mm-1039 === Subject: are C_r fields CÕ_r? any progress? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Some time around Ō52, Serge Lang generalized ArtinÕs notion of quasi-algebraically closed fields (C_1) and defined C_r fields: A field k is said to satisfy property (C_r) when any homogeneous polynomial of degree d with coefficients in k in n+1 variables satisfying n >= d^r has a non-trivial zero. (Or, in arithmetic geometry terminology, any hypersurface in P^n over k with degree d has a k-point if n >= d^r.) (Note that a C_0 field is merely an algebraically closed field. Finite fields are C_1 by a theorem of Chevalley, the field of rational functions, or Laurent series, in r variables over an algebraically closed fields, are C_r by results of Tsen and Lang. The field Q_p of p-adic numbers is not C_2, contrary to what was believed for some time: Terjanian gave some counterexamples to that effect.) Under a certain technical assumption (viz., the existence of normic forms of order r and of all degrees), Lang proved that the existence of a non-trivial zero generalizes to families of polynomials, provided the inequality is satisfied for the sum of the degrees-to-the-r. Precisely, let us define: A field k is said to satisfy property (CÕ_r) when any s homogeneous polynomials of degrees d_1,...,d_s with coefficients in k in n+1 variables satisfying n >= d_1^r + ... + d_s^r have a common non-trivial zero. (Or, in arithmetic geometry terminology, any intersection of s hypersurfaces in P^n over k with degrees d_1,...,d_s has a k-point if n >= d_1^r + ... + d_s^r.) (IÕm not sure that my notation is perfectly standard. Perhaps what other people call CÕ_r is not exactly what is written above. But let us continue with this notation.) So Lang proves that, under a certain technical assumption which I wonÕt recall, a C_r field is actually CÕ_r. Later (around Ō57), Nagata showed that the technical assumption is not necessary provided all the d_j are equal. However, the question of whether the technical assumption is necessary in full generality remained open (as far as I know). My question is: has any progress been made on this question since then? Is there now a known example of a C_r field which is not CÕ_r, or a proof that all C_r fields are CÕ_r? Perhaps if we restrict to the (most interesting) r=1 case? At any rate, what would the educated guess be? Does it help in any way if we assume the (hypersurfaces defined by the) f_j to be in complete intersection (I canÕt see a way to reduce the general problem to that case, but it seems like a reasonable assumption to make)? (Basically, IÕd like to formulate the conjecture that, over a C_1 field k, any smooth projective (geometrically) separably rationally connected variety has a k-point: this is known when k is finite or when it is the field of rational functions over an algebraically closed field, by results of Esnault on the one hand, and Graber, Harris, de Jong and Starr on the other. At the very least, it is necessary, for the conjecture to be sensible, for the field to be CÕ_1, so it would be embarrassing if there were already a known example of a C_1 field that is not CÕ_1.) of Maths) and Nagata (in Kyoto University something-or-other) if necessary. -- David A. Madore (david.madore@ens.fr, http://www.dma.ens.fr/~madore/ ) === Subject: Paper published by Geometry and Topology Received-SPF: Received-SPF: pass (mailbox2.ucsd.edu: domain of gt@maths.warwick.ac.uk designates 137.205.233.100 as permitted sender) receiver=mailbox2.ucsd.edu; client_ip=137.205.233.100; envelope-from=gt@maths.warwick.ac.uk; Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The following paper has been published: URL: http://www.maths.warwick.ac.uk/gt/GTVol8/paper37.abs.html Title: Commensurations of the Johnson kernel Author(s): Tara E Brendle, Dan Margalit Abstract: Let K be the subgroup of the extended mapping class group, Mod(S), generated by Dehn twists about separating curves. Assuming that S is a closed, orientable surface of genus at least 4, we confirm a conjecture of Farb that Comm(K), Aut(K) and Mod(S) are all isomorphic. More generally, we show that any injection of a finite index subgroup of K into the Torelli group I of S is induced by a homeomorphism. In particular, this proves that K is co-Hopfian and is characteristic in I. Further, we recover the result of Farb and Ivanov that any injection of a finite index subgroup of I into I is induced by a homeomorphism. Our method is to reformulate these group theoretic statements in terms of maps of curve complexes. Secondary: 20F38, 20F36 Keywords: Torelli group, mapping class group, Dehn twist Proposed: Walter Neumann Seconded: Shigeyuki Morita, Joan Birman Author(s) address(es): Department of Mathematics, Cornell University 310 Malott Hall, Ithaca, NY 14853, USA and Department of Mathematics, University of Utah 155 S 1440 East, Salt Lake City, UT 84112, USA Email: brendle@math.cornell.edu, margalit@math.utah.edu === Subject: Teiji Takagi and principalization Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I couldnÕt get an answer on sci.math, so I hope this is suitable for here: Some sources say that Takagi proved not only that the maximal unramified extension of a L/K number field K has a Galois group corresponding to the class group, but that principalization occurs in this field--all the ideals of K extend to principal ideals of L. Others say it had to wait for Artin and Furtwangler to get the proof of this. Does anyone have the straight dope? === Subject: Re: Teiji Takagi and principalization Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > I couldnÕt get an answer on sci.math, so I hope this is suitable for here: > Some sources say that Takagi proved not only that the maximal > unramified extension of a L/K number field K has a Galois group > corresponding to the class group, but that principalization occurs in > this field--all the ideals of K extend to principal ideals of L. > Others say it had to wait for Artin and Furtwangler to get the proof > of this. Does anyone have the straight dope? in Cassells and Froelich, _Algebraic Number Theory_ (Thompson, 1967). He says (p. 273): With the help of this [general reciprocity] law, Artin could also reduce the principal divisor theorem, enunciated by Hilbert and not yet proved by Takagi, to a pure group-theoretical proposition, which was then proved by Furtwaengler. IÕve looked at ArtinÕs paper, Idealklassen in Oberkoerpern und allgemeines Reziprozitaetsgesetz (Collected Papers 159-164), and itÕs clear that he did not know of any earlier proof. William C. Waterhouse Penn State === Subject: Normal subgroups of surface groups? Epigone-thread: thulgonstrix Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Let G be a surface group, and H be a non-trivial normal subgroup of G. How can I prove that H is of finite index in G ? (This is a conjecturally true for any one-relator group) === Subject: Re: Normal subgroups of surface groups? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Let G be a surface group, and H be a non-trivial normal subgroup of G. > How can I prove that H is of finite index in G ? > (This is a conjecturally true for any one-relator group) You canÕt, because this is not true. For instance, any surface group (other than the 2-sphere or projective plane) has a surjection to the integers, whose kernel is of infinite index. The simplest of course would be the torus, where the obvious Z subgroup is normal, and of infinite index. DR === Subject: Re: Normal subgroups of surface groups? Epigone-thread: thulgonstrix Originator: bergv@math.uiuc.edu (Maarten Bergvelt) YouÕre right. Here is the restatement of the question: Show that every finitely generated normal subgroup of a non-abelian surface group (with or without boundary) is of finite index. JJ >> Let G be a surface group, and H be a non-trivial normal subgroup of G. >> How can I prove that H is of finite index in G ? >> (This is a conjecturally true for any one-relator group) >You canÕt, because this is not true. For instance, any surface group >(other than the 2-sphere or projective plane) has a surjection to the >integers, whose kernel is of infinite index. The simplest of course >would be the torus, where the obvious Z subgroup is normal, and of >infinite index. === Subject: rapidly converging rational sqrt Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Below is a description of an algorithm which, with each iteration, will double the number of significant digits in the computation of a rational square root approximation. I do not know if this algorithm is new, but I found it interesting nonetheless. Lisp code for implementing this algorithm can be found at: http://thegreves.com/david/sqrt/sqrt.html If by convention we say that: s = isqrt(C) d = C - s^2 Then the square root of C can be expressed as as the infinite continued fraction: d s + ----------------- d 2s + ------------ d 2s + ------- 2s + .. We designate the tail of this continued fraction using e(n) = nth error term and we say that the nth error term of the continued fraction representation has the form: d e(n) = ---------- 2s + e(n+1) Although we sometimes drop the subscript on the error term for notational convenience. A pretty good first order approximation for a square root can be computed as follows (even allowing e(1) to be zero): d sqrt(C) ~= s + -------- 2s + e(1) Without proof, we claim that a generalized expression for a partial evaluation of our continued fraction can be represented as: A + Be sqrt(C) = s + ------- C + De It is easy to see that the first order approximation given above is an instance of this expression when A = d, B = 0, C = 2s, D = 1. The generalized representation is useful for representing the result of evaluating some number of sucessive terms in the continued fraction representation. Assuming that the above representation is the result of evaluating n terms of the continued fraction for sqrt(C), then the n+1 term would be computed by substituting the next error term into the error expression in the representation. A + B(d/(2s + e)) s + --------------- C + D(d/(2s + e)) A(2s + e) + Bd s + -------------- C(2s + e) + Dd (A2s + Bd) + Ae s + ----------------- (C2s + Dd) + Ce Which, we observe, is once again in the general representational form. If the evaluation of the first n terms produced A + Be ------ C + De and the evaluation of the next m terms were to produce W + Xe ------ Y + Ze Then the evaluation of the first n+m terms would be: W + Xe A + B(------) Y + Ze A(Y + Ze) + B(W + Xe) ------------ = --------------------- W + Xe C(Y + Ze) + D(W + Xe) C + D(------) Y + Ze (AY + BW) + (AZ + BX)e ---------------------- (CY + DW) + (CZ + DX)e Because the continued fraction representation of the square root is uniform, the evaluation of any n sucessive terms will always produce the same result. We can take advantage of this fact to refine our first order approximation by substituting A,B,C, and D in for W,X,Y and Z in the above expression. The will double the number of terms we have evaluated. Of course, this procedure can be repeated again and again, with each iteration of the algorithm doubling the number of significant digits in our representation. Here is an example run computing the sqrt of 1973 for 1 to 10 iterations of the algorithm. Note that after iteration 6 we have more than 100 significant digits. 1 : 44.41845521141241485670222336460609176198432078139056676519727 541447114766739 49363834982650044981364863 2 : 44.41846462881467219505665982727838995343272680250852496023827 788068578570854 49370627285274658896791048 3 : 44.41846462902561876427524312325572040240985914842970903422305 486056596616470 16608600579504981095676558 4 : 44.41846462902561876438107965740906053956833272941354961112468 508573375443791 08155113103260155775597834 5 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 26805283703947239542091268 6 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 43686917147360058911830087 7 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 43686917147360058911830087 8 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 43686917147360058911830087 9 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 43686917147360058911830087 10 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 43686917147360058911830087 Dave === Subject: Re: rapidly converging rational sqrt Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Below is a description of an algorithm which, with > each iteration, will double the number of significant > digits in the computation of a rational square root > approximation. > Here is an example run computing the sqrt of 1973 for 1 to 10 > iterations of the algorithm. Note that after iteration 6 we have > more than 100 significant digits. > 1 : 44.4184 | 55211412414856702223364606091761984320781390566765197275414471 147667394936383 4982650044981364863 > 2 : 44.41846462 | 88146721950566598272783899534327268025085249602382778806857857 085449370627285 274658896791048 > 3 : 44.418464629025618764 | 27524312325572040240985914842970903422305486056596616470166086 005795049810956 76558 > 4 : 44.4184646290256187643810796574090605395 | 68332729413549611124685085733754437910815511310326015577559783 4 > 5 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 | 26805283703947239542091268 > 6 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 43686917147360058911830087 > 7 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 43686917147360058911830087 > 8 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 43686917147360058911830087 > 9 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 43686917147360058911830087 > 10 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 43686917147360058911830087 I donÕt wish to denigrate your algorithm unduly, but square root algorithms with this rate of convergence have been known for millennia. The iteration x_{n+1} = 1/2 (x_n + C/x_n), which is believed to have been known to the ancient Babylonians, yields the following output if we take C = 1973 and start it at 44 (since your algorithm presupposes we know isqrt(C) this seems a fair comparison.) I have added vertical bars to indicate the correct portion of each decimal expansion; I have done the same for the quoted output from your algorithm above. 1 : 44.4 | 20454545454545454545454545454545454545454545454545454545454545 454545454545454 54545454545454545455 2 : 44.4184646 | 73597060396753412870066745738272983092630061164213121235377566 920160933975208 72578432056559 3 : 44.418464629025618786 | 74355237890368384233529296754366053971048032852669727540929455 186628564646861 810 4 : 44.4184646290256187643810796574090605 | 45224167043182294394699285292428539812171115734428794096905254 72 5 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066 | 216106505407447827420345844 6 : 44.41846462902561876438107965740906053959497442704659903610246 205761940066180 436869171473600589118301... So like your algorithm this also produces 100 digits in 6 iterations, and it is easy to prove (since this is a special case of NewtonÕs method for a general function) that convergence is quadratic in general. In short, your algorithm is interesting but it doesnÕt outperform the standard algorithms for square roots. Yours, David Loefßer (student, Trinity College, University of Cambridge, UK) === Subject: LaplaceÕs method and a Double Integral Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hello all, I wonder if anyone can show me how to evaluate the asymptotics (as N) gets large of this integral: int_0^R int_{-u}^0 r^3 [1-c_1 (c_2 n - c_3 r^2)^2]^N dn dr IÕve looked in several books on LaplaceÕs method (Erdelyi, Wong, Bleistein). If we define f(n,r) = -log[1-c_1 (c_2 n - c_3 r^2)^2] Then the integral is of the form int int r^3 exp(-N f(n,r)) dn dr where f takes on its mininum value of 0 at the pt (0,0). Unfortunately the Hessian at that pt is zero....because f_{rr}(0,0) = 0. All the explicit expansions I have seen require that the Hessian be non-zero. Does anyone know of another reference to try that might have this worked out? Jim PS: IÕm not a mathematician....just a humble plodding engineer! === Subject: Re: LaplaceÕs method and a Double Integral Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Hello all, >I wonder if anyone can show me how to evaluate the asymptotics (as N) >gets large of this integral: >int_0^R int_{-u}^0 r^3 [1-c_1 (c_2 n - c_3 r^2)^2]^N dn dr >IÕve looked in several books on LaplaceÕs method (Erdelyi, Wong, Bleistein). >If we define >f(n,r) = -log[1-c_1 (c_2 n - c_3 r^2)^2] >Then the integral is of the form int int r^3 exp(-N f(n,r)) dn dr >where f takes on its mininum value of 0 at the pt (0,0). Unfortunately >the Hessian at that pt is zero....because f_{rr}(0,0) = 0. All the >explicit expansions I have seen require that the Hessian be non-zero. Do the substitutions t=r^2, s=-n help? The integral becomes (1/2)int_0^{R^2} int_0^u t [1 - c_1(c_2 s + c_3 t)^2]^N ds dt For the corresponding f(s,t), a quick calculation gave me f_{tt}(0,0) not= 0, but you should check that yourself. Dan -- Dan Luecking Department of Mathematical Sciences University of Arkansas Fayetteville, Arkansas 72701 To reply by email, change Look-In-Sig to luecking === Subject: Re: LaplaceÕs method and a Double Integral Originator: israel@math.ubc.ca (Robert Israel) DanÕs substitutions are OK, a nice trick, but I wonder if the origin (0,0) is the main contributing point. I expect also contributions from the boundaries, but this depends on the values of c_1, c_2, c_3, u and R. Nico M. Temme, http://homepages.cwi.nl/~nicot/ C W I: Centrum voor Wiskunde en Informatica Kruislaan 413, NL-1098 SJ Amsterdam Tel +31 20 592 4240 P.O. Box 94079, NL-1090 GB Amsterdam Fax +31 20 592 4199 === > Subject: Re: LaplaceÕs method and a Double Integral >Hello all, >I wonder if anyone can show me how to evaluate the asymptotics (as N) >gets large of this integral: >int_0^R int_{-u}^0 r^3 [1-c_1 (c_2 n - c_3 r^2)^2]^N dn dr >IÕve looked in several books on LaplaceÕs method (Erdelyi, Wong, Bleistein). >If we define >f(n,r) = -log[1-c_1 (c_2 n - c_3 r^2)^2] >Then the integral is of the form int int r^3 exp(-N f(n,r)) dn dr >where f takes on its mininum value of 0 at the pt (0,0). Unfortunately >the Hessian at that pt is zero....because f_{rr}(0,0) = 0. All the >explicit expansions I have seen require that the Hessian be non-zero. > Do the substitutions t=r^2, s=-n help? The integral becomes > (1/2)int_0^{R^2} int_0^u t [1 - c_1(c_2 s + c_3 t)^2]^N ds dt > For the corresponding f(s,t), a quick calculation gave me > f_{tt}(0,0) not= 0, but you should check that yourself. > Dan > -- > Dan Luecking Department of Mathematical Sciences > University of Arkansas Fayetteville, Arkansas 72701 > To reply by email, change Look-In-Sig to luecking === Subject: Re: LaplaceÕs method and a Double Integral Originator: israel@math.ubc.ca (Robert Israel) Re: LaplaceÕs method and a Double Integral Since Nico thinks that the asymptotic form of the integral has also contributions not only from the neighborhood of (0,0), I derived an asymptotic approximation given in the following. It shows that only the neighborhood of (0,0) is important. Ciao Karl Breitung Schellingstr. 21 D-80799 Munich, Germany AN ASYMPTOTIC APPROXIMATION FOR THE TWO-DIMENSIONAL INTEGRAL: ,- R ,- 0 3 [ 2 2 ]N I(N)= | | r [ 1-a(bn-cr ) ] dn dr, N --> oo -0 --u Here we write instead of the original form: a=c , b=c , c=c . 1 2 3 1/2 2 1/2 Making the substitutions r --> v=a cr and n --> z=-a bn transforms this into: 2 ,- cR ,- bu / v 3/2 [ 2 ]N d r d n I(N)=- | | | ----- | [ 1-(z+v) ] --- ---- dz dv -0 -0 | 1/2 | d v d z a c / 2 ,- cR ,- bu / v 3/2 [ 2 ]N 1 1/2 -1/2 1 = | | | ----- | [ 1-(z+v) ] -(a cv) ----- dz dv= -0 -0 | 1/2 | 2 1/2 a c / a b 1/2 2 1/2 ,- a cR ,- a bu [ 2 ]N K | | v [ 1-(z+v) ] dz dv -0 -0 3/2 2 -1 with K=(2a bc ) . Now we will consider only the integral over a triangle 1/2 2 1/2 (0,0), (d,0) and (0,d) with 00 and K >0 are constants. Then: 1 2 ,- d ,- d-v [ 2 ]N I(N) sim K | | v [ 1-(z+v) ] dz dv -0 -0 In this triangle, we make the variable transformation (v,z) --> (w,y) with w=z+v, y=z-v. Then: ,- d [ ,- w w-y [ 2 ]N ] I(N) sim K | | | --- [ 1-w ] |det (J(w,y))| dy | dw= -0 [ --w 2 ] ,- d [ ,- w w-y [ 2 ]N ] K | | | --- [ 1-w ] dy | dw (*) -0 [ --w 2 ] J(w,y) is the Jacobian of the inverse transformation with its determinant equal to 1/2. The integral in the brackets is: ,- w w-y [ 2 ]N [ 2 ]N ,- w / w y [ 2 ]N ,- w w | --- [ 1-w ] dy=[ 1-w ] | | - - - | dy=[ 1-w ] | - dy= --w 2 --w 2 2 / --w 2 [ 2 ]N 2 [ 1-w ] w 2 Inserting this into equ. (*) and writing f(w)=log (1-w ) gives: ,- d 2 I(N) sim K | w exp (Nf(w)) dw -0 This we can evaluate using the generalized Laplace method (derived in [2], p. 37, see also [1], p. 48). Here we derive the result directly by approximating 2 2 f(w) by its second order Taylor expansion at zero, i.e. f(w)=-2w +o(w ) and 2 then making the substitution w --> x=2N w . This gives: ,- dN x -x dw ,- dN x -x 1 -1/2 I(N)sim K | -- e --dx= K | -- e ------ x dx = -0 2N dx -0 2N +--+ 2|2N -3/2 K ,- dN 1/2 -x N ----- | x e dx +-+ -0 4|2 For this we get the asymptotic form replacing dN by oo: -3/2 K ,- oo 1/2 -x -3/2 K I(N)sim N ----- | x e dx sim N ----- Gamma(3/2)= +-+ -0 +-+ 4|2 4|2 +---+ +---+ -3/2 K |pi -3/2 |pi N --- ----- = N ---------------- , N --> oo +-+ 2 +-+ 3/2 2 4|2 16|2 a bc If necessary this approximation can be refined by deriving a second term in the asymptotic expansion of I(N). Bibliography: [1] K. Breitung. Asymptotic Approximations for Probability Integrals. Springer, Berlin, 1994. Lecture Notes in Mathematics, Nr.1592. [2] A. ErdÕelyi. Asymptotic Expansions. Dover, New York, 1956. === Subject: groups Epigone-thread: geikreizhan Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Let $G$ be a finitely generated subgroup of Homeo(E), where $E$ is an arbitrary metric space, such that all orbits of $G$ are finite. We suppose that there exists $pin N$ such that for all $gin G$ we have $g^p = 1$. The group $G$ is it finite? === Subject: Re: groups Epigone-thread: geikreizhan Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Let $G$ be a finitely generated subgroup of Homeo(E), where $E$ is an >arbitrary metric space, such that all orbits of $G$ are finite. >We suppose that there exists $pin N$ such that for all $gin G$ we >have $g^p = 1$. The group $G$ is it finite? The answer is no. This was essentially known as the Burnside problem, and solved in the negative by Novikov and Adjan in 1968: there exist infinite two-generator groups identically satisfying x^n=1 with n any sufficiently large odd integer. (The additional condition stipulating that G should be a subgroup of Homeo(E) adds nothing to the problem, for every group can be so represented: choose E = G, with the discrete metric where distinct points always have distance 1, and let G operate on itself by left translations; these, of course, are homeomorphisms in the given metric.) === Subject: Re: groups Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >>Let $G$ be a finitely generated subgroup of Homeo(E), where $E$ is >>an arbitrary metric space, such that all orbits of $G$ are finite. >>We suppose that there exists $pin N$ such that for all $gin G$ we >>have $g^p = 1$. The group $G$ is it finite? >The answer is no. This was essentially known as the Burnside >problem, and solved in the negative by Novikov and Adjan in 1968: >there exist infinite two-generator groups identically satisfying >x^n=1 with n any sufficiently large odd integer. (The additional >condition stipulating that G should be a subgroup of Homeo(E) adds >nothing to the problem, for every group can be so represented: choose >E = G, with the discrete metric where distinct points always have >distance 1, and let G operate on itself by left translations; these, >of course, are homeomorphisms in the given metric.) But among the hypotheses you have that the orbits of G are finite. This implies that G is residually finite and thus, by ZelmanovÕs positive solution to the restricted Burnside problem, G is indeed finite. Andreas === Subject: Re: groups Epigone-thread: geikreizhan Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Sorry for my oversight! Of course, Andreas is right; I overlooked the requirement that the orbits should be finite. Regretfully, Peter === Subject: Re: groups Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >>Let $G$ be a finitely generated subgroup of Homeo(E), where $E$ is an >>arbitrary metric space, such that all orbits of $G$ are finite. >>We suppose that there exists $pin N$ such that for all $gin G$ we >>have $g^p = 1$. The group $G$ is it finite? >The answer is no. This was essentially known as the Burnside >problem, and solved in the negative by Novikov and Adjan in 1968: >there exist infinite two-generator groups identically satisfying x^n=1 >with n any sufficiently large odd integer. (The additional condition >stipulating that G should be a subgroup of Homeo(E) adds nothing to >the problem, for every group can be so represented: choose E = G, with >the discrete metric where distinct points always have distance 1, and >let G operate on itself by left translations; these, of course, are >homeomorphisms in the given metric.) But the orbits of G will not be finite in this example. G having a faithful representation as mappings of a set with all orbits finite means that G has a collection of normal subgroups of finite index whose intersection is the identity. I donÕt know if this is true for any of the known examples in the Burnside problem. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Fall Pacific NW Geometry Seminar at U of Oregon Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Final announcement: PACIFIC NORTHWEST GEOMETRY SEMINAR University of Oregon Eugene, OR SCHEDULE Saturday, November 6 10:30 - 11:00 Morning Reception 11:00 - 12:15 Charles Doran (University of Washington) Mirror Symmetry, K-Theory, and Toric Geometry 12:15 - 2:00 Lunch 2:00 - 3:15 Mutao Wang (Columbia) Mean curvature ßows of Lagrangian submanifolds 3:15 - 4:00 Break 4:00 - 5:15 Lei Ni (UCSD) Ancient Solutions of the K.8ahler-Ricci Flow 7:00 Party at BotvinnikÕs Sunday, November 7 8:30 - 9:00 Morning Reception 9:00 - 10:15 John Lott (University of Michigan) Ricci curvature for metric-measure spaces 10:15 - 10:45 Break 10:45 - 12:00 David Auckly (Kansas State University) The structure of maps into homogenous spaces and the Faddeev and Skyrme models Note: Each speakerÕs time allotment includes 15 minutes for a discussion of Open Problems related to his topic. The talks will be in 110 Fenton Hall (D-7 on the campus map). The receptions and breaks will be right outside 110 Fenton. ----------------------------------------------------------- For general information about the PNGS, visit the PNGS web site: http://www.math.washington.edu/~lee/PNGS It contains up-to-date information about this meeting, travel and lodging information, general information about the PNGS, and a historical record of all PNGS meetings and speakers. ----------------------------------------------------------- For more information about this meeting, contact the organizers: Boris Botvinnik (botvinn@math.uoregon.edu) Jim Isenberg (jim@newton.uoregon.edu) === Subject: Non linear hyperbolic PDEs : ill posedness ? Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hi everybody I am studying the well-posedness of the Cauchy problem for systems of PDE of the form : d U/dt = A d U/dx, with unknown vector U(x,t) In the linear case (A is a matrix function of (x,t)), it is well known that the problem is well-posed (existence of a unique solution depending continuously on the initial data) iff A is diagonalisable with real eigenvalues for all x and t. Here is my question : In the non linear case (A is a function of U, x and t) does one know such a system where the matrix A is not always diagonalisable but wich is still well-posed ? Michael === Subject: Partitioning 4 space with ultraskew lines, and the three body problem. Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Suppose we are in 4 space. Two lines are skew if they do not lie in the same plane(2 space). Skewness is a binary relation on lines. Now two skew lines will lie in the same 3 space. But it is possible for three lines to not lie in the same 3 space. This is a ternary relation on lines. I donÕt know if thereÕs a name for this relation, so letÕs call it ultraskewness until we find out. 1) Can one foliate, or at least partition, 4 space with lines that are tripletwise ultraskew? I mean EACH 3 line subset of the partition must not lie in the same 3 space. 2) In the three-body problem, one could approximate the trajectory of each of the 3 masses as a straight lines until they became close enough to each other for their gravitation to have an appreciable effect. I think much of the work on this problem assumes all three trajectories lie in the same plane.(The restricted 3-body problem.) But some work has been where the trajectories are not coplanar. WouldnÕt it be fun to explore the three-body problem when the trajectories are not cospatial? Richard Peterson, CSU Sacramento === Subject: Re: Partitioning 4 space with ultraskew lines, and the three body problem. ath: nntpswitch.com Originator: israel@math.ubc.ca (Robert Israel) > Suppose we are in 4 space. Two lines are skew if they do not lie in > the same plane(2 space). Skewness is a binary relation on lines. Now > two skew lines will lie in the same 3 space. But it is possible for > three lines to not lie in the same 3 space. This is a ternary relation > on lines. I donÕt know if thereÕs a name for this relation, so letÕs > call it ultraskewness until we find out. > 1) Can one foliate, or at least partition, 4 space with lines that > are tripletwise ultraskew? I mean EACH 3 line subset of the partition > must not lie in the same 3 space. > 2) In the three-body problem, one could approximate the trajectory > of each of the 3 masses as a straight lines until they became close > enough to each other for their gravitation to have an appreciable > effect. I think much of the work on this problem assumes all three > trajectories lie in the same plane.(The restricted 3-body problem.) > But some work has been where the trajectories are not coplanar. > WouldnÕt it be fun to explore the three-body problem when the > trajectories are not cospatial? > Richard Peterson, CSU Sacramento I donÕt know about the three-body problem, or about foliations, but I think one can partition 4-space into ultraskew lines without much trouble. The proof is via a transfinite induction of c (the cardinality of the continuum) many steps; at step k one considers the k-th point p in a fixed enumeration of 4-space, and one has already constructed a collection L of |k|-many (fewer than c) lines. If p is in the union of L there is nothing to do. Otherwise one need only find a unit tangent vector u at p so that the line through p in direction u is (a) disjoint from each line in L and (b) ultraskew to every pair of lines in L. Since p lies on no line in L (a) is satisfied so long as u does not lie in any plane containing both p and a line in L. Also, (b) is satisfied so long as u does not lie in any translate containing p of an (affine) 3-space generated by a pair of lines in L. So we need a point on the unit 3-sphere in R^4 not lying in any of a collection of fewer than c many subspaces of R^4 of dimension at most 3. Without loss of generality we may assume all the subspaces have dimension 3, so each has a perp that meets the 3-sphere in at most 2 points. As we have fewer than c subspaces, there is a point w on the 3-sphere not in the perp of any of them. Then the perp P of the line spanned by w is a 3-dimensional space meeting each of the spaces we want to avoid in a subspace of dimension at most 2. Thus it suffices to find a point on the intersection of the unit 3-sphere with P (i.e., a unit 2-sphere) not in any of a collection of fewer than c subspaces of P of dimension at most 2. By a similar argument we can drop the dimension once again, and then we need to find a point on the unit circle avoiding fewer than c lines through the origin. As the circle has c points, and each line meets it in two point, that is easy. Well, itÕs late and IÕm hurrying, but I think this argument holds water. Bob Beaudoin === Subject: Integral recurrence relation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) I have encountered the following integral in some research in the physical sciences int |u-A|^(2a) |u-B|^(2b) Exp[-|u|^2] du where u, A and B are cartesian vectors in 3 dimensions and the integral is to performed over all space. This seems like quite a straightforward integral but the best I can do is to write it as a triple infinite series in A^2, B^2 and |A-B|^2 (which quickly truncates, depending on the values of a and b). I was wondering if anyone has any suggestions as how I might produce a more useful formulation. Even more useful would be a suggestion as to how I might derive a recurrence relation to generate integrals of higher values of a and b or if it is possible to prove or disprove the existence of such a relation. Darragh === Subject: Re: Integral recurrence relation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >I have encountered the following integral in some research in the >physical sciences > int |u-A|^(2a) |u-B|^(2b) Exp[-|u|^2] du >where u, A and B are cartesian vectors in 3 dimensions and the >integral is to performed over all space. This seems like quite a >straightforward integral but the best I can do is to write it as a >triple infinite series in A^2, B^2 and |A-B|^2 (which quickly >truncates, depending on the values of a and b). I was wondering if >anyone has any suggestions as how I might produce a more useful >formulation. Let your integral be F(a,b) (for nonnegative integers a,b). Consider the exponential generating function f(s,t) = sum_{a=0}^infinity sum_{b=0}^infinity F(a,b) s^a t^b/(a! b!) = int_{R^3} exp(s |u-A|^2) exp(t |u-B|^2) exp(-|u|^2) du = int_{R^3} exp(-(1-s-t) |u|^2 - 2 u.(sA+tB) + s|A|^2 + t|B|^2) du = exp(s|A|^2 + t|B|^2 + |sA+tB|^2/(1-s-t)) int_{R^3} exp(-(1-s-t) |u-(sA+tB)/sqrt(1-s-t)|^2) du = exp(s|A|^2 + t|B|^2 + |sA+tB|^2/(1-s-t)) (pi/(1-s-t))^(3/2) for |s|+|t| < 1. Then F(a,b) can be obtained from the coefficients of the bivariate Taylor series for f(s,t) around (0,0). