punch line, which was a factorization > of a kind that he had said was impossible. Which, Iım sure you noticed, I recently more or less reposted in Jamesıs Point of logic thread (November 11). I didnıt 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 beneŜts of those who are new to the Harris Magnum Opus. Rick === Subject: Re: JSH: Simple proof > Why donıt you simply and brieŝy 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 coefŜcients. 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 coefŜcients 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 Ŝnd that if the uıs 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. Itıs not complicated. Basically you canıt 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?<< Iım not discussing that, other posters made a big deal out of it. > here? The notion of a polynomial is deŜned since a long time and > can be found in every introductory book on algebra. So? > If we consistently use the common deŜnitions 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. Iıve seen posters come and go, and every once in a while thereıs a poster like you who claims to care about working things out. When it turns out that Iım 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. Iıve seen it lots of times. Do yourself a favor, and just walk away now. James Harris === Subject: Re: JSH: Simple proof >>Why donıt you simply and brieŝy 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 coefŜcients. Thatıs the deŜnition. It is not a property of a ring. What is the problem with the ring? What prevents it from being everything thatıs 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 coefŜcients 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 deŜned. 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 isnıt a unit, then its multiplicative inverse fails to be in the ring. Thereıs no ambiguity here. How about providing a deŜnition 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 Ŝnd that if the uıs 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? Hereıs 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, letıs look at a speciŜc example: k = 5 25 - 12 = 13 u = (5 +/- sqrt(13))/6 Iıll let the root with + be u, and the root with - be ubar. Note that you canıt have *both* solutions in your ring: u ubar = (25 - 13)/36 = 12/36 = 1/3 Considering your recent rant about how you canıt get rid of the evil ambiguity of root extractions, how do you determine *which* of these roots is in your ring? Note that the Ŝelds 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 Iıve called u, or what Iıve called ubar? You canıt have both, and you canıt 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, youıll 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 Rı that contains R. It is also not difŜcult 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. > Itıs not complicated. Basically you canıt just rely on whether or not > some number is in the ring of algebraic integers when considering > factors of roots of a polynomial. Thatıs what you say. However, every speciŜc example youıve proposed that is supposed to highlight the ŝaws 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?<< > Iım not discussing that, other posters made a big deal out of it. >>here? The notion of a polynomial is deŜned since a long time and >>can be found in every introductory book on algebra. > So? >>If we consistently use the common deŜnitions 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. > Iıve seen posters come and go, and every once in a while thereıs a > poster like you who claims to care about working things out. > When it turns out that Iım right, you go over to the other side, and > either run away, or turn to bizarre behavior. It has not turned out youıre 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. > Iıve seen it lots of times. Do yourself a favor, and just walk away > now. I think youıre 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@ıELIi $t^ VcLWP@J5p^rst0+(Œ>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 coefŜcients >> 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 deŜned. > 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 isnıt a unit, then its multiplicative > inverse fails to be in the ring. > Thereıs no ambiguity here. > How about providing a deŜnition 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 youıd be considering 1/2 in the Ŝrst 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 donıt you simply and brieŝy 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 coefŜcients. > 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 coefŜcients 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 Ŝrst question irrelevant.) -- Come on people!!! The US just blew up a lot of people in Iraq, donıt you realize that a person with my exposure might just end up dead, by mysterious circumstances? --James Harris, on the dangers of proving Fermatıs last theorem === Subject: Re: JSH: Simple proof !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ıELIi $t^ VcLWP@J5p^rst0+(Œ>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Why donıt you simply and brieŝy 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 coefŜcients. Well, thatıs the deŜnition. Thatıs what makes them interesting in the Ŝrst place. If you donıt like the particular class of numbers deŜned by this property, you are free to deŜne a different class of numbers deŜned by different properties. You just canıt 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 speciŜed ring. > but because their multiplicative inverse is not the root of some > monic polynomial with integer coefŜcients 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. > Iıve 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 donıt you simply and brieŝy 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 coefŜcients. > 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 coefŜcients they are not > units in the ring of algebraic integers. Has it occurred to you that the problem might be with your undeŜned 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 Ŝnd that if the uıs 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 uıs. 