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Paper published by Geometry and Topology Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The following paper has been published: URL: http://www.maths.warwick.ac.uk/gt/GTVol8/paper38.abs.html Title: Noncommutative localisation in algebraic K-theory I Author(s): Amnon Neeman, Andrew Ranicki Abstract: associative rings, a long exact sequence in algebraic K-theory. The main result goes as follows. Let A be an associative ring and let A-->B be the localisation with respect to a set sigma of maps between finitely generated projective A-modules. Suppose that Tor_n^A(B,B) vanishes for all n>0. View each map in sigma as a complex (of length 1, meaning one non-zero map between two non-zero objects) in the category of perfect complexes D^perf(A). Denote by the thick subcategory generated by these complexes. Then the canonical functor D^perf(A)-->D^perf(B) induces (up to direct factors) an equivalence D^perf(A)/--> D^perf(B). As a consequence, one obtains a homotopy fibre sequence K(A,sigma)-->K(A)-->K(B) (up to surjectivity of K_0(A)-->K_0(B)) of Waldhausen K-theory spectra. consequences of the main theorem in a form more suitable for the applications to surgery. For example if, in addition to the vanishing of Tor_n^A(B,B), we also assume that every map in sigma is a monomorphism, then there is a description of the homotopy fiber of the map K(A)-->K(B) as the Quillen K-theory of a suitable exact category of torsion modules. Secondary: 19D10, 55P60 Keywords: Noncommutative localisation, $K$--theory, triangulated category Proposed: Bill Dwyer Seconded: Thomas Goodwillie, Gunnar Carlsson Author(s) address(es): Centre for Mathematics and its Applications The Australian National University Canberra, ACT 0200, Australia and School of Mathematics, University of Edinburgh Edinburgh EH9 3JZ, Scotland, UK Email: Amnon.Neeman@anu.edu.au, a.ranicki@ed.ac.uk === Subject: monoidal enriched natural transformations Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hi. I have a problem which, it seems to me, requires the notion of enriched natural transformation between enriched monoidal functors, but I havenÕt been able to find a good reference for it. IÕve taken a look at the books of Borceux (Handbook of categorical algebra), Kelly (Basic notions of enriched category) and the appendix in LevineÕs Mixed motives, but they all end just where I need them, or even long before that point. More precisely, the situation IÕve encountered seems to be the following. I have: - Two monoidal (symmetric) categories: V and W . - A couple of monoidal (symmetric) functors between them: S, T : V ---> W - A monoidal natural transformation between S and T : And here is where my problems begin. There is a well-known notion of what is a V-functor between V-categories and what a V-natural transformation is. My first need is to understand what an S-functor between a V-category C and a W-category D should be: F : C ---> D I didnÕt find this thing in the literature, but I expect it ought to be something like a V-functor, but with a family of morphisms in W lambda_{XY} : S[X,Y] ---> [FX,FY] for every pair of objects X, Y in C . (Here the square brackets [,] stand for the objects in V and W of morphisms of C and D , and IÕm leaving aside units and commutative isomorphisms for the moment.) This seems reasonable to me, since (a) is the situation I have in the real world and (b) if I put S = id_V , I find the definition of a V-functor. Next, I would need the notion of an omega - natural transformation and I think this should be something like a V-natural transformation between an S-functor F : C ---> D and a T-functor G: C ---> D, but placing at the beginning of the commutative diagram which defines a V-natural transformation an arrow like omega_{[X,Y]} : S[X,Y] ---> T[X,Y] . Assuming that this is ok, I should also need to understand what might be the definition of a monoidal omega - natural transformation. That is to say, C is a monoidal V-category, D is a monoidal W-category, F is a monoidal S-functor and G a monoidal T-functor: what is a monoidal natural transformation between F and G , over IÕve drawn a couple of commutative diagrams that should appear in the definition of such a construct, but I feel I could be forgetting a dozen more. Any references for it? Unfortunately for me, KellyÕs book ends before this point: it explicitely says: is our decision not to discuss the Ōchange of base-categorygiven by a symmetric monoidal functor V ---> W. Has someone else done the job after KellyÕs book? Agust.92 Roig === Subject: Re: monoidal enriched natural transformations Epigone-thread: tixslongcax Content-Length: 6836 Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Hi. >I have a problem which, it seems to me, requires the notion of >enriched natural transformation between enriched monoidal functors, >but I havenÕt been able to find a good reference for it. >IÕve taken a look at the books of Borceux (Handbook of categorical >algebra), Kelly (Basic notions of enriched category) and the appendix >in LevineÕs Mixed motives, but they all end just where I need them, >or even long before that point. >More precisely, the situation IÕve encountered seems to be the >following. I have: >- Two monoidal (symmetric) categories: > V and W . >- A couple of monoidal (symmetric) functors between them: > S, T : V ---> W >- A monoidal natural transformation between S and T : >And here is where my problems begin. >There is a well-known notion of what is a V-functor between >V-categories and what a V-natural transformation is. >My first need is to understand what an S-functor between a >V-category C and a W-category D should be: > F : C ---> D >I didnÕt find this thing in the literature, but I expect it ought to >be something like a V-functor, but with a family of morphisms in W > lambda_{XY} : S[X,Y] ---> [FX,FY] >for every pair of objects X, Y in C . (Here the square brackets >[,] stand for the objects in V and W of morphisms of C and D >, and IÕm leaving aside units and commutative isomorphisms for the >moment.) >This seems reasonable to me, since (a) is the situation I have in the >real world and (b) if I put S = id_V , I find the definition of a >V-functor. >Next, I would need the notion of an omega - natural transformation >and I think this should be something like a V-natural transformation >between an S-functor F : C ---> D and a T-functor G: C ---> D, >but placing at the beginning of the commutative diagram which defines >a V-natural transformation an arrow like > omega_{[X,Y]} : S[X,Y] ---> T[X,Y] . >Assuming that this is ok, I should also need to understand what might >be the definition of a monoidal omega - natural transformation. >That is to say, C is a monoidal V-category, D is a monoidal >W-category, F is a monoidal S-functor and G a monoidal T-functor: >what is a monoidal natural transformation between F and G , over >IÕve drawn a couple of commutative diagrams that should appear in the >definition of such a construct, but I feel I could be forgetting a >dozen more. Any references for it? >Unfortunately for me, KellyÕs book ends before this point: it >explicitely says: is our decision not to discuss the Ōchange of >base-categorygiven by a symmetric monoidal functor V ---> W. Has >someone else done the job after KellyÕs book? If you havenÕt done so already, I recommend that you get on the categories mailing list: categories@mta.ca where you would surely get a reply and advice about the literature. Alternatively, you might write Max Kelly (at the University of Sydney) or Ross Street (Macquarie University) directly. I wish I had suitable references at hand, but here are some remarks on your query. First, a monoidal functor S: V --> W induces a 2-functor S_{*}: V-Cat --> W-Cat, making straightforward use of the monoidal structure on S. If C is a V-category and D is a W-category, then what you call an S-functor is undoubtably the same as a W-functor of the form F: S_{*}C --> D. Next, a monoidal natural transformation omega: S --> T induces a 2-natural transformation between 2-functors omega_{*}: S_{*} --> T_{*} and in particular provides, for each V-category C, a W-functor of the form omega_{*}(C): S_{*}C --> T_{*}C This too is straightforward, using just the data and equations for an m.n.t. Then what you call an omega-natural transformation from F to G is undoubtably the same as a W-transformation of the form F --> G(omega_{*}(C)) where F: S_{*}C --> D and G: T_{*}C --> D are W-functors, and the target on the right is a composite of W-functors. Now to define a monoidal omega-natural transformation, you want to do a jazzed-up version of the above definitions. HereÕs what youÕll need (minimally): -- V, W braided monoidal categories -- S, T braided monoidal functors of the form V --> W -- omega a m.n.t. of the form S --> T (You can of course replace braided by symmetric, but you lose some generality in doing so.) Since V is braided monoidal, V-Cat is a monoidal 2-category, and monoidal V-categories are the same as (pseudo-)monoids in V-Cat as a monoidal 2-category. Indeed, the 2-category Mon(V-Cat) whose objects are monoidal V-categories, whose objects are monoidal V-functors, and whose 2-cells are monoidal V-transformations, is definable purely in terms of the monoidal 2-category structure on V-Cat, and therefore, the desired change of base induced by S: V --> W, Mon(V-Cat) --> Mon(W-Cat), requires only a monoidal 2-functor S_{*}: V-Cat --> W-Cat to get off the ground. The point of demanding that S be braided monoidal is so that the 2-functor S_{*} is in fact monoidal. So: under these hypotheses, a monoidal S-functor (to use your terminology) should be the same as a monoidal W-functor of the form F: S_{*}C --> D where C is a monoidal V-category and D is a monoidal W-category. Finally, if omega: S --> T is an m.n.t., there is a monoidal W-functor omega_{*}(C): S_{*}C --> T_{*}D and what you call a monoidal omega-natural transformation should just be a monoidal W-transformation of the form F --> G(omega_{*}(C)). Notice that it is unnecessary to write down a whole bunch of commutative diagrams to define these constructs: the data and axioms inherent in ordinary enriched notions and in braided monoidal notions do the work for you. However, this approach does involve some machinery of higher-dimensional categories (monoidal 2-categories, monoidal 2-functors), which is probably why Kelly didnÕt touch this in his book -- the relevant notions hadnÕt yet been formulated properly. If you want to follow up on this machinery, you might want to look at Coherence for Tricategories by Gordon, Power & Street. The full-ßedged definition of monoidal 2-category can be found there, and is shown to be equivalent in an appropriate sense to so-called Gray-monoids, which are much simpler (indeed, for V braided monoidal, V-Cat *is* a Gray-monoid). Also look at the havenÕt already done so. Todd Trimble === Subject: Re: monoidal enriched natural transformations Originator: bergv@math.uiuc.edu (Maarten Bergvelt) vas dir: >>Assuming that this is ok, I should also need to understand what might >>be the definition of a monoidal omega - natural transformation. [...] >>Unfortunately for me, KellyÕs book ends before this point: it >>explicitely says: is our decision not to discuss the Ōchange of >>base-categorygiven by a symmetric monoidal functor V ---> W. Has >>someone else done the job after KellyÕs book? >If you havenÕt done so already, I recommend that you get on the >categories mailing list: > categories@mta.ca >where you would surely get a reply and advice about the literature. >Alternatively, you might write Max Kelly (at the University of >Sydney) or Ross Street (Macquarie University) directly. >I wish I had suitable references at hand, but here are some remarks >on your query. [...] >Todd Trimble Agust.92 Roig === Subject: weighted tree generation Originator: bergv@math.uiuc.edu (Maarten Bergvelt) For a given undirected graph G(V,E) I would like to generate all possible rooted trees. How many will there be? If there are a lot, at least I would like to generate a set of different ones or make a single change to a given tree. Secondly, is there any standard method to assign weights to an undirected graph such that for a given root, the all shortest paths algorithm (e.g. dijkstra) will yield a given tree? (I guess that you could assign low values to the links that are elements of the tree and high values for all others, but can it be proven that it will always yield the given tree?) Diego diego at aulignac dot com www.aulignac.com === Subject: fundamental bounds on the elements of covariance matrices Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Let x and y be n-tuplets of normally distributed random variables, and let C = cov(x,yÕ) be the nxn covariance matrix of x and y. Let z be some fixed n-tuplet whose elements are all strictly positive. 1) Are all the elements of the n-vector, Cz, non-negative? 2) Is zÕCz > 0? Where can I read more about bounds on the elements of covariance matrices? === Subject: Re: fundamental bounds on the elements of covariance matrices Originator: bergv@math.uiuc.edu (Maarten Bergvelt) See Morris L. EatonÕs book with multivariate in the title. I was about to answer that any matrix whatsoever could be in the role of C, but I find IÕm a bit rusty. -- Mike Hardy > Let x and y be n-tuplets of normally distributed random variables, and let > C = cov(x,yÕ) be the nxn covariance matrix of x and y. > Let z be some fixed n-tuplet whose elements are all strictly positive. > 1) Are all the elements of the n-vector, Cz, non-negative? > 2) Is zÕCz > 0? > Where can I read more about bounds on the elements of covariance matrices? === Subject: Re: binary vector packings Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Could someone please help me identify the following problem? > Consider binary arrays {u_i} of length n with k 1Õs (and n-k 0Õs). How > many can we choose such that all pairwise inner products > sum_i u_i v_i < t ? > Equivalently, what is the maximum number of k-subsets of the n-set with > pairwise intersections less than t elements? > Does this problem, or some equivalent, have a name? Any references? Looks like your problem is more or less equivalent to finding the maximum size of a certain constant weight code (aka fixed weight code). Searching with those buzzwords should lead you to sources of known results, tables of upper bounds etc. Jyrki Lahtonen, Turku, Finland === Subject: two fibrations Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hallo, consider an isometric action of a compact Lie group G on a connected Riemannian manifold M with a fixed point p. I am interested in the topology (especially the homology) of the space of paths in M, starting at p and ending in some fixed orbit N=Gq, denoted by P(M,ptimes N). Even for arbitrary p and submanifolds N, the end point map P(M,ptimes N)to N; c mapsto c(1) is a fibration with fibre P(M,ptimes q), the space of paths from p to q (which ist homotopy equivalent to the space of loops on M). So I can deduce some information on the homology from this fibration, e.g. by using the Leray-Serre spectral sequence. Furthermore, now restricting to the case of N being some orbit, we have P(M,ptimes Gq)=P(M,ptimes q)times_{G_q} G (twisted product - G_q is the isotropy group at q), which can be easily seen by regarding the mapping P(M,ptimes q)times Gto P(M,ptimes Gq); (c,g)mapsto gc (well-defined since p is fixed). Summarizing, we have two fibrations: i) P(M,ptimes q)to P(M,ptimes Gq)to Gq and ii) G_qto P(M,ptimes q)times G to P(M,ptimes Gq). Now my question: Does the existence of such two fibrations give any new relation between the topology of these spaces? Is there some method of extracting information from such two similar-looking fibrations? Oliver Goertsches === Subject: RA Positions at UNR Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The Computer Vision Laboratory (CVL) at the University of Nevada, Reno (UNR) invites applications for research assistant positions starting in Spring2005/Fall 2005. Preference will be given to students who want to pursue a PhD degree in Computer Vision. Active research areas within CVL include object recognition, visual motion analysis, face detection and recognition, biometrics, tracking and pose estimation of human body/head/hand/eye-gaze, surveillance and activity recognition. CVL is currently funded by NSF, NASA, ONR, and Ford Motor Company. We are also collaborating with several government and industry laboratories. For more information, please visit http://www.cs.unr.edu/CVL Requirements: You must have a first degree in either an Engineering subject, in Mathematics, in Physics, or in Computer Science. Good Mathematical background, programming skills in C or C++, and familiarity with Unix/Linux/Windows are necessary. Prior familiarity with Image Processing, Computer Vision, Pattern Recognition, and Machine Learning is desirable. Good communication and writing skills in English are essential. Interested students should send their CV by regular mail, e-mail, or fax to Dr. George Bebis (bebis@cs.unr.edu) or Dr. Mircea Nicolescu (mircea@cs.unr.edu) Dr. George Bebis Department of Computer Science & Engineering University of Nevada Reno, NV 89557, USA phone: (775) 784-6463 email: bebis@cs.unr.edu http://www.cs.unr.edu/~bebis Dr. Mircea Nicolescu Department of Computer Science & Engineering University of Nevada Reno, NV 89557, USA phone: (775) 784-4356 email: mircea@cs.unr .edu http://www.cs.unr.edu/~mircea === Subject: A type of regular graph Originator: bergv@math.uiuc.edu (Maarten Bergvelt) For a graph with verticies the integers mod n, we may draw an edge between a and b iff b=a+i for some i in a set of residues mod n. Clearly such a graph is a regular graph, is it possible to characterize it further with known graph-theoretic properties? === Subject: Re: A type of regular graph 3QLpj-NoP*NzsIC,boYU]bQ]HÕy<#4ga3$21: Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > For a graph with verticies the integers mod n, we may draw an edge > between a and b iff b=a+i for some i in a set of residues mod n. > Clearly such a graph is a regular graph, is it possible to > characterize it further with known graph-theoretic properties? These graphs are known as circulants, e.g. see . That doesnÕt answer your question, but it should at least help in searching for an answer. -- David Eppstein Computer Science Dept., Univ. of California, Irvine http://www.ics.uci.edu/~eppstein/ === Subject: building 3d shapes Originator: bergv@math.uiuc.edu (Maarten Bergvelt) hi world, iÕm looking for toolkits of segments / magnets / plastic caps to build 3d shapes like the 1- or 2-skeleta of tetrahedra, octahedra, etc. i want to play around with the cell decomposition of the lattice a_3, so iÕll need quite a bunch of these. i quickly searched on the net for some way of buying these online, with no success. does anybody know where i can find them? tia, laurent -- Laurent Bartholdi laurent.bartholdiepßch EPFL, IGAT, B.89timent BCH T.8el.8ephone: +41 21-6930380 CH-1015 Lausanne, Switzerland Fax: +41 21-6930385 === Subject: Re: building 3d shapes Originator: israel@math.ubc.ca (Robert Israel) > hi world, > iÕm looking for toolkits of segments / magnets / plastic caps to build 3d > shapes like the 1- or 2-skeleta of tetrahedra, octahedra, etc. > i want to play around with the cell decomposition of the lattice a_3, so > iÕll need quite a bunch of these. > i quickly searched on the net for some way of buying these online, with no > success. does anybody know where i can find them? One possibility: http://www.zometool.com/ -- http://hertzlinger.blogspot.com === Subject: Re: building 3d shapes Originator: israel@math.ubc.ca (Robert Israel) > hi world, > iÕm looking for toolkits of segments / magnets / plastic caps to > build 3d shapes like the 1- or 2-skeleta of tetrahedra, octahedra, > etc. > i want to play around with the cell decomposition of the lattice a_3, > so iÕll need quite a bunch of these. > i quickly searched on the net for some way of buying these online, > with no success. does anybody know where i can find them? > tia, laurent Or try Geomag and Supermag (donÕt know if these are the same), e.g. at http://www.toymagnets.com/geomag/index.cfm or http://www.geomags.com/. IÕve seen them sold in Swiss toyshops as well. Christian Graf === Subject: Re: building 3d shapes Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > hi world, > iÕm looking for toolkits of segments / magnets / plastic caps to build 3d > shapes like the 1- or 2-skeleta of tetrahedra, octahedra, etc. > i want to play around with the cell decomposition of the lattice a_3, so > iÕll need quite a bunch of these. > i quickly searched on the net for some way of buying these online, with no > success. does anybody know where i can find them? > tia, laurent > -- > Laurent Bartholdi laurent.bartholdiepßch > EPFL, IGAT, B.89timent BCH T.8el.8ephone: +41 21-6930380 > CH-1015 Lausanne, Switzerland Fax: +41 21-6930385 For magnets look at http://www.supermagnete.ch/magnets.php?at=Z Hugo Pfoertner === Subject: Re: building 3d shapes Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > .... > iÕm looking for toolkits of segments / magnets / plastic caps to build 3d > shapes like the 1- or 2-skeleta of tetrahedra, octahedra, etc. > i want to play around with the cell decomposition of the lattice a_3, so > iÕll need quite a bunch of these. > i quickly searched on the net for some way of buying these online, with no > success. does anybody know where i can find them? .... Is http://www.korthalsaltes.com/ any use to you? Ken Pledger. === Subject: natural numbers as coequalizer (Re: Those Naughty Category Theorists) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) begin{quote} Incidentally for those who think that by defining numbers as lengths of strings IÕm making some sort of obscure automata theoretic point with my computer science hat on, let me just point out that not only can this representation be made entirely respectable mathematically, but via a slicker mathematical name than any other IÕve seen proposed in this thread so far: the free monoid (N,+,0) on one generator. Just as punchy, and as clear to a category theorist as free monoid is to an algebraist: the coequalizer of the elements of (the ordinal) *2. The set-theoretic explanation of this is a bit clumsy, but here goes. The scene is Cat, the category of all small categories. The diagram to be coequalized is the left half of 0 ----> F *1 *2 ----> Coeq(0,1) ----> 1 with the coequalizer Coeq(0,1) and its coequalizing arrow F shown on the right. The ordinal *2 is the one-nonidentity-arrow category {0->1}, the ordinal *1 is the evident {0}, the elements 0,1 form the set Hom(*1,*2). The coequalizer of 0 and 1 creates Coeq(0,1), a copy of *2 which identifies 0 and 1. This has the side effect of looping the nonidentity arrow back on itself. Since we are in Cat, we now have to specify a composition law for this arrow with itself in the least constraining way, i.e. Coeq(0,1) has to be universal. Clearly we need all composites f, ff, fff, etc. Identifying any two of these is an unwanted constraint, so we leave them all unidentified. We now have a monoid whose arrows 1,f,ff,fff,... represent the natural numbers 0,1,2,3,..., composition represents addition, and the identity arrow represents 0. (Represent is meaningful only in this set-theoretic view.) The functor F takes both objects of *2 to the object of Coeq(0,1), and takes the nonidentity arrow of *2 to f or 1, the generator of (N,+,0). This construction may seem a bit contrived until you look at how category theoretic foundations are typically organized. (Good reading: McLarty, Axiomatizing a Category of Categories, J.Symbolic Logic, 56:4(DecÕ91).) Ordinal constructions involving the four ordinals up to *3, along with the product *2 x *2, are at the heart of this organization, and the above construction of the natural numbers as a monoid in Cat is not only slick but very natural and in that setting. end{quote} Can we construct in a similar manner integers, rational, reals and complex numbers? David, === Subject: Re: SchlomilchÕs series Epigone-thread: yendwholgrimp Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Hi--IÕm a newbie to special functions, I hope somebody can help me out. IÕm doing a problem in scattering, and it would be really helpful if I could sum the following series: sum_{n=1}^inf J_0(n x + b) where b is a real number. If b=0 then this is a bread & butter Schlomilch series (see e.g. Gradshteyn and Ryzhik 5th ed. 8.521). I tried expanding J_0(n x + b) via the addition theorem. But I didnÕt get too far. I was wondering if this was a known series...hoping the experts have some help! :) === Subject: Re: SchlomilchÕs series Epigone-thread: yendwholgrimp Content-Length: 782 Originator: rusin@vesuvius There is a sort of obvious way to do it if b is an integer...take your original sum, replace x by 2x and then subtract the two infinite sums, etc. For arbitrary b IÕm really not sure. >Hi--IÕm a newbie to special functions, I hope somebody can help me >out. IÕm doing a problem in scattering, and it would be really >helpful if I could sum the following series: > sum_{n=1}^inf J_0(n x + b) >where b is a real number. >If b=0 then this is a bread & butter Schlomilch series (see e.g. >Gradshteyn and Ryzhik 5th ed. 8.521). >I tried expanding J_0(n x + b) via the addition theorem. But I didnÕt >get too far. >I was wondering if this was a known series...hoping the experts have >some help! :) === Subject: Re: SchlomilchÕs series Content-Length: 1302 Originator: rusin@vesuvius Please donÕt top-post. IÕm putting the original question here where it belongs: >>Hi--IÕm a newbie to special functions, I hope somebody can help me >>out. IÕm doing a problem in scattering, and it would be really >>helpful if I could sum the following series: >> sum_{n=1}^inf J_0(n x + b) >There is a sort of obvious way to do it if b is an integer...take >your original sum, replace x by 2x and then subtract the two infinite >sums, etc. Sorry, I donÕt understand this. You seem to be saying take the original sum, J_0(x+b) + J_0(2x+b) + J_0(3x+b) + ... and subtract J_0(2x+b) + J_0(4x+b) + J_0(6x+b) + .... But that will just give you J_0(x+b) + J_0(3x+b) + J_0(5x+b) + ... and I donÕt see how that helps, or what difference b being an integer makes. On the other hand, if b is an integer multiple of x, you can say something: if F(x,b) is the sum, F(x,b+x) = F(x,b) - J_0(x+b) so F(x, kx) = F(x,0) - sum_{j=1}^k J_0(jx) for positive integers k Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: On a property of Bernoulli numbers Epigone-thread: clephoystald Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Let A(n,k)=(n^(2k)-1)*B(2k) where B(2k) denotes the 2k-th Bernoulli >number. Then I suspect the existence of a minimal positive rational >value, depending on n, r(n)=P(n)/Q(n) with (P(n),Q(n))=1 and such >that : >for any k>0 r(n)*A(n,k) is an integer value. >r(2)=2, r(3)=3/4 .... >P(n) appears to be the largest square-free divisor of n but I didnÕt >observation when n is a power of 2 : >for p prime, if 2^p-1 and (2^p+1)/3 are both primes then >Q(2^p)=(4^p-1)/3 (converse doesnÕt hold). >Can anyone confirm theorically the existence of r(n) and the formula >for P(n)? If so, what is the formula for Q(n)? Update : I found that r(n)=rad(n^3-n)/(n^2-1) where rad(n) is the square-free kernel of n, the largest square-free divisor of n. This explains why P(n)=rad(n) and we have Q(n)=(n^2-1)/rad(n^2-1). I canÕt say if this property of BernoulliÕs numbers is known. Studying r(n) I came across something looking as an integer formulation of AgohÕs conjecture : p is prime iff p divides (p^p-p)*B(p-1)-1 and I unearthed this amusing connection with 3-smooth numbers (numbers of form 2^i*3^j i,j>=0) : fractional part of ((n^(2k)-1)*B(2k)) is constant for any k>0 iff n is a 3-smooth number. Benoit Cloitre === Subject: Paper published by Algebraic and Geometric Topology Originator: bergv@math.uiuc.edu (Maarten Bergvelt) The following paper has been published: Algebraic and Geometric Topology URL: http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-42.abs.html Title: A class of tight contact structures on Sigma_2 x I Author(s): Tanya Cofer Abstract: We employ cut and paste contact topological techniques to classify some tight contact structures on the closed, oriented genus-2 surface times the interval. A boundary condition is specified so that the Euler class of the of the contact structure vanishes when evaluated on each boundary component. We prove that there exists a unique, non-product tight contact structure in this case. Secondary: 53C15 Keywords: Tight, contact structure, genus-2 surface Author(s) address(es): Department of Mathematics, Northeastern Illinois University 5500 North St Louis Avenue, Chicago, IL 60625-4699, USA Email: T-Cofer@neiu.edu URL: http://www.neiu.edu/~tcofer/ === Subject: This week in the mathematics arXiv (18 Oct - 22 Oct) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Here are this weekÕs titles in the mathematics arXiv, available at: http://front.math.ucdavis.edu/ http://front.math.ucdavis.edu/submissions This week in the mathematics arXiv may be freely redistributed with attribution and without modification. Titles in the mathematics arXiv (18 Oct - 22 Oct) ------------------------------------------------- AC: Commutative Algebra ----------------------- math.AC/0410375 Kiran S. Kedlaya: Finite automata and algebraic extensions of function fields math.AC/0410340 Claudia Polini, Bernd Ulrich: A formula for the core of an ideal AG: Algebraic Geometry ---------------------- math.AG/0410469 FrÕedÕeric Campana: Fibres multiples des surfaces math.AG/0410458 Samuel Boissiere: Chern classes of the tangent bundle on the Hilbert scheme of points on the affine plane math.AG/0410444 S. Kaplan, E. Liberman, M. Teicher: Braid Monodromy Computation of Real Singular Curves math.AG/0410442 Nickolas Michelacakis, Apostolos Thoma: On the geometry of complete intersection toric varieties math.AG/0410432 Qi Zhang: On projective varieties with nef anticanonical divisors hep-th/0410055 Volker Braun, Burt A. Ovrut, Tony Pantev, Rene Reinbacher: Elliptic Calabi-Yau Threefolds with Z_3 x Z_3 Wilson Lines math.AG/0410408 Ivan Cheltsov: Double cubics and double quartics hep-th/0410018 A. Klemm, M. Kreuzer, E. Riegler, E. Scheidegger: Topological String Amplitudes, Complete Intersection Calabi-Yau Spaces and Threshold Corrections math.AG/0410401 GÕabor SzÕekelyhidi: Extremal metrics and K-stability math.AG/0410394 Ana Cristina Lopez: Relative Jacobians of elliptic fibrations with reducible fibers math.AG/0410393 Ana Cristina Lopez: Simpson Jacobians of reducible curves math.AG/0410392 variety math.AG/0410388 M. E. Kazaryan, S. K. Lando: Towards the Intersection Theory on Hurwitz Spaces math.AG/0410383 Philibert Nang, Kiyoshi Takeuchi: Addendum to the paper Characteristic Cycles of Perverse Sheaves and Milnor Fibers math.AG/0410379 Seongchun Kwon: Transversality properties on the moduli space of genus 0 stable maps to a smooth rational projective surface and their real enumerative implications math.AG/0410378 Silvano Baggio: Equivariant K-Theory of Smooth Toric Varieties math.AG/0410360 Tyler J. Jarvis, William E. Lang, Nansen Petrosyan, Gretchen Rimmasch, Julie Rogers, Erin D. Summers: Classification of Singular Fibres on Rational Elliptic Surfaces in Characteristic Three hep-th/0410170 Bjorn Andreas, Daniel Hernandez Ruiperez: U(n) Vector Bundles on Calabi-Yau Threefolds for String Theory Compactifications math.AG/0410349 Igor Burban, Bernd Kreussler: On a relative Fourier-Mukai transform on genus one fibrations math.AG/0410346 Oliver Lorscheid: Completeness and compactness for varieties over local fields AP: Analysis of PDEs -------------------- math.AP/0410475 Zhongwei Shen: Bounds of Riesz Transforms on $L^p$ Spaces for Second Order Elliptic Operators math.AP/0410462 Fernando Cardoso, Georgi Vodev: Weighted L^p decay estimates of solutions to the wave equation with a potential math.AP/0410452 A.G.Ramm: Existence of a solution to a nonlinear equation math.AP/0410451 A.G.Ramm: A singular perturbation problem math.AP/0410443 Arnaud Debussche, Cyril Odasso: Ergodicity for the weakly damped stochastic non-linear Schrodinger equations math.AP/0410441 Giuseppe Da Prato, Arnaud Debussche, Luciano Tubaro: Coupling for some partial differential equations driven by white noise math.AP/0410431 Burak Erdogan, Wilhelm Schlag: Dispersive estimates for Schr{o}dinger operators in the presence of a resonance and/or an eigenvalue at zero energy in dimension three: I math.AP/0410416 Dian Palagachev, Lubomira Softova: Fine regularity for elliptic systems with discontinuous ingredients math.AP/0410415 Dian K. Palagachev, Lubomira G. Softova: AÕpriori estimates and precise regularity for parabolic systems with discontinuous data math.AP/0410380 Fabian Waleffe: On some dyadic models of the Euler equations math.AP/0410344 Isabelle Gallagher, Thierry Gallay, Pierre-Louis Lions: On the uniqueness of the solution of the two-dimensional Navier-Stokes equation with a Dirac mass as initial vorticity AT: Algebraic Topology ---------------------- math.AT/0410405 Scott O. Wilson: Partial Algebras Over Operads of Complexes and Applications math.AT/0410398 R. Brown, H.K. Kamps, T. Porter: A homotopy double groupoid of a Hausdorff space II: a van Kampen theorem math.AT/0410374 Norio Iwase, Donald Stanley, Jeffrey Strom: Implications of the Ganea Condition math.AT/0410367 Z. Fiedorowicz, R. M. Vogt: Topological Hochschild Homology of $E_n$-Ring Spectra math.AT/0410363 A.D.R. Choudary, A. Dimca, S. Papadima: Some Analogs of Zariski Theorem on Nodal Line Arrangements math.AT/0410342 Nicholas J. Kuhn: Goodwillie towers and chromatic homotopy: an overview CA: Classical Analysis and ODEs ------------------------------- math.CA/0410439 JosÕe L. LÕopez, Nico M. Temme: Convergent Asymptotic Expansions of Charlier, Laguerre and Jacobi Polynomials math.CA/0410436 JosÕe L. LÕopez, Nico M. Temme: Multi-point Taylor Expansions of Analytic Functions math.CA/0410395 Sever Silvestru Dragomir: Some Inequalities for Functions of Bounded Variation with Applications to Landau Type Results CO: Combinatorics ----------------- math.CO/0410471 Michiel Hazewinkel: Word Hopf algebras math.CO/0410466 Charles F. Dunkl: Hook-lengths and Pairs of Compositions math.CO/0410455 David E Speyer: Tropical Linear Spaces math.CO/0410429 Jens Christian Claussen: Time-evolution of the Rule 150 cellular automaton activity from a Fibonacci iteration math.CO/0410425 Joseph E. Bonin, Omer Gimenez: Multi-Path Matroids math.CO/0410410 Annalies Vuong, M. Ian Wyckoff: Conditions for Weighted Cover Pebbling of Graphs math.CO/0410404 F. Bonetto, H. Matzinger: Fluctuations of the Longest Common Subsequence in the Asymmetric Case of 2- and 3-Letter Alphabets math.CO/0410382 B.M. Kim, Y. Rho: Van der WaerdenÕs Theorem on Homothetic copies of {1,1+s, 1+s+t} math.CO/0410373 Ira M. Gessel, Louis H. Kalikow: Hypergraphs and a functional equation of Bouwkamp and de Bruijn math.CO/0410366 Michiel Hazewinkel: Explicit polynomial generators for the ring of quasi-symmetric functions over the integers math.CO/0410361 Howard Kleiman: The Floyd-WarshallAlgorithm and the Asymmetric TSP nlin.AO/0407024 Fatihcan M. Atay, Tuerker Biyikoglu, Juergen Jost: On the synchronization of networks with prescribed degree distributions math.CO/0410347 Svante Linusson, Johan Waestlund: Completing a k-1 assignment math.CO/0410345 Svante Linusson, John Shareshian, Volkmar Welker: Complexes of graphs with bounded matching size CT: Category Theory ------------------- math.CT/0410412 Dominic Verity: Complicial Sets CV: Complex Variables --------------------- math.CV/0410445 P. Ebenfelt, L. P. Rothschild: Transversality of CR mappings math.CV/0410420 Rostyslav O. Hryniv, Yaroslav V. Mykytyuk: Asymptotics of zeros for some entire functions math.CV/0410399 Vladimir V. Kisil, Debapriya Biswas: Elliptic, Parabolic and Hyperbolic Analytic Function Theory--0: Geometry of Domains math.CV/0410390 Franc Forstneric, Joerg Winkelmann: Holomorphic discs with dense images math.CV/0410386 Franc Forstneric, Christine Laurent-Thiebaut: Stein compacts in Levi-ßat hypersurfaces math.CV/0410376 Bertrand Deroin: Laminations dans les esapces projectifs complexes math.CV/0410362 Young-Heon Kim: Holomorphic extensions of determinants of Laplacians math.CV/0410353 J. J. Kohn: Superlogarithmic estimates on pseudoconvex domains and CR manifolds math.CV/0410343 Mikhail Sodin: Zeroes of Gaussian analytic functions math.CV/0410341 Fedor Nazarov, Mikhail Sodin: Coarse equidistribution of the argument of entire functions of finite order DG: Differential Geometry ------------------------- math.DG/0410461 Josef Janyv{s}ka: Natural connections given by general linear and classical connections math.DG/0410460 Tom Mestdag, Bavo Langerock: A Lie algebroid framework for non-holonomic systems math.DG/0410456 Mikhail G. Katz, Yuli B. Rudyak: Lusternik-Schnirelmann category and systolic category of low dimensional manifolds nlin.SI/0407057 Paolo Lorenzoni, Marco Pedroni: On the bi-Hamiltonian structures of the Camassa-Holm and Harry Dym equations math.DG/0410435 Isabel Fernandez, Francisco J. Lopez: Relative parabolicity of zero mean curvature surfaces in $R^3$ and $R_1^3$ math.DG/0410434 Michael Schulze: On the resolvent of the Laplacian on functions for degenerating surfaces of finite geometry math.DG/0410418 Jian Song, Ben Weinkove: On the convergence and singularities of the J-ßow with applications to the Mabuchi energy math.DG/0410413 2-surfaces of prescribed mean curvature DS: Dynamical Systems --------------------- math.DS/0410464 I.Dynnikov, S.Novikov: Topology of quasiperiodic functions on the plane physics/0410160 R. Ball: The case of the trapped singularities math.DS/0410417 Charles Favre, Mattias Jonsson: Eigenvaluations math.DS/0410384 N. Haydn, Y. Lacroix & S. Vaienti: Hitting and return times in ergodic dynamical systems math.DS/0410355 Marco Lenci: Typicality of recurrence for Lorentz gases nlin.CD/0410019 Sylvie Oliffson Kamphorst, Sonia Pinto de Carvalho: The First Birkhoff Coefficient and the Stability of 2-Periodic Orbits on Billiards FA: Functional Analysis ----------------------- math.FA/0410427 W. B. Johnson, N. L. Randrianarivony: $ell_p$ (p>2) does not coarsely embed into a Hilbert space math.FA/0410422 Assaf Naor, Yuval Peres, Oded Schramm, Scott Sheffield: Markov chains in smooth Banach spaces and Gromov hyperbolic metric spaces math.FA/0410403 Stefan Bildea, Dorin Ervin Dutkay, Gabriel Picioroaga: MRA Super-wavelets math.FA/0410391 Roberto GiambÕo, Fabio Giannoni, Paolo Piccione: Orthogonal Geodesic Chords, Brake Orbits and Homoclinic Orbits in Riemannian Manifolds math.FA/0410351 FrÕedÕeric Bayart, Catherine Finet, Daniel Li, HervÕe QueffÕelec: Composition operators on the Wiener-Dirichlet algebra math.FA/0410348 Wojciech Czaja: Remarks on NaimarkÕs duality GM: General Mathematics ----------------------- math.GM/0410377 Jacky Cresson: Non-differentiable variational principles GT: Geometric Topology ---------------------- math.GT/0410476 J.-F. Lafont: Strong Jordan separation and applications to rigidity math.GT/0410474 Brent Everitt, John Ratcliffe, Steven Tschantz: The smallest hyperbolic 6-manifolds math.GT/0410433 Carlo Petronio: Complexity of 3-orbifolds math.GT/0410381 Suhyoung Choi: Drilling cores of hyperbolic 3-manifolds to prove tameness math.GT/0410370 Jerzy Dydak, Michael Levin: Extension of maps to the projective plane math.GT/0410369 Michael Levin: Rational acyclic resolutions math.GT/0410368 Michael Levin: Universal acyclic resolutions for arbitrary coefficient groups math.GT/0410358 Robion Kirby, Paul Melvin: Local surgery formulas for quantum invariants and the Arf invariant math.GT/0410356 Eaman Eftekhary: Filtration of Heegaard Floer homology and gluing formulas HO: History and Overview ------------------------ math.HO/0410411 Tommaso Toffoli: MaxwellÕs daemon, the Turing machine, and Jaynesrobot math.HO/0410397 V.G.Gurzadyan: Kolmogorov and Aleksandrov in Sevan Monastery, Armenia, 1929 MG: Metric Geometry ------------------- math.MG/0410440 Andreas Balser, Alexander Lytchak: Centers of convex subsets of buildings math.MG/0410437 Andreas Balser, Alexander Lytchak: Building-like spaces math.MG/0410421 Alexander Lytschak, Viktor Schroeder: Affine functions on CAT(kappa) spaces MP: Mathematical Physics ------------------------ quant-ph/0410131 Xiong-Jun Liu, Hui Jing, Xin Liu, Mo-Lin Ge: Dynamical Symmetry and Its Applications In Electromagnetically Induced Transparency math-ph/0410046 Michiel Hazewinkel, Hugo H Torriani: Coherence and uniqueness theorems for averaging processes in statistical mechanics hep-th/0410199 A.P. Balachandran, A. Pinzul: On Time-Space Noncommutativity for Transition Processes and Noncommutative Symmetries hep-th/0008117 M. Hssaini, M. Kessabi, B. Maroufi, M.B.Sedra: Central extended D=2 N=4 SU(2) Liouville self interacting model and explicit hyperkahler metric math-ph/0410045 Alexei F. Cheviakov: Plasma equilibrium equations in coordinates connected with magnetic surfaces. Exact equilibrium solutions gr-qc/0410069 Antonio Lopez-Pinto: Nonstandard spin 2 field theory physics/0410127 A. Figotin, J. H. Schenker: Hamiltonian treatment of time dispersive and dissipative media within the linear response theory math-ph/0410044 Daniel Peralta-Salas: A geometric approach to the equilibrium shapes of self-gravitating ßuids hep-th/0410013 Patrick Dorey, Adam Millican-Slater, Roberto Tateo: Beyond the WKB approximation in PT-symmetric quantum mechanics cond-mat/0410435 F. Guerra: Mathematical aspects of mean field spin glass theory math-ph/0410043 Volodymyr Sushch: On some discrete model of the magnetic Laplacian math-ph/0410042 Jochen Bruening, Vladimir Geyler, Konstantin Pankrashkin: Continuity of integral kernels related to Schrodinger operators on manifolds math-ph/0410041 O.M. Kiselev, S.G. Glebov, V.A. Lazarev: Resonant pumping in nonlinear Klein-Gordon equation and solitary packets of waves math-ph/0410040 G.Giachetta, L.Mangiarotti, G.Sardanashvily: Geometric and Algebraic Topological Methods in Quantum Mechanics hep-th/0408241 S. Meljanac, A. Samsarov: Matrix oscillator and Calogero-type models math-ph/0410039 Nasser Saad, Richard L. Hall, Qutaibeh D. Katatbeh: Study of anharmonic singular potentials NT: Number Theory ----------------- math.NT/0410428 L.A.Gutnik: On the difference equation of the Poincare type math.NT/0410409 A. Agboola: Galois modules and p-adic representations math.NT/0410387 C. S. Rajan: Recovering modular forms and representations from tensor and symmetric powers math.NT/0410372 Mark van Hoeij: Solving conics over Q(t1,..,tk) OA: Operator Algebras --------------------- math.OA/0410449 Kenneth Davidson, Elias Katsoulis: Nest representations of directed graph algebras math.OA/0410426 BenjamÕ{i}n ItzÕa-Ortiz: Eigenvalues, K-theory and Minimal Flows math.OA/0410400 Marius Dadarlat: On the topology of the Kasparov groups and its applications OC: Optimization and Control ---------------------------- math.OC/0410467 Antonios Armaou, Ioannis G. Kevrekidis: Equation-free optimal switching policies for bistable reacting systems using coarse time-steppers PR: Probability --------------- math.PR/0410465 Federico Camia: Scaling Limit and Critical Exponents for Two-Dimensional Bootstrap Percolation math.PR/0410459 Florent Benaych-Georges: Taylor expansions of R-transforms, application to supports and moments math.PR/0410457 Catherine Donati-Martin: Large deviations for Wishart processes math.PR/0410453 Patrick Cheridito, Freddy Delbaen, Michael Kupper: Dynamic monetary risk measures for bounded discrete-time processes math.PR/0410447 Michail Loulakis: On the Symmetry of the Diffusion Coefficient in Asymmetric Simple Exclusion math.PR/0410430 Yuval Peres, David Revelle: Scaling limits of the uniform spanning tree and loop-erased random walk on finite graphs math.PR/0410414 Robert C. Dalang, Carl Mueller, Lorenzo Zambotti: Hitting properties of s.p.d.e.Õs with reßection math.PR/0410402 David J. Aldous, Lea Popovic: A critical branching process model for biodiversity math.PR/0410371 Harry Kesten, Vladas Sidoravicius: A phase transition in a model for the spread of an infection math.PR/0410359 Bela Bollobas, Oliver Riordan: A short proof of the Harris-Kesten Theorem cond-mat/0410309 V. Sood, S. Redner, D. ben-Avraham: First Passage Properties of the Erdos-Renyi Random Graph QA: Quantum Algebra ------------------- math.QA/0410470 Michiel Hazewinkel: Symmetric functions, noncommutative symmetric functions, and quasisymmetric functions II math.QA/0410468 Michiel Hazewinkel: Symmetric functions, noncommutative symmetric functions, and quasisymmetric functions math.QA/0410463 S. SinelÕshchikov, A. Stolin, L. Vaksman: A Quantum Analogue of the Bernstein Functor math.QA/0410450 J.E. McClure: On the chain-level intersection pairing for PL manifolds math.QA/0410448 Christian Blohmann: Reconstruction of universal Drinfeld twists from representations math.QA/0410446 Haisheng Li, Gaywalee Yamskulna: On certain vertex algebras and their modules associated with vertex algebroids math.QA/0410407 K. Szlachanyi: Monoidal Morita equivalence math.QA/0410396 S. SinelÕshchikov, L. Vaksman: Quantum groups and non-commutative complex analysis math.QA/0410389 Harald Grosse, Stefan Schraml: The Eigenfunctions of the q-Harmonic Oscillator on the Quantum Line math.QA/0410365 Michiel Hazewinkel: The primitives of the Hopf algebra of noncommutative symmetric functions math.QA/0410364 Michiel Hazewinkel: Hopf algebras of endomorphisms of Hopf algebras math.QA/0410350 Henrique Bursztyn, Stefan Waldmann: Hermitian star products are completely positive deformations RA: Rings and Algebras ---------------------- math.RA/0410473 Gizem Karaali: A New Lie Bialgebra Structure on sl(2,1) math.RA/0410406 Michael Pinsker: The number of unary clones containing the permutations on an infinite set RT: Representation Theory ------------------------- math.RT/0410472 Paolo Bravi, Guido Pezzini: Wonderful varieties of type D math.RT/0410454 Francois Digne, Jean Michel, Raphael Rouquier: Cohomologie des varietes de Deligne-Lusztig math.RT/0410423 Calin Chindris: Quivers, long exact sequences and Horn type inequalities math.RT/0410357 Helmer Aslaksen, Mong Lung Lang: Extending $pi$-systems to bases of root systems SG: Symplectic Geometry ----------------------- math.SG/0410352 Eaman Eftekhary: Embedded curves and Gromov-Witten invariants of three-folds SP: Spectral Theory ------------------- math.SP/0410438 M.A. Kaashoek, A.L. Sakhnovich: Discrete skew selfadjoint canonical systems and the isotropic Heisenberg magnet model ST: Statistics -------------- math.ST/0410424 George Kahrimanis, Daniel Berleant: Direct pivotal predictive inference math.ST/0410419 Grace Wahba: An introduction to (smoothing spline) ANOVA models in RKHS with examples in geographical data, medicine, atmospheric science and machine learning math.ST/0410385 Mario Ruetti, Matthias Troyer, Wesley P. Petersen: A Generic Random Number Generator Test Suite math.ST/0410354 Stephane Gaiffas: Rates of convergence for pointwise curve estimation with a degenerate design -- / Greg Kuperberg (UC Davis) / Home page: http://www.math.ucdavis.edu/~greg/ / Visit the Math ArXiv Front at http://front.math.ucdavis.edu/ / * All the math thatÕs fit to e-print * === Subject: Question about Gomory-Cut Originator: israel@math.ubc.ca (Robert Israel) for an application I have to find an integer solution. I use the simplex-algorithm and the Gomory-cut. It works fine, but it is to slowly. There are multiply opportunities to make a Gomory-cut. Which Gomory-cuts should I take, to get a solution in shortest time ? Where can I get information about that (in the internet) ? Ulrich === Subject: Combinatorial inequality Epigone-thread: wunloifo Originator: israel@math.ubc.ca (Robert Israel) Can anyone help me in deriving this inequality which appears to be true? {C(n,k)/C(a,k)}*(m/n)^k <={((n-m)/(n-a))^(n-a)}*{(n/a)^a} where k=(a-m)/(1-m/n); and 0Can anyone help me in deriving this inequality which appears to be >true? > {C(n,k)/C(a,k)}*(m/n)^k <={((n-m)/(n-a))^(n-a)}*{(n/a)^a} > where k=(a-m)/(1-m/n); and 1 Here C(n,k) means the binomial coefficient and ^ means >exponentiation,* denotes ordinary multiplication. === Subject: Conformal Mapping Question Originator: israel@math.ubc.ca (Robert Israel) IÕd like to find the explicit formula of a bijective conformal mapping from the following region, say K, K={ x+iy in C mid x>0, y > arccos(e^{-x}) } to the interior of the unit disk (K is just the subregion in the first quadrant of the complex plane bounded below by the graph of e^{x}cos(y)=1). IÕve tried basic ones and looked into Dictionary of Conformal Representations by H. Kober, but so far nothing works for me. Any suggestions are greatly appreciated. -- So Okada Ph.D. Student of Math at UMass Amherst okada@math.umass.edu === Subject: This week in the mathematics arXiv (25 Oct - 29 Oct) Originator: israel@math.ubc.ca (Robert Israel) Here are this weekÕs titles in the mathematics arXiv, available at: http://front.math.ucdavis.edu/ http://front.math.ucdavis.edu/submissions This week in the mathematics arXiv may be freely redistributed with attribution and without modification. Titles in the mathematics arXiv (25 Oct - 29 Oct) ------------------------------------------------- AC: Commutative Algebra ----------------------- math.AC/0410598 Kamran Divaani-Aazar, Amir Mafi: Associated primes of local cohomology module math.AC/0410585 Nicholas Baeth: A Krull-Schmidt Theorem for One-dimensional Rings of Finite Cohen-Macaulay Type math.AC/0410535 Anurag K. Singh, Uli Walther: On the arithmetic rank of certain Segre products math.AC/0410497 Juan C. Migliore, Uwe Nagel, Tim Romer: The Multiplicity Conjecture in low codimensions math.AC/0410478 Carlos DÕAndrea, Laurent Buse: Properness and inversion problems by means of matrices AG: Algebraic Geometry ---------------------- math.AG/0410604 Elizabeth S. Allman, John A. Rhodes: Phylogenetic ideals and varieties for the general Markov model math.AG/0410602 E. Carlini: Codimension one decompositions and Chow varieties math.AG/0410600 J. C. Sierra, L. Ugaglia: On double Veronese embeddings in the Grassmannian G(1,N) math.AG/0410584 Carolina Araujo: Rational curves of minimal degree and characterizations of ${mathbb P}^n$ math.AG/0410572 Israel Moreno MejÕ{i}a: The trace of an automorphism on H^0(J,O(nTheta)) math.AG/0410558 Ivan Cheltsov: Birationally superrigid cyclic triple spaces math.AG/0410554 Meirav Amram, David Goldberg: Higher degree Galois covers of CP^1 x T math.AG/0410547 D. A. Stepanov: Non-rational divisors over non-Gorenstein terminal singularities math.AG/0410540 Pan Peng: A simple proof of Gopakumar-Vafa conjecture for local toric Calabi-Yau manifolds math.AG/0410537 Eduardo Esteves, Steven Kleiman: The compactified Picard scheme of the compactified Jacobian math.AG/0410527 Cristiano Bocci: Special effect varieties and (-1)-curves math.AG/0410526 Shihoko Ishii: Arcs, valuations and the Nash map math.AG/0410524 Boris E. Kunyavskii, Louis H. Rowen, Sergey V. Tikhonov, Vyacheslav I. Yanchevskii: Division algebras that ramify only on a plane quartic curve math.AG/0410520 Laurent Manivel, Emilia Mezzetti: On linear spaces of skew-symmetric matrices of constant rank math.AG/0410518 Elena Drozd: Curves on a nonsingular Del Pezzo Surface in $P^4_k$ math.AG/0410513 Kalle Karu: The cd-index of fans and lattices AP: Analysis of PDEs -------------------- math.AP/0410581 G. Olafsson, A. Pasquale: Support properties and HolmgrenÕs uniqueness theorem for differential operators with hyperplane singularities math.AP/0410564 James Nolen, Jack Xin: A Variational Principle Based Study of KPP Minimal Front Speeds in Random Shears math.AP/0410546 Plamen Stefanov, Gunther Uhlmann: Stable determination of generic simple metrics from the hyperbolic Dirichlet-to-Neumann map math.AP/0410538 J. Colliander, W. Staubach: $L^2$ blowup solutions of cubic NLS on $R^2$ concentrate a fixed amount of mass math.AP/0410525 Gianni Dal Maso, Rodica Toader: On a notion of unilateral slope for the Mumford-Shah functional math.AP/0410499 Hans Lindblad, Jacob Sterbenz: Global Stability for Charge Scalar Fields on Minkowski Space AT: Algebraic Topology ---------------------- math.AT/0410589 Nora Ganter: Smash products of E(1)-local spectra at an odd prime math.AT/0410552 Javier Turiel: Polynomials Maps and Even Dimensional Spheres math.AT/0410503 Kathryn Hess, Ran Levi: An algebraic model for the loop homology of a homotopy fiber CA: Classical Analysis and ODEs ------------------------------- math.CA/0410548 Marc Artzrouni: A new family of periodic functions as explicit roots of a class of polynomial equations math.CA/0410542 Projections And Universal Encoding Strategies math.CA/0410508 Stephen Semmes: Potpourri, 8 math.CA/0410490 Stephen Semmes: Potpourri, 7 math.CA/0410489 Stephen Semmes: Potpourri, 6 math.CA/0410483 A. A. Bolibruch, S. Malek, C. Mitschi: On the generalized Riemann-Hilbert problem with irregular singularities CO: Combinatorics ----------------- math.CO/0410592 S. Ole Warnaar: Hall--Littlewood functions and the A_2 Rogers--Ramanujan identities quant-ph/0410226 P. Blasiak, A. Horzela, K. A. Penson, A. I. Solomon: Deformed Bosons: Combinatorics of Normal Ordering math.CO/0410550 Ewa Krot: Further develpoements in finite fibonomial calculus math.CO/0410529 math.CO/0410482 Michael Anshelevich: Orthogonal polynomials with a resolvent-type generating function CT: Category Theory ------------------- math.CT/0410555 Alan Robinson: Partition complexes, duality and integral tree representations gr-qc/0410104 J. Daniel Christensen, Louis Crane: Causal sites as quantum geometry CV: Complex Variables --------------------- math.CV/0410599 Laurent Gendre: Inegalites de Markov tangentielles locales sur les courbes algebriques singulieres de R^n math.CV/0410578 Dmitri Prokhorov, Alexander VasilÕev: Optimal control in BombieriÕs and TammiÕs conjectures math.CV/0410509 David E. Barrett: A ßoating body approach to FeffermanÕs hypersurface measure DG: Differential Geometry ------------------------- math.DG/0410610 Francisco Martin Cabrera: SU(3)-structures on hypersurfaces of manifolds with $G_2$-structure hep-th/0410183 Anton Alekseev, Thomas Strobl: Current Algebras and Differential Geometry math.DG/0410579 Jorge Lauret: A canonical compatible metric for geometric structures on nilmanifolds math.DG/0410575 Joseph H.G. Fu: Structure of the unitary valuation algebra math.DG/0410561 math.DG/0410559 Tomasz S. Mrowka, Yann Rollin: Legendrian knots and monopoles math.DG/0410557 A. V. Kiselev, G. Manno: On the symmetry structure of the minimal surface equation math.DG/0410553 Anton Deitmar: A prime geodesic theorem for higher rank II: singular geodesics math.DG/0410551 Eduardo Martinez: Classical field theory on Lie algebroids: Variational aspects math.DG/0410512 Maks A. Akivis, Vladislav V. Goldberg, Arto V. Chakmazyan: Induced connections on submanifolds in spaces with fundamental groups math.DG/0410511 Maks A. Akivis, Vladislav V. Goldberg: Dually degenerate varieties and the generalization of a theorem of Griffiths--Harris math.DG/0410498 Boris S. Kruglikov, Vladimir S. Matveev: Strictly non-proportional geodesically equivalent metrics have $h_text{top}(g)=0$ math.DG/0410494 George Papadopoulos: Spin Cohomology math.DG/0410493 Abdenago Barros G. Pacelli Bessa: Estimates of the first eigenvalue of minimal hypersurfaces of $mathbb{S}^{n+1} math.DG/0410487 Hiroshi Iritani: Quantum D-modules and equivariant Floer theory for free loop spaces math.DG/0410484 Aleksis Raza: An application of Guillemin-Abreu theory to a non-abelian group action DS: Dynamical Systems --------------------- math.DS/0410580 I. Binder, M. Braverman, M. Yampolsky: Filled Julia sets with empty interior are computable math.DS/0410517 Le Van Hien: Stability of Solutions of Fuzzy Differential Equations math.DS/0410507 Topologies on the group of homeomorphisms of a Cantor set math.DS/0410506 Topologies on the group of Borel automorphisms of a standard Borel space math.DS/0410505 Sergey Bezuglyi, Anthoni H. Dooley, Konstantin Medynets: The Rokhlin lemma for homeomorphisms of a Cantor set math.DS/0410504 Sergey Bezuglyi, Konstantin Medynets: Smooth automorphisms and path-connectedness in Borel dynamics math.DS/0410500 Ara Basmajian, Mahmoud Zeinalian: Maximal Convergence Groups and Rank One Symmetric Spaces math.DS/0410481 Pavlos B. Konstadinidis: The Real 3x+1 Problem FA: Functional Analysis ----------------------- math.FA/0410596 Ralf Meyer: Embeddings of derived categories of bornological modules math.FA/0410573 Jorge Antezana, Gustavo Corach, Demetrio Stojanoff: Spectral shorted operators math.FA/0410571 Massimo Fornasier, Holger Rauhut: Continuous Frames, Function Spaces, and the Discretization Problem math.FA/0410567 A. Brudnyi: Contractibility of Maximal Ideal Spaces of Certain Algebras of Almost Periodic Functions math.FA/0410549 Massimo Fornasier: Banach frames for alpha-modulation spaces math.FA/0410501 V.Yaskin: The Busemann-Petty problem in hyperbolic and spherical spaces math.FA/0410496 A.Koldobsky, V.Yaskin, M.Yaskina: Modified Busemann-Petty problem on sections of convex bodies math.FA/0410491 T. Banks, T. Constantinescu, Nermine El-Sissi: Tensor algebras and displacement structure. IV. Invariant kernels math.FA/0410479 A.G.Ramm: Dynamical systems method (DSM) for nonlinear equations in Banach spaces GM: General Mathematics ----------------------- math.GM/0410556 Joao R. Cardoso: An Explicit Formula for the Matrix Logarithm GR: Group Theory ---------------- math.GR/0410593 Henrik Baarnhielm: The Schreier-Sims algorithm for matrix groups math.GR/0410590 Edith Adan-Bante: Products of characters with few irreducible constituents math.GR/0410583 Edith Adan-Bante: Products of characters and derived length II math.GR/0410582 Edith Adan-Bante: Squares of characters and groups of odd order math.GR/0410539 Daniel Farley, Lucas Sabalka: Discrete Morse theory and graph braid groups math.GR/0410533 Stephen DeBacker: Parametrizing nilpotent orbits via Bruhat-Tits theory math.GR/0410516 J.Mostovoy, J.M. PÕerez-Izquierdo: Dimension filtration on loops math.GR/0410515 Jacob Mostovoy: On the notion of lower central series for loops GT: Geometric Topology ---------------------- math.GT/0410606 Greg Friedman: Knot spinning math.GT/0410603 R. C. Penner, Dennis Sullivan: The Structure and Singularities of Arc Complexes math.GT/0410595 Pascal Hubert, Samuel Lelievre: Noncongruence subgroups in H(2) math.GT/0410570 Andras Nemethi: On the Heegaard Floer homology of S^3_{-p/q}(K) math.GT/0410565 Brooke Brennan, Thomas W. Mattman, Roberto Raya, Dan Tating: Ribbonlength of torus knots math.GT/0410541 Ensil Kang, J. Hyam Rubinstein: Ideal triangulations of 3--manifolds I: spun normal surface theory math.GT/0410495 Dror Bar-Natan: KhovanovÕs Homology for Tangles and Cobordisms KT: K-Theory and Homology ------------------------- math.KT/0410597 Ralf Meyer: Combable groups have group cohomology of polynomial growth LO: Logic --------- math.LO/0410523 Fredrik Engstrom: Expansions, omitting types, and standard systems MG: Metric Geometry ------------------- math.MG/0410566 Piotr W. Nowak: On coarse embeddability into $ell_p$-spaces and a conjecture of Dranishnikov MP: Mathematical Physics ------------------------ quant-ph/0410201 Kazuyuki Fujii: Jaynes-Cummings Model and a Non-Commutative Geometry : A Few Problems Noted math-ph/0410062 David Damanik, Daniel Lenz, Gunter Stolz: Lower Transport Bounds for One-Dimensional Continuum Schrodinger Operators math-ph/0410061 P. Di Francesco, P. Zinn-Justin: Razumov-Stroganov sum rule: a proof based on multi-parameter generalizations math-ph/0410060 A.W.Beckwith: How false vacuum synthesis of a universe sets initial conditions which permit the onset of variations of a nucleation rate per Hubble volume per Hubble time math-ph/0410059 Manfred Requardt: Supersymmetry on Graphs and Networks math-ph/0410058 Vitaly V. Bulatov, Yuriy V. Vladimirov, Vasily A. Vakorin: Weak Singularity for Two-Dimensional Nonlinear Equations of Hydrodynamics and Propagation of Shock Waves math-ph/0410057 Joseph V. Pule, Andre F. Verbeure, Valentin A. Zagrebnov: Models with Recoil for Bose-Einstein Condensation and Superradiance math-ph/0410056 Petko Nikolov, Tihomir Valchev: Description of all conformally invariant differential operators, acting on scalar functions astro-ph/0404408 Ing-Guey Jiang, Li-Chin Yeh: On the Chaotic Orbits of Disc-Star-Planet Systems math-ph/0410055 A. A. Hujeirat: Problem-orientable numerical algorithm for modelling multi-dimensional radiative MHD ßows in astrophysics -- the hierarchical solution scenario math-ph/0410054 A.A. Hujeirat: A method for enhancing the stability and robustness of explicit schemes in astrophysical ßuid dynamics hep-th/0410172 Beatriz Gato-Rivera: The Adapted Ordering Method in Representation Theory math-ph/0410053 Alexander Rybko, Senya Shlosman: Poisson Hypothesis for Information Networks II. Cases of Violations and Phase Transitions math-ph/0410052 A.C.D.van Enter, E.A.Verbitskiy: On the Variational Principle for Generalized Gibbs Measures math-ph/0410051 Xavier Gracia, Ruben Martin: Time-dependent singular differential equations math-ph/0410050 Nicolae Cotfas: Systems of orthogonal polynomials defined by hypergeometric type equations math-ph/0410049 Joachim Kupsch, Subhashish Banerjee: Ultracoherence and Canonical Transformations math-ph/0410048 Sebastian Bauer: Post-Newtonian approximation of the Vlasov-Nordstrom system hep-th/0410212 C. Chryssomalakos, E. Okon: Generalized Quantum Relativistic Kinematics: a Stability Point of View cond-mat/0410424 Sandeep Tyagi: New series representation for Madelung constant quant-ph/0410151 S. Twareque Ali, F. Bagarello: Some Physical Appearances of Vector Coherent States and CS Related to Degenerate Hamiltonians math-ph/0410047 Volodymyr Sushch: Discrete model of Yang-Mills equations in Minkowski space hep-th/0410109 S. Odake, R. Sasaki: Equilibrium Positions, Shape Invariance and Askey-Wilson Polynomials hep-th/0410102 S. Odake, R. Sasaki: Shape Invariant Potentials in Discrete Quantum Mechanics NA: Numerical Analysis ---------------------- math.NA/0410488 Paul Sablonniere: Recent Results on Near-Best Spline Quasi-Interpolants NT: Number Theory ----------------- math.NT/0410563 Dragos Ghioca: The Mordell-Lang Theorem for Drinfeld modules math.NT/0410536 construction of some Galois modules math.NT/0410531 Takashi Taniguchi: A mean value theorem for the square of class numbers of quadratic fields math.NT/0410522 A. Ivic, E. Kratzel, M. Kuhleitner, W.G. Nowak: Lattice points in large regions and related arithmetic functions: Recent developments in a very classic topic math.NT/0410519 Howard Kleiman: Bounds for the Solutions of Cubic Diophantine Equations math.NT/0410502 Avner Ash, David Pollack, Warren Sinnott: A_6-extensions of Q and the mod p cohomology of GL(3,Z) math.NT/0410477 Bernd C. Kellner: Some remarks on KurepaÕs left factorial OA: Operator Algebras --------------------- math.OA/0410607 P. S. Muhly, M. Skeide, B. Solel: Representations of B^a(E) math.OA/0410601 Marek Bozejko, Wlodzimierz Bryc: On a class of free Levy laws related to a regression problem math.OA/0410594 Kenley Jung: Some free entropy dimension inequalities for subfactors math.OA/0410587 Wei Wu: Non-commutative metric topology on matrix state space math.OA/0410534 Todd Kemp: Strong hypercontractivity in non-commutative holomorphic spaces math.OA/0410492 Gelu Popescu: Unitary invariants in multivariable operator theory math.OA/0410480 Marius Ionescu, Yasuo Watatani: $C^{ast}$-Algebras associated with Mauldin-Williams Graphs PR: Probability --------------- math.PR/0410569 Aaron Abrams, Henry Landau, Zeph Landau, James Pommersheim, Eric Zaslow: Random Multiplication Approaches Uniform Measure in Finite Groups math.PR/0410560 Elchanan Mossel, Ryan OÕDonnell, Oded Regev, Jeffrey Steif, Benjamin Sudakov: Non-interactive correlation distillation, inhomogeneous Markov chains, and the reverse Bonami-Beckner inequality math.PR/0410545 Ravi Montenegro: Vertex and edge expansion properties for rapid mixing math.PR/0410544 Frederik Herzberg: The fairest price of an asset in an environment of temporary arbitrage math.PR/0410543 Frederik Herzberg: On measures of unfairness and an optimal currency transaction tax math.PR/0410532 Anna Rudas: Random tree growth with general weight function math.PR/0410514 Jianjun Tian, Xiao-Song Lin: Colored Genealogical Trees and Coalescent Theory math.PR/0410510 Anna Karczewska: Properties of convolutions arising in stochastic Volterra equations math.PR/0410485 Jacques Franchi, Yves Le Jan: Relativistic Diffusions and Schwarzschild Space cond-mat/0410543 Federico Camia: A Note on EdwardsHypothesis for Zero-Temperature Ising Dynamics QA: Quantum Algebra ------------------- math.QA/0410605 D. Shklyarov, S. SinelÕshchikov, L. Vaksman: Fock Representations and Quantum Matrices math.QA/0410562 Vasiliy Dolgushev, Pavel Etingof: Hochschild cohomology of quantized symplectic orbifolds and the Chen-Ruan cohomology math.QA/0410530 S. SinelÕshchikov, L. Vaksman: Quantum Groups and Bounded Symmetric Domains math.QA/0410528 Michel Van den Bergh: Double Poisson algebras math.QA/0410486 Vladimir D. Lyakhovsky: On a class of skew classical r-matrices with large carrier RA: Rings and Algebras ---------------------- math.RA/0410591 Aaron Lauve: NSym into Q_{infty} is not a Hopf Map math.RA/0410576 Friedrich Wehrung, Jiri Tuma: Congruence lifting of diagrams of finite Boolean semilattices requires large congruence varieties math.RA/0410521 Jerzy Matczuk: Ore Extensions over Duo Rings RT: Representation Theory ------------------------- math.RT/0410588 Konstantin Styrkas: Regular representation on the big cell and big projective modules in the category O SG: Symplectic Geometry ----------------------- math.SG/0410609 Joa Weber: Noncontractible periodic orbits in cotangent bundles and Floer homology math.SG/0410608 Weimin Chen: Pseudoholomorphic curves in four-orbifolds and some applications math.SG/0410568 Eugene Lerman: Gradient ßow of the norm squared of a moment map SP: Spectral Theory ------------------- math.SP/0410577 Sergio Albeverio, Alexander K. Motovilov: Operator integrals with respect to a spectral measure and solutions to some operator equations ST: Statistics -------------- math.ST/0410586 Rasa Karapandza, Milos Bozovic: You Can Fool Some People Sometimes math.ST/0410574 Igor Podlubny: A note on comparison of scientific impact expressed by number of citations in different fields of science -- / Greg Kuperberg (UC Davis) / Home page: http://www.math.ucdavis.edu/~greg/ / Visit the Math ArXiv Front at http://front.math.ucdavis.edu/ / * All the math thatÕs fit to e-print * === Subject: 3rd cohomology of loop group Originator: israel@math.ubc.ca (Robert Israel) IÕd like to know the 3rd cohomology group of the loop group OmegaE_8 over E_8, H^3(OmegaE_8,H) , with H some abelian group. Does anyone know where I could find respective information? === Subject: Re: Open questions related to periodic continued fractions Originator: israel@math.ubc.ca (Robert Israel) Diana Mecum asked: > I am starting research for a thesis on continued fractions, and want > to look at open questions related to periodic continued fractions. Is > anyone aware of current open questions of interest? If n is not a perfect square, let p(n) be the length of the period of the s.c.f. of sqrt(n). In terms of n, how large can p(n) be? ItÕs been shown that p(n) = O(sqrt(n) log(n)). It follows from some generalization of the Riemann hypothesis that p(n) = O(sqrt(n) log(log(n))). It seems likely that p(n)/sqrt(n) is unbounded, but I donÕt think itÕs even been shown that it doesnÕt tend to 0. The first 23 record-setting values of p(n)/sqrt(n) are shown below: n p(n) p(n)/sqrt(n) 2 1 0.70711 3 2 1.15470 7 4 1.51186 43 10 1.52499 46 12 1.76930 211 26 1.78991 331 34 1.86881 631 48 1.91085 919 60 1.97922 1726 88 2.11818 4846 152 2.18349 7606 194 2.22445 10399 228 2.23583 10651 234 2.26736 10774 238 2.29292 18379 322 2.37517 19231 332 2.39407 32971 438 2.41217 48799 544 2.46260 61051 614 2.48497 78439 696 2.48510 82471 716 2.49323 111094 834 2.50219 See http://www.research.att.com/projects/OEIS?Anum=A003285 for some references. Dean Hickerson dean@math.ucdavis.edu === Subject: Re: Open questions related to periodic continued fractions > I am starting research for a thesis on continued fractions, and want > to look at open questions related to periodic continued fractions. Is > anyone aware of current open questions of interest? As you may know, the outstanding problem in this area is to improve on known conditions for the length of the period of the continued fraction expansion of sqrt{D} to be odd (D > 0, D not a square, equivalent to x^2 - Dy^2 = -1 having solutions). The following might be a current summary of known conditions: B. D. Beach and H. C. Williams, A Numerical Investigation of the Diophantine Equation x^2 - dy^2 = -1, Congressus Numerantium VI, Proceedings of the Third Southeastern Conference on Combinatorics, Graph Theory and Computing, Utilitas Mathematica Publishing Inc., Winnipeg, Canada, 1972, pages 37 to 52. A less well known problem is as follows. As far as I know, this is an open problem, but I cannot be sure. For D > 0 not a square, D congruent to 1 modulo 4, let L1(D) and L4(D) denote the lengths of the periods of the continued fraction expansions of sqrt{D} and (1+sqrt{D})/2 respectively. The question is whether it is possible to have L1(D) = L4(D) + 4 when L4(D) == 3 (mod 6). As far as I can tell, this is not prohibited by results in the literature. I have empirical evidence that it is not possible based on testing those D up to 30 billion that have L4(D) <= 255. It is not hard to show this for one particular case, namely if L4(D) = 3 then L1(D) cannot be 7. See discussion under the heading ``Periods of Continued FractionsÕin April and May of 2000 in the archives of the Number Theory Listserver at http://listserv.nodak.edu/archives/nmbrthry.html for related comments and some references that might be of interest. John Robertson === Subject: Re: Open questions related to periodic continued fractions Received-SPF: Received-SPF: none (mailbox4.ucsd.edu: domain of news@newsread1.news.pas.earthlink.net does not designate permitted sender hosts) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > I am starting research for a thesis on continued fractions, and want > to look at open questions related to periodic continued fractions. Is > anyone aware of current open questions of interest? > As you may know, the outstanding problem in this area is to improve on known > conditions for the length of the period of the continued fraction expansion of > sqrt{D} to be odd (D > 0, D not a square, equivalent to x^2 - Dy^2 = -1 having > solutions). The following might be a current summary of known conditions: > B. D. Beach and H. C. Williams, A Numerical Investigation of the > Diophantine Equation x^2 - dy^2 = -1, Congressus Numerantium VI, > Proceedings of the Third Southeastern Conference on Combinatorics, > Graph Theory and Computing, Utilitas Mathematica Publishing Inc., > Winnipeg, Canada, 1972, pages 37 to 52. > A less well known problem is as follows. As far as I know, this is an open > problem, but I cannot be sure. For D > 0 not a square, D congruent to 1 modulo > 4, let L1(D) and L4(D) denote the lengths of the periods of the continued > fraction expansions of sqrt{D} and (1+sqrt{D})/2 respectively. The question > is whether it is possible to have L1(D) = L4(D) + 4 when L4(D) == 3 (mod 6). > As far as I can tell, this is not prohibited by results in the literature. I > have empirical evidence that it is not possible based on testing those D up to > 30 billion that have L4(D) <= 255. It is not hard to show this for one > particular case, namely if L4(D) = 3 then L1(D) cannot be 7. > See discussion under the heading ``Periods of Continued FractionsÕin April > and May of 2000 in the archives of the Number Theory Listserver at > http://listserv.nodak.edu/archives/nmbrthry.html > for related comments and some references that might be of interest. > John Robertson === Subject: Re: Open questions related to periodic continued fractions Epigone-thread: drermestrin Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Regarding the maximal element in the CF of sqrt(n) (say M(n)) . Recently I noticed M(n)-2*sqrtint(n) = 0 or 1 and I suspect this behaviour : sum_{i non square <= n} 2*sqrt(i)-M(i) = c*n+o(n) where c is >0 and = 0.8... I donÕt know references on this subject. Should exist some. B. Cloitre === Subject: Re: Open questions related to periodic continued fractions Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Regarding the maximal element in the CF of sqrt(n) (say M(n)) . > Recently I noticed M(n)-2*sqrtint(n) = 0 or 1 and I suspect this > behaviour : > sum_{i non square <= n} 2*sqrt(i)-M(i) = c*n+o(n) where c is >0 > and = 0.8... > I donÕt know references on this subject. Should exist some. IÕm not sure what ``sqrtint(n)Õis. The maximum partial quotient in the continued fraction expansion of sqrt(n) (for n not a square) is 2 times the integer part of the square root of n. This is proved in most references that consider the continued fraction expansion of sqrt(n), e.g., MollinÕs Fundamental Number Theory with Applications, or Niven, Zuckerman, and Montgomery. Also, (1/n) Sum_{i=1}^{n} [sqrt(i) - int(sqrt(i))] tends to 1/2. John Robertson === Subject: Re: Open questions related to periodic continued fractions Epigone-thread: drermestrin Originator: bergv@math.uiuc.edu (Maarten Bergvelt) >Regarding the maximal element in the CF of sqrt(n) (say M(n)) . >Recently I noticed M(n)-2*sqrtint(n) = 0 or 1 and I suspect this >behaviour : >sum_{i non square <= n} 2*sqrt(i)-M(i) = c*n+o(n) where c is >0 and = >0.8... >I donÕt know references on this subject. Should exist some. >B. Cloitre May be the following remarks are all well known: observe for instance sqrt(41) giving sequence (6;1/2,1/2,1/12,...) the period (1/2,1/2,1/12 ..)is linked to function (1/2,1/2,1/12,x) or (5x+62)/(2x+25) with two fixed points ,the positive -6+sqrt(41) is related to our continuous fraction. f(x)=(5x+62)/(2x+25) is easily iterated (sci.math 20 oct), Friendly yours,Alain. === Subject: Re: Open questions related to periodic continued fractions Epigone-thread: drermestrin Content-Length: 2687 Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Diana, in case it can help. Is anything known about the continued fraction of log_2(3)? (that is, of the logarithm of 3, base 2) Is it bounded,...!?? Maybe there are some results on the continued fractions of logarithms.. I conjecture, with the standard notations, that a_{n+1}/q_n <= 1/4, for that number. (One more thing, is log_2(3) algebraic or trascendent?..!) Jose >> I am starting research for a thesis on continued fractions, and want >> to look at open questions related to periodic continued fractions. Is >> anyone aware of current open questions of interest? >> As you may know, the outstanding problem in this area is to improve on >known >> conditions for the length of the period of the continued fraction >expansion of >> sqrt{D} to be odd (D > 0, D not a square, equivalent to x^2 - Dy^2 = -1 >having >> solutions). The following might be a current summary of known conditions: >> B. D. Beach and H. C. Williams, A Numerical Investigation of the >> Diophantine Equation x^2 - dy^2 = -1, Congressus Numerantium VI, >> Proceedings of the Third Southeastern Conference on Combinatorics, >> Graph Theory and Computing, Utilitas Mathematica Publishing Inc., >> Winnipeg, Canada, 1972, pages 37 to 52. >> A less well known problem is as follows. As far as I know, this is an >open >> problem, but I cannot be sure. For D > 0 not a square, D congruent to 1 >modulo >> 4, let L1(D) and L4(D) denote the lengths of the periods of the continued >> fraction expansions of sqrt{D} and (1+sqrt{D})/2 respectively. The >question >> is whether it is possible to have L1(D) = L4(D) + 4 when L4(D) == 3 (mod >6). >> As far as I can tell, this is not prohibited by results in the literature. >> have empirical evidence that it is not possible based on testing those D >up to >> 30 billion that have L4(D) <= 255. It is not hard to show this for one >> particular case, namely if L4(D) = 3 then L1(D) cannot be 7. >> See discussion under the heading ``Periods of Continued FractionsÕÕ in >April >> and May of 2000 in the archives of the Number Theory Listserver at >> http:// listserv.no dak.edu/archives/nmbrthry.html for related comments and some references that might be of interest. >> John Robertson === Subject: Re: Open questions related to periodic continued fractions Received-SPF: Received-SPF: none (mailbox3.ucsd.edu: domain of news@nntp.itservices.ubc.ca does not designate permitted sender hosts) Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Diana, > in case it can help. > Is anything known about the continued fraction of log_2(3)? (that >is, of the logarithm of 3, base 2) > Is it bounded,...!?? > Maybe there are some results on the continued fractions of >logarithms.. I doubt that this will help. Very little is known about the continued fractions of closed-form numbers apart from rationals and quadratic irrationals, and AFAIK there is no prospect, with currently available mathematical techniques, of being able to prove whether the continued fraction of a number such as this has bounded elements. IMHO this is not a problem to give to a student starting her thesis. > I conjecture, with the standard notations, that a_{n+1}/q_n <= 1/4, >for that number. > (One more thing, is log_2(3) algebraic or trascendent?..!) Transcendental, by the Gelfond-Schneider Theorem. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Open questions related to periodic continued fractions Originator: bergv@math.uiuc.edu (Maarten Bergvelt) > Diana, > in case it can help. > Is anything known about the continued fraction of log_2(3)? (that > is, of the logarithm of 3, base 2) > Is it bounded,...!?? No, itÕs unbounded. But, so far, nobody knows how to prove that. > Maybe there are some results on the continued fractions of > logarithms.. > I conjecture, with the standard notations, that a_{n+1}/q_n <= 1/4, > for that number. How does that compare to almost all continued fractions? > (One more thing, is log_2(3) algebraic or trascendent?..!) It is transcendental. -- G. A. Edgar edgar at math.ohio-state.edu === Subject: Re: Galois group of a given quartic equation Epigone-thread: twomprendlex Originator: israel@math.ubc.ca (Robert Israel) The pending issue is whether the quartic Q(x) = 0 may have the Galois group isomorphic to Z4 for a and b different. No progress has been posted. One way to rule out D4 in the Z4/D4 case is given below. (If Q(x) is irreducible over Z, the discriminant is not a square and the cubic resolvent R(t) = 0 of the quartic has one and only one integral root t0, then the Galois group G is either Z4 or D4.) Let R(t) = (t - t0) r(t) where r(t) is a monic irreducible quadratic with integral coefficients. Further, let the discriminant of r(t) be D. Let d be the squarefree part of D and E the splitting field of R(t) = 0. Then E = Q[Sqrt[d]] and therefore easy to determine. Then put E to use as follows: If Q(x) is reducible over E then G is Z4; otherwise G is D4. May be this can be used to settle the case whether G may be Z4 or not. Kent Holing === Subject: matrix derivative Epigone-thread: zoypryrwhou Originator: israel@math.ubc.ca (Robert Israel) I am currently having a matrix derivative problem What is derivative of trace{(A+FÕ*B*F)^(-1)} with respect to matrix F where ()is the transpose of a matrix. A and B is diagonal matrices. I searched online and was only able to find the derivative of d trace{(FÕ*B*F)^(-1)}/d F =-2*B*F*(FÕ*B*F)^(-2) without knowing how they get it. Moreover, all these matrix derivative problems seem to be difficult forme. Could anyone be kind enough to give me some good references on this topic. === Subject: Re: matrix derivative Originator: israel@math.ubc.ca (Robert Israel) > I am currently having a matrix derivative problem > What is derivative of > trace{(A+FÕ*B*F)^(-1)} > with respect to matrix F > where ()is the transpose of a matrix. > A and B is diagonal matrices. > I searched online and was only able to find the derivative of > d trace{(FÕ*B*F)^(-1)}/d F > =-2*B*F*(FÕ*B*F)^(-2) > without knowing how they get it. > Moreover, all these matrix derivative problems seem to be > difficult forme. Could anyone be kind enough to give me > some good references on this topic. d A denotes the matrix differential for the matrix A. x = trace{ (A+FÕ*B*F)^(-1) } d x = trace{ d ( (A+FÕ*B*F)^(-1) ) } = trace{ - (A+FÕ*B*F)^(-1) ( d (A+FÕ*B*F) ) (A+FÕ*B*F)^(-1) } = trace{ - (A+FÕ*B*F)^(-2) ( d (FÕ*B*F) ) } = trace{ - (A+FÕ*B*F)^(-2) ( d F*B*F + FÕ*B* d F ) } = - ( trace{ (A+FÕ*B*F)^(-2) d F*B*F } + trace{ (A+FÕ*B*F)^(-2) FÕ*B* d F } ) = - trace{ B * F * (A+FÕ*B*F)^(-2) d F} - trace{ (A+FÕ*B*F)^(-2) FÕ*B* d F } = - trace{ (A+FÕ*B*F)^(-2) * F* B * d F} - trace{ (A+FÕ*B*F)^(-2) FÕ*B* d F } (Note that B is symmetric, and B=BÕ; also, (A+FÕ*B*F) is also symmetric) = -2 trace{ (A+FÕ*B*F)^(-2) * F* B * d F} Therefore, by the first identification theorem, the differential is as given above. Recall that trace( (d x / d F)* d F ) = d x I find working with matrix differential much easier than other alternatives. The book recommended by Peter has a great account on how to manipulate matrix differential. It also provides the theorectical justification. Hope that helps. === Subject: Re: matrix derivative Originator: israel@math.ubc.ca (Robert Israel) >I am currently having a matrix derivative problem >What is derivative of >trace{(A+FÕ*B*F)^(-1)} >with respect to matrix F >where ()is the transpose of a matrix. >A and B is diagonal matrices. >I searched online and was only able to find the derivative of >d trace{(FÕ*B*F)^(-1)}/d F >=-2*B*F*(FÕ*B*F)^(-2) >without knowing how they get it. >Moreover, all these matrix derivative problems seem to be >difficult forme. Could anyone be kind enough to give me >some good references on this topic. Derivatives are not appropriate for functions of more than one variable, and you are asking for derivatives with respect to a matrix. Differentials satisfy the usual properties, but the failure of commutativity is quite important. So if q is a differential function of an argument in a locally ßat space, the Frechet derivative is dq(x, m) = lim ((q(x+em) - q(x))/e) as e -> 0. Using this, and the result that d(X^(-1)) = - X^(-1) dX X^(-1), we get that the differential you seek is trace (-2*(A+FÕ*B*F)^(-1)*FÕ*B*dF*(A+F[CapitalOTilde ]*B*F)^(-1)); this uses the results about trace being invariant under permutation and transposition. -- 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: matrix derivative Originator: israel@math.ubc.ca (Robert Israel) >I am currently having a matrix derivative problem >What is derivative of >trace{(A+FÕ*B*F)^(-1)} >with respect to matrix F >where ()is the transpose of a matrix. >A and B is diagonal matrices. >I searched online and was only able to find the derivative of >d trace{(FÕ*B*F)^(-1)}/d F >=-2*B*F*(FÕ*B*F)^(-2) >without knowing how they get it. >Moreover, all these matrix derivative problems seem to be >difficult forme. Could anyone be kind enough to give me >some good references on this topic. some reference which might help: Magnus, J.R., Neudecker, H. Matrix Differential Calculus with Applications in Statistics and Econometrics. paul fackler has some notes at http://www4.ncsu.edu/~pfackler/MATCALC.ps hth peter === Subject: standard probability spaces Originator: israel@math.ubc.ca (Robert Israel) let $(Omega,F)$ be a standard probability space and $X:[0,t]times Omegato E$ a stochastic process with values in a Polish space $E$ and RCLL trajectories. Is it true that $(Omega,F^X)$ is standard where $F^X$ denotes the $sigma$-algebra generated by $X$? J. p.s.: Please reply to email as well. Thx. === Subject: Re: Multivariable calculus help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAF2RA514573; >I have two questions. Any help would be greatly appreciated. >Find the volume of the solid spherical cap cut from the sphere x^2 + >y^2 +z^2 =8 by the plane y=2. I know I need to use spherical >coordinates but I am having trouble figuring out the different limits >of integration. >Consider the triple integral: > Integral from (0 to 1), (x to 1), (0 to 1-y^2) of >(e^z)/(z-1)dzdydx. >--Sketch the three dimensional region of integration and describe the >surfaces that form its boundary. >--Evaluate the integral by changing the order of integration to >dxdydz. It is better to use cylindrical coordinates x = r*cos(f) y = y z = r*sin(f) R = sqrt(8) = 2*sqrt(2) is the radius of the sphere. R 2pi sqrt(8-y^2) V = S S S r drdfdy = 2 0 2 (integrate over r) R 2pi sqrt(8-y^2) = S S [r^2/2] dfdy = 2 0 2 (and substitute limits for r) R 2pi = 1/2 * S S [8 - y^2 - (8 - 2^2)] dfdy = 2 0 R 2pi = 1/2 * S S (4 - y^2) dfdy = 2 0 (integrate over f) R 2pi = 1/2 * S (4 - y^2) [f] dy = 2 0 (and substitute limits for f) R = 1/2*2pi * S (4 - y^2) dy = 2 (integrate over y) R = pi * [4y - y^3/3] = 2 (and substitute limits for y) = pi * (4*2*sqrt(2) - 8*2*sqrt(2)/3 - 4*2 + 8/3) = = pi/3 *(8*sqrt(2) - 8) = 8/3*pi*(sqrt(2) - 1) ----------------------------------------------------------- 1 1 1-y^2 S S S e^z/(z-1) dzdydx = 0 x 0 (0 <= x <= 1 and x <= y <= 1 is equivalent to 0 <= y <= 1 and 0 <= x <= y) 1 y 1-y^2 = S S S e^z/(z-1) dzdxdy = 0 0 0 (integration over x and z can be swapped, because both limits depend only on y, the outermost integration) 1 1-y^2 y = S S S e^z/(z-1) dxdzdy = 0 0 0 (0 <= y <= 1 and 0 <= z <= 1-y^2 is equivalent to 0 <= z <= 1 and 0 <= y <= sqrt(1-z) 1 sqrt(1-z) y = S S S e^z/(z-1) dxdydz = 0 0 0 (integrate over x) 1 sqrt(1-z) y = S S e^z/(z-1) * [x] dydz = 0 0 0 (and substitute limits for x) 1 sqrt(1-z) = S S e^z/(z-1) * y dydz = 0 0 (integrate over y) 1 sqrt(1-z) = S e^z/(z-1) * [y^2/2] dz = 0 0 (and substitute limits for x) 1 = S e^z/(z-1) * (1-z)/2 dz = 0 (integrate over z) 1 1 = -1/2 * S e^z dz = -1/2 * [e^z] = 0 0 (and substitute limits for z) = -1/2*(e - 1) = (1 - e)/2 === Subject: finding inverse of matrix using linear equation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAF2RAb14595; HI, Could someone help me solve this. I tried a lot but avain. Help me out plz. Given is the system of linear equations A x = b with -1 1 -2 1 A = 2 -1 1 b = 2 -3 0 -1 , -5 1 Determine the inverse matrix B(inverse of A) and the solution x of the system of linear equations === Subject: Re: finding inverse of matrix using linear equation B = [ {1/4,1/4,-1/4},{-1/4,-5/4,-3/4},{-3/4,-3/4,-1/4}] there is a method to find B using gaussian matrices, anyway i used my TI-83 calculator. then x = B b, > HI, > Could someone help me solve this. I tried a lot but avain. Help me out > plz. > Given is the system of linear equations A x = b with > -1 1 -2 1 > A = 2 -1 1 b = 2 > -3 0 -1 , -5 > Determine the inverse matrix B(inverse of A) and the solution x of > the system of linear equations === Subject: Re: finding inverse of matrix using linear equation > B = [ {1/4,1/4,-1/4},{-1/4,-5/4,-3/4},{-3/4,-3/4,-1/4}] > there is a method to find B using gaussian matrices, anyway i used my > TI-83 calculator. > then x = B b, yes you can encapsulate the rules of gaussian elimination in Ōmatrix formÕ, and so boil down the process of gaussian elimination to multiplying your matrix by other matrices which represent implementations of GE. but our friends answer is to use gaussian row operations to operate on the augmented matrix consisting of the original matrix and the identity matrix >> HI, >> Could someone help me solve this. I tried a lot but avain. Help me out >> plz. >> Given is the system of linear equations A x = b with >> -1 1 -2 1 >> A = 2 -1 1 b = 2 >> -3 0 -1 , -5 >> 1 >> Determine the inverse matrix B(inverse of A) and the solution x of >> the system of linear equations === Subject: Proving G or ~G is Not Planar Hey All, Let G be a simple graph with 11 or more vertices. Show that either G or ~G is not planar. Induction seems to be in order. If G is not planar, IÕm done, so letÕs assume G is planar. For the basis step, I figured it would be best to construct G with 11 vertices and the largest number of possible edges keeping G planar. Then from this contruct ~G and show that it is not planar. Any subgraph of G with the same amount of vertices would Ōsufferthe same thing (i.e. ~G would be unplanar) thus completing the basis step. However, when I realized how complex it turned out to be when I started drawing G and soon gave up. I canÕt think of any other strategy at the moment for this. Any tips? Bernd === Subject: Re: Proving G or ~G is Not Planar > Hey All, > Let G be a simple graph with 11 or more vertices. Show that either G > or ~G is not planar. Induction seems to be in order. If G is not > planar, IÕm done, so letÕs assume G is planar. For the basis step, I > figured it would be best to construct G with 11 vertices and the > largest number of possible edges keeping G planar. Then from this > contruct ~G and show that it is not planar. Any subgraph of G with the > same amount of vertices would Ōsufferthe same thing (i.e. ~G would > be unplanar) thus completing the basis step. However, when I realized > how complex it turned out to be when I started drawing G and soon gave > up. I canÕt think of any other strategy at the moment for this. Any > tips? > Bernd First, by ~G I am assuming you mean the graph formed by the edges that are not in G, the complement of G. If you mean something else, the problem might be a little more difficult (or maybe easier, I donÕt know). Okay, here is a really, *really* big hint for this problem. Think about the degrees of the vertices for planar and nonplanar graphs. Oh, and every planar graph on 9 vertices has a nonplanar complement (Battle et al. 1962) and any nonplanar graph on 9 vertices, well, is already nonplanar. Ergo, for any simple graph G with 9 vertices either G or ~G is nonplanar. The real question is does the property hold for all graphs on 10 vertices? - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Proving G or ~G is Not Planar > First, by ~G I am assuming you mean the graph formed by the edges that are > not in G, the complement of G. If you mean something else, the problem > might be a little more difficult (or maybe easier, I donÕt know). Yes, when I write ~G, I mean the complement of G. > Okay, here is a really, *really* big hint for this problem. Think about > the degrees of the vertices for planar and nonplanar graphs. If a graph is not planar, then it has no K[5] or K[3,3] embeddings which means there canÕt be five vertices with degree 4 or greater, and there canÕt be 6 vertices with degree 3 or greater. So, given a planar graph G with 11 vertices, it must satisfy the aformentioned property. Hence, ~G will have at least 7 vertices with degree 7 or greater I think. Is this right? > Oh, and every planar graph on 9 vertices has a nonplanar complement (Battle > et al. 1962) and any nonplanar graph on 9 vertices, well, is already > nonplanar. Ergo, for any simple graph G with 9 vertices either G or ~G is > nonplanar. The real question is does the property hold for all graphs on > 10 vertices? Interesting... === Subject: Re: Proving G or ~G is Not Planar > If a graph is not planar, then it has no K[5] or K[3,3] embeddings > which means there canÕt be five vertices with degree 4 or greater, > and there canÕt be 6 vertices with degree 3 or greater. So, given a > planar graph G with 11 vertices, it must satisfy the aformentioned > property. Hence, ~G will have at least 7 vertices with degree 7 or > greater I think. Is this right? You are very much on the right track. There is a small problem with the second part of your argument, and there canÕt be 6 vertices with degree 3 or greater Look at the following graph (use a monospaced font): o--*--*--o | | | | *--*--*--o | | | o--*--o There are exactly 6 vertices, the *Õs, with degree 3 or greater. Yet it is planar. The complement is not planar, however, so the theorem is still true. You just need to tighten things up a bit. Do you know any theorems relating planarity to vertex degree? - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Proving G or ~G is Not Planar > If a graph is not planar, then it has no K[5] or K[3,3] embeddings > which means there canÕt be five vertices with degree 4 or greater, > and there canÕt be 6 vertices with degree 3 or greater. So, given a > planar graph G with 11 vertices, it must satisfy the aformentioned > property. Hence, ~G will have at least 7 vertices with degree 7 or > greater I think. Is this right? > > You are very much on the right track. There is a small problem with the > second part of your argument, and there canÕt be 6 vertices with degree 3 > or greater Look at the following graph (use a monospaced font): > o--*--*--o > | | | | > *--*--*--o > | | | > o--*--o > There are exactly 6 vertices, the *Õs, with degree 3 or greater. Yet it is > planar. The complement is not planar, however, so the theorem is still > true. > You just need to tighten things up a bit. Do you know any theorems > relating planarity to vertex degree? The only theorems I know relating to planarity are KuratowskiÕs Theorem and EurlerÕs Formula. I was trying to apply KuratowskiÕs Theorem in the above but I guess it didnÕt work very well. I canÕt think of anything else. Bernd === Subject: Re: Proving G or ~G is Not Planar > The only theorems I know relating to planarity are KuratowskiÕs > Theorem and EurlerÕs Formula. I was trying to apply KuratowskiÕs > Theorem in the above but I guess it didnÕt work very well. I canÕt > think of anything else. > Bernd By EulerÕs Formula, I assume you mean for a graph G with n vertices, e edges and f faces, n - e + f = 2 or f = 2 + e - n Most texts list a corollary or another theorem to this formula. A quick statement of the theorem and a proof follows. If G is a simple, planar graph with n vertices (n>=3) and e edges, then e <= 3n - 6 If also G contains no triangles, then e <= 2n - 4 Proof: Assume G is connected. Count the edges in two ways, directly and as boundaries of faces. Let fi denote the number of edges forming the boundary of a face i. Note that a face must have at least three boundaries. Then we have 2e = sum(fi, over all faces i) >= 3f [*] Substitute into EulerÕs formula: 2e >= 3f = 6 + 3e - 3n 2e >= 6 + 3e - 3n -e >= 6 - 3n e <= 3n - 6 If there are no triangles, we end up with [*] as 2e >= 4f. The first part of the theorem is the result that will make your life a little easier, e <= 3n - 6 for planar graphs. You know have a formula that relates edges, vertices (and implicitly degrees of vertices) to planarity. - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Proving G or ~G is Not Planar > By EulerÕs Formula, I assume you mean for a graph G with n vertices, e > edges and f faces, > n - e + f = 2 > or > f = 2 + e - n Yes, exactly. > Most texts list a corollary or another theorem to this formula. A quick > statement of the theorem and a proof follows. > If G is a simple, planar graph with n vertices (n>=3) and e edges, then > e <= 3n - 6 > If also G contains no triangles, then > e <= 2n - 4 > Proof: Assume G is connected. Count the edges in two ways, directly and > as boundaries of faces. Let fi denote the number of edges forming the > boundary of a face i. Note that a face must have at least three > boundaries. Then we have > 2e = sum(fi, over all faces i) >= 3f [*] > Substitute into EulerÕs formula: > 2e >= 3f = 6 + 3e - 3n > 2e >= 6 + 3e - 3n > -e >= 6 - 3n > e <= 3n - 6 > If there are no triangles, we end up with [*] as 2e >= 4f. Funny. This theorem wasnÕt given in the chapterÕs content but rather as an exercise. Funny enough, I was having trouble solving this problem too, i.e. proving e <= 3v - 6, v >= 3. Interesting proof. Never would have thought of it. > The first part of the theorem is the result that will make your life a > little easier, e <= 3n - 6 for planar graphs. You know have a formula that > relates edges, vertices (and implicitly degrees of vertices) to planarity. I was thinking of using this for the problem at hand by using the following argument. Let E be the number of edges in K[11], e the number of edges in G and ethe number of edges in ~G. By definition, E = e + eÕ. If G is planar then e <= 3(11) - 6 = 27. E = 10 + 9 + 8 + ... + 1 = 55 I think so that means e>= 55 - 27 = 28. Hence, by the corollary stated above, ~G canÕt be planar. This should be correct. Tell me otherwise. Bernd === Subject: Re: Proving G or ~G is Not Planar >> If G is a simple, planar graph with n vertices (n>=3) and e edges, >> then >> e <= 3n - 6 >> If also G contains no triangles, then >> e <= 2n - 4 > I was thinking of using this for the problem at hand by using the > following argument. Let E be the number of edges in K[11], e the > number of edges in G and ethe number of edges in ~G. By definition, > E = e + eÕ. If G is planar then e <= 3(11) - 6 = 27. E = 10 + 9 + 8 + > ... + 1 = 55 I think so that means e>= 55 - 27 = 28. Hence, by the > corollary stated above, ~G canÕt be planar. > This should be correct. Tell me otherwise. Looks pretty good. You might want to state why e>= 28 makes ~G not planar (e>= 28 = 3*(11) - 5 > 3*(11) - 6). Now, rewrite the same proof changing G for ~G and ~G for G to show if ~G is planar, then G is not. Okay, it might not need to be that extreme, but you should at least say something like, The proof that ~G is planar implies G is not planar is similar. You need to make sure you cover both cases. You might also want to make the algebra a little easier to follow. > Bernd Using the above theorem, you can easily prove K[5] and K[3,3] are not planar. In fact, the proof in this method is almost trivial. In K[5] there are 10 edges and 5 vertices. Plugging in gives 10 <= 9. Contradiction. K[5] is not planar. Likewise, K[3,3] contains no triangles, 6 vertices, and 9 edges. Plugging in gives 9 <= 8. Contradiction. K[3,3] is not planar. Probably an easier proof than any other you have seen to show it. In fact, using the above theorem is usually easier to try and prove if a graph is planar than most any other method. Kuratowski is nice, if you have time to search for the K[5] or K[3,3]. Now use this technique to figure out whether the theorem is true for 10 vertices (i.e. either G or ~G is not planar on 10 vertices). - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Proving G or ~G is Not Planar >> If G is a simple, planar graph with n vertices (n>=3) and e edges, >> then >> e <= 3n - 6 >> If also G contains no triangles, then >> e <= 2n - 4 > > I was thinking of using this for the problem at hand by using the > following argument. Let E be the number of edges in K[11], e the > number of edges in G and ethe number of edges in ~G. By definition, > E = e + eÕ. If G is planar then e <= 3(11) - 6 = 27. E = 10 + 9 + 8 + > ... + 1 = 55 I think so that means e>= 55 - 27 = 28. Hence, by the > corollary stated above, ~G canÕt be planar. > This should be correct. Tell me otherwise. WasnÕt the original problem to show that the result holds for any G with at least 11 vertices ?? Your proof seems to handle only the case of a G with _exactly_ 11 vertices ... > Looks pretty good. You might want to state why e>= 28 makes ~G not > planar (e>= 28 = 3*(11) - 5 > 3*(11) - 6). Now, rewrite the same > proof changing G for ~G and ~G for G to show if ~G is planar, then G is > not. Okay, it might not need to be that extreme, but you should at > least say something like, The proof that ~G is planar implies G is not > planar is similar. You need to make sure you cover both cases. You > might also want to make the algebra a little easier to follow. ThereÕs a slightly different approach that handles G and ~G somewhat more symmetrically. Using the above result, itÕs easy to see that the average degree of a planar graph is < 6. More explicitly, if G is a graph with V vertices, E edges and F faces, define d(G) = (sum_{v a vertex of G} deg(v))/V. Then d(G) V = 2E <= 6V - 12 < 6V and it follows that d(G) < 6. Now, note that (obviously) d(G) + d(~G) = d(K[V]) = V-1 . If V > 11, the result weÕre trying to prove follows immediately, since at least one of d(G), d(~G) must then be >= 6. Unfortunately, this argument does _not_ handle the V=11 case -- for that one, we need to refine the bound on d(G) just the teensiest bit: we _really_ know that d(G) <= 6 - (12/V) which, if V happens to be 11, shows that d(G) < 5. And we get the result in this case just as before ... > > Bernd > Using the above theorem, you can easily prove K[5] and K[3,3] are not > planar. In fact, the proof in this method is almost trivial. In K[5] > there are 10 edges and 5 vertices. Plugging in gives 10 <= 9. > Contradiction. K[5] is not planar. Likewise, K[3,3] contains no > triangles, 6 vertices, and 9 edges. Plugging in gives 9 <= 8. > Contradiction. K[3,3] is not planar. *Definitely* the nicest approach to this ... :-) > Probably an easier proof than any other you have seen to show it. In > fact, using the above theorem is usually easier to try and prove if a > graph is planar than most any other method. Kuratowski is nice, if you > have time to search for the K[5] or K[3,3]. > Now use this technique to figure out whether the theorem is true for 10 > vertices (i.e. either G or ~G is not planar on 10 vertices). > - Tim === Subject: Re: Proving G or ~G is Not Planar > WasnÕt the original problem to show that the result holds for > any G with at least 11 vertices ?? Your proof seems to handle > only the case of a G with _exactly_ 11 vertices ... Easy enough. You can use induction on the number of vertices v with the basis step being when v = 11. For the inductive step, if G is planar, then G contains a planar subgraph of 11 vertices whose complement is not planar (by the basis step) and is a subgraph of ~G. So ~G is not planar. For the case of a simple planar graph G with 10 vertices: Let E be the number of edges in K[10], e the number of edges in G and ethe nubmer of edges in ~G. E = 45 and E = e + eÕ. Since e <= 3v - 6 = 30 - 6 = 24, then e>= 21. Hmm...but this implies that ~G is planar. I think the problem here is that the theorem e <= 3v - 6 only works for simple planar *connected* graphs. Interesting... === Subject: Re: Proving G or ~G is Not Planar > For the case of a simple planar graph G with 10 vertices: > Let E be the number of edges in K[10], e the number of edges in G > and ethe nubmer of edges in ~G. E = 45 and E = e + eÕ. Since > e <= 3v - 6 = 30 - 6 = 24, then e>= 21. Hmm...but this implies that > ~G is planar. I think the problem here is that the theorem e <= 3v - 6 > only works for simple planar *connected* graphs. Interesting... Yes, connectivity is one of the problems. The other problem is that it says every planar graph must have this property and thus any graph without the property is not planar. It does *not* say that any graph with this property is planar. In fact, as I stated earlier, Battle et.al. showed that for any graph G on 9 vertices, either G or ~G is not planar. This formula will not work for that proof (right? I didnÕt check it myself). And that is one of the beauties (or curses) of combinatorics: a proof technique does not work in every case. This is why there are so many open problems in Combinatorics. It can be proven for certain constructions, but not all. Ramsey numbers is a classic example. There are probably as many different proofs for Ramsey numbers as there are known Ramsey numbers. - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Proving G or ~G is Not Planar ... of course, I wasnÕt careful enough ... > >> If G is a simple, planar graph with n vertices (n>=3) and e edges, >> then >> e <= 3n - 6 >> If also G contains no triangles, then >> e <= 2n - 4 > > I was thinking of using this for the problem at hand by using the > following argument. Let E be the number of edges in K[11], e the > number of edges in G and ethe number of edges in ~G. By definition, > E = e + eÕ. If G is planar then e <= 3(11) - 6 = 27. E = 10 + 9 + 8 + > ... + 1 = 55 I think so that means e>= 55 - 27 = 28. Hence, by the > corollary stated above, ~G canÕt be planar. This should be correct. Tell me otherwise. > WasnÕt the original problem to show that the result holds for > any G with at least 11 vertices ?? Your proof seems to handle > only the case of a G with _exactly_ 11 vertices ... > > Looks pretty good. You might want to state why e>= 28 makes ~G not > planar (e>= 28 = 3*(11) - 5 > 3*(11) - 6). Now, rewrite the same > proof changing G for ~G and ~G for G to show if ~G is planar, then G is > not. Okay, it might not need to be that extreme, but you should at > least say something like, The proof that ~G is planar implies G is not > planar is similar. You need to make sure you cover both cases. You > might also want to make the algebra a little easier to follow. > > ThereÕs a slightly different approach that handles G and ~G > somewhat more symmetrically. Using the above result, itÕs > easy to see that the average degree of a planar graph is < 6. > More explicitly, if G is a graph with V vertices, E edges and > F faces, define d(G) = (sum_{v a vertex of G} deg(v))/V. Then > d(G) V = 2E <= 6V - 12 < 6V and it follows that d(G) < 6. Now, > note that (obviously) > d(G) + d(~G) = d(K[V]) = V-1 . > If V > 11, the result weÕre trying to prove follows immediately, > since at least one of d(G), d(~G) must then be >= 6. Unfortunately, > this argument does _not_ handle the V=11 case -- for that one, > we need to refine the bound on d(G) just the teensiest bit: we > _really_ know that > d(G) <= 6 - (12/V) > which, if V happens to be 11, shows that d(G) < 5. And we get > the result in this case just as before ... You also need the more refined estimate to handle the V=12 case -- there it says d(G) <= 5 for a planar G with 12 vertices ... > > Bernd > > Using the above theorem, you can easily prove K[5] and K[3,3] are not > planar. In fact, the proof in this method is almost trivial. In K[5] > there are 10 edges and 5 vertices. Plugging in gives 10 <= 9. > Contradiction. K[5] is not planar. Likewise, K[3,3] contains no > triangles, 6 vertices, and 9 edges. Plugging in gives 9 <= 8. > Contradiction. K[3,3] is not planar. > > *Definitely* the nicest approach to this ... :-) > Probably an easier proof than any other you have seen to show it. In > fact, using the above theorem is usually easier to try and prove if a > graph is planar than most any other method. Kuratowski is nice, if you > have time to search for the K[5] or K[3,3]. > > Now use this technique to figure out whether the theorem is true for 10 > vertices (i.e. either G or ~G is not planar on 10 vertices). > > - Tim === Subject: The kids of today are too dumb for math ! ! ! Their brains are full of music, videogame rubbish and parties. There is no room for math anymore in their little heads, so they will end up as soldiers in the US army and fight in some war in 2010-2015. === Subject: Re: The kids of today are too dumb for math ! ! ! > Their brains are full of music, But not, unfortunately, what any normal person would call music. === Subject: Re: The kids of today are too dumb for math ! ! ! > Their brains are full of music, > But not, unfortunately, what any normal person would call music. Umm, guys - why are you replying to can.sun-stroke???? Hans-Marc Olsen appears to be trolling just to trick people into ßooding can.sun-stroke... Mike. === Subject: Re: The kids of today are too dumb for math ! ! ! > Their brains are full of music, videogame rubbish and parties. > There is no room for math anymore in their little heads, so they will > end up as soldiers in the US army and fight in some war in 2010-2015. come on dont be so harsh on the little bighters.... we were the same at their age :) besides videogames are great for developing IQ. kids IQs have been rising quite steadily since the 70s, a rise put down to videogame playing. i put the decline more down to the great growth of new subjects recently being taught in schools, and also the difficulty of employing good math teachers (in the UK anyway) === Subject: Re: The kids of today are too dumb for math ! ! ! > come on dont be so harsh on the little bighters.... we were the same at > their age :) besides videogames are great for developing IQ. kids IQs > have been rising quite steadily since the 70s, a rise put down to videogame > playing. IÕve never believed that IQ measures anything important... now IÕm sure of it. === Subject: alarm clock probablility by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAFIeZm01550; Can you guys help me with this... it seems so simple but the answer just doesnt seem right. heres the problem. Unfortunetely, you do NOT have a very reliable alarm clock. In fact, the three alarm clocks you have only ring some of the time. The first clock only rings 20% of the time, the second clock only rings 30% of the time, and the third alarm clock only rings 50% of the time. If you will be alerted to get up at the correct time.. and the answer choices are A. 0% b. 3% c. 100% d. cannot determine from the given info. e.none of the above. Now i would think it would be b. 3% , by just multiplying the three out. but it doesnt seem right... anyone??? === Subject: Re: alarm clock probablility by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAFKSMo10979; >Can you guys help me with this... it seems so simple but the answer >just doesnt seem right. >heres the problem. >Unfortunetely, you do NOT have a very reliable alarm clock. In fact, >the three alarm clocks you have only ring some of the time. The first >clock only rings 20% of the time, the second clock only rings 30% of >the time, and the third alarm clock only rings 50% of the time. If >you will be alerted to get up at the correct time.. >and the answer choices are >A. 0% >b. 3% >c. 100% >d. cannot determine from the given info. >e.none of the above. >Now i would think it would be b. 3% , by just multiplying the three >out. but it doesnt seem right... anyone??? Look at it this way: whatÕs the probability that none will work? ThereÕs an 80% chance the first wonÕt work, a 70% chance the second wonÕt work, and a 50% chance the third wonÕt work. The probability that the first AND the second AND the third wonÕt work is (.8)(.7)(.5) = .28. The probability that at least one will work is therefore 1 - .28 = .72. I would therefore choose e. (assuming of course that doesnÕt work very well means that correct time is kept, but the ringer is faulty). Your answer of 3% corresponds to the event 1st clock works and 2nd clock works and 3rd clock works, but you need only one to work, so for the event you want, change the ands to ors. My solution above got at this using the complementary event. Hope this makes sense to you, Todd Trimble === Subject: Re: alarm clock probablility alt.math.undergrad: >Unfortunetely, you do NOT have a very reliable alarm clock. In fact, >the three alarm clocks you have only ring some of the time. The first >clock only rings 20% of the time, the second clock only rings 30% of >the time, and the third alarm clock only rings 50% of the time. If >you will be alerted to get up at the correct time.. >and the answer choices are >A. 0% >b. 3% >c. 100% >d. cannot determine from the given info. >e.none of the above. >Now i would think it would be b. 3% , by just multiplying the three >out. but it doesnt seem right... anyone??? Think for a minute about why you multiplied the three probabilities of the three clocks working, and what that would mean. Is that the probability you are awakened at the correct time? No, it is the probability that all three clocks fire at the correct time. That will wake you up, all right, but itÕs more than sufficient. What is sufficient? For _at_least_one_ clock to fire when needed. How can you compute that probability? (Hint: In many problems, including this one, itÕs easier to compute the probability of the event you _donÕt_ want, and subtract it from 1.) -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youÕre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: alarm clock probablility > Can you guys help me with this... it seems so simple but the answer > just doesnt seem right. > heres the problem. > Unfortunetely, you do NOT have a very reliable alarm clock. In fact, > the three alarm clocks you have only ring some of the time. The first > clock only rings 20% of the time, the second clock only rings 30% of > the time, and the third alarm clock only rings 50% of the time. If > you will be alerted to get up at the correct time.. > and the answer choices are > A. 0% > b. 3% > c. 100% > d. cannot determine from the given info. > e.none of the above. > Now i would think it would be b. 3% , by just multiplying the three > out. but it doesnt seem right... anyone??? By multiplying the probabilities of events together you are calculating the probability that ALL of the events occur (given the assumption that the events are independent of each other, which seems plausible for your alarm clocks). So you have calculated the probability that all three alarm clocks ring at the right time... What you actually want is for at least one alarm clock to go off. The way to approach this is that if this doesnÕt happen, then ALL three of the following independent events must occur: a) First alarm clock FAILS to go off b) Second alarm clock FAILS to go off c) Third alarm clock FAILS to go off Hope this points you in the right direction! Mike. === Subject: stock probability by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAFLRxS16034; heres another one i have... You recently purchased three different stocks (AAA, BBB, CCC), with the anticipation that the price per share will go up and your investment will increase in value. Given the following probability distributions below, what is the probability that all of the three stocks will AT LEAST retain the value of your initial purchase price? DOWN) AAA 0.30 0.30 0.40 BBB 0.20 0.70 0.10 CCC 0.40 0.10 0.50 A. .21 B. .27 C. .95 D. 1.00 E. None of the above I come out to 0.27 but not sure im doin it right. well i used a decision tree but still not quite sure.. === Subject: Re: stock probability > heres another one i have... > You recently purchased three different stocks (AAA, BBB, CCC), with > the anticipation that the price per share will go up and your > investment will increase in value. Given the following probability > distributions below, what is the probability that all of the three > stocks will AT LEAST retain the value of your initial purchase price? > DOWN) > AAA 0.30 0.30 0.40 > BBB 0.20 0.70 0.10 > CCC 0.40 0.10 0.50 > A. .21 > B. .27 > C. .95 > D. 1.00 > E. None of the above > I come out to 0.27 but not sure im doin it right. well i used a > decision tree but still not quite sure.. I got the same thing. Bill === Subject: Re: stock probability > heres another one i have... > You recently purchased three different stocks (AAA, BBB, CCC), with > the anticipation that the price per share will go up and your > investment will increase in value. Given the following probability > distributions below, what is the probability that all of the three > stocks will AT LEAST retain the value of your initial purchase price? > DOWN) > AAA 0.30 0.30 0.40 > BBB 0.20 0.70 0.10 > CCC 0.40 0.10 0.50 > A. .21 > B. .27 > C. .95 > D. 1.00 > E. None of the above > I come out to 0.27 but not sure im doin it right. well i used a > decision tree but still not quite sure.. > I got the same thing. > Bill This is becoming a familiar refrain in recent probability-related questions, but itÕs probably worth pointing out again that the question of the INDEPENDENCE (or otherwise) of the distributions is crucial. The answer of 0.27 is true if the distributions are independent. But in the real world the movements of stock prices are often correlated (maybe highly so, depending on the stocks), which would alter the result. === Subject: traffic light by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAFLbuD16814; Heres a good one On your way to a particular location that you regularly travel, there are three intersections with traffic lights. Each of the three lights has a different stop and go pattern: Light one is green 10% of the time, light two is green 40 % of the time, and light three is green 25% of the time. Assuming that you are a very safe driver and only drive through an intersection when the light is green, what is the probability that you will arrive at your destination without having to stop( that is, hitting all three traffic lights green)? I got 1% using a decision tree .... thats a low number, so im not too sure.. === Subject: Re: traffic light alt.math.undergrad: >Heres a good one >On your way to a particular location that you regularly travel, there >are three intersections with traffic lights. Each of the three lights >has a different stop and go pattern: Light one is green 10% of the >time, light two is green 40 % of the time, and light three is green >25% of the time. Assuming that you are a very safe driver and only >drive through an intersection when the light is green, what is the >probability that you will arrive at your destination without having to >stop( that is, hitting all three traffic lights green)? >I got 1% using a decision tree .... thats a low number, so im not too >sure.. YouÕre quite correct, assuming that the lights are independent (different stop and go pattern). The probability of green on the first two is .1*.4 = .04.The probability of also getting green on the third is .04*.25 = .01. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youÕre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: traffic light > Heres a good one > On your way to a particular location that you regularly travel, there > are three intersections with traffic lights. Each of the three lights > has a different stop and go pattern: Light one is green 10% of the > time, light two is green 40 % of the time, and light three is green > 25% of the time. Assuming that you are a very safe driver and only > drive through an intersection when the light is green, what is the > probability that you will arrive at your destination without having to > stop( that is, hitting all three traffic lights green)? > I got 1% using a decision tree .... thats a low number, so im not too > sure.. The correct answer for the problem _as_stated_ is that the desired probability canÕt be determined. Even given the stated probabilities that the lights are green, the lights could still be set so that itÕs 1 (100%) if you drive exactly at the speed limit, or so that itÕs 0 if you drive exactly at the speed limit, or anything in between. But if you assume that the lights are independent, then .1 * .4 * .25 = .01, or 1%, is correct. Brian === Subject: Re: traffic light > .., what is the > probability that you will arrive at your destination without having to > stop( that is, hitting all three traffic lights green)? > I got 1% using a decision tree .... thats a low number, so im not too 1% is not a probability. Probabilities are real numbers x in the range 0 <= x <= 1. === Subject: Re: traffic light alt.math.undergrad: >1% is not a probability. Probabilities are real numbers x in the >range 0 <= x <= 1. Uh, 1% = 1/100 = .01 _is_ in the interval you specify. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youÕre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: traffic light > alt.math.undergrad: >1% is not a probability. Probabilities are real numbers x in the >range 0 <= x <= 1. > Uh, 1% = 1/100 = .01 _is_ in the interval you specify. 1% isnÕt even a real number. x% means x parts out of a hundred _of something_. === Subject: Re: traffic light >> alt.math.undergrad: >>1% is not a probability. Probabilities are real numbers x in the >>range 0 <= x <= 1. >> Uh, 1% = 1/100 = .01 _is_ in the interval you specify. > 1% isnÕt even a real number. CÕmon George weÕve all seen many of your posts so we KNOW that you know better! Maybe someone else is using your account? 1% is the same real number as 1/100 is, as .01 is. Just written differntly, thatÕs all. 100% is the same real number as 1 is. Percent is defined such that these are implications. Look at a definition. HavenÕt you ever converted fractions or decimals into percents and vice versa? Percents are often _used_ in contexts where there is some other non-number entity being described, eg. a 30% chance of rain, but so what, so are other forms of real numbers. Numbers (in various forms) are used all the time to to quantify things, eg. 4 inches of rain, 79 dollars, 5 books, half of a pizza, etc. Surely youÕre not going to suggest rain and dollar and book are real numbers. Of course these are not the numbers. 4 and 79 and 5 are. That doesnÕt mean the numbers 4, 79, and 5 are meaningless outside of the context of quantifying something specific, like rain, dollars, or books. Likewise, when you see 1% _of something_ it is not necessarily the _something_ that is a real number (though it may be, of course, eg 40% of 57) but the 1% part definitely IS a real number. By definition a percent (a implying 1) is simply a way to express a ratio in parts of 100. 1% is the _exact_ same real number 1/100 is. > x% means x parts out of a hundred _of > something_. Think about what youÕre saying. WhatÕs 50% of 100? Is it 50 inches of rain, 50 dollars, 50 books, or is it just the number 50? IOW, if you want to insist that it must be x parts out of a hundred of something, donÕt you see that the something is just a whole. A whole is the real number 1. Just like the fraction 1/2 means one half of 1. -- Darrell === Subject: Re: traffic light > ... > x% means x parts out of a hundred _of > something_. > Think about what youÕre saying. WhatÕs 50% of 100? Is it 50 inches of > rain, 50 dollars, 50 books, or is it just the number 50? IOW, if you want > to insist that it must be x parts out of a hundred of something, I make no such claim. An example of x% of something may indeed be x% of 100. I donÕt know why you are adding another of something. > donÕt you > see that the something is just a whole. A whole is the real number 1. > Just like the fraction 1/2 means one half of 1. > -- > Darrell === Subject: Re: traffic light > > alt.math.undergrad: >1% is not a probability. Probabilities are real numbers x in the >range 0 <= x <= 1. > > Uh, 1% = 1/100 = .01 _is_ in the interval you specify. > 1% isnÕt even a real number. x% means x parts out of a hundred _of > something_. And sometimes the _of something_ is 1. Like when someone says a fair coin has a 50% chance of coming up heads. Only the most pedantic would deny the speaker means a probability of 0.5. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: traffic light >> .., what is the >> probability that you will arrive at your destination without having to >> stop( that is, hitting all three traffic lights green)? >> I got 1% using a decision tree .... thats a low number, so im not too > 1% is not a probability. Probabilities are real numbers x in the > range 0 <= x <= 1. 1% _is_ a real number between 0 and 1. -- Darrell === Subject: Re: traffic light > Heres a good one > On your way to a particular location that you regularly travel, there > are three intersections with traffic lights. Each of the three lights > has a different stop and go pattern: Light one is green 10% of the > time, light two is green 40 % of the time, and light three is green > 25% of the time. Assuming that you are a very safe driver and only > drive through an intersection when the light is green, what is the > probability that you will arrive at your destination without having to > stop( that is, hitting all three traffic lights green)? > I got 1% using a decision tree .... thats a low number, so im not too > sure.. Looks OK to me. Though there are other complications the problem does not address - like traffic backups, which might make you close to 100% of hitting a light. Bill === Subject: Re: traffic light >> Heres a good one >> On your way to a particular location that you regularly travel, there >> are three intersections with traffic lights. Each of the three lights >> has a different stop and go pattern: Light one is green 10% of the >> time, light two is green 40 % of the time, and light three is green >> 25% of the time. Assuming that you are a very safe driver and only >> drive through an intersection when the light is green, what is the >> probability that you will arrive at your destination without having to >> stop( that is, hitting all three traffic lights green)? >> I got 1% using a decision tree .... thats a low number, so im not too >> sure.. >Looks OK to me. Though there are other complications the problem does not >address - like traffic backups, which might make you close to 100% of hitting >a light. >Bill Also unaddressed is whether the three lights are timed independently of each other. In many real world situations, the lights are in fact timed with each other in some fashion, therefore definitely not independent. --Lynn === Subject: Re: traffic light > Heres a good one > On your way to a particular location that you regularly travel, there > are three intersections with traffic lights. Each of the three lights > has a different stop and go pattern: Light one is green 10% of the > time, light two is green 40 % of the time, and light three is green > 25% of the time. Assuming that you are a very safe driver and only > drive through an intersection when the light is green, what is the > probability that you will arrive at your destination without having to > stop( that is, hitting all three traffic lights green)? > I got 1% using a decision tree .... thats a low number, so im not too > sure.. >>Looks OK to me. Though there are other complications the problem does not >>address - like traffic backups, which might make you close to 100% of >>hitting >>a light. >>Bill > Also unaddressed is whether the three lights are timed independently > of each other. In many real world situations, the lights are in fact > timed with each other in some fashion, therefore definitely not > independent. > --Lynn You could be right. But I took a different stop and go pattern to imply independence. Bill === Subject: Re: Population Models and Generalised Geometric Means by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAGDegU31176; Hi Everyone! Much has happened since the I posted the original message above. Have had to create a MULTIPLICATIVE version of infinitesimal calculus (Multigral Calculus)which can be used to estimate long-term modes of STOCHASTIC RECURSIVE EQUATIONS like the popultation model P--> (1+b-d)*P where P the popultaion level, b=stochastic birth rate and d= stochastic death rate. See the website www.geocities.com/multigrals2000 for details. Surprisingly, if b=1.243865664..*ran# and d=ran# (where ran#= a random number between 0 and 1 and b,d are independent) then mode(P)-->initial P as n-->infinity even though E(P)--> infinity. True. Check out website for details. So the standard treatment of Stochastic Population models using means is mathematically correct but inappropriate when modelling a SINGLE population. Truly. Yours, D === Subject: Solve 2x /(3x +1) = 2g(x) / (3g(x)+1) , generalize by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAGDefx31122; Why wouldnÕt you try solving 2x /(3x +1) = 2g(x) / (3g(x)+1); g is R*R->R ,continuous and one-to-one , unknown . Generalization is possible : f and g are R*R->R ,continuous and one-to-one . See you soon,Alain. === Subject: Re: Solve 2x /(3x +1) = 2g(x) / (3g(x)+1) , generalize by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAGEZLW04482; >Why wouldnÕt you try solving 2x /(3x +1) = 2g(x) / (3g(x)+1); >g is R*R->R ,continuous and one-to-one , unknown . >Generalization is possible : f and g are R*R->R ,continuous > and one-to-one . >See you soon,Alain. I meant 2x /(3x +1)*g(x) = 2g(x) / (3g(x)+1); sorry,Alain. === Subject: Re: Solve 2x /(3x +1) = 2g(x) / (3g(x)+1) , generalize by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAID3vh11772; >>Why wouldnÕt you try solving 2x /(3x +1) = 2g(x) / (3g(x)+1); >>g is R*R->R ,continuous and one-to-one , unknown . >>Generalization is possible : f and g are R*R->R ,continuous >> and one-to-one . >>See you soon,Alain. >I meant 2x /(3x +1)*g(x) = 2g(x) / (3g(x)+1); >sorry,Alain. Of course there is a direct computing but let us write f(x)=2x /(3x +1), Here we may observe the Ōformf(x)*g(x)=f(g(x)) ; bijectivity of g and f ->f(g^-1(x))*g(g^-1(x))=f(g(g^-1(x))) or f(g^-1(x))*x = f(x) , then g^-1(x) = f^-1(f(x)/x); this formula gives after inversion g(x). An other generalization is possible:g unknown f(x)*p(g(x))=f(g(x)) ; g^-1(x) = f^-1(f(x)/p(x)); Alain. === Subject: slope by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAGG1g813602; i need help on x and y intercept, please. i have a quiz today === Subject: Re: slope by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAGGGnK15119; >i need help on x and y intercept, please. i have a quiz today i need help soooo badly, please help me. === Subject: Re: slope >>i need help on x and y intercept, please. i have a quiz today >i need help soooo badly, please help me. Try opening your textbook. Your question is so vague and general that the only possible answer is a long essay on intercepts (or maybe slopes), which would only duplicate whatÕs in your book, less successfully. Usenet is like God: it helps those who help themselves. Ask a _specific_ question and you may get a useful answer. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youÕre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: slope >i need help on x and y intercept, please. i have a quiz today >>i need help soooo badly, please help me. An x-intercept for a equation (graph) is the value of x when y=0. The y-intercept is the value of y when x=0. All you have to remember is to set the *other* one equal to 0 and solve. A simple example: Find the x and y intercepts of 2x + 7y = 4 2x + 0 = 4 2x = 4 x = 2 ...so the point (2,0) is the x-intercept. y-intercept: 2(0) + 7y = 4 0 + 7y = 4 7y = 4 y = 4/7 ...so the y-intercept is the point (0, 4/7). > Try opening your textbook. Your question is so vague and general > that the only possible answer is a long essay on intercepts (or > maybe slopes), which would only duplicate whatÕs in your book, less > successfully. A very long essay, that was. > Usenet is like God: it helps those who help themselves. No, Stan is like God (or so he thinks.) His way or the highway, is his philosophy. Even when he has no control over the road, he still dogs everyone as if he does. Notice he did not address the question at all, other than to essentially say heÕs _not_ going to address it. WhatÕs the matter, Stan, are your students stressing you out so much you have to take it out on innocent Usenet posters, again? Get a grip. Take baby steps, if you must, for ex. you can start out promising yourself you wonÕt bitch and moan unless you actually address the question posed. === Subject: Re: slope by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAGITmQ28403; >>i need help on x and y intercept, please. i have a quiz today >i need help soooo badly, please help me. Assume you are given the equation of a line ax + by + c = 0, where a not= 0, b not= 0, c not= 0. Rearrange the equation to look like x/p + y/q = 1 Then p is the x-axis intercept and q is the y-axis intercept. ax + by = -c -a/c*x - b/c*y = 1 x/(-c/a) + y/(-c/b) = 1 p = -c/a is the x-axis intercept, q = -b/c is the y-axis intercept. This is true even if c = 0. The line then passes through the origin and both intercepts are 0. Example: 2x + 3y - 6 = 0 2x + 3y = 6 2/6*x + 3/6*y = 1 1/3*x + 1/2*y = 1 x/3 + y/2 = 1 === Subject: Re: Solve 2x /(3x +1) = 2g(x) / (3g(x)+1) , generalize by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAGG1gw13606; simplify the problem === Subject: trig questions.. help I have a final tomorrow and IÕm going over some homework problems that may be on the test. If anyone knows how to solve these please explain and show KuLL@cox-internet.com jkc014@latech.edu x = theta Write cos (4x) as a trigonometric function of the single angle x. (I know this has to do with double angle identities, but still can not figure it out. 12.5sin{pi/6 (t-5)} + 58 in simplest form using an identity. Proving an identity: cos (2a) = cot(a)-tan(a)/csc(a)sec(a) when i work this out i end up with -tan(a)/1/sinAcosA.. so I donÕt know what IÕm doing and lastly, Find sinA given that cos(2a)= -2/squareroot of 7 2a is in quad II. tomorrow. === Subject: Re: trig questions.. help >Write cos (4x) as a trigonometric function of the single angle x. (I know >this has to do with double angle identities, but still can not figure it >out. Hint: 4x = 2*2x. So first write cos(4x) as functions of 2x, then apply double-angle identities again. >12.5sin{pi/6 (t-5)} + 58 in simplest form using an identity. I canÕt figure out what you mean here. Is t-5 supposed to be part of numerator or denominator? >Proving an identity: >cos (2a) = cot(a)-tan(a)/csc(a)sec(a) >when i work this out i end up with -tan(a)/1/sinAcosA.. so I donÕt know what >IÕm doing Hint: When having trouble proving any identity, try rewriting all other functions in terms of sin x and cos x. >Find sinA given that cos(2a)= -2/squareroot of 7 2a is in quad II. Hint: (sin x)^2 + (cos x)^2 = 1. ThatÕs the master identity, which should be graven on your cortex. It means that if you know sine of an angle you know cosine of the same angle and vice versa -- the only question is whether it takes a + or - sign, and you know that if you know what quadrant itÕs in. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youÕre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: trig questions.. help Cc: jkc014@latech.edu > x = theta > Write cos (4x) as a trigonometric function of the single angle x. (I know > this has to do with double angle identities, but still can not figure it > out. Look at what the double-angle identity for cosine says: cos(2t) = [cos(t)]^2 - [sin(t)]^2 Let t=2x [thatÕs the important step] and you immediately get cos(4x) = [cos(2x)]^2 - [sin(2x)]^2 Then apply the double angle formulas again to turn those 2xÕs into xÕs. Or, if you had some complex numbers and have learned e^(it) = cos(t) + i sin(t) consider what happens if you raise each side of that equation to the 4th power. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: trig questions.. help > x = theta > > Write cos (4x) as a trigonometric function of the single angle x. (I know > this has to do with double angle identities, but still can not figure it > out. This sort of problem is, in general, most easily approached by using De MoivreÕs theorem, that, in trigonometric form, says that For i^2 = -1 and any positive integer n, cos(n*x) + i*sin(n*x) = ( cos(x) + i*sin(x) )^n One can then separate the real and imaginary parts of this equations into two eparate equations, when the right hand side is expanded, to get expressions for cos(n*x) and sin(n*x) in terms of cos(x) and sin(x). === Subject: Stats help I have some answers but i am unsure that they are correct. 1) find theprobability of finding a difference between the sample means X1 and X2 of more than 1.0. My data is as the fallowing.. N= 64 for both populations; first pop. Mu = 45 and sigma = 5; second pop. Mu = 47 and sigma = 8. 2) i have a large container with 5000 red and 10000 green marbles. The container is mixed and i draw out 100 marbles. What is the probability that the sample has more than 40% red marbles? 3) The average guest bill at one of the managers hotel is atleast $250. To test this she obtains a sample of 60 bills and finds, for her sample, Xbar = $325 and s = 50. Do the data disagree with her hypothesis? L.O.S = .01. Find a P-Value for your test. Jimi === Subject: Re: Stats help > ... To test this she obtains a sample of 60 bills and finds, How did she obtain these bills? Is this not a case of customersÕ confidentiality being violated? === Subject: Re: Stats help Please DO NOT post the same query multiple times, and not to multiple newsgroups either. I spent considerable time answering your questions in sci.math -- it is rather irritating to see the same questions posted here a day later. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youÕre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: Stats help > Please DO NOT post the same query multiple times, and not to > multiple newsgroups either. > I spent considerable time answering your questions in sci.math -- it > is rather irritating to see the same questions posted here a day > later. Here let me just quote you, Sir.. Please note that homework questions might be better in alt.math.undergrad. looks like thats what i did! Jimi === Subject: JSH: Legacy of error Basically the latter 1800Õs and early 1900Õs represent a period when a substantial amount of errors, mathematical and logical entered into the math field, and then, in what has been a covering action, the concept of pure math was emphasized, which helped to hide the errors. There is more than just the problem with the ring of algebraic integers that IÕve discussed so much. Like, consider the Riemann Hypothesis. I find it hard to face the issue myself at times, but my find of a three dimensional formula that counts primes gives a rather clear and readily understandable explanation for the closeness of the prime distribution to continuous functions. ItÕs just simple. But you have these complex ideas that got a lot of attention, get a lot of attention, and I know that nothing will come of them. If you look at RiemannÕs actual work, youÕll find a section where he talks about certain terms looking like theyÕll tend to zero, as they look like they should balance out. But has anyone ever proven that they will? Goldstone had what he thought was a proof shot down for using the same type reasoning just recently. So why wouldnÕt anyone go back to RiemannÕs work with the same analysis? Yet, why is my work still just supposedly the rantings of some crank when all of it is now rigorously proven? You are in a field with a legacy of error. Andrew Wiles didnÕt prove FermatÕs Last Theorem. RiemannÕs Hypothesis can be shown to probably be wrong on some very basic grounds, but no one will admit it. And thatÕs not all of the error. I have research I donÕt bother even talking about in public, as whatÕs the point? Actually itÕs just freaking depressing. The mistakes cluster around 1900, both before and after it, and itÕs just weird looking at them, wondering. Like think about this algebraic integer thing. Why wouldnÕt it occur to people that maybe they might want to consider roots of non-monic polynomials irreducible over Q that might correlate to integer examples like 3x^2 + 4x + 1 = (3x + 1)(x + 1)? I mean, itÕs not like it takes brilliance to consider that maybe some non-monics--even if they were irreducible over Q--might have roots that have integral properties despite being irrational. It doesnÕt take brilliance to consider possibilities. And as for counting primes, why not have a p(x,y) function that gives a count with p(x,sqrt(x))? Why couldnÕt the answer to linking the prime distribution to continuous functions like x/ln x come from multi-variable functions? Why not ask why? Have you seen my derivation of the prime counting function? Such a simple derivation could have been done a thousand years ago. Maybe it was and lost. If I knew a few years ago what I know now, IÕd have never made these discoveries--if I could have stopped myself. Ignorance is bliss. There is no profit in it after all, and IÕm not talking about money. I never really was. James Harris === Subject: Re: JSH: Legacy of error > Like think about this algebraic integer thing. Why wouldnÕt it occur > to people that maybe they might want to consider roots of non-monic > polynomials irreducible over Q that might correlate to integer > examples like > 3x^2 + 4x + 1 = (3x + 1)(x + 1)? > I mean, itÕs not like it takes brilliance to consider that maybe some > non-monics--even if they were irreducible over Q--might have roots > that have integral properties despite being irrational. Ah, but people HAVE considered algebraic integer roots to non-monic polynomials. The result is that a non-monic polynomial with at least one algebraic integer root is reducible with at least one of the polynomial factors being monic. Like in your example, 3x^2 + 4x + 1 is reducible and the (at least) one monic polynomial factor is (x + 1). > It doesnÕt take brilliance to consider possibilities. > And as for counting primes, why not have a p(x,y) function that gives > a count with p(x,sqrt(x))? Why couldnÕt the answer to linking the > prime distribution to continuous functions like x/ln x come from > multi-variable functions? > Why not ask why? > Have you seen my derivation of the prime counting function? > Such a simple derivation could have been done a thousand years ago. > Maybe it was and lost. Most likely algorithms very much like yours have been rediscovered many times, it is just that the algorithm is so grossly inefficient compared to other algorithms that there is no interest. Especially when you consider that the prime factorization is all about speed to computer scientists/cryptographers. === Subject: Re: JSH: Legacy of error > > Like think about this algebraic integer thing. Why wouldnÕt it occur > to people that maybe they might want to consider roots of non-monic > polynomials irreducible over Q that might correlate to integer > examples like > > 3x^2 + 4x + 1 = (3x + 1)(x + 1)? > > I mean, itÕs not like it takes brilliance to consider that maybe some > non-monics--even if they were irreducible over Q--might have roots > that have integral properties despite being irrational. > Ah, but people HAVE considered algebraic integer roots to non-monic > polynomials. The result is that a non-monic polynomial with at least > one algebraic integer root is reducible with at least one of the > polynomial factors being monic. Like in your example, 3x^2 + 4x + 1 > is reducible and the (at least) one monic polynomial factor is (x + > 1). DonÕt just say reducible, say reducible over Q, or over rationals. But how about? 3x^2 + Kx + 1 = (3x + u_1)(x + u_2) where u_1 and u_2 are properly units for some integer K? Now the wrong answer is to say that Galois Theory shows that if u_1 and u_2 are rational, then ok, they can both be units, like with K=4, but if they are irrational, then NO MATTER what supposedly thereÕs never another unit case? ThatÕs just stupid. When you think about it, why should there not be another unit case just because u_1 and u_2 arenÕt algebraic integers? ItÕs just DUMB, and defending it, especially after IÕve proven that such cases must exist, is just an exercise in futility. > > It doesnÕt take brilliance to consider possibilities. > > And as for counting primes, why not have a p(x,y) function that gives > a count with p(x,sqrt(x))? Why couldnÕt the answer to linking the > prime distribution to continuous functions like x/ln x come from > multi-variable functions? > > Why not ask why? > > Have you seen my derivation of the prime counting function? > > Such a simple derivation could have been done a thousand years ago. > Maybe it was and lost. > Most likely algorithms very much like yours have been rediscovered > many times, it is just that the algorithm is so grossly inefficient > compared to other algorithms that there is no interest. Especially > when you consider that the prime factorization is all about speed to > computer scientists/cryptographers. However, Gauss and others noticed a peculiar closeness between the prime distribution and continuous functions like x/ln x, and later Li(x). The three-dimensional function I found, which when properly constrained counts prime numbers, shows WHY that closeness exists in a very simple manner. It is in fact what Riemann was looking for--an explanation--when he made his famous hypothesis, which is why your posting is worse than misleading, itÕs anti-mathematical. The real point of my discovery of a partial difference equation to count primes is that it gives THE ANSWER to the why of the prime distributions closeness, and it probably means that the Riemann Hypothesis itself is false. Now supposedly the Riemann Hypothesis is a big deal, but itÕs all politics in todayÕs math world, or there would be people all over my work, either showing what I just said, or explaining why itÕs not true as thereÕs just NOT THAT MUCH NEW in this area, and what I have is distinctive. Or, are you going to try to claim another partial difference equation exists that can be used to count primes, which can be mapped to a partial differential equation which integrates closely to the prime distribution and to x/ln x and Li(x)? If not, why donÕt you come back and freaking apologize for your b.s. post? You wonÕt, now will you? Because you are an anti-mathematician. James Harris === Subject: Re: JSH: Legacy of error > > Like think about this algebraic integer thing. Why wouldnÕt it occur > to people that maybe they might want to consider roots of non-monic > polynomials irreducible over Q that might correlate to integer > examples like > > 3x^2 + 4x + 1 = (3x + 1)(x + 1)? > > I mean, itÕs not like it takes brilliance to consider that maybe some > non-monics--even if they were irreducible over Q--might have roots > that have integral properties despite being irrational. > > Ah, but people HAVE considered algebraic integer roots to non-monic > polynomials. The result is that a non-monic polynomial with at least > one algebraic integer root is reducible with at least one of the > polynomial factors being monic. Like in your example, 3x^2 + 4x + 1 > is reducible and the (at least) one monic polynomial factor is (x + > 1). > > DonÕt just say reducible, say reducible over Q, or over rationals. > But how about? > 3x^2 + Kx + 1 = (3x + u_1)(x + u_2) > where u_1 and u_2 are properly units for some integer K? You are putting the cart before the horse. The factorization you propose is not correct for all K. If you were to define K in terms of u_1 and u_2 (ie let K = 3*u_2 + u_1) then there is no problem. But by defining u_1 and u_2 in terms of K, you make an assumption that (3x + u_1)(x + u_2) is the correct factorization. In fact I think that is the core error in all of your algebraic integer work. You are defining an arbitrary factorization and blaming algebraic integers when it turns out that every equation can not be factored in that way. Consider for example 6x^2 + Ax + 1. If I were to say 6x^2 + Ax + 1 = (3x + v_1)(2x + v_2), I would quickly run into problems trying to factor when A=7. The scenario you never seem to consider is to say the factorization (3x + v_1)(2x + v_2) is wrong for this case and that there exists some other factorization. Indeed there is: 6x^2 + Ax + 1 = (6x + v_1)(x + v_2). Your problem is that you are so busy trying to fit all cases into a single factorization that that you do not ever try to look at the possibility of different factorizations. > Now the wrong answer is to say that Galois Theory shows that if u_1 > and u_2 are rational, then ok, they can both be units, like with K=4, > but if they are irrational, then NO MATTER what supposedly thereÕs > never another unit case? > ThatÕs just stupid. When you think about it, why should there not be > another unit case just because u_1 and u_2 arenÕt algebraic integers? Again, there is another case, it is simply a different factorization from what you started with. > ItÕs just DUMB, and defending it, especially after IÕve proven that > such cases must exist, is just an exercise in futility. > > It doesnÕt take brilliance to consider possibilities. > > And as for counting primes, why not have a p(x,y) function that gives > a count with p(x,sqrt(x))? Why couldnÕt the answer to linking the > prime distribution to continuous functions like x/ln x come from > multi-variable functions? > > Why not ask why? > > Have you seen my derivation of the prime counting function? > > Such a simple derivation could have been done a thousand years ago. > Maybe it was and lost. > > Most likely algorithms very much like yours have been rediscovered > many times, it is just that the algorithm is so grossly inefficient > compared to other algorithms that there is no interest. Especially > when you consider that the prime factorization is all about speed to > computer scientists/cryptographers. > However, Gauss and others noticed a peculiar closeness between the > prime distribution and continuous functions like x/ln x, and later > Li(x). Different issue. I am addressing the slowness of your formula, not is correctness. > The three-dimensional function I found, which when properly > constrained counts prime numbers, shows WHY that closeness exists in a > very simple manner. > It is in fact what Riemann was looking for--an explanation--when he > made his famous hypothesis, which is why your posting is worse than > misleading, itÕs anti-mathematical. > The real point of my discovery of a partial difference equation to > count primes is that it gives THE ANSWER to the why of the prime > distributions closeness, and it probably means that the Riemann > Hypothesis itself is false. > Now supposedly the Riemann Hypothesis is a big deal, but itÕs all > politics in todayÕs math world, or there would be people all over my > work, either showing what I just said, or explaining why itÕs not true > as thereÕs just NOT THAT MUCH NEW in this area, and what I have is > distinctive. > Or, are you going to try to claim another partial difference equation > exists that can be used to count primes, which can be mapped to a > partial differential equation which integrates closely to the prime > distribution and to x/ln x and Li(x)? I am not claiming your function is wrong. I am claiming that your function is a slow, inefficient, redundant method of counting primes. A claim which can be verified by holding a speed test. You can post your code for the algorithm and it can be run by anyone. Anyone then can compare the result to their own algorithm (or anyone elseÕs) run on the same machine, written in the same programming language, and decinde which one is faster. If I am not mistaken, that has already been done and your algorithm did rather poorly, especially compared to Christian BauÕs algorithm. > If not, why donÕt you come back and freaking apologize for your b.s. > post? Why do I need to apologize for stating my opinion of your work. You posted looking for feedback and I gave you feedback. I would have thought by now that you would have learned to handle negative feedback gracefully. > You wonÕt, now will you? Because you are an anti-mathematician. Wow, from the way you talk, you are the archetype of anti-mathematicians. > James Harris === Subject: Re: JSH: Legacy of error > > Like think about this algebraic integer thing. Why wouldnÕt it occur > to people that maybe they might want to consider roots of non-monic > polynomials irreducible over Q that might correlate to integer > examples like > > 3x^2 + 4x + 1 = (3x + 1)(x + 1)? > > I mean, itÕs not like it takes brilliance to consider that maybe some > non-monics--even if they were irreducible over Q--might have roots > that have integral properties despite being irrational. > > Ah, but people HAVE considered algebraic integer roots to non-monic > polynomials. The result is that a non-monic polynomial with at least > one algebraic integer root is reducible with at least one of the > polynomial factors being monic. Like in your example, 3x^2 + 4x + 1 > is reducible and the (at least) one monic polynomial factor is (x + > 1). > > DonÕt just say reducible, say reducible over Q, or over rationals. > But how about? > 3x^2 + Kx + 1 = (3x + u_1)(x + u_2) > where u_1 and u_2 are properly units for some integer K? What does it mean for a number to be a unit Ōproperly? -- Larry Lard Replies to group please === Subject: Re: JSH: Legacy of error >[...] >However, Gauss and others noticed a peculiar closeness between the >prime distribution and continuous functions like x/ln x, and later >Li(x). >The three-dimensional function I found, which when properly >constrained counts prime numbers, shows WHY that closeness exists in a >very simple manner. You keep saying this, but youÕve never explained HOW your work shows HOW this is true. >It is in fact what Riemann was looking for--an explanation--when he >made his famous hypothesis, which is why your posting is worse than >misleading, itÕs anti-mathematical. >The real point of my discovery of a partial difference equation to >count primes is that it gives THE ANSWER to the why of the prime >distributions closeness, and it probably means that the Riemann >Hypothesis itself is false. Only in your imagination. >Now supposedly the Riemann Hypothesis is a big deal, but itÕs all >politics in todayÕs math world, or there would be people all over my >work, either showing what I just said, or explaining why itÕs not true This is a classic. People have been explaining why essentially everything you say is not true for years, since you started posting on usenet. Now you say you must be right, otherwise people would be explaining why youÕre wrong. A new one - exactly the sort of thing that keeps your fans reading >as thereÕs just NOT THAT MUCH NEW in this area, and what I have is >distinctive. Ok. I think we can all agree that distinctive is right... >Or, are you going to try to claim another partial difference equation >exists that can be used to count primes, which can be mapped to a >partial differential equation which integrates closely to the prime >distribution and to x/ln x and Li(x)? Why have you never _domeonstrated_ that this thing integrates closely to x/ln x and Li(x)? Herea hint: YouÕve just given away the fact that your amazing result is nothing at all - even if everything youÕve said so far is true, it follows, _from what you just said_, that your amazing result is a huge step _backwards_ from existing knowledge. HereÕs why: As is well known, x/ln x gives an approximation to pi(x) and Li(x) gives a _better_ approximation. Your thing integrates closely to _both_, evidently the approximation you get is much worse than whatÕs known. (ThatÕs all assuming everything youÕve said is true, which of course is not true. The only reason you have for thinking that the solution to the continuous equation approximates pi(x) is a vague and _incorrect_ feeling that it must be so - the example I keep posting and you keep ignoring shows that thereÕs simply no reason the solutions to the continuous version of the difference equation should be close to pi(x) (unless of course weÕre counting e^x as a good approximation to 2^x.) >If not, why donÕt you come back and freaking apologize for your b.s. >post? >You wonÕt, now will you? Because you are an anti-mathematician. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Legacy of error > > Like think about this algebraic integer thing. Why wouldnÕt it occur > to people that maybe they might want to consider roots of non-monic > polynomials irreducible over Q that might correlate to integer > examples like > > 3x^2 + 4x + 1 = (3x + 1)(x + 1)? > > I mean, itÕs not like it takes brilliance to consider that maybe some > non-monics--even if they were irreducible over Q--might have roots > that have integral properties despite being irrational. > > Ah, but people HAVE considered algebraic integer roots to non-monic > polynomials. The result is that a non-monic polynomial with at least > one algebraic integer root is reducible with at least one of the > polynomial factors being monic. Like in your example, 3x^2 + 4x + 1 > is reducible and the (at least) one monic polynomial factor is (x + > 1). > > DonÕt just say reducible, say reducible over Q, or over rationals. > But how about? > 3x^2 + Kx + 1 = (3x + u_1)(x + u_2) > where u_1 and u_2 are properly units for some integer K? > Now the wrong answer is to say that Galois Theory shows that if u_1 > and u_2 are rational, then ok, they can both be units, like with K=4, > but if they are irrational, then NO MATTER what supposedly thereÕs > never another unit case? OOPS! There is an irrational unit case, which is with K=0. So, if you believe the way Galois Theory is taught, then you can have u_1 and u_2 units ONLY when K=0, K=-1, or K=1, and for every other case--according to those beliefs--neither u_1 and u_2 are units just because theyÕre not units in the ring of algebraic integers. > ThatÕs just stupid. When you think about it, why should there not be > another unit case just because u_1 and u_2 arenÕt algebraic integers? ItÕs always funny to come back to a post where you said something was stupid when you made a mistake. Ok, so I was stupid, but at least I can admit my error... James Harris === Subject: Re: JSH: Legacy of error > > Like think about this algebraic integer thing. Why wouldnÕt it occur > to people that maybe they might want to consider roots of non-monic > polynomials irreducible over Q that might correlate to integer > examples like > > 3x^2 + 4x + 1 = (3x + 1)(x + 1)? > > I mean, itÕs not like it takes brilliance to consider that maybe some > non-monics--even if they were irreducible over Q--might have roots > that have integral properties despite being irrational. > > Ah, but people HAVE considered algebraic integer roots to non-monic > polynomials. The result is that a non-monic polynomial with at least > one algebraic integer root is reducible with at least one of the > polynomial factors being monic. Like in your example, 3x^2 + 4x + 1 > is reducible and the (at least) one monic polynomial factor is (x + > 1). > > > DonÕt just say reducible, say reducible over Q, or over rationals. > > But how about? > > 3x^2 + Kx + 1 = (3x + u_1)(x + u_2) > > where u_1 and u_2 are properly units for some integer K? > > Now the wrong answer is to say that Galois Theory shows that if u_1 > and u_2 are rational, then ok, they can both be units, like with K=4, > but if they are irrational, then NO MATTER what supposedly thereÕs > never another unit case? > OOPS! There is an irrational unit case, which is with K=0. Damn. Original was right, but the correction I made this morning was wrong. ThatÕs annoying. Guess IÕm too worried here that I screwed up somewhere. Now IÕm correcting what I think are mistakes that are right only to have to correct the correction which is wrong. > So, if you believe the way Galois Theory is taught, then you can have > u_1 and u_2 units ONLY when K=0, K=-1, or K=1, and for every other > case--according to those beliefs--neither u_1 and u_2 are units just > because theyÕre not units in the ring of algebraic integers. > ThatÕs just stupid. When you think about it, why should there not be > another unit case just because u_1 and u_2 arenÕt algebraic integers? > > ItÕs always funny to come back to a post where you said something was > stupid when you made a mistake. Ok, so I was stupid, but at least I > can admit my error... And just weird to come back to the come back post and note that it was wrong. Twilight Zone. But itÕs been that for a while. James Harris === Subject: Re: JSH: Legacy of error >> [...] >> OOPS! There is an irrational unit case, which is with K=0. >Damn. Original was right, but the correction I made this morning was >wrong. >ThatÕs annoying. Not at all, itÕs hilarious. > Guess IÕm too worried here that I screwed up >somewhere. YouÕre worried you screwed up somewhere? Why would that be? I mean ok, itÕs true that for many years every non-trivial new and significant fact youÕve announced here has turned out to be wrong. But I donÕt see why that would make you worry that you screwed up here, after all the _current_ post is never wrong. >Now IÕm correcting what I think are mistakes that are right only to >have to correct the correction which is wrong. ************************ David C. Ullrich === Subject: Re: JSH: Legacy of error >>Like think about this algebraic integer thing. Why wouldnÕt it occur >>to people that maybe they might want to consider roots of non-monic >>polynomials irreducible over Q that might correlate to integer >>examples like >>3x^2 + 4x + 1 = (3x + 1)(x + 1)? >>I mean, itÕs not like it takes brilliance to consider that maybe some >>non-monics--even if they were irreducible over Q--might have roots >>that have integral properties despite being irrational. >Ah, but people HAVE considered algebraic integer roots to non-monic >polynomials. The result is that a non-monic polynomial with at least >one algebraic integer root is reducible with at least one of the >polynomial factors being monic. Like in your example, 3x^2 + 4x + 1 >is reducible and the (at least) one monic polynomial factor is (x + >1). >>DonÕt just say reducible, say reducible over Q, or over rationals. >>But how about? >>3x^2 + Kx + 1 = (3x + u_1)(x + u_2) >>where u_1 and u_2 are properly units for some integer K? So what? TheyÕre units in any ring where the above factorization occurs. TheyÕre integral over Z[1/3]. WhatÕs your problem? >>Now the wrong answer is to say that Galois Theory shows that if u_1 >>and u_2 are rational, then ok, they can both be units, like with K=4, >>but if they are irrational, then NO MATTER what supposedly thereÕs >>never another unit case? Why should you care more about units than about the ring? Is it more important to have lots of units, or to have a ring that allows you to understand the integers? > OOPS! There is an irrational unit case, which is with K=0. When K=0, you have (over the algebraic integers): 3 x^2 + 1 = (sqrt(3) x + i)(sqrt(3) x - i) Note how the factorization isnÕt necessarily what you wanted. You could, of course, divide and multiply by sqrt(3) to get 3 x^2 + 1 = (3 x + sqrt(3)i)(x - i/sqrt(3)) but that requires 3 to be a unit in the ring of coefficients. Are you sure thatÕs the way you want things? I mean, youÕre all het up over the concept of units, and youÕre also pretty keen on the only integers that can be units being +/- 1. You canÕt have your factorization *and* the restriction on which integers can be units. > So, if you believe the way Galois Theory is taught, then you can have > u_1 and