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 ŝaw 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. > Itıs not complicated. Basically you canıt 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. Hereıs 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 X-RFC2646: Format=Flowed; Original > 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 coefŜcients 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 coefŜcients 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. Itıs sort of like complaining that thereıs no solution to 3x - 1 = 0 in the ring of integers and then going on to conclude that thereıs 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@ıELIi $t^ VcLWP@J5p^rst0+(Œ>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 coefŜcients 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. > Itıs sort of like complaining that thereıs no solution to 3x - 1 = 0 > in the ring of integers and then going on to conclude that thereıs a > fundamental problem with the integers. Oh, but there is. Thatıs why we have rationals in the Ŝrst 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 Ŝelds (uh, pun gone bad) interesting in the Ŝrst place. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Simple proof X-RFC2646: Format=Flowed; Original > 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. > Thatıs 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 speciŜc way. So for example: x^2 + 5x + 6 => (x+2)(x+3) Youıre 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 80ıs. 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. >> Iım 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. >> Iıve 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 youıre 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. Letıs 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 sufŜciently 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 sufŜciently > 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 itıs 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 Œdensityı function Iım 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 sufŜciently > 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 itıs 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 Œdensityı function Iım 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 sufŜciently > 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 itıs 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 Œdensityı function Iım 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 sufŜciently > 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 itıs 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 Œdensityı function Iım 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 youıre 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 youıre 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 itıs usually more accurate to say that the >sentences are understood as relevant implications. Theyıre not >really if and only if; theyıre just not truth functional at all. >> Sure, but thatıs on the natural language side. Suppose a person, >> during normal discourse, tells his son: If you eat your vegetables, >> Iıll give you some ice cream. Conversationally implied is the >> sentence If you donıt eat your vegetables, I wonıt give you ice >> cream. > No, I donıt really think it is. I think thatıs an oversimpliŜcation. > Whatıs understood conversationally is that thereıs 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 wouldnıt have put it as he did. But > if he caves in at the end and allows ice cream with no vegetables, > itıs not generally considered that heıs been inaccurate. > These intensional relationships are much more difŜcult to > model mathematically than truth-functional relations such > as material implication. > I think this sums it up very well. > After all, the father doesnıt 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 > shouldnıt conclude that the kid still gets the ice cream after eating > his veggies and killing his sister. Depending, of course, on Dadıs > opinions of sis. > These arenıt 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 arenıt problems for natural language *speakers*, just people who wish to translate a natural language into the FOL. >If theyıre 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 doesnıt 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, youıve shown the existence of two classes of conversational conditionals. This isnıt 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 itıs worth, my view is that our natural language abilities combined with rough, though strong intuitions of truth and falsity constitute most peopleıs 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. Œcid Œooh === Subject: Re: logic is innate? > Granting this, your example of the non-monotonicity of this > conversational conditional doesnıt 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, youıve shown the existence of two classes of > conversational conditionals. Can you explain why you think this example involves two different kinds of conditional? ŒDifferent contexts generatedı 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 doesnıt 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, youıve shown the existence of two classes of > conversational conditionals. > Can you explain why you think this example involves two different kinds > of conditional? ŒDifferent contexts generatedı is almost always what is > going on in this kind of examples of non-monotony. Well, letıs 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 Ŝrst makes intuitive sense--a father and son (and daughter) eating dinner. Now, in context, the Ŝrst 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 (Iım not too happy with it as an argument, since things arenıt actually equal -- one *could* argue that the conditionals are in fact the same and that itıs 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 speciŜcity, 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) Œcid Œooh === Subject: Re: logic is innate? >Granting this, your example of the non-monotonicity of this >conversational conditional doesnıt 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, youıve shown the existence of two classes of >conversational conditionals. >>Can you explain why you think this example involves two different kinds >>of conditional? ŒDifferent contexts generatedı is almost always what is >>going on in this kind of examples of non-monotony. > Well, letıs 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 Ŝrst makes intuitive > sense--a father and son (and daughter) eating dinner. Now, in > context, the Ŝrst 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 (Iım not too > happy with it as an argument, since things arenıt actually equal -- > one *could* argue that the conditionals are in fact the same and that > itıs 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 speciŜcity, 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) > Œcid Œooh Hmmm. Iıd say: we have here two meanings for the conditional connective. And the example shows that both these conditionals are non-monotonic. Thereıs just one additional aspect of this example that may obfuscate that non-monotony, though: the Ŝrst 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 theyıre 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? Letıs 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 didnıt disagree with your claim that this example is better understood as a biconditional, but even then, Iıd 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 doesnıt really show what you think it > does. Iım not sure what youıre correcting here. I didnıt want to claim that the fact natural language conditionals arenıt truth-functional *proves* logic is not innate. But if natural language gives us any hints on that question, Iıd guess it pushes us towards the negative at least as strongly as to the positive. To be honest, though, itıs 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, youıve shown the existence of two classes of > conversational conditionals. This isnıt 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 didnıt 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 Ŝnd 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 theyıre 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? Letıs 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). Iıll 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 didnıt disagree with your claim that this example is better > understood as a biconditional, but even then, Iıd 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 doesnıt really show what you think it > does. > Iım not sure what youıre correcting here. Well, you claimed that _The_ conditional operator of natural language is non-monotonic, whereas your example shows that it isnıt unique. (emph mine) > I didnıt want to claim that > the fact natural language conditionals arenıt truth-functional > *proves* logic is not innate. But if natural language gives us any > hints on that question, Iıd guess it pushes us towards the negative at > least as strongly as to the positive. I donıt have any strong opinions (that would settle the issue, anyway). Like Iıve said before, in my experience -- say, during freshman year analysis, I would reason by internally thinking things along the lines of So ŒXı is true, and ŒX->Yı 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 satisŜability problem -- it would be difŜcult to say that this person wasnıt using logic, even though all theyıve 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 *Ŝnding* 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 Ŝnd peopleıs rough, intuitive notions of how logic works somewhere between the two. But this isnıt philosophical or empirical. Itıs downright ŝakey. > To be honest, though, itıs 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. Œcid Œooh === 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 doesnıt really show what you think it >> does. >> Iım not sure what youıre correcting here. > Well, you claimed that _The_ conditional operator of natural language > is non-monotonic, whereas your example shows that it isnıt unique. > (emph mine) -- Jesse F. Hughes LOL. How arrogant you are. Now when you realize that I DID prove Goldbachıs conjecture and that I proved Fermatıs Last Theorem as well, how are you going to feel then? -- James Harris === Subject: Complex analysis power series question Iım rather new to complex analysis, so Iım 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 X-RFC2646: Format=Flowed; Original > Iım rather new to complex analysis, so Iım 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 > Iım rather new to complex analysis, so Iım 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 Liouvilleıs 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 Liouvilleıs 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 Cauchyıs 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 Liouvilleıs 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 Cauchyıs estimates to prove >something interesting about c_m for large m. Given the way the hypothesis was stated you can in fact use Cauchyıs 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. Iım 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. Isnıt 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 Ŝelds 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. >Iım 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. Isnıt 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 Ŝelds 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. > Iım 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. Isnıt 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 Ŝelds 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 Ŝles, or Word Ŝles, etc. TeX is plain text, at least the way I understand the term. Occasionally I post something here that Iıve cut Œnı pasted from another source, where it was in TeX. Going through the TeX, especially if itıs at all long, and manually editing it into a form youıd approve is not an option; your options are, I post the TeX, or I donıt 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 Iıve cut Œnı pasted from > another source, where it was in TeX. Going