u_2 units ONLY when K=0, K=-1, or K=1, and for every other > case--according to those beliefs--neither u_1 and u_2 are units just > because theyÕre not units in the ring of algebraic integers. This statement is just silly. In the first case, you presume to know the way Galois Theory is taught. This is simply a presumption on your part, and a fairly outlandish one at that, since itÕs abundantly clear that you donÕt know the first thing about Galois Theory. In the second case, no one in mathematics is really concerned about belief. To be sure, people will use phrases akin to I believe X when they should be using phrases like I suspect X, that is, to express some sort of hunch that they have if they have some evidence, but no proof, of the statement X. However, itÕs clear that this is not that youÕre talking about. Statements are proven or they are not proven. The results of Galois Theory are *proven*, not just *believed*. Further, the proofs are revisited thousands, if not millions, of times, every semester, as large numbers of students take higher-level courses in algebra; this has been going on for a long time, and to my knowledge, no mistake has been found in any of the proofs that support the theory. Finally, one cannot reasonably discuss numbers being (or not being) units without having the context of a ring in which they are (or are not) invertible. If one is working in the ring of algebraic integers, then your polynomial may not factor in the way you want. One problem is seen in the above example with K=0, which suggests that you may need to have units in your ring that youÕd prefer werenÕt there, just to get that factorization. >>ThatÕs just stupid. When you think about it, why should there not be >>another unit case just because u_1 and u_2 arenÕt algebraic integers? No one says that there isnÕt a ring in which the uÕs are units. Of course there is. Why do you refuse to understand this? Find a ring in which the polynomial factors in the fashion youÕve chosen. Any ring where it factors that way, will then have the uÕs as units. For instance, the ring A[1/3], where A is the ring of algebraic integers. Spending all this effort whining about numbers failing to be units, when theyÕre not even in the ring youÕre complaining about, is not very different from complaining that your inßatable sweetie isnÕt Family Reunions, that sort of thing]. In fact, such a tasteless analogy deserves to be stretched further; the resolution in each case is the same: find a context in which the offense is removed. In the algebraic case, go to a ring where the factorization exists; in the other, either avail yourself of a more presentable paramour, or find a circle of jerks who share the same interest [either your balloon babe, or each his own]. > ItÕs always funny to come back to a post where you said something was > stupid when you made a mistake. Ok, so I was stupid, but at least I > can admit my error... ThatÕs what we can always count on. Your abiding faithfulness to Truth. > James Harris Dale === Subject: Re: JSH: Legacy of error > In fact, such a tasteless analogy deserves to be stretched further... Ouch!! Buy a bigger size next time! === Subject: Re: JSH: Legacy of error > > Like think about this algebraic integer thing. Why wouldnÕt it occur > to people that maybe they might want to consider roots of non-monic > polynomials irreducible over Q that might correlate to integer > examples like > > 3x^2 + 4x + 1 = (3x + 1)(x + 1)? > > I mean, itÕs not like it takes brilliance to consider that maybe some > non-monics--even if they were irreducible over Q--might have roots > that have integral properties despite being irrational. > > Ah, but people HAVE considered algebraic integer roots to non-monic > polynomials. The result is that a non-monic polynomial with at least > one algebraic integer root is reducible with at least one of the > polynomial factors being monic. Like in your example, 3x^2 + 4x + 1 > is reducible and the (at least) one monic polynomial factor is (x + > 1). > > DonÕt just say reducible, say reducible over Q, or over rationals. > But how about? > 3x^2 + Kx + 1 = (3x + u_1)(x + u_2) > where u_1 and u_2 are properly units for some integer K? What exactly does properly units mean ? You could call them perfect units or pling-plong units and no-one would be any closer to understanding what you might mean. If you have a clear conception of what you are talking about a clear statement of your insights should be possible. > Now the wrong answer What is the question that is being asked ? Without knowing the question it is impossible to know whether a particular answer is right or wrong >is to say that Galois Theory shows that if u_1 > and u_2 are rational, then ok, they can both be units, like with K=4, > but if they are irrational, then NO MATTER what supposedly thereÕs > never another unit case? There are plenty of algebraic numbers that are integers and units i, 1+sqrt(2),... what has their rationality or irrationality to do with it ? What relevance has Galois theory ? > ThatÕs just stupid. When you think about it, why should there not be > another unit case just because u_1 and u_2 arenÕt algebraic integers? What does unit case mean precisely. There are plenty of numbers that give 1 when multiplied together. So what ? > ItÕs just DUMB, and defending it, especially after IÕve proven that > such cases must exist, is just an exercise in futility. What is dumb and what precisely is being defended or attacked ? Unless you say precisely what you were intending to prove instead of keeping it a personal secret, couched in undefined terminology, then it is impossible to determine whether your proof is valid or not. It would be fascinating if you could explain in detail, assuming that we know what the intgers, Z, are, what it means to call a number that is not in Z integral or integr-like or even an integer. How do you distingusih between numbers that are integers in some sense and those that are not ? === Subject: Re: JSH: Legacy of error > However, Gauss and others noticed a peculiar closeness between the prime > distribution and continuous functions like x/ln x, and later Li(x). ItÕs not all that peculiar. I donÕt know if it still there in the current edition, but there used to be in Courant and RobbinsWhat is Mathematics? a simple heuristic argument as to why pi(x) ~= x/ln x (or was it Li(x)?). It is hard to rigourously *prove* this, but to see why it is so is not hard. ... > The real point of my discovery of a partial difference equation to count > primes is that it gives THE ANSWER to the why of the prime distributions > closeness, and it probably means that the Riemann Hypothesis itself is > false. > Now supposedly the Riemann Hypothesis is a big deal, but itÕs all politics > in todayÕs math world, or there would be people all over my work, either > showing what I just said, or explaining why itÕs not true as thereÕs just > NOT THAT MUCH NEW in this area, and what I have is distinctive. People canÕt be all over your work, because you HAVENÕT DEMONSTRATED ANY WORK YET. All youÕve done is give a prime counting formula that works but no one (but you) sees as providing any new insights, and have stated that it probably means that RH is false. Drop the Anne Elk impersonation, and actually present real results, instead of just talking about them. -- --Tim Smith === Subject: Re: JSH: Legacy of error > However, Gauss and others noticed a peculiar closeness between the prime > distribution and continuous functions like x/ln x, and later Li(x). > ItÕs not all that peculiar. I donÕt know if it still there in the current > edition, but there used to be in Courant and RobbinsWhat is Mathematics? > a simple heuristic argument as to why pi(x) ~= x/ln x (or was it Li(x)?). > It is hard to rigourously *prove* this, but to see why it is so is not > hard. So? The important fact is that Gauss and others noticed a peculiar closeness between the prime distribution and continuous functions like x/ln x and Li(x). Riemann was actually looking for THE function that linked them, and thought heÕd found it, and called it R(x). He was wrong, but thatÕs a sidepoint. > ... > The real point of my discovery of a partial difference equation to count > primes is that it gives THE ANSWER to the why of the prime distributions > closeness, and it probably means that the Riemann Hypothesis itself is > false. > > Now supposedly the Riemann Hypothesis is a big deal, but itÕs all politics > in todayÕs math world, or there would be people all over my work, either > showing what I just said, or explaining why itÕs not true as thereÕs just > NOT THAT MUCH NEW in this area, and what I have is distinctive. > People canÕt be all over your work, because you HAVENÕT DEMONSTRATED ANY > WORK YET. All youÕve done is give a prime counting formula that works but > no one (but you) sees as providing any new insights, and have stated that it > probably means that RH is false. Hey, IÕm posting on Usenet. Usenet is not important. Math journals are important. The real deal goes on behind closed doors where IÕm learning a system that I thought IÕd escape by going directly to the people via the Internet. The Internet doesnÕt have that kind of power, yet. You donÕt matter. IÕm just here wasting time while I wait on the people who do. > Drop the Anne Elk impersonation, and actually present real results, instead > of just talking about them. I present the real results to the journals. I talk on Usenet for entertainment purposes. You are just here for my amusement. Nothing more, so you have no basis for any demands. You just arenÕt important enough to make demands, and donÕt bother shifting now to saying I should not post. Usenet is a public forum. And besides, what makes you think that demand would work any better than others. This is entertainment. Get with the program. James Harris === Subject: Re: JSH: Legacy of error > This is entertainment. === Subject: Re: JSH: Legacy of error >Basically the latter 1800Õs and early 1900Õs represent a period when a >substantial amount of errors, mathematical and logical entered into >the math field, and then, in what has been a covering action, the >concept of pure math was emphasized, which helped to hide the >errors. >There is more than just the problem with the ring of algebraic >integers that IÕve discussed so much. >Like, consider the Riemann Hypothesis. You donÕt know what RH even _says_. >I find it hard to face the >issue myself at times, but my find of a three dimensional formula that >counts primes gives a rather clear and readily understandable >explanation for the closeness of the prime distribution to continuous >functions. And although you keep saying this, youÕve never _given_ that explanation. Over and over you talk about connections between your work and RH, but you never say what the connection actually is. For new readers: He has a recurrence that counts primes. HeÕs written down the corresponding partial differential equation (itÕs not quite a pde, but never mind.) But heÕs never shown that the solution to the pde has anything whatever to do with counting primes. And he has many times simply ignored simple examples like the following: The solution to y= y, y(0) = 1 is y(t) = e^t. The solution to the analogous recurrence, a[n+1] - a[n] = a[n], a[0] = 1 is a[n] = 2^n. By the argument heÕs using above it follows that 2^n is a good approximation to e^n. >ItÕs just simple. But you have these complex ideas that got a lot of >attention, get a lot of attention, and I know that nothing will come >of them. >If you look at RiemannÕs actual work, youÕll find a section where he >talks about certain terms looking like theyÕll tend to zero, as they >look like they should balance out. Guffaw. YouÕve looked at RiemannÕs work, right? You can tell us where you found it, and what those terms are? >But has anyone ever proven that they will? No. RH is an open question. >Goldstone had what he thought was a proof shot down for using the same >type reasoning just recently. So why wouldnÕt anyone go back to >RiemannÕs work with the same analysis? >Yet, why is my work still just supposedly the rantings of some crank >when all of it is now rigorously proven? Guffaw. >You are in a field with a legacy of error. >Andrew Wiles didnÕt prove FermatÕs Last Theorem. Guffaw. >RiemannÕs Hypothesis >can be shown to probably be wrong on some very basic grounds, but no >one will admit it. And thatÕs not all of the error. >I have research I donÕt bother even talking about in public, as whatÕs >the point? >Actually itÕs just freaking depressing. The mistakes cluster around >1900, both before and after it, and itÕs just weird looking at them, >wondering. >Like think about this algebraic integer thing. Why wouldnÕt it occur >to people that maybe they might want to consider roots of non-monic >polynomials irreducible over Q that might correlate to integer >examples like >3x^2 + 4x + 1 = (3x + 1)(x + 1)? >I mean, itÕs not like it takes brilliance to consider that maybe some >non-monics--even if they were irreducible over Q--might have roots >that have integral properties despite being irrational. >It doesnÕt take brilliance to consider possibilities. You got that one right. btw, it also doesnÕt take brilliance to post crap on usenet. >And as for counting primes, why not have a p(x,y) function that gives >a count with p(x,sqrt(x))? Why couldnÕt the answer to linking the >prime distribution to continuous functions like x/ln x come from >multi-variable functions? >Why not ask why? >Have you seen my derivation of the prime counting function? >Such a simple derivation could have been done a thousand years ago. >Maybe it was and lost. Or maybe (hint) it _was_ done a few centuries ago and not lost. (Again, for new readers: HeÕs told us about an un-named top mathematician who told _him_ that 20% of the grad students working in the field come up with something similar to his recurrence on their own. Why he told us this nobody knows - how it can be that it has no effect on his conviction that heÕs made an earth-shamking discovery nobody knows.) >If I knew a few years ago what I know now, IÕd have never made these >discoveries--if I could have stopped myself. Ignorance is bliss. >There is no profit in it after all, and IÕm not talking about money. >I never really was. Except for the times youÕve been talking about money, you mean. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Legacy of error >>If you look at RiemannÕs actual work, youÕll find a section where he talks >>about certain terms looking like theyÕll tend to zero, as they look like >>they should balance out. > Guffaw. YouÕve looked at RiemannÕs work, right? You can tell us where you > found it, and what those terms are? Finding a copy of RiemannÕs work is pretty easy, actually. ThereÕs a Dover edition of EdwardÕs RiemannÕs Zeta Function available in many Barnes & Noble bookstores for about $15, and from Amazon for $10. HereÕs DoverÕs description of the book: Superb, high-level study of one of the most inßuential classics in mathematics examines landmark 1859 publication entitled .89´[Thorn]ÕOn the Number of Primes Less Than a Given MagnitudeÕ,.89´Ų and traces developments in theory inspired by it. Topics include RiemannÕs main formula, the prime number theorem, the Riemann-Siegel formula, large-scale computations, Fourier analysis, and other related topics. English translation of RiemannÕs original document appears in the Appendix Book News, Inc, says this: Edwards elaborates on Bernard RiemannÕs eight-page paper ŌOn the Number of Primes Less Than a Given MagnitudeÕ, published in German in 1859. His goal is not to supplant the classic work, but to provide mathematics students access to it. Indeed an English translation of the original is appended ItÕs funny what you can find in mainstream bookstores. There are very few people in the small town/rural county I live in who have anywhere near the background for that book, but there it was, in my local B&N. And when I bought it, they restocked it within a few days. My copy of Misner, Thorne, and WheelerÕs Gravitation was picked up at a Waldenbooks in a mall. :-) -- --Tim Smith === Subject: Re: JSH: Legacy of error >>If you look at RiemannÕs actual work, youÕll find a section where he talks >>about certain terms looking like theyÕll tend to zero, as they look like >>they should balance out. > > Guffaw. YouÕve looked at RiemannÕs work, right? You can tell us where you > found it, and what those terms are? > Finding a copy of RiemannÕs work is pretty easy, actually. ThereÕs a Dover > edition of EdwardÕs RiemannÕs Zeta Function available in many Barnes & > Noble bookstores for about $15, and from Amazon for $10. HereÕs DoverÕs > description of the book: > Superb, high-level study of one of the most inßuential classics in > mathematics examines landmark 1859 publication entitled .89??ÕOn the Number > of Primes Less Than a Given MagnitudeÕ,.89?Ų and traces developments in > theory inspired by it. Topics include RiemannÕs main formula, the prime > number theorem, the Riemann-Siegel formula, large-scale computations, > Fourier analysis, and other related topics. English translation of > RiemannÕs original document appears in the Appendix > Book News, Inc, says this: > Edwards elaborates on Bernard RiemannÕs eight-page paper ŌOn the Number > of Primes Less Than a Given MagnitudeÕ, published in German in 1859. His > goal is not to supplant the classic work, but to provide mathematics > students access to it. Indeed an English translation of the original is > appended > ItÕs funny what you can find in mainstream bookstores. There are very few > people in the small town/rural county I live in who have anywhere near the > background for that book, but there it was, in my local B&N. And when I > bought it, they restocked it within a few days. > My copy of Misner, Thorne, and WheelerÕs Gravitation was picked up at a > Waldenbooks in a mall. :-) Oh, yeah, I have that, and noticed itÕs a favorite of math folks who think they want to learn some physics. I discovered that a few years back when I went in to talk to a math professor in Washington state, and he had a copy. ItÕs like the one area of physics where a lot of math types seem to get interested, and then they get mixed up on the physics as they forget that in the real world, what actually happens rules, not what they THINK the math says. You see, the book is NOT REALLY pure math but about the real world, so math types get lost at the end. I, on the other hand, am a physics major. So I understand what the real world is. James Harris === Subject: Re: JSH: Legacy of error > I, on the other hand, am a physics major. So I understand what the > real world is. The fact that you may have lied, cheated and/or bribed your way into a diploma doesnÕt mean you really know anything about physics. You have demonstrated repeatedly that what goes on in your mind has, at best, a tenuous connection with the real world. -- Wayne Brown (HPCC #1104) | When your tailÕs in a crack, you improvise fwbrown@bellsouth.net | if youÕre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Legacy of error > Basically the latter 1800Õs and early 1900Õs represent a period when a > substantial amount of errors, mathematical and logical entered into > the math field, and then, in what has been a covering action, the > concept of pure math was emphasized, which helped to hide the > errors. Ah, now you are a math historian as well? This is interesting. Are you saying that guys like Hilbert et. al. realized that the entire enterprise was a fraud, and decided to hide the fact by confusing everyone with abstraction? ThatÕs an interesting thesis. You have evidence, of course. > If you look at RiemannÕs actual work, youÕll find a section where he > talks about certain terms looking like theyÕll tend to zero, as they > look like they should balance out. So that I can follow along, could you please provide a citation to the Riemann paper to which you refer? > I have research I donÕt bother even talking about in public, as whatÕs > the point? Well youÕve been amusing the heck out of me for several years. And many others, as you know. > Actually itÕs just freaking depressing. The mistakes cluster around > 1900, both before and after it, and itÕs just weird looking at them, > wondering. Well I looked up 1900 at http://www.andibradley.com/whatya/othertrivia_1801_1900.htm. It turns out that 1900 was the year that The Wonderful Wizard of Oz was first published. And coincidentally -- or maybe NOT a coincidence -- Max Planck discovered energy comes in packets called quanta, giving birth to quantum theory. Pay no attention to the man behind the curtain! > Like think about this algebraic integer thing. Why wouldnÕt it occur > to people that maybe they might want to consider roots of non-monic > polynomials irreducible over Q that might correlate to integer > examples like > 3x^2 + 4x + 1 = (3x + 1)(x + 1)? To be honest, every time you write an incoherent run-on sentence like that I find it very frustrating. I am really trying to follow your argument, but I donÕt have the patience of some others. I wish you would state your argument more clearly. Tell me WHAT exactly is wrong with the ring of algebraic integers? Does it somehow violate the ring axioms? IÕm being serious, I would really like to understand what you are trying to get at. > If I knew a few years ago what I know now, IÕd have never made these > discoveries--if I could have stopped myself. Ignorance is bliss. > There is no profit in it after all, and IÕm not talking about money. > I never really was. IÕm glad you actually understand that. You know that making money by posting mathematical discoveries to Usenet is a real longshot for making anyone any money. === Subject: Re: interesting problems i am not able to count yet by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAH43SF13615; >Hello everyone, first thing i want to say is that i am from Slovakia ( >somewhere in middle of Europe ) so please excuse my english. Well.. >here are the problems: >1. How many of (A,B) - (?ordered pairs? in english?) do we have if >A is subset of B & B is subset of {1,2,...n} Yes, ordered pairs in English. You can think of it like this: color the elements of A white, color the elements of B which are not in A red, and color all the other elements blue. How many ways can you color {1, 2, ..., n} in this way? >2.(this one is tuff) Lets say that q is prime number. How many >matrixes has determinant dividable by q? ( or how many matrixes above >GF(q) is singular? ) matrices not matrixes divisible not dividable Count instead the nonsingular n x n matrices by considering the columns as forming a basis of F^n; count the number of bases by using the fact that the (k+1)st column must be linearly independent of the first k columns (which span a k-dimensional space). Hence the number of 1st columns is q^n - 1, the number of 2nd columns is q^n - q, number of 3rd is q^n - q^2, etc. You should be able to solve the problem from here. Todd Trimble === Subject: SORRY, I corrected the equation Find a cubic fcn f(x) = ax^3 +bx^2+cx+d that has a local max value of 3 at -2 and a local min value of 0 at 1. === Subject: Re: SORRY, I corrected the equation > Find a cubic fcn f(x) = ax^3 +bx^2+cx+d that has a local max value of 3 > at -2 > and a local min value of 0 at 1. === Subject: Re: SORRY, I corrected the equation max/min at -2,+1 implies constant times (x+2)(x-1)=0=dy/dx Integrate, interpret this constant as 3a and last as d, to give form y= a[x^3 +(3/2)x^2 -6x] +d which leads to 2 eqns with 2 unknowns. 1,0] 0=(-7/2)a +d -2,3] 3= 10a +d, so a =2/9, d=3-20/9=7/9 and the required cubic is y= (2/9)x^3+(1/3)x^2 -(4/3)x +7/9. Check it and see. Hope this helps Ian Hutcheson === Subject: Re: SORRY, I corrected the equation hgl escribi.97: > Find a cubic fcn f(x) = ax^3 +bx^2+cx+d that has a local max value of > 3 at -2 > and a local min value of 0 at 1. Well, you know that f(-2) = 3 and f(1) = 0. Do you know the value of fÕ(x) at -2 an 1? You have a system of 4 equations in 4 indetermined, it is easy. To check your result, 9*f(x) has all coefficients integers. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: SORRY, I corrected the equation just to check ur answer a = 2/9 b = 1/3 c = -4/3 d = 7/9 > hgl escribi.97: >> Find a cubic fcn f(x) = ax^3 +bx^2+cx+d that has a local max value of >> 3 at -2 >> and a local min value of 0 at 1. > Well, you know that f(-2) = 3 and f(1) = 0. Do you know the value of fÕ(x) > at -2 an 1? > You have a system of 4 equations in 4 indetermined, it is easy. To check > your result, 9*f(x) has all coefficients integers. > -- > Ignacio Larrosa Ca.96estro > A Coru.96a (Espa.96a) > ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: A Vectors question I am doing some pre-calculus for a university algebra & calc course. Currently I am looking at vectors, I am hoping I could ask a few questions. (I am assuming u and v are common terms when dealing with vectors) I am trying to understand the theorem: u.v = |u| |v| cos@ I am confused about what |u| really means. I understand that when used with vectors it does not really mean Ōabsolute valueÕ, and that some textbooks will actually use the following symbols instead: u.v = ||u|| ||v|| cos@ So what does the symbol mean? Should u.v=|u||v|? Could anyone please assist, Cassandra Thompson === Subject: Re: A Vectors question alt.math.undergrad: >I am confused about what |u| really means. It means the length of the vector u. > u.v = |u| |v| cos@ The dot product (scalar product) of two vectors equals the length of the first times the length of the second times of the cosine of the angle between them. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youÕre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: A Vectors question > I am doing some pre-calculus for a university algebra & calc course. > Currently I am looking at vectors, I am hoping I could ask a few questions. > (I am assuming u and v are common terms when dealing with vectors) I am > trying to understand the theorem: > u.v = |u| |v| cos@ > I am confused about what |u| really means. I understand that when used > with vectors it does not really mean Ōabsolute valueÕ, and that some > textbooks will actually use the following symbols instead: > u.v = ||u|| ||v|| cos@ > So what does the symbol mean? > Should u.v=|u||v|? No. > Could anyone please assist, > Cassandra Thompson Usually, |u| usually refers to the length of the vector u, which is also often indicated by ||u||. Since there is no other reasonable meaning for |u| than ||u||, the symbolism should not be ambiguous. === Subject: Re: A Vectors question >>I am doing some pre-calculus for a university algebra & calc course. >>Currently I am looking at vectors, I am hoping I could ask a few questions. >>(I am assuming u and v are common terms when dealing with vectors) I am >>trying to understand the theorem: >> u.v = |u| |v| cos@ >>I am confused about what |u| really means. I understand that when used >>with vectors it does not really mean Ōabsolute valueÕ, and that some >>textbooks will actually use the following symbols instead: >> u.v = ||u|| ||v|| cos@ >>So what does the symbol mean? >>Should u.v=|u||v|? > No. >>Could anyone please assist, >>Cassandra Thompson > Usually, |u| usually refers to the length of the vector u, which is also > often indicated by ||u||. Since there is no other reasonable meaning for > |u| than ||u||, the symbolism should not be ambiguous. Someonekicked, Virgil, you please correct me if I still seem as though I donÕt understand: Q) Find the angle between the vetors u = 2i+3j and v = -7i+j. A) u.v = |u||v| cos@ u.v = (2i + 3j) . (-7i + j) = -14i + 3j = -11 (Actually I got this last line from the example, I donÕt understand how -14i + 3j becomes -11. I do understand that i and j = 1, but I didnÕt realise you could just multiple them out?!?) |u| = sqrt(2^2 + 3^2) = sqrt(4 + 9) = sqrt(13) |v| = sqrt(-7^2 + 1^2) = sqrt(49 + 1) = sqrt(50) |u||v| = sqrt(13) . 5 . sqrt(2) = 5 . sqrt(26) cos@ = u.v / |u||v| = -11 / sqrt(650) = -0.43145... @ = 1/cos(-0.43145..) = 2.01690 (I am still hazy on by radians, and when they should be written as a value of pi, ie pi/2, as opposed to 2.016...) === Subject: Re: A Vectors question alt.math.undergrad: >u.v = (2i + 3j) . (-7i + j) > = -14i + 3j No, the dot product is formed from the components, and does not use the unit vectors i and j. The dot product is also called the scalar product because it contains only numbers, no vectors. The dot product of your two vectors is defined (in rectangular coordinates) as 2*(-7) + 3*1 = -14+3 = 11. >I do understand that i and j = 1, No, i does not equal 1. Its _length_ is 1, but it is a vector not a scalar. It is the unit vector in the positive x direction (east). Similarly j does not equal 1; it is the vector in the positive y direction (north) whose length is 1. [snipped: lengths are sqrt(13) and sqrt(50)] >cos@ = u.v / |u||v| > = -11 / sqrt(650) > = -0.43145... Okay. >@ = 1/cos(-0.43145..) No, you need the angle whose cosine is -0.431456. True, thatÕs often written cos(superscript -1)(-.43146), but thatÕs inverse cosine, not 1 over cosine. It appears you actually did that operation, but please use correct notation. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youÕre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: A Vectors question I suggest you learn about cross posting into a few newsgroups at once, instead of one at a time. > Q) Find the angle between the vectors u = 2i+3j and v = -7i+j. > A) > u.v = |u||v| cos@ > u.v = (2i + 3j) . (-7i + j) > = -14i + 3j > = -11 (Actually I got this last line from the example, I donÕt > understand how -14i + 3j becomes -11. I do understand that i and j = 1, > but I didnÕt realise you could just multiple them out?!?) Go back to the definitions. When u = (u1,u2), v = (v1,v2) the dot product of u and v, u.v = u1 u2 + v1 v2 which is a number. However for u and v being the sum of two vectors, use distributivity. (2i + 3j).(-7i + j) = 2i.(-7i + j) + 3j(-7i + j) = 2i.(-7i) + 2i.j + 3j.(-7i) + 3j.j = (2*-7)i.i + 2(i.i) + (3*-7)j.i + 3(j.j) = -14 + 3 = -11 because i.i = 1 = j.j and i.j = 0 = j.i > |u| = sqrt(2^2 + 3^2) = sqr(u.u) as noted other newsgroup > = sqrt(4 + 9) > = sqrt(13) > |v| = sqrt(-7^2 + 1^2) No. sqr((-7)^2 + 1^2) > = sqrt(49 + 1) > = sqrt(50) > |u||v| = sqrt(13) . 5 . sqrt(2) > = 5 . sqrt(26) As you are already using . for vector dot product. it is better to use * for multiplying numbers. Also . for multiplication makes trouble 10 = 2*5 = 2.5 = 5/2. > cos@ = u.v / |u||v| > = -11 / sqrt(650) > = -0.43145... > @ = 1/cos(-0.43145..) > = 2.01690 (I am still hazy on by radians, and when they should be > written as a value of pi, ie pi/2, as opposed to 2.016...) No no. @ /= 1/cos(-0.43) @ = arccos(-0.43) The answer needs be presented 2.017 radians or degrees depending if the arccos function you used was set to return radians or degrees. Radians are like degrees, like real life solutions to polynomials. Instead of easy practice problem with nice cute answers like pi/2 radians, pi/4 radians, 90 deg, 45 deg, 4, sqr 15, etc. You have to get real like engineer calculating satellite orbit. 2.06 radians, 118.03 degrees, 4.01323, 3.8729, etc. === Subject: Re: A Vectors question > I suggest you learn about cross posting into a few newsgroups at once, > instead of one at a time. Yep, I get the cross-posting thing, just only discovered this newsgroup *after* I had posted to the other group... >>Q) Find the angle between the vectors u = 2i+3j and v = -7i+j. >>A) >>u.v = |u||v| cos@ >>u.v = (2i + 3j) . (-7i + j) >> = -14i + 3j >> = -11 (Actually I got this last line from the example, I donÕt >>understand how -14i + 3j becomes -11. I do understand that i and j = 1, >>but I didnÕt realise you could just multiple them out?!?) > Go back to the definitions. When u = (u1,u2), v = (v1,v2) > the dot product of u and v, > u.v = u1 u2 + v1 v2 > which is a number. I see, I had read it as meaning u.v = (u1+v1, u2+v2), ie that it multiplied to become another vector. You are saying it is not a vector, just a number. Also, did you make a mistake in your little example or am I incorrect. Should you have written: u.v = u1.v1 + u2.v2 ? > However for u and v being the sum of two vectors, use distributivity. > (2i + 3j).(-7i + j) > = 2i.(-7i + j) + 3j(-7i + j) > = 2i.(-7i) + 2i.j + 3j.(-7i) + 3j.j > = (2*-7)i.i + 2(i.i) + (3*-7)j.i + 3(j.j) > = -14 + 3 = -11 > because i.i = 1 = j.j and i.j = 0 = j.i >>|u| = sqrt(2^2 + 3^2) > = sqr(u.u) as noted other newsgroup I donÕt understand that. If |u| means the magnitude (length) of the vector u, then this is calculated using the pythagorus formula. On closer inspection I have it written in my textbook. ie. |v| = sqrt(v1^2 + v2^2) You are disagreeing with this formula, or are you saying my workings do not reßect this forumula. >> = sqrt(4 + 9) >> = sqrt(13) >>|v| = sqrt(-7^2 + 1^2) > No. sqr((-7)^2 + 1^2) Yes, well from my workings below you can see that that is what I was implying. Sorry if you found that confusing... >> = sqrt(49 + 1) >> = sqrt(50) >>|u||v| = sqrt(13) . 5 . sqrt(2) >> = 5 . sqrt(26) > As you are already using . for vector dot product. > it is better to use * for multiplying numbers. > Also . for multiplication makes trouble 10 = 2*5 = 2.5 = 5/2. >>cos@ = u.v / |u||v| >> = -11 / sqrt(650) >> = -0.43145... >>@ = 1/cos(-0.43145..) >> = 2.01690 (I am still hazy on by radians, and when they should be >>written as a value of pi, ie pi/2, as opposed to 2.016...) > No no. @ /= 1/cos(-0.43) Are you saying Ōno nobecause you are annoyed that I havenÕt learnt the acceptable way to write mathematical formulaes in ASCII yet? @ = cos^-1(-0.43) Does that make you happier? Or do you prefer I write it some other way. Does it all mean the same thing??? > @ = arccos(-0.43) > The answer needs be presented 2.017 radians or degrees depending > if the arccos function you used was set to return radians or degrees. > Radians are like degrees, like real life solutions to polynomials. > Instead of easy practice problem with nice cute answers like > pi/2 radians, pi/4 radians, 90 deg, 45 deg, 4, sqr 15, etc. > You have to get real like engineer calculating satellite orbit. > 2.06 radians, 118.03 degrees, 4.01323, 3.8729, etc. I see what you are saying, when doing assignments, or taking exams any solution that does not produce a Ōcuteanswer is a good indication that it is wrong (sad but true...they like to keep things simple). So 1.5708 radians is another way of writing pi/2. Is there a symbol/abbreviation or should I write ŌradiansÕ after the result? ie Which is correct: 1.5708 or 1.5708 radians I would imagine that I should always write ŌradiansÕ, however the answers in my textbook never seem to do this. === Subject: Re: A Vectors question Q) Find the angle between the vectors u = 2i+3j and v = -7i+j. >>A) >>u.v = |u||v| cos@ >>u.v = (2i + 3j) . (-7i + j) >> = -14i + 3j >> = -11 (Actually I got this last line from the example, I donÕt >>understand how -14i + 3j becomes -11. I do understand that i and j = 1, >>but I didnÕt realise you could just multiple them out?!?) > Go back to the definitions. When u = (u1,u2), v = (v1,v2) > the dot product of u and v, > u.v = u1 u2 + v1 v2 > which is a number. > I see, I had read it as meaning > u.v = (u1+v1, u2+v2), ie that it multiplied to become another vector. > You are saying it is not a vector, just a number. Yup. > Also, did you make a mistake in your little example or am I incorrect. > Should you have written: > u.v = u1.v1 + u2.v2 ? Yes. > However for u and v being the sum of two vectors, use distributivity. > (2i + 3j).(-7i + j) > = 2i.(-7i + j) + 3j(-7i + j) > = 2i.(-7i) + 2i.j + 3j.(-7i) + 3j.j > = (2*-7)i.i + 2(i.i) + (3*-7)j.i + 3(j.j) > = -14 + 3 = -11 > because i.i = 1 = j.j and i.j = 0 = j.i >>|u| = sqrt(2^2 + 3^2) > = sqr(u.u) as noted other newsgroup > I donÕt understand that. If |u| means the magnitude (length) of the > vector u, then this is calculated using the pythagorus formula. On > closer inspection I have it written in my textbook. > ie. > |v| = sqrt(v1^2 + v2^2) Indeed, indeed. > You are disagreeing with this formula, or are you saying my workings do > not reßect this forumula. Neither, just commenting how thatÕs an example of |u| = sqr(u.u) >> = sqrt(4 + 9) >> = sqrt(13) >>|v| = sqrt(-7^2 + 1^2) > No. sqr((-7)^2 + 1^2) > Yes, well from my workings below you can see that that is what I was > implying. Sorry if you found that confusing... You made two mutually correcting errors. >> = sqrt(49 + 1) >> = sqrt(50) >>|u||v| = sqrt(13) . 5 . sqrt(2) >> = 5 . sqrt(26) >>cos@ = u.v / |u||v| >> = -11 / sqrt(650) >> = -0.43145... >>@ = 1/cos(-0.43145..) >> = 2.01690 (I am still hazy on by radians, and when they should be >>written as a value of pi, ie pi/2, as opposed to 2.016...) > No no. @ /= 1/cos(-0.43) > Are you saying Ōno nobecause you are annoyed that I havenÕt learnt the > acceptable way to write mathematical formulaes in ASCII yet? No, because itÕs wrong. > @ = cos^-1(-0.43) > Does that make you happier? Or do you prefer I write it some other way. > Does it all mean the same thing??? No, it makes it correct. Yes cos^2 x = (cos x)^2 and cos^-1 x = arccos x instead of 1/cos x is just one of those historical hangovers of mathematic notations. There are many. > @ = arccos(-0.43) > The answer needs be presented 2.017 radians or degrees depending > if the arccos function you used was set to return radians or degrees. > Radians are like degrees, like real life solutions to polynomials. > Instead of easy practice problem with nice cute answers like > pi/2 radians, pi/4 radians, 90 deg, 45 deg, 4, sqr 15, etc. > You have to get real like engineer calculating satellite orbit. > 2.06 radians, 118.03 degrees, 4.01323, 3.8729, etc. > I see what you are saying, when doing assignments, or taking exams any > solution that does not produce a Ōcuteanswer is a good indication that > it is wrong (sad but true...they like to keep things simple). > So 1.5708 radians is another way of writing pi/2. ItÕs another way of writing pi/2 radians. Yes 1.5708 radians is the for sure way of keeping things clear even with the convention angles are measure in radians where pi/2 is implied to mean pi/2 radians. After all would ever one see pi/2 degrees for 1.5708 deg? > Is there a symbol/abbreviation or should I write ŌradiansÕ after the result? > ie DonÕt know and as for ascii, thereÕs degrees or deg. There could be rad. but thatÕs a unit of radioactivity. > Which is correct: > 1.5708 pi/2 with understanding all angles are measured in radians 1.5708 is ok under that convention, however > or > 1.5708 radians is best for it wonÕt loose track of the convention. For example you work a problem out in radians and give answer to carpenture to build complicated box. Of course he thinks in degrees. Well for him, you better convert it to degrees as he might miss the radians. However for an engineer, you best make your results explicit, degrees or radians, no matter what convention you use or even what the shop may use. For example, NASA uses metric and those engineers think metric. Lockeed Marting apparently uses English. They build Mars satellight and gave it to NASA with lots of nice numbers for this and that. Come landing time, whoop satellight missed Mars and a few days later NASA had to admit, they missed a 3,000 mile target because somebodies didnÕt every time put ft, lb and cm, kg etc on every number. A $100 million dollar throwaway, that nobody even got to see where it went, for lack of a ft-lb/sec booster. > I would imagine that I should always write ŌradiansÕ, however the > answers in my textbook never seem to do this. all itÕs units. ItÕs is exacting method, but it pays off in correct answers. Yes, I suggest for all but pi/4 etc, that you explicate. Then no matter where you go or to whom you talk, youÕll be appreciated for your exactness and clarity. I would convert 5 ft = 5 ft * 12in/ft = 60 in = 60 in * 2.54 cm/in = 152.4 cm Then when I went over my work, it was all clearly laid out to check step by step until I found where it was I was getting absurd answers like, cut taxes + spend money = balanced budget, but excuse me, now IÕve digressed. ;-) === Subject: Re: A Vectors question >>|u||v| = sqrt(13) . 5 . sqrt(2) >> = 5 . sqrt(26) >>cos@ = u.v / |u||v| >> = -11 / sqrt(650) >> = -0.43145... >>@ = 1/cos(-0.43145..) >> = 2.01690 (I am still hazy on by radians, and when they should be >>written as a value of pi, ie pi/2, as opposed to 2.016...) >No no. @ /= 1/cos(-0.43) >>Are you saying Ōno nobecause you are annoyed that I havenÕt learnt the >> acceptable way to write mathematical formulaes in ASCII yet? > No, because itÕs wrong. >>@ = cos^-1(-0.43) >>Does that make you happier? Or do you prefer I write it some other way. >>Does it all mean the same thing??? > No, it makes it correct. Yes cos^2 x = (cos x)^2 > and cos^-1 x = arccos x instead of 1/cos x is just one of > those historical hangovers of mathematic notations. There are many. I am still really confused about why the notation/meaning is wrong. I know that I can just start using cos^-1 instead of 1/cos, however I would really like to know why it is important to do so. cos^-1 x = 1/cos x However my understanding is that you are saying the above is incorrect. I donÕt understand why. 1/cos x /= 1/(cos x) (/= does not equal, am not sure what conventions you would use, and years of programming has left me with lots of variations. ie != <> /= etc) I know this is a basic math question, my apologies for that. === Subject: Re: A Vectors question and cos^-1 x = arccos x instead of 1/cos x is just one of > those historical hangovers of mathematic notations. There are many. > However my understanding is that you are saying the above is incorrect. > I donÕt understand why. > 1/cos x /= 1/(cos x) I see you have already been given a worthy answer by another. > (/= does not equal, am not sure what conventions you would use, and > years of programming has left me with lots of variations. ie != <> /= etc) Indeed and all common ascii. /= is best rendition of = with line thru it. <> is same an /= only when you are considering linear or total orders. If you have a partially ordered structured, like a tree, then a <> b means a is above or below b. Two leaves on the same stem have not a <> b and a /= b > I know this is a basic math question, my apologies for that. No need to appology for asking questions about ascii notation conventions or basic math. The people weÕd like to have appologise are those bone head egos that canÕt ever accept that their fantastic ideas have been rejected as nonsense by the hard cold reality of mathematical review. Your welcome, happy to hear that my iota of insight or whatever has contributed to your learning. It is the apt student who can ask a coherent question pin pointing what needs be explained. IÕve see some school kids in these forums who canÕt even ask a coherent question. === Subject: Re: A Vectors question > I am still really confused about why the notation/meaning is wrong. I > know that I can just start using cos^-1 instead of 1/cos, however I > would really like to know why it is important to do so. > cos^-1 x = 1/cos x > However my understanding is that you are saying the above is incorrect. > I donÕt understand why. > 1/cos x /= 1/(cos x) > (/= does not equal, am not sure what conventions you would use, and > years of programming has left me with lots of variations. ie != <> /= etc) > I know this is a basic math question, my apologies for that. The standard notation for one divide by the cosine of x is 1/cos x. The standard notation for inverse cosine of x is cos^-1 x. You could, of course, say that when you write 1/cos that means the inverse cosine and that when you want to write 1 divided by cos x you will write 1/(cos x) but thatÕs simply not what everyone else is used to, so even if you know that, in your notation, 1/cos x != 1/(cos x), everyone else will interpret 1/cos x as 1/(cos x), not as cos^-1 x. meeroh -- If this message helped you, consider buying an item from my wish list: === Subject: Re: A Vectors question >>I am still really confused about why the notation/meaning is wrong. I >>know that I can just start using cos^-1 instead of 1/cos, however I >>would really like to know why it is important to do so. >>cos^-1 x = 1/cos x >>However my understanding is that you are saying the above is incorrect. >>I donÕt understand why. >>1/cos x /= 1/(cos x) >>(/= does not equal, am not sure what conventions you would use, and >>years of programming has left me with lots of variations. ie != <> /= etc) >>I know this is a basic math question, my apologies for that. > The standard notation for one divide by the cosine of x is 1/cos x. The > standard notation for inverse cosine of x is cos^-1 x. You could, of course, > say that when you write 1/cos that means the inverse cosine and that when you > want to write 1 divided by cos x you will write 1/(cos x) but thatÕs simply > not what everyone else is used to, so even if you know that, in your notation, > 1/cos x != 1/(cos x), everyone else will interpret 1/cos x as 1/(cos x), not as > cos^-1 x. > meeroh Meeroh, I see now, you are correct, I was writing 1/cos x and seeing it as cos^-1 x, everyone else was seeing (per standard ASCII notation) (cos x)^1, and I was couldnÕt understand why. Much appreciated. === Subject: Re: A Vectors question > snip > I donÕt understand that. If |u| means the magnitude (length) of the > vector u, then this is calculated using the pythagorus formula. On > closer inspection I have it written in my textbook. > ie. > |v| = sqrt(v1^2 + v2^2) > snip v1^2+v2^2 is v.v so |v| = sqrt(v1^2 + v2^2) = sqrt (v.v) === Subject: Re: A Vectors question >>snip >>I donÕt understand that. If |u| means the magnitude (length) of the >>vector u, then this is calculated using the pythagorus formula. On >>closer inspection I have it written in my textbook. >>ie. >>|v| = sqrt(v1^2 + v2^2) >>snip > v1^2+v2^2 is v.v > so |v| = sqrt(v1^2 + v2^2) = sqrt (v.v) Duh, of course...sorry forest and trees... === Subject: Re: A Vectors question > I am doing some pre-calculus for a university algebra & calc course. > Currently I am looking at vectors, I am hoping I could ask a few > questions. > (I am assuming u and v are common terms when dealing with vectors) I > am trying to understand the theorem: > u.v = |u| |v| cos@ > I am confused about what |u| really means. I understand that when used > with vectors it does not really mean Ōabsolute valueÕ, and that some > textbooks will actually use the following symbols instead: > u.v = ||u|| ||v|| cos@ > So what does the symbol mean? > Should u.v=|u||v|? > Could anyone please assist, > Cassandra Thompson It means the norm of a vector (or matrix). A norm has to satisfy a few properties. one of which is that it is always nonnegative, and only zero when what is inside the bars is zero. For what you are doing, and unless otherwise specified, is the 2-norm or the Euclidean norm (though not the Euclidean norm also called the Frobenius norm). This uses the distance formula from geometry. That is, if x = then |x| = sqrt(x1^2 + x2^2 + x3^2 + ... + xn^2) The use of the bars is probably because absolute value is a norm for one dimensions. Then there are a few more conventions if you are dealing with complex numbers, but you can ask about them if you need to. - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: A Vectors question |u| or ||u|| means the length of a vector, ( remember vector definition based on length and direction ) if u.v = |u| |v|, then one or more of the following are true: 1) u = 0 ( here 0 vector ) 2) v = 0 ( here 0 vector) 3) the two vectors are of the same direction, so cos@ = 1; >I am doing some pre-calculus for a university algebra & calc course. >Currently I am looking at vectors, I am hoping I could ask a few questions. > (I am assuming u and v are common terms when dealing with vectors) I am > trying to understand the theorem: > u.v = |u| |v| cos@ > I am confused about what |u| really means. I understand that when used > with vectors it does not really mean Ōabsolute valueÕ, and that some > textbooks will actually use the following symbols instead: > u.v = ||u|| ||v|| cos@ > So what does the symbol mean? > Should u.v=|u||v|? > Could anyone please assist, > Cassandra Thompson === Subject: Info for Field Theory.... i would like to know if there are any books on-line about Field theory (pdf,ps,or something like that!) thank you in advance....! === Subject: Re: Info for Field Theory.... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAIMKu130585; You might try the following: http://www.scienceoxygen.com/math/284.html >i would like to know if there are any books on-line >about Field theory (pdf,ps,or something like that!) >thank you in advance....! === Subject: Can anyone help me prove this closed and bounded sequence problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAID3ss11592; The question is Let the sequence {An} be nonempty, closed and of bounded intervals such that A(n+1) is contained in An for all ng greater than or equal to 1. Note for An here n is the subscript. Prove that the intersection of An from n=1 to infinity is nonempty. === Subject: Re: Can anyone help me prove this closed and bounded sequence problem > The question is Let the sequence {An} be nonempty, closed and of > bounded intervals such that A(n+1) is contained in An for all ng > greater than or equal to 1. Note for An here n is the subscript. > Prove that the intersection of An from n=1 to infinity is nonempty. Write An = [an,bn]. Then a1 <= a2 <= ... and b1 >= b2 >= ... The endpoint sequences are therefore bounded and monotone, hence converge to limits a and b resp., and you can show a <= b. It seems likely that [a,b] is the intersection of all An, doesnÕt it? === Subject: Re: Can anyone help me prove this closed and bounded sequence problem Just an interesting note this made me think of, although not entirely neccesary for this problem: In fact, you can make a much stronger argument here, that all of these intervals have only ONE point in common, and that point is also the limit of the an sequence, and of the bn sequence. Proof: If every bn is greater than an, then an is bounded from above, so it must have a least upper bound. If every an is smaller than bn, then bn is bounded from below, which means bn has a greatest lower bound. Now, because an is monotone increasing, its l.u.b is also its limit. And, because bn is monotone decreasing, its g.l.b is also its limit. Now, the distance between an and bn keeps getting smaller, So: lim(an - bn) = 0 - which means lim(an) - lim(bn) = 0, which means lim(an) = lim(bn) - and that number is also the only one which is common to all the intervals... > The question is Let the sequence {An} be nonempty, closed and of > bounded intervals such that A(n+1) is contained in An for all ng > greater than or equal to 1. Note for An here n is the subscript. > Prove that the intersection of An from n=1 to infinity is nonempty. > Write An = [an,bn]. Then a1 <= a2 <= ... and b1 >= b2 >= ... The endpoint > sequences are therefore bounded and monotone, hence converge to limits a > and b resp., and you can show a <= b. It seems likely that [a,b] is the > intersection of all An, doesnÕt it? === Subject: Re: Can anyone help me prove this closed and bounded sequence problem alt.math.undergrad: > Just an interesting note this made me think of, although not entirely > neccesary for this problem: > In fact, you can make a much stronger argument here, that all of these > intervals have only ONE point in common, and that point is also the limit of > the an sequence, and of the bn sequence. Not without a hypothesis to the effect that the lengths of the intervals converge to 0. > Proof: > If every bn is greater than an, then an is bounded from above, so it must > have a least upper bound. > If every an is smaller than bn, then bn is bounded from below, which means > bn has a greatest lower bound. > Now, because an is monotone increasing, its l.u.b is also its limit. > And, because bn is monotone decreasing, its g.l.b is also its limit. > Now, the distance between an and bn keeps getting smaller, So: lim(an - bn) > = 0 DoesnÕt follow. Let each a_n = 0, and let b_n = 1 + 1/2^n. The intersection is [0, 1]. [...] Brian === Subject: Re: Can anyone help me prove this closed and bounded sequence problem OK, yourÕe right - you have to assume that an and bn are not constant, meaning, theyÕre really increasing and decreasing, and then my assertion is true. BTW, I didnÕt think of that myself, itÕs CantorÕs lemma. > alt.math.undergrad: > Just an interesting note this made me think of, although not entirely > neccesary for this problem: > In fact, you can make a much stronger argument here, that all of these > intervals have only ONE point in common, and that point is also the limit of > the an sequence, and of the bn sequence. > Not without a hypothesis to the effect that the lengths of > the intervals converge to 0. > Proof: > If every bn is greater than an, then an is bounded from above, so it must > have a least upper bound. > If every an is smaller than bn, then bn is bounded from below, which means > bn has a greatest lower bound. > Now, because an is monotone increasing, its l.u.b is also its limit. > And, because bn is monotone decreasing, its g.l.b is also its limit. > Now, the distance between an and bn keeps getting smaller, So: lim(an - bn) > = 0 > DoesnÕt follow. Let each a_n = 0, and let b_n = 1 + 1/2^n. > The intersection is [0, 1]. > [...] > Brian === Subject: Re: Can anyone help me prove this closed and bounded sequence problem > OK, yourÕe right - you have to assume that an and bn are not constant, > meaning, theyÕre really increasing and decreasing, and then my assertion is > true. No, it isnÕt. Try a_n = -1/2^n, b_n = 1 + 1/2^n. [...] Brian === Subject: Re: Can anyone help me prove this closed and bounded sequence problem OK, Cantor DID have another condition, saying lim(a_n - b_n) = 0 - without > OK, yourÕe right - you have to assume that an and bn are not constant, > meaning, theyÕre really increasing and decreasing, and then my assertion is > true. > No, it isnÕt. Try a_n = -1/2^n, b_n = 1 + 1/2^n. > [...] > Brian === Subject: Re: Can anyone help me prove this closed and bounded sequence problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAJDJx304959; >OK, yourÕe right - you have to assume that an and bn are not constant, >meaning, theyÕre really increasing and decreasing, and then my assertion is >true. No, itÕs still wrong: consider a_n = 1/2 - 1/2^n and b_n = 1 + 1/2^n. >BTW, I didnÕt think of that myself, itÕs CantorÕs lemma. ?? >> <> alt.math.undergrad: >> Just an interesting note this made me think of, although not entirely >> neccesary for this problem: >> In fact, you can make a much stronger argument here, that all of these >> intervals have only ONE point in common, and that point is also the >limit of >> the an sequence, and of the bn sequence. >> Not without a hypothesis to the effect that the lengths of >> the intervals converge to 0. >> Proof: >> If every bn is greater than an, then an is bounded from above, so it >must >> have a least upper bound. >> If every an is smaller than bn, then bn is bounded from below, which >means >> bn has a greatest lower bound. >> Now, because an is monotone increasing, its l.u.b is also its limit. >> And, because bn is monotone decreasing, its g.l.b is also its limit. >> Now, the distance between an and bn keeps getting smaller, So: lim(an - >bn) >> = 0 >> DoesnÕt follow. Let each a_n = 0, and let b_n = 1 + 1/2^n. >> The intersection is [0, 1]. >> [...] >> Brian === Subject: Re: Can anyone help me prove this closed and bounded sequence problem > OK, yourÕe right - you have to assume that an and bn are not constant, > meaning, theyÕre really increasing and decreasing, and then my assertion is > true. Not quite - you have to assume (b_n - a_n) ---> 0 E.g. think of [a_n, b_n] = [-1/n, 1 + 1/n] ... > BTW, I didnÕt think of that myself, itÕs CantorÕs lemma. I expect Cantor included a condition (b_n - a_n) ---> 0 ? > alt.math.undergrad: > Just an interesting note this made me think of, although not entirely > neccesary for this problem: > In fact, you can make a much stronger argument here, that all of these > intervals have only ONE point in common, and that point is also the > limit of > the an sequence, and of the bn sequence. > Not without a hypothesis to the effect that the lengths of > the intervals converge to 0. > Proof: > If every bn is greater than an, then an is bounded from above, so it > must > have a least upper bound. > If every an is smaller than bn, then bn is bounded from below, which > means > bn has a greatest lower bound. > Now, because an is monotone increasing, its l.u.b is also its limit. > And, because bn is monotone decreasing, its g.l.b is also its limit. > Now, the distance between an and bn keeps getting smaller, So: lim(an - > bn) > = 0 > DoesnÕt follow. Let each a_n = 0, and let b_n = 1 + 1/2^n. > The intersection is [0, 1]. > [...] > Brian === Subject: Re: Can anyone help me prove this closed and bounded sequence problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAIF7GY22737; >The question is Let the sequence {An} be nonempty, closed and of >bounded intervals such that A(n+1) is contained in An for all ng >greater than or equal to 1. Note for An here n is the subscript. >Prove that the intersection of An from n=1 to infinity is nonempty. Have you studied compact sets? If you have, think of all the sets An as inside the compact set A1, and consider their complements as open sets. If the intersection in your question were empty, what would this mean for its complement, and how could you use that together with compactness of A1 to reach a contradiction? ItÕs a little tricky, but I hope youÕll see it. Todd Trimble === Subject: Now see the b.s. zone Some of the sci.math posters who like to track me--throwing up smoke--have finally tracked over to this newsgroup. ThatÕs what they do. They will make a lot of posts, make a lot of noise, and basically just disagree. ItÕs actually a waste of time as what happens here on Usenet isnÕt really all that big of a deal, while part of the reason I post here is that itÕs a gateway to the Internet. I post here and watch the posts move about on Yahoo! and Google. ItÕs mildly entertaining. The real deal is the paper at the math journal. So why do those other people bother? Well, partly because theyÕre losers. Well, I think mostly because theyÕre losers, and replying to my posts is the closest theyÕll ever get to any kind of real fame. Pathetic, eh? But I invented extreme mathematics--posting your ideas as you get them and daring the rest of the world to catch you--and with that invention, comes attention parasites: hangerÕs on, trying to be, what they will never be. James Harris === Subject: Re: Now see the b.s. zone > Some of the sci.math posters who like to track me--throwing up > smoke--have finally tracked over to this newsgroup. > ThatÕs what they do. > They will make a lot of posts, ... YouÕre the one who makes lots of posts. The number of posts about you and your concerns that are _not_ replies to your posts is tiny. What do you do when someone makes a negative comment on your work? Do you reply to it and refute them or agree with them? No. You start another thread. ItÕs pots and kettles again. === Subject: Re: Now see the b.s. zone > Some of the sci.math posters who like to track me--throwing up > smoke--have finally tracked over to this newsgroup. Throwing up smoke would seem to be an activity more associated with a trackEE than a trackER, but youÕve never been much good with metaphor. > ThatÕs what they do. > They will make a lot of posts, make a lot of noise, and basically just > disagree. > ItÕs actually a waste of time as what happens here on Usenet isnÕt > really all that big of a deal So why do you get so worked up about it? > , while part of the reason I post here is > that itÕs a gateway to the Internet. > I post here and watch the posts move about on Yahoo! and Google. Uh, yeah... > ItÕs mildly entertaining. > The real deal is the paper at the math journal. Ah yes. Remind us, which journal? And how long have they had this paper? > So why do those other people bother? > Well, partly because theyÕre losers. Well, I think mostly because > theyÕre losers, and replying to my posts is the closest theyÕll ever > get to any kind of real fame. > Pathetic, eh? But I invented extreme mathematics--posting your ideas > as you get them and daring the rest of the world to catch you--and > with that invention, comes attention parasites: hangerÕs on, trying to > be, what they will never be. Your English is almost as bad as your maths. And you certainly didnÕt invent Ōposting to Usenet whatever comes into your headÕ. People have been doing that for almost as long as thereÕs been a Usenet. -- Larry Lard Replies to group please === Subject: Re: Now see the b.s. zone >Some of the sci.math posters who like to track me--throwing up >smoke--have finally tracked over to this newsgroup. >ThatÕs what they do. >They will make a lot of posts, make a lot of noise, and basically just >disagree. >ItÕs actually a waste of time as what happens here on Usenet isnÕt >really all that big of a deal, while part of the reason I post here is >that itÕs a gateway to the Internet. >I post here and watch the posts move about on Yahoo! and Google. >ItÕs mildly entertaining. >The real deal is the paper at the math journal. >So why do those other people bother? >Well, partly because theyÕre losers. Hmm, thereÕs a little inconsistency here. All that matters is math journals, IÕve published 30 or so papers in math journals, the journals continually reject your submissions, but IÕm the loser and youÕre not? DoesnÕt add up, quite - I think you dropped a minus sign somewhere or something. >Well, I think mostly because >theyÕre losers, and replying to my posts is the closest theyÕll ever >get to any kind of real fame. >Pathetic, eh? But I invented extreme mathematics--posting your ideas >as you get them and daring the rest of the world to catch you Sorry, but the idea of making hundreds of posts to usenet full of total crap is not something that originated with you either - thatÕs not quite as old as your prime counting function, but it was around long before you were. >--and >with that invention, comes attention parasites: hangerÕs on, trying to >be, what they will never be. >James Harris ************************ David C. Ullrich === Subject: Planar Graphs & Euler Cycles I will quote the problem from my book: Prove that if a planar graph has an Euler cycle, it has an Euler cycle with no crossings. A path P in a planar graph has a crossing if a vertex v appears at least twice in P and P crosses itself at v; that is P = (...,a,v,b,...c,v,d,...), where the vertices are arranged so that a,v,b crosses c,v,d at v in the following figure. c | | | | a----------v----------b | | | | d This problem is bogus. Consider the following planar graph c | | | a---v---b | | | d There is absolutely no Euler cycle that doesnÕt contain a crossing at v. None. What gives!? Bernd === Subject: Re: Planar Graphs & Euler Cycles > I will quote the problem from my book: > Prove that if a planar graph has an Euler cycle, it has an > Euler cycle with no crossings. A path P in a planar graph has > a crossing if a vertex v appears at least twice in P and P > crosses itself at v; that is > P = (...,a,v,b,...c,v,d,...), > where the vertices are arranged so that a,v,b crosses c,v,d > at v in the following figure. > c > | > | > a----------v----------b > | > | > d > This problem is bogus. Consider the following planar graph > c > | > | > | > a---v---b > | > | > | > d The Eulerain cycle a v b c v d a has two crossings > There is absolutely no Euler cycle that doesnÕt contain a > crossing at v. None. What gives!? The Eulerian cycle a d v b c v a has no crossings. === Subject: Re: Planar Graphs & Euler Cycles > I will quote the problem from my book: > Prove that if a planar graph has an Euler cycle, it has an > Euler cycle with no crossings. A path P in a planar graph has > a crossing if a vertex v appears at least twice in P and P > crosses itself at v; that is > P = (...,a,v,b,...c,v,d,...), > where the vertices are arranged so that a,v,b crosses c,v,d > at v in the following figure. > c > | > | > a----------v----------b > | > | > d > This problem is bogus. Consider the following planar graph > c > | > | > | > a---v---b > | > | > | > d > The Eulerain cycle > a v b c v d a > has two crossings > There is absolutely no Euler cycle that doesnÕt contain a > crossing at v. None. What gives!? > The Eulerian cycle > a d v b c v a > has no crossings. I see now. OK. Suppose C is a Euler cycle and contains a crossing, i.e. C = (w...,a,v,b,...c,v,d,...w). Consider the following cycle C= (w...,a,v,c,...b,v,d,...w). This cycle has no crossings. Is that it!? === Subject: Re: Planar Graphs & Euler Cycles >> Prove that if a planar graph has an Euler cycle, it has an >> Euler cycle with no crossings. A path P in a planar graph has >> a crossing if a vertex v appears at least twice in P and P >> crosses itself at v; that is >> P = (...,a,v,b,...c,v,d,...), >> This problem is bogus. Consider the following planar graph >> c >> | >> | >> | >> a---v---b >> | >> | >> | >> d >> The Eulerain cycle >> a v b c v d a >> has two crossings >> The Eulerian cycle >> a d v b c v a >> has no crossings. > I see now. OK. Suppose C is a Euler cycle and contains a crossing, > i.e. C = (w...,a,v,b,...c,v,d,...w). Consider the following cycle > C= (w...,a,v,c,...b,v,d,...w). This cycle has no crossings. Is > that it!? Okay, but now what happens if in making this change, you introduced a new crossing and also that your new cycle will still work? I will try to draw a graph that might demonstrate what I mean, but it probably wonÕt come out nicely. --a-- --b-- / / / --c-- / --d-- | / v / | e< >f< >g | / ^ / / --h-- / --i-- / / / --j-- --k-- Imagine two large diamonds, eafj and fbgk. They share the common point f. Also, there are two smaller diamonds, one inside each of the larger diamonds, ecfh and fdgi. One Eulerian cycle is like a double figure eight: e-a-f-k-g-b-f-j-e-c-f-i-g-d-f-h-e You cross the vertex f four times. If you try to rearrange incorrectly by simply reversing the sequence from the first f to the last f you might end up with e-a-f-d-g-i-f-c-e-j-f-b-g-k-f-h-e This still contains a crossing, three I think. If you start with a cycle that only contains two crossings, such as: e-a-f-b-g-k-f-j-e-c-f-i-g-d-f-h-e You might introduce a new crossing while removing the old if you change it by reversing at each occurence of f to e-a-f-k-g-b-f-j-e-c-f-d-g-i-f-h-e This creates a new crossing while removing the old crossing. How can you be sure you do not allow this to occur? - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Planar Graphs & Euler Cycles @agora.rdrop.com: >> I will quote the problem from my book: >> Prove that if a planar graph has an Euler cycle, it has an >> Euler cycle with no crossings. A path P in a planar graph has >> a crossing if a vertex v appears at least twice in P and P >> crosses itself at v; that is >> P = (...,a,v,b,...c,v,d,...), >> This problem is bogus. Consider the following planar graph >> c >> | >> | >> | >> a---v---b >> | >> | >> | >> d > The Eulerain cycle > a v b c v d a > has two crossings >> There is absolutely no Euler cycle that doesnÕt contain a >> crossing at v. None. What gives!? > The Eulerian cycle > a d v b c v a > has no crossings. I havenÕt seen this problem before. But I think the above is your hint to the proof. You assumed the following cycle: a -> d -> v -> c -> b -> v -> a or something similar. Notice how the example given above does not contain a crossing: a -> d -> v -> b -> c -> v -> a Look very closely at these two cycles. Figure out how the crossing disappeared. Therein lies the proof. - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: how do you show that a function has a sup or infimum? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAJ2x6k21273; >I really having trouble with this in my class. >How do you Prove that the function g attains its infimum when: > --g: Real --> Real is continuous and > --the limit of g(x) as x ---> infinity is infinity and the limit of >g(x) as x---> negative infinity is also infinity? Let g(0) = A. By the limit statements, there exists N > 0 such that g(x) > A whenever |x| > N. So {x: g(x) <= A} is bounded (why?), and also closed (why?). Can you finish it from here? Todd Trimble === Subject: Re: how do you show that a function has a sup or infimum? >I really having trouble with this in my class. >How do you Prove that the function g attains its infimum when: > --g: Real --> Real is continuous and > --the limit of g(x) as x ---> infinity is infinity and the limit of >g(x) as x---> negative infinity is also infinity? > Let g(0) = A. By the limit statements, there exists N > 0 > such that g(x) > A whenever |x| > N. So {x: g(x) <= A} is > bounded (why?), and also closed (why?). Can you finish it > from here? More directly, the set [-N, N] is bounded and closed... Mike. > Todd Trimble === Subject: Re: how do you show that a function has a sup or infimum? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iANLaT914845; >IÕm sorry but I still donÕt see how this set [-N,N] is bounded and >closed leads to g attaining its infimum. >>I really having trouble with this in my class. >How do you Prove that the function g attains its infimum when: > --g: Real --> Real is continuous and > --the limit of g(x) as x ---> infinity is infinity and the limit >g(x) as x---> negative infinity is also infinity? > Let g(0) = A. By the limit statements, there exists N > 0 > such that g(x) > A whenever |x| > N. So {x: g(x) <= A} is > bounded (why?), and also closed (why?). Can you finish it > from here? >>More directly, the set [-N, N] is bounded and closed... >>Mike. > Todd Trimble ThereÕs a famous theorem in calculus (maybe called the extreme value theorem) which for your problem says that if f: [-N, N] --> R is continuous, then there exist numbers c, d in [-N, N] such that f(c) <= f(x) <= f(d) for all x in [-N, N] (or in different language, that an absolute min of f occurs at x = c and an absolute max at x = d, for some c and d). Show that an absolute min for the restriction of g to [-N, N] is an absolute min for g on the entire domain R (hint: how was N chosen?). In first courses in real analysis, one studies the notion of compactness, which is relevant here. Is this something youÕve studied? Todd Trimble === Subject: Re: how do you show that a function has a sup or infimum? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAJ5oVL01609; >>I really having trouble with this in my class. >>How do you Prove that the function g attains its infimum when: >> --g: Real --> Real is continuous and >> --the limit of g(x) as x ---> infinity is infinity and the limit of >>g(x) as x---> negative infinity is also infinity? >Let g(0) = A. By the limit statements, there exists N > 0 >such that g(x) > A whenever |x| > N. So {x: g(x) <= A} is >bounded (why?), and also closed (why?). Can you finish it > from here? >Todd Trimble IÕm a little confused, how did you go from g(x) > A whenever |x| > N to {g(x) <= A} is bounded and closed? === Subject: Monotone sequence by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAJ6lYk05919; Suppose that S is nonempty is bounded above, K = sup(S) but K is not in S. Prove that there exists a strictly monotone sequence {x_n} such that x_n --> K and x_n is in S for all n=>1. === Subject: Re: Monotone sequence > Suppose that S is nonempty is bounded above, K = sup(S) but K is not > in S. Prove that there exists a strictly monotone sequence {x_n} such > that x_n --> K and x_n is in S for all n=>1. We know that (K-a, K) intersect S is non-empty for all a > 0. So choose x1 in (K-1, K) intersect S. For n >= 1, choose x_(n+1) in ((K+x_n)/2, K) intersect S. Then (x_n) is a strictly increasing sequence in S that converges to K. === Subject: Re: Monotone sequence Because K = sup(S), for all epsilon > 0, there is an x1 in S so that: K - epsilon < x1. Let epsilon = K - x1. So there is an x2 in S so that x1 = K - (K - x1) < x2. Let epsilon = K - x2. So there is an x3 in S so that x2 = K - (K - x2) < x3. We can continue this, until we get to x_n in S, so that x_(n-1) = K - (K - x_(n-1)) < x_n. So x_(n-1) < x_n, and This means that {x_n} is a strictly monotone sequence such that every x_n is in S for all n>=1. Now to prove that lim(x_n) = k: As we proved, for every epsilon > 0, there is an x_N in S, so that K - epsilon < x_N < K (N is a natural number). Because {x_n} is strictly monotone increasing, for all n > N, x_n > x_N, so we get: K - epsilon < x_N < x_n < K < K + epsilon. (The last < is true because epsilon > 0) => K - epsilon < x_n < K + epsilon => abs(K - x_n) < epsilon. (for all n > N) => By definition, K=lim(x_n). > Suppose that S is nonempty is bounded above, K = sup(S) but K is not > in S. Prove that there exists a strictly monotone sequence {x_n} such > that x_n --> K and x_n is in S for all n=>1. === Subject: Re: Monotone sequence > Suppose that S is nonempty is bounded above, K = sup(S) but K is not > in S. Prove that there exists a strictly monotone sequence {x_n} such > that x_n --> K and x_n is in S for all n=>1. I presume S subset reals. For each n in N, show K_n = S / (K-1/n, K) is nonnul Pick any x1 out of K_1. Eventually thereÕs some n1 for which x1 not in K_n1. Pick any x2 out K_n1. Repeat x3, x4, ... to create quested sequence. === Subject: Re: Monotone sequence > Suppose that S is nonempty is bounded above, K = sup(S) but K is not > in S. Prove that there exists a strictly monotone sequence {x_n} such > that x_n --> K and x_n is in S for all n=>1. Use the definition of K = sup S. Since K is not in S, there cannot be a finite number of points in S or else there would be a positive distance between S and K. Define the sequence {y_n}: Let y_1 be any point in S such that y_1 > K - 1/1. Assume you have picked points {y_1,...,y_k}. Pick y_{k+1} in S{y_1,...,y_k} such that y_{k+1} > K - 1/(k+1). Now define another sequence {x_n}. Let x_1 = inf {y_n}. Assume you have picked points {x_1, ..., x_k}. Pick x_{k+1} = inf {y_n}{x_1, ..., x_k}. I think this {x_n} should do the trick. Kira. === Subject: Understanding the mistake with algebraic integers It turns out that the math showing that the current teaching on the ring of algebraic integers is wrong, with horrible implications for much of what was thought to be mathematics in algebraic number theory, is harder to show than explaining the problem and how it entered the field. It goes back to irreducibility over rationals typically shortened to the math slang, irreducible over Q, or just irreducible. Look at x^2 + 3x + 2, and use the quadratic formula (though I know, itÕs not necessary with something that easy) and you get x = (-3 +/- sqrt(9 - 8))/2, and with the convention that you take the positive square root you can use that recipe to get an answer, as x = (-3 +/- 1)/2 = -1 or -2. Now thatÕs trivial but what happens with x^2 + 5x + 2? Well you use the recipe again and get x = (-5 +/- sqrt(25 - 8))/2 and youÕre stopped because now you have x = (-5 +/- sqrt(17))/2 and the recipe says to take the square root of 17, but you donÕt have an natural number for it like before, so you stop. Now, what if *one* of the xÕs STILL has only 2 as a factor while the other is really like a unit? Can you tell? Well, people playing around with algebra for quite a while COULD NOT tell. Try as they might, no matter what they did, there didnÕt seem to be any way you could answer that question. Notice that, yeah, sure, you can use approximate solutions to try and actually say you solved for x, but that tells you NOTHING about whether or not x has 2 as a factor for either solution. You might just go ahead and say, let x = 2y, and substitute to get 4y^2 + 10y + 2 = 0, which is 2y^2 + 5y + 1 = 0, and solve for y, but you still get stuck with your recipe. And here something interesting happened. Try that with x^2 + 3x + 2, and you get 4y^2 + 6y + 2 = 0, which is 2y^2 + 3y + 1 = 0 so you get a non-monic there as well. If you solve it, you find that letting x=2y, means you divide *BOTH* roots by 2, which is important so hang on to that information. When algebraic integers came into the picture--remember complex numbers were already well-known by that time--it seemed that a complete understanding of the basic properties of numbers could be had. And it *SEEMED* that algebraic integers covered a lot of territory, as any algebraic number can be written as a ratio of algebraic integers. BUT, it turns out you can prove that the roots of a non-monic polynomial with rational coefficients irreducible over Q could not be algebraic integers! Suddenly it looked like there was an answer! There was no way to look at, pry at, or otherwise fiddle with (-5 +/- sqrt(17))/2 to find out if either of the roots had 2 itself as a factor, and it seemed reasonable to suppose that meant the question was *undeterminable*, but the number is an algebraic integer, while if you try to let one of its roots have 2 as a factor you get 2y^2 + 5y + 1 = 0 which is non-monic and irreducible over Q, meaning it doesnÕt have algebraic integer roots! It seemed *reasonable* to suppose that if you canÕt determine whether or not either root has a 2 as a factor, but can determine that they are not algebraic integers that the roots are highly symmetric in some way, and *neither* has 2 itself as a factor. So, consider the reasoning, as crucial to it is the inability to check the roots of x^2 + 5x + 2 to see if either has 2 itself as a factor like with x^2 + 3x + 2 where one does, as you have an irreducible and your recipe for solving it gets you stuck with a square root that isnÕt rational. To your knowledge THERE IS NO WAY TO BREAK THROUGH and find out if 2 itself is a factor! But you have that both roots are algebraic integers, and neither can have 2 as a factor in that ring, so it seems REASONABLE TO SUPPOSE that neither has 2 itself as a factor. After all, if you have a negative result with algebraic integers, and an inability to otherwise determine, why would you assume otherwise? But thatÕs where my research comes in, as IÕve found that at least with some expressions there are in fact tools for determining factors even with irreducibles. Worse, IÕve found that being or not being in the ring of algebraic integers tells you no more than that a number is or is not the root of some monic polynomial with integer coefficients!!! So what seemed reasonable a long