through the TeX, especially > if itıs at all long, and manually editing it into a form youıd approve > is not an option; your options are, I post the TeX, or I donıt post > at all. Then donıt post it. Unless itıs simple Tex, likely I may never read it. Of the many many problems in the 7 groups I read, Iıve little time to translate hard to read stuff. So if youıve 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 Ŝrst. === Subject: Re: The joy of plain text > Occasionally I post something here that Iıve cut Œnı pasted from > another source, where it was in TeX. Going through the TeX, especially > if itıs at all long, and manually editing it into a form youıd approve > is not an option; your options are, I post the TeX, or I donıt post > at all. > Then donıt post it. Unless itıs simple Tex, likely I may never read it. > Of the many many problems in the 7 groups I read, Iıve little time to > translate hard to read stuff. So if youıve 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 Ŝrst. I see that I didnıt make myself clear. Most of what I post here is not problems but answers to other peopleıs problems. Iıve been providing, not asking for, the free tutoring for over a decade now (for the most part), though I guess itıs too much to expect that even regulars contributors such as yourself would have noticed. The typical situation in which Iıll post in TeX is when someone asks a research-level question, and I happen to know that thereıs a paper on the topic already in the literature, and Iıll go cut Œnı paste the review from Math Reviews, which review will be written in TeX. If the person who asked the question canıt 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 Iıll survive. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: The joy of plain text >The typical situation in which Iıll post in TeX is when someone >asks a research-level question, and I happen to know that thereıs >a paper on the topic already in the literature, and Iıll go cut >ını paste the review from Math Reviews, which review will be written >in TeX. For Williamıs sake, perhaps you could use Zentralblatt instead. Of course, then youıd have to schneidınıklebb. Lee Rudolph === Subject: Re: The joy of plain text >> Occasionally I post something here that Iıve cut Œnı pasted from >> another source, where it was in TeX. Going through the TeX, especially >> if itıs at all long, and manually editing it into a form youıd approve >> is not an option; your options are, I post the TeX, or I donıt post >> at all. >Then donıt post it. [...] No, please do. Derek Holtıs post says it all, so I wonıt labour the point. I just wanted to add another vote for sanity (not that Iım claiming to be sane). -- Angus Rodgers (angus_prune@ eats spam; reply to angusrod@) Contains mild peril === Subject: Re: The joy of plain text >> Occasionally I post something here that Iıve cut Œnı pasted from >> another source, where it was in TeX. Going through the TeX, especially >> if itıs at all long, and manually editing it into a form youıd approve >> is not an option; your options are, I post the TeX, or I donıt post >> at all. >Then donıt post it. Unless itıs 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 simpliŜed 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 difŜcult 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, Iıve little time to >translate hard to read stuff. So if youıve 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 Ŝrst. === 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 simpliŜed 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 difŜcult 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 (Œı), if it looks like that wonıt 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 Œfracı or Œoverı (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-Ŝle I had written, it probably wouldnıt have that extra space surrounding Œplusı and Œequal toı 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 Iım in a hurry and donıt 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 difŜcult 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 wasnıt complaining that you noted it at all. My soul was suffused with the light that you brought unto me. I wasnıt 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 ŝeeting Ŝt 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 persiŝage ? 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 wasnıt 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. Weıre already ing impressed. At least I know I am. -- Jesse F. Hughes To be honest, I donıt 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 Ŝfth-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 Ŝfth-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 donıt think so!How did you get this? I donıt 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. Thatıs the way I see it, FWIW. --OL === Subject: Re: Proposed deŜnition for comparing the sizes of two sets X-Spam-This: SpamCopies@YahooGroups.Com > Do you have any inŜnite sets? Please state what you mean when you use the word have. The ordinary meaning, synonym with own, isnıt applicable, because sets are abstract mathematical ideas which canıt be owned by any one person. > Please name an inŜnite 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, doesnıt seem useful in this context. So thereıs this inŜnite 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 Ŝnite 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 itıs 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 inŜnity (aleph_null factorial, which equals aleph_one) of different ordering relations. > completely intuitive natural counting numbers. The natural numbers arenıt completely intuitive. Only the numbers from one up to about four or Ŝve 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 difŜcult. > the set of even integers within the integers, is deŜned 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 arenıt deŜned in terms of something else such as cardinality of sets whereby you can use that deŜnition to generate arithmetic properties, there is no way whatsoever to deŜne which are even and which arenıt even except by an inŜnitely 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 deŜnition: The empty set, and any set containing exactly one element which is an integer. Thus {} {{}} {{{}}} etc. are the consecutive integers. Even can be deŜned recursively like this: N is even iff N = {M} and M is not even. This works because integers arenıt just abstract elements, but are actually constructed via set theory in a way such that even can be deŜned from that. (2) Arithmetic deŜnition: Start with Peanoıs postulates, in particular the successor function S, with natural numbers deŜned recursively as 1, and any S(N) where N is an integer. Even can be deŜned 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 donıt have an inŜnite amount of time to type them all: {apat, isa, lima, delawa, tatlo}. Unless you know that Iıve used Tagalog names for those Ŝve integers, and unless you actually know what those Ŝve Tagalog words mean, and unless Iım actually using the corect Tagalog names instead of shufŝing them, you canıt Ŝgure 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 arenıt 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 arenıt per my deŜnition unless I simply tell you the answer and promise not to change my deŜnition to pull a trick on you? (4) In all the above, I had some sort of name for each integer. But suppose I donıt 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 inŜnite 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 arenıt? No matter how you guess which are even and which arenıt, 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 deŜne even vs. not-even for all the asterisks? Well because they are located in a lattice, you could perhaps make up a deŜnition that is based on the location within the pattern. It wouldnıt match my deŜnition, but at least you *could* deŜne what is meant by even on that set of integers. But that would not be deŜning even on the elements themselves, rather youıd be deŜnining even based on their position within some sort of structue, in this case a regular lattice. But suppose they werenıt 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, thereıs no way to deŜne 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 arenıt? No, the best you can do is make arbitrary decisions, which wouldnıt be consistent from one run of the demo to another. Can you deŜne a predicate: boolean isEven() which is well deŜned, using *only* calls to the API Iıve speciŜed above? No, you canıt. The best you can do is randomly decide for each new integerObject whether itıs even or odd, and keep a cache of such decisions so you donıt contradict yourself later, but thatıs not a deŜnition, thatıs a random sampler, and again, just like the manual demo, youıd get different results with each run of the program, so your function/method doesnıt *deŜne* a function, it merely generates a new random sample each time itıs run. By the way, with *all* the integers, not just the positive integers, even if you know the total ordering, thatıs still not enough to deŜne what is even and what isnıt 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 itıs 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 donıt know whether two integers are adjacent or not you canıt in a Ŝnite number of steps determine the two alternating sets. > The set of all sets ... Thereıs no such thing. If you understood proof-by-contradiction I could present you a very simple proof to that fact, but you donıt 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 thereıs a corresponding even integer, and vice versa. What you said is equivalent to saying thereıs a bijection between the Ŝngers on my left hand and the Ŝngers on my right hahd, which shows there are twice as many Ŝngers on my left hand as on my right hand. === Subject: Re: Proposed deŜnition for comparing the sizes of two sets <41AAD2D3.301BBFC3@tiki-lounge.com> X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark GrifŜth X-Treme: C&C,DWS at 07:03 PM, rem642b@Yahoo.Com (tinyurl.com/uh3t) said: >Thereıs no such thing. There are set theories that include a set of all sets. They donıt, 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 reply to spamtrap@library.lspace.org === Subject: Re: Proposed deŜnition for comparing the sizes of two sets X-Spam-This: SpamCopies@YahooGroups.Com > There are set theories that include a set of all sets. > They donıt, however, satisfy, e.g., the axiom of foundation. Do they include an axiom that allows subsets of a given set to be deŜned in terms of a property (boolean-valued function) deŜned on that set, i.e.: if S is already known to be a set, and P is a property deŜned on that set, do they allow a new set to be deŜned as: {x in S such that P(x)} Do they include an axiom that allows ordered pairs to be deŜned 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 deŜne 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 deŜned 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 Russelıs paradox? === Subject: Re: Proposed deŜnition for comparing the sizes of two sets <41b24220$12$fuzhry+tra$mr2ice@news.patriot.net> X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark GrifŜth X-Treme: C&C,DWS 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 >deŜned in terms of a property (boolean-valued function) deŜned on >that set, i.e.: if S is already known to be a set, and P is a >property deŜned on that set, do they allow a new set to be deŜned >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 deŜned >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 deŜned 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 Russelıs 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 reply to spamtrap@library.lspace.org === Subject: Re: Proposed deŜnition for comparing the sizes of two sets > Do you have any inŜnite sets? > Please state what you mean when you use the word have. The ordinary > meaning, synonym with own, isnıt applicable, because sets are > abstract mathematical ideas which canıt be owned by any one person. > Please name an inŜnite 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, doesnıt seem useful in this context. So thereıs this inŜnite > 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 Ŝnite 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 inŜnite 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 sufŜciently exact. > and there is an ordering relation on each of those sets of things, > That is such a gross understatment that itıs 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 > inŜnity (aleph_null factorial, which equals aleph_one) of different > ordering relations. Itıs deŜnitely not a lie. There are obviously many ordering relations, pick one. > completely intuitive natural counting numbers. > The natural numbers arenıt completely intuitive. Only the numbers from > one up to about four or Ŝve 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 difŜcult. 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 deŜned 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 arenıt > deŜned in terms of something else such as cardinality of sets whereby > you can use that deŜnition to generate arithmetic properties, there is > no way whatsoever to deŜne which are even and which arenıt even except > by an inŜnitely 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 deŜnition: The empty set, and any set containing exactly > one element which is an integer. Thus {} {{}} {{{}}} etc. are the > consecutive integers. Even can be deŜned recursively like this: N is > even iff N = {M} and M is not even. This works because integers arenıt > just abstract elements, but are actually constructed via set theory in > a way such that even can be deŜned from that. > (2) Arithmetic deŜnition: Start with Peanoıs postulates, in particular > the successor function S, with natural numbers deŜned recursively as > 1, and any S(N) where N is an integer. Even can be deŜned 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 donıt have an inŜnite amount of time to type them > all: {apat, isa, lima, delawa, tatlo}. Unless you know that Iıve used > Tagalog names for those Ŝve integers, and unless you actually know > what those Ŝve Tagalog words mean, and unless Iım actually using the > corect Tagalog names instead of shufŝing them, you canıt Ŝgure 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 arenıt > 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 arenıt per my deŜnition unless I simply tell you the > answer and promise not to change my deŜnition 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 donıt 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 inŜnite 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 arenıt? I donıt care. I wouldnıt. > No matter how you guess which are even and which arenıt, 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 deŜne even > vs. not-even for all the asterisks? Well because they are located in a > lattice, you could perhaps make up a deŜnition that is based on the > location within the pattern. It wouldnıt match my deŜnition, but at > least you *could* deŜne what is meant by even on that set of > integers. But that would not be deŜning even on the elements > themselves, rather youıd be deŜnining even based on their position > within some sort of structue, in this case a regular lattice. But > suppose they werenıt 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, thereıs > no way to deŜne 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 arenıt? No, the best you can do is make > arbitrary decisions, which wouldnıt 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 deŜne a predicate: > boolean isEven() > which is well deŜned, using *only* calls to the API Iıve speciŜed above? > No, you canıt. The best you can do is randomly decide for each new > integerObject whether itıs even or odd, and keep a cache of such > decisions so you donıt contradict yourself later, but thatıs not a > deŜnition, thatıs a random sampler, and again, just like the manual > demo, youıd get different results with each run of the program, so your > function/method doesnıt *deŜne* a function, it merely generates a new > random sample each time itıs run. Thatıs 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, thatıs still not enough to deŜne > what is even and what isnıt 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 itıs 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 donıt know whether two integers are adjacent or not you > canıt in a Ŝnite number of steps determine the two alternating sets. > The set of all sets ... > Thereıs 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 donıt 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, youıre 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 thereıs a > corresponding even integer, and vice versa. Youıre wrong, it shows exactly that thing. > What you said is equivalent to saying thereıs a bijection between the > Ŝngers on my left hand and the Ŝngers on my right hahd, which shows > there are twice as many Ŝngers on my left hand as on my right hand. No, it doesnıt. 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 setıs 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 youıve been about all this? Half of the integers are even, true or false? Itıs true. If itıs 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 deŜnition 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 Iım correct, and letıs 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 aıs 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 aıs are not polynomials, then how do you deŜne 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 speciŜc numbers, speciŜcally 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 aıs have 7 as a factor, and Iıve > arbitrarily selected a_1(x) and a_2(x). Iım 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 coefŜcients, then the aıs 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 coefŜcients? 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 Iım correct, and letıs 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 aıs 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 aıs are not polynomials, then how do you deŜne 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 speciŜc numbers, speciŜcally 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