time ago just turned out to be wrong. Key is the lack of the tool. The tool involves non-polynomial factorization, and itÕs not even really hard to understand. No one peer review (according to the math journal Southwest Journal of Pure and Applied Mathematics which *claims* it does formal peer review) and to date no one has been able to show a ßaw in the technique or the reasoning. The tool works. So why all the arguing? Why donÕt people just accept what I have if itÕs mathematically correct? Well, basically, the results I have mean that many of your textbooks on algebraic number theory are wrong. It means that much of what youÕre being taught about Galois Theory is wrong. And it means that famous proofs like WilesÕs of FLT, are wrong. ThatÕs why. ItÕs extraordinary, but a simple assumption made a long time ago, lead to a lot of research based on a rotten foundation. And mathematics is not gracious about such things. Now human nature being what it is, people can argue with me for a while. But just because they disagree it doesnÕt change the truth. ItÕs like in the book 1984 when the Ministry of Truth makes up things for people to know, and they get Winston and work to convince him that 2+2=5 in that famous example. And for a while, social pressure will probably work. But as time passes and the information is absorbed and digested...slowly, over time, itÕs more likely that the old ways will simply pass away, and the math field will adjust. In the meantime, you get arguing, as people fight to maintain their old ways and old beliefs, just like how you see it in the rest of the world. People resist change. James Harris http://mathforprofit.blogspot.com/ === Subject: Re: Understanding the mistake with algebraic integers > It turns out that the math showing that the current teaching on the > ring of algebraic integers is wrong, with horrible implications for > much of what was thought to be mathematics in algebraic number theory, > is harder to show than explaining the problem and how it entered the > field. > It goes back to irreducibility over rationals typically shortened to > the math slang, irreducible over Q, or just irreducible. > Look at x^2 + 3x + 2, and use the quadratic formula (though I know, > itÕs not necessary with something that easy) and you get > x = (-3 +/- sqrt(9 - 8))/2, > and with the convention that you take the positive square root you can > use that recipe to get an answer, as > x = (-3 +/- 1)/2 = -1 or -2. > Now thatÕs trivial but what happens with x^2 + 5x + 2? > Well you use the recipe again and get > x = (-5 +/- sqrt(25 - 8))/2 > and youÕre stopped because now you have > x = (-5 +/- sqrt(17))/2 > and the recipe says to take the square root of 17, but you donÕt have > an natural number for it like before, so you stop. > Now, what if *one* of the xÕs STILL has only 2 as a factor while the > other is really like a unit? Good question, and it should be pretty simple to answer, as soon as you identify in which ring. Assuming you mean in the ring of algebraic integers and that one has 2 as a factor and the other does not, simply divide each of your solutions by 2, construct the minimal polynomial itÕs a root of and determine if that is a monic polynomial with integer coefficients. > Can you tell? Well, people playing around with algebra for quite a > while COULD NOT tell. Try as they might, no matter what they did, > there didnÕt seem to be any way you could answer that question. What on earth makes you say that? Observe: The two solutions are x = (-5 +/- sqrt(17))/2 So, the question becomes: is (-5 + sqrt(17))/4 or (-5 - sqrt(17))/4 an algebraic integer? x = (-5 +/- sqrt(17))/4 can be rewritten as 16x^2 + 40x + 8 = 0, or 2x^2 + 5x + 1 = 0. This is clearly an irreducible polynomial, and is not monic. It seems safe to say that 2 was not a factor of either of (-5 +/- sqrt(17))/2 in the ring of algebraic integers. Any objections? > Notice that, yeah, sure, you can use approximate solutions to try and > actually say you solved for x, but that tells you NOTHING about > whether or not x has 2 as a factor for either solution. Why would you mention approximations? > You might just go ahead and say, let x = 2y, and substitute to get > 4y^2 + 10y + 2 = 0, which is 2y^2 + 5y + 1 = 0, > and solve for y, but you still get stuck with your recipe. > And here something interesting happened. Try that with > x^2 + 3x + 2, and you get 4y^2 + 6y + 2 = 0, which is > 2y^2 + 3y + 1 = 0 > so you get a non-monic there as well. If you solve it, you find that > letting x=2y, means you divide *BOTH* roots by 2, which is important > so hang on to that information. But, 2y^2 + 3y + 1 factors as (2y + 1)(y + 1), one of which *is* a monic polynomial. > When algebraic integers came into the picture--remember complex > numbers were already well-known by that time--it seemed that a > complete understanding of the basic properties of numbers could be > had. > And it *SEEMED* that algebraic integers covered a lot of territory, as > any algebraic number can be written as a ratio of algebraic integers. > BUT, it turns out you can prove that the roots of a non-monic > polynomial with rational coefficients irreducible over Q could not be > algebraic integers! > Suddenly it looked like there was an answer! IÕve now completely lost the thread of what youÕre trying to say. Do you have a point? If so, how is the above related to it? > There was no way to look at, pry at, or otherwise fiddle with > (-5 +/- sqrt(17))/2 > to find out if either of the roots had 2 itself as a factor, and it > seemed reasonable to suppose that meant the question was > *undeterminable*, but the number is an algebraic integer, while if you > try to let one of its roots have 2 as a factor you get > 2y^2 + 5y + 1 = 0 > which is non-monic and irreducible over Q, meaning it doesnÕt have > algebraic integer roots! Look at what I did above, though your approach was more elegant. Either way, we *both* showed that (-5 +/- sqrt(17))/2 do not have 2 as a factor in the algebraic integers. > It seemed *reasonable* to suppose that if you canÕt determine whether > or not either root has a 2 as a factor, but can determine that they > are not algebraic integers that the roots are highly symmetric in some > way, and *neither* has 2 itself as a factor. Or you could just prove it, as above. > So, consider the reasoning, as crucial to it is the inability to check > the roots of > x^2 + 5x + 2 > to see if either has 2 itself as a factor like with > x^2 + 3x + 2 > where one does, as you have an irreducible and your recipe for solving > it gets you stuck with a square root that isnÕt rational. But both of us did. We both provided the evidence needed to say no. > To your knowledge THERE IS NO WAY TO BREAK THROUGH and find out if 2 > itself is a factor! Huh? > But you have that both roots are algebraic integers, and neither can > have 2 as a factor in that ring, so it seems REASONABLE TO SUPPOSE > that neither has 2 itself as a factor. But you showed that they donÕt! > After all, if you have a negative result with algebraic integers, and > an inability to otherwise determine, why would you assume otherwise? > But thatÕs where my research comes in, as IÕve found that at least > with some expressions there are in fact tools for determining factors > even with irreducibles. There are clearly some tools around. > Worse, IÕve found that being or not being in the ring of algebraic > integers tells you no more than that a number is or is not the root of > some monic polynomial with integer coefficients!!! By definition. And? > So what seemed reasonable a long time ago just turned out to be wrong. Perhaps what you say seemed reasonable was never viewed as reasonable. Can you document your historical claims? > Key is the lack of the tool. The tool involves non-polynomial > factorization, and itÕs not even really hard to understand. No one > peer review (according to the math journal Southwest Journal of Pure > and Applied Mathematics which *claims* it does formal peer review) and > to date no one has been able to show a ßaw in the technique or the > reasoning. The tools exist. > The tool works. > So why all the arguing? Why donÕt people just accept what I have if > itÕs mathematically correct? Because what you submitted had numerical counter-examples proving that its conclusions were wrong. > Well, basically, the results I have mean that many of your textbooks > on algebraic number theory are wrong. > It means that much of what youÕre being taught about Galois Theory is > wrong. Where is the tie-in to Galois Theory in any of the above? > And it means that famous proofs like WilesÕs of FLT, are wrong. Where is the connection? > ThatÕs why. > ItÕs extraordinary, but a simple assumption made a long time ago, lead > to a lot of research based on a rotten foundation. And mathematics is > not gracious about such things. > Now human nature being what it is, people can argue with me for a > while. > But just because they disagree it doesnÕt change the truth. > ItÕs like in the book 1984 when the Ministry of Truth makes up things > for people to know, and they get Winston and work to convince him that > 2+2=5 in that famous example. > And for a while, social pressure will probably work. But as time > passes and the information is absorbed and digested...slowly, over > time, itÕs more likely that the old ways will simply pass away, and > the math field will adjust. > In the meantime, you get arguing, as people fight to maintain their > old ways and old beliefs, just like how you see it in the rest of the > world. > People resist change. > James Harris > http://mathforprofit.blogspot.com/ James, please pick up a book on Abstract Algebra and read it. John B. FraleighÕs A First Course in Abstract Algebra is quite accessible. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Understanding the mistake with algebraic integers Discussion, linux) > This is clearly an irreducible polynomial, and is not monic. It seems > safe to say that 2 was not a factor of either of (-5 +/- sqrt(17))/2 in > the ring of algebraic integers. Yeah, yeah, but was it *properly* a factor? Get with the program. Geez. -- Jesse F. Hughes Depression hits more people than thought. --headline in Lexington, KY newspaper, as reported on NPRÕs Morning Edition === Subject: Re: Understanding the mistake with algebraic integers > It turns out that the math showing that the current teaching on the > ring of algebraic integers is wrong, with horrible implications for > much of what was thought to be mathematics in algebraic number theory, > is harder to show than explaining the problem and how it entered the > field. > It goes back to irreducibility over rationals typically shortened to > the math slang, irreducible over Q, or just irreducible. > Look at x^2 + 3x + 2, and use the quadratic formula (though I know, > itÕs not necessary with something that easy) and you get > x = (-3 +/- sqrt(9 - 8))/2, > and with the convention that you take the positive square root you can > use that recipe to get an answer, as > x = (-3 +/- 1)/2 = -1 or -2. > Now thatÕs trivial but what happens with x^2 + 5x + 2? > Well you use the recipe again and get > x = (-5 +/- sqrt(25 - 8))/2 > and youÕre stopped because now you have > x = (-5 +/- sqrt(17))/2 > and the recipe says to take the square root of 17, but you donÕt have > an natural number for it like before, so you stop. What is the problem? Your first polynomial is reducible, but your second is IRREDUCIBLE. That is the reason that neither of the two factors are divisible by 2. Instead, since the polynomial is irreducible, the two roots are defined as algebraic integer factors of 2. Your problem is trying to force all possibilites to fit in a single case. But the two polynomials you use are different cases: reducible and irreducible. You are trying to take a irreducible polynomial and force it to fit into the reducible case. > Now, what if *one* of the xÕs STILL has only 2 as a factor while the > other is really like a unit? If the polynomial is irreducible, then there is nothing preventing one of the roots to be a multiple of 2 and the other to be coprime. Likewise there is nothing preventing both roots to be neither multiples nor factors of 2. > Can you tell? Well, people playing around with algebra for quite a > while COULD NOT tell. Try as they might, no matter what they did, > there didnÕt seem to be any way you could answer that question. > Notice that, yeah, sure, you can use approximate solutions to try and > actually say you solved for x, but that tells you NOTHING about > whether or not x has 2 as a factor for either solution. Actually the structure of the polynomial can tell about the nature of the roots. If the polynomial is irreducible, none of the roots will be multiples of the constant term, unless the constant term is -1, 0, or 1 > You might just go ahead and say, let x = 2y, and substitute to get > 4y^2 + 10y + 2 = 0, which is 2y^2 + 5y + 1 = 0, Out of scope. The polynomial now has at least one root which is an algebraic number but not an algebraic integer. The ring in which the roots exist now expands outside the scope of algebraic integers and any conclusions made in that ring do not need to apply to algebraic integers. > and solve for y, but you still get stuck with your recipe. > And here something interesting happened. Try that with > x^2 + 3x + 2, and you get 4y^2 + 6y + 2 = 0, which is > 2y^2 + 3y + 1 = 0 > so you get a non-monic there as well. If you solve it, you find that > letting x=2y, means you divide *BOTH* roots by 2, which is important > so hang on to that information. That information is important. It means that the polynomial and its roots are outside the scope of algebraic integers. Again, any conclusions drawn from the larger ring which the roots reside in do not need to apply to algebraic integers. > When algebraic integers came into the picture--remember complex > numbers were already well-known by that time--it seemed that a > complete understanding of the basic properties of numbers could be > had. > And it *SEEMED* that algebraic integers covered a lot of territory, as > any algebraic number can be written as a ratio of algebraic integers. > BUT, it turns out you can prove that the roots of a non-monic > polynomial with rational coefficients irreducible over Q could not be > algebraic integers! That is part of the distinction between algebraic integers and algebraic numbers. If a number is proven to be a root of an irreducible non-monic polynomial then it means that the scope of the ring it resides in extends outside the algebraic integers and any conclusions drawn from the larger ring do not need to apply to algebraic integers. > Suddenly it looked like there was an answer! > There was no way to look at, pry at, or otherwise fiddle with > (-5 +/- sqrt(17))/2 > to find out if either of the roots had 2 itself as a factor, Yes there is a way, look at the polynomial (-5 +/- sqrt(17))/2 are roots of, in this case x^2 + 5x + 2. The polynomial is irreducible therefore the roots are non-unit factors of 2, which are not multiples of 2. > and it > seemed reasonable to suppose that meant the question was > *undeterminable*, but the number is an algebraic integer, while if you > try to let one of its roots have 2 as a factor you get > 2y^2 + 5y + 1 = 0 > which is non-monic and irreducible over Q, meaning it doesnÕt have > algebraic integer roots! Well, duh. Being non-monic and irreducible over Q means that the polynomial has no algebraic integer roots. Part of the definition of algebraic number. > It seemed *reasonable* to suppose that if you canÕt determine whether > or not either root has a 2 as a factor, but can determine that they > are not algebraic integers that the roots are highly symmetric in some > way, and *neither* has 2 itself as a factor. > So, consider the reasoning, as crucial to it is the inability to check > the roots of > x^2 + 5x + 2 > to see if either has 2 itself as a factor like with > x^2 + 3x + 2 > where one does, as you have an irreducible and your recipe for solving > it gets you stuck with a square root that isnÕt rational. > To your knowledge THERE IS NO WAY TO BREAK THROUGH and find out if 2 > itself is a factor! Again, I did determine the relationship between the roots of x^2 + 5x + 2 and 2. Neither root is a multiple of 2, but both are factors of 2. > But you have that both roots are algebraic integers, and neither can > have 2 as a factor in that ring, so it seems REASONABLE TO SUPPOSE > that neither has 2 itself as a factor. > After all, if you have a negative result with algebraic integers, and > an inability to otherwise determine, why would you assume otherwise? > But thatÕs where my research comes in, as IÕve found that at least > with some expressions there are in fact tools for determining factors > even with irreducibles. > Worse, IÕve found that being or not being in the ring of algebraic > integers tells you no more than that a number is or is not the root of > some monic polynomial with integer coefficients!!! Why is that worse? By the DEFINITION of algebraic integers, being or not being in the ring of algebraic integers tells you no more than that a number is or is not the root of some monic polynomial with integer coefficients > So what seemed reasonable a long time ago just turned out to be wrong. So, are you saying that the definition is wrong? > Key is the lack of the tool. The tool involves non-polynomial > factorization, and itÕs not even really hard to understand. No one > peer review (according to the math journal Southwest Journal of Pure > and Applied Mathematics which *claims* it does formal peer review) and > to date no one has been able to show a ßaw in the technique or the > reasoning. > The tool works. Um, no. If fails miserably. I have read your papers where you start with an arbitrary factorization and blame algebraic integers when it turns out that every equation can not be factored in that way. You never even look at the possibility that the factorization you started with was the wrong one. Consider for example 6x^2 + Ax + 1. If I were to say 6x^2 + Ax + 1 = (3x + v_1)(2x + v_2), I would quickly run into problems trying to factor when A=7. But if I take a step back and look for an alternate factorization then I find one that does work: 6x^2 + Ax + 1 = (6x + v_1)(x + v_2). You never take that big step all the way back and look for a different factorization. > So why all the arguing? Why donÕt people just accept what I have if > itÕs mathematically correct? The problem is that it is not correct. You start with a ßawed premise, see above. > Well, basically, the results I have mean that many of your textbooks > on algebraic number theory are wrong. Or it means that all the textbooks are right and you are wrong. > It means that much of what youÕre being taught about Galois Theory is > wrong. > And it means that famous proofs like WilesÕs of FLT, are wrong. You donÕt even know what is being taught about Galois Theory or anything about the Wiles proof. Unless you took a class which covers Galois Theory, you have no standing to criticize what is being taught. > ThatÕs why. > ItÕs extraordinary, but a simple assumption made a long time ago, lead > to a lot of research based on a rotten foundation. And mathematics is > not gracious about such things. I am going to quote you on that. It perfectly describes your entire algebraic integer/core error campaign. > Now human nature being what it is, people can argue with me for a > while. > But just because they disagree it doesnÕt change the truth. > ItÕs like in the book 1984 when the Ministry of Truth makes up things > for people to know, and they get Winston and work to convince him that > 2+2=5 in that famous example. > And for a while, social pressure will probably work. But as time > passes and the information is absorbed and digested...slowly, over > time, itÕs more likely that the old ways will simply pass away, and > the math field will adjust. > In the meantime, you get arguing, as people fight to maintain their > old ways and old beliefs, just like how you see it in the rest of the > world. > People resist change. You more than most. You absolutely resist any change involving your work. Especially when change means admitting that your previous work was wrong. > James Harris > http://mathforprofit.blogspot.com/ === Subject: Re: Understanding the mistake with algebraic integers > It turns out that the math showing that the current teaching on the > ring of algebraic integers is wrong, with horrible implications for > much of what was thought to be mathematics in algebraic number theory, > is harder to show than explaining the problem and how it entered the > field. > It goes back to irreducibility over rationals typically shortened to > the math slang, irreducible over Q, or just irreducible. > Look at x^2 + 3x + 2, and use the quadratic formula (though I know, > itÕs not necessary with something that easy) and you get > x = (-3 +/- sqrt(9 - 8))/2, > and with the convention that you take the positive square root you can > use that recipe to get an answer, as > x = (-3 +/- 1)/2 = -1 or -2. > Now thatÕs trivial but what happens with x^2 + 5x + 2? > Well you use the recipe again and get > x = (-5 +/- sqrt(25 - 8))/2 > and youÕre stopped because now you have > x = (-5 +/- sqrt(17))/2 > and the recipe says to take the square root of 17, but you donÕt have > an natural number for it like before, so you stop. > Now, what if *one* of the xÕs STILL has only 2 as a factor while the > other is really like a unit? > Can you tell? Well, people playing around with algebra for quite a > while COULD NOT tell. Try as they might, no matter what they did, > there didnÕt seem to be any way you could answer that question. Both roots belong to the field Q(sqrt(17)), that is, the field of numbers of the form r + s*sqrt(17) where r and s are rational. The members of that field are algebraic numbers; the algebraic integers in the field are of the form a + b(1 + sqrt(17))/2 where a and b are integers. The set of all these algebraic integers is a ring, for which the notation R(sqrt(17)) is found in at least one book[1]. Within this ring, both values of x are prime, that is to say: if x = y*z where y and z are in R(sqrt(17)), then y or z is a unit. In particular, neither value of x has 2 as a factor, that is, x/2 is not an algebraic integer. Also, neither value of x is a unit, that is, 1/x is not an algebraic integer. > Notice that, yeah, sure, you can use approximate solutions to try and > actually say you solved for x, but that tells you NOTHING about > whether or not x has 2 as a factor for either solution. > You might just go ahead and say, let x = 2y, and substitute to get > 4y^2 + 10y + 2 = 0, which is 2y^2 + 5y + 1 = 0, > and solve for y, but you still get stuck with your recipe. > And here something interesting happened. Try that with > x^2 + 3x + 2, and you get 4y^2 + 6y + 2 = 0, which is > 2y^2 + 3y + 1 = 0 > so you get a non-monic there as well. If you solve it, you find that > letting x=2y, means you divide *BOTH* roots by 2, which is important > so hang on to that information. > When algebraic integers came into the picture--remember complex > numbers were already well-known by that time--it seemed that a > complete understanding of the basic properties of numbers could be > had. > And it *SEEMED* that algebraic integers covered a lot of territory, as > any algebraic number can be written as a ratio of algebraic integers. > BUT, it turns out you can prove that the roots of a non-monic > polynomial with rational coefficients irreducible over Q could not be > algebraic integers! If the non-monic polynomial is irreducible, then its roots are not algebraic integers. If it is reducible, then the roots could be a mixed bag: some algebric integers, and some not. Example: 2y^2 + 3y + 1 which has roots -1 and -1/2. > Suddenly it looked like there was an answer! > There was no way to look at, pry at, or otherwise fiddle with > (-5 +/- sqrt(17))/2 > to find out if either of the roots had 2 itself as a factor, and it > seemed reasonable to suppose that meant the question was > *undeterminable*, but the number is an algebraic integer, while if you > try to let one of its roots have 2 as a factor you get > 2y^2 + 5y + 1 = 0 > which is non-monic and irreducible over Q, meaning it doesnÕt have > algebraic integer roots! OK, this settles the question. y is not an algebraic integer; x does not have 2 as a factor. Neither of the values of x given above has 2 as a factor. Actually, 2 has x as a factor. Check it out. The two xÕs are xp = (-5 + sqrt(17))/2 and xm = (-5 - sqrt(17))/2. xp * xm = (25 - 17)/4 = 8/4 = 2. But 2 is prime! Well yes, 2 is prime in the ring of rational integers; But, in a larger ring, such as R(sqrt(17)), it has nontrivial factors. Prime is a concept that is relative to a given ring. Unit is relative to a given ring also; if z is a member of a ring, then z is a unit in that ring if 1/z is also a member of that ring. We get away with a little bit of sloppiness by implicitly supposing unit to mean unit in the ring of algebraic integers. -- Chris Henrich === Subject: Re: Understanding the mistake with algebraic integers > It turns out that the math showing that the current teaching on the > ring of algebraic integers is wrong, with horrible implications for > much of what was thought to be mathematics in algebraic number theory, > is harder to show than explaining the problem and how it entered the > field. > > It goes back to irreducibility over rationals typically shortened to > the math slang, irreducible over Q, or just irreducible. > > Look at x^2 + 3x + 2, and use the quadratic formula (though I know, > itÕs not necessary with something that easy) and you get > > x = (-3 +/- sqrt(9 - 8))/2, > > and with the convention that you take the positive square root you can > use that recipe to get an answer, as > > x = (-3 +/- 1)/2 = -1 or -2. > > Now thatÕs trivial but what happens with x^2 + 5x + 2? > > Well you use the recipe again and get > > x = (-5 +/- sqrt(25 - 8))/2 > > and youÕre stopped because now you have > > x = (-5 +/- sqrt(17))/2 > > and the recipe says to take the square root of 17, but you donÕt have > an natural number for it like before, so you stop. > > Now, what if *one* of the xÕs STILL has only 2 as a factor while the > other is really like a unit? > > Can you tell? Well, people playing around with algebra for quite a > while COULD NOT tell. Try as they might, no matter what they did, > there didnÕt seem to be any way you could answer that question. > Both roots belong to the field Q(sqrt(17)), that is, the field of > numbers of the form r + s*sqrt(17) where r and s are rational. > The members of that field are algebraic numbers; the algebraic integers > in the field are of the form a + b(1 + sqrt(17))/2 where a and b are > integers. > The set of all these algebraic integers is a ring, for which the > notation R(sqrt(17)) is found in at least one book[1]. > Within this ring, both values of x are prime, that is to say: if x = > y*z where y and z are in R(sqrt(17)), then y or z is a unit. > In particular, neither value of x has 2 as a factor, that is, x/2 is > not an algebraic integer. > Also, neither value of x is a unit, that is, 1/x is not an algebraic > integer. That highlights the logical leap thatÕs made. What can actually be proven is that you donÕt have 2 as a factor in the ring of algebraic integers. The reason is that an algebraic integer cannot be the root of a non-monic polynomial with integer coefficients irreducible over Q. However, in the ring of integers you trivially have the example (3x + 1)(x + 1) = 3x^2 + 4x + 1 and the assertion is that there is no irrational corrollary with (3x + u_1)(x + u_2) = 3x + Kx + 1, where K is an integer other than 4. But the reasoning is that since u_2 cannot be a unit as it is not the root of a monic polynomial with integer coefficients as it is a root of a non-monic polynomial with integer coefficients irreducible over Q that you have proven that u_2 is never a unit. In actuality though all youÕve proven is that itÕs never a unit in the ring of algebraic integers. For a field where precision matters itÕs important to focus on the specifics to understand the reasoning. > > Notice that, yeah, sure, you can use approximate solutions to try and > actually say you solved for x, but that tells you NOTHING about > whether or not x has 2 as a factor for either solution. > > You might just go ahead and say, let x = 2y, and substitute to get > > 4y^2 + 10y + 2 = 0, which is 2y^2 + 5y + 1 = 0, > > and solve for y, but you still get stuck with your recipe. > > And here something interesting happened. Try that with > > x^2 + 3x + 2, and you get 4y^2 + 6y + 2 = 0, which is > > 2y^2 + 3y + 1 = 0 > > so you get a non-monic there as well. If you solve it, you find that > letting x=2y, means you divide *BOTH* roots by 2, which is important > so hang on to that information. > > When algebraic integers came into the picture--remember complex > numbers were already well-known by that time--it seemed that a > complete understanding of the basic properties of numbers could be > had. > > And it *SEEMED* that algebraic integers covered a lot of territory, as > any algebraic number can be written as a ratio of algebraic integers. > > BUT, it turns out you can prove that the roots of a non-monic > polynomial with rational coefficients irreducible over Q could not be > algebraic integers! > If the non-monic polynomial is irreducible, then its roots are not > algebraic integers. If it is reducible, then the roots could be a mixed > bag: some algebric integers, and some not. > Example: > 2y^2 + 3y + 1 > which has roots -1 and -1/2. Yup. And notice that here you have a fraction paired with an integer for the *rational* case, but supposedly that never occurs with irrationals, if the polynomial has integer coefficients. Obviously if you have irrational coefficients you can force such a pairing, but the idea is that when the coefficients are rational, if you donÕt have algebraic integers then you canÕt have a pairing of a number with integral properties with one that does not. ThatÕs been seen as a fundamental property of numbers for over a hundred years, and forms the basis for quite a bit of mathematical arguments ßoated over the years. Trouble is, itÕs *provably* false, and I emphasize because IÕm not just asking you to believe, but talking about the implications of the non-polynomial factorization techniques IÕve found. > > Suddenly it looked like there was an answer! > > There was no way to look at, pry at, or otherwise fiddle with > > (-5 +/- sqrt(17))/2 > > to find out if either of the roots had 2 itself as a factor, and it > seemed reasonable to suppose that meant the question was > *undeterminable*, but the number is an algebraic integer, while if you > try to let one of its roots have 2 as a factor you get > > 2y^2 + 5y + 1 = 0 > > which is non-monic and irreducible over Q, meaning it doesnÕt have > algebraic integer roots! > OK, this settles the question. y is not an algebraic integer; x does > not have 2 as a factor. Neither of the values of x given above has 2 > as a factor. Actually, 2 has x as a factor. But why should that settle the question? All it tells you is that y is not an algebraic integer. It turns out that it canÕt be because itÕs not the root of a monic polynomial with integer coefficients. ThatÕs what you know. For over a hundred years though given that information and the presumption that there was no other way to find out, it seemed reasonable to conclude that 2 just couldnÕt be a factor. My work changes that by showing there is a technique for probing irrational numbers which is to use non-polynomial factorizations. > Check it out. The two xÕs are > xp = (-5 + sqrt(17))/2 > and > xm = (-5 - sqrt(17))/2. > xp * xm = (25 - 17)/4 = 8/4 = 2. > But 2 is prime! Well yes, 2 is prime in the ring of rational integers; > But, in a larger ring, such as R(sqrt(17)), it has nontrivial factors. > Prime is a concept that is relative to a given ring. > Unit is relative to a given ring also; if z is a member of a ring, > then z is a unit in that ring if 1/z is also a member of that ring. We > get away with a little bit of sloppiness by implicitly supposing unit > to mean unit in the ring of algebraic integers. And that closes the circle on the circular reasoning. Supposedly the ring of algebraic integers is all you need. So when it turns out that youÕre outside of the ring of algebraic integers, then you must have proven something, but I have yet another example. x^2 + 2^{1/2}x + 1 has roots that are NOT ALGEBRAIC INTEGERS!!! Now then, itÕs roots are NOT UNITS in that ring as theyÕre not even in the ring. ThatÕs because 2^{1/2} is a transcendental number, and you canÕt get to integer coefficients because it is. So, just like that, I can show you reason to wonder about numbers outside of the ring of algebraic integers. Over a hundred years ago some people found they couldnÕt do something: probe irrational roots. But I found, over a hundred years later, that you *can* probe irrational roots with an advanced idea, which is non-polynomial factorization. It should be one of the hottest topics in mathematics now but politics. You know? James Harris http://mathforprofit.blogspot.com/ === Subject: Re: Understanding the mistake with algebraic integers So when it turns out that youÕre outside of the ring of algebraic > integers, then you must have proven something, but I have yet another > example. > x^2 + 2^{1/2}x + 1 > has roots that are NOT ALGEBRAIC INTEGERS!!! OOPS! That should be x^2 + 2^{sqrt(2)}x + 1, as 2^{sqrt(2)}) is a transcendental number while sqrt(2), which I put instead is not. Was just thinking of one thing and put another. > Now then, itÕs roots are NOT UNITS in that ring as theyÕre not even in > the ring. > ThatÕs because 2^{1/2} is a transcendental number, and you canÕt get > to integer coefficients because it is. Should be 2^{sqrt(2)} is a transcendental number. James Harris === Subject: Re: Understanding the mistake with algebraic integers > 2^{sqrt(2)} is a transcendental number PI and E are transcendental numbers too, but none of these three numbers has anything to do with the topic of this thread, so why did you bother to mention such a number here? Really you shoot yourself in the foot when you mention something totally off-topic like that. By the way, in one of your Web pages I happened to find when doing a Google search, you were talking about algebraic integers, or at least you claimed to be talking about them, when suddenly you said 7 is prime, which is blatantly false, showing you donÕt even know the definition of prime in a ring. In the ring of algebraic integers, (2*sqrt(2) + 1) and (2*sqrt(2) - 1) are both non-units, yet their product is 7, showing 7 to be composite, not prime, in that ring. Since your claim that 7 was prime was a necessary step in your proof, that simple mistake invalidated your entire proof. === Subject: Re: Understanding the mistake with algebraic integers You are so back in my killfile. When will you ever learn? James Harris schreef in bericht > So when it turns out that youÕre outside of the ring of algebraic > integers, then you must have proven something, but I have yet another > example. > x^2 + 2^{1/2}x + 1 > has roots that are NOT ALGEBRAIC INTEGERS!!! > OOPS! That should be x^2 + 2^{sqrt(2)}x + 1, as 2^{sqrt(2)}) is a > transcendental number while sqrt(2), which I put instead is not. > Was just thinking of one thing and put another. > Now then, itÕs roots are NOT UNITS in that ring as theyÕre not even in > the ring. > ThatÕs because 2^{1/2} is a transcendental number, and you canÕt get > to integer coefficients because it is. > Should be 2^{sqrt(2)} is a transcendental number. > James Harris === Subject: Re: Understanding the mistake with algebraic integers <10q5revegbm4s0e@corp.supernews.com> Discussion, linux) > Actually it has been a little better this time around, as besides > the sci.math people coming over, I havenÕt seen a lot of what IÕd > deem local chatter. The perfect conversation is one in which youÕre the only speaker, isnÕt it? -- Jesse F. Hughes There are VERY FEW real mathematicians and I am one of them. Few of you can handle the pressure of real mathematics, like being wrong, while I demonstrably can. -- James S. Harris === Subject: Re: Understanding the mistake with algebraic integers >> Look at x^2 + 3x + 2, and use the quadratic formula (though I know, > itÕs not necessary with something that easy) and you get > > x = (-3 +/- sqrt(9 - 8))/2, > > and with the convention that you take the positive square root you can > use that recipe to get an answer, as > > x = (-3 +/- 1)/2 = -1 or -2. > > Now thatÕs trivial but what happens with x^2 + 5x + 2? > > Well you use the recipe again and get > > x = (-5 +/- sqrt(25 - 8))/2 > > and youÕre stopped because now you have > > x = (-5 +/- sqrt(17))/2 > > and the recipe says to take the square root of 17, but you donÕt have > an natural number for it like before, so you stop. > > Now, what if *one* of the xÕs STILL has only 2 as a factor while the > other is really like a unit? > > Can you tell? Well, people playing around with algebra for quite a > while COULD NOT tell. Try as they might, no matter what they did, > there didnÕt seem to be any way you could answer that question. > Both roots belong to the field Q(sqrt(17)), that is, the field of > numbers of the form r + s*sqrt(17) where r and s are rational. > > The members of that field are algebraic numbers; the algebraic integers > in the field are of the form a + b(1 + sqrt(17))/2 where a and b are > integers. > > The set of all these algebraic integers is a ring, for which the > notation R(sqrt(17)) is found in at least one book[1]. > > Within this ring, both values of x are prime, that is to say: if x = > y*z where y and z are in R(sqrt(17)), then y or z is a unit. > > In particular, neither value of x has 2 as a factor, that is, x/2 is > not an algebraic integer. > > Also, neither value of x is a unit, that is, 1/x is not an algebraic > integer. > That highlights the logical leap thatÕs made. > What can actually be proven is that you donÕt have 2 as a factor in > the ring of algebraic integers. > The reason is that an algebraic integer cannot be the root of a > non-monic polynomial with integer coefficients irreducible over Q. > However, in the ring of integers you trivially have the example > (3x + 1)(x + 1) = 3x^2 + 4x + 1 (x+1) is a monic polynomial with integer coefficients. -1 is the root of this monic polynomial with integer coefficients therefore -1 is an algebraic integer. Is there another algebraic integer Z such that -1z = 1. Yes (-1)(-1)Z = =1 => z = -1 => z+1 =0 So -1 is a unit. > and the assertion is that there is no irrational corrollary with What is an irrational corrollary ? > (3x + u_1)(x + u_2) = 3x + Kx + 1, where K is an integer other than 4. > But the reasoning is that since u_2 cannot be a unit as it is not the > root of a monic polynomial with integer coefficients as it is a root > of a non-monic polynomial with integer coefficients irreducible over Q > that you have proven that u_2 is never a unit. No, take (x -(1+ sqrt(2))(x - (1-sqrt(2)) =0 => (x -(1+ sqrt(2)) =0 or (x - (1-sqrt(2)) =0 In neither case is x the root of a monic polynomoal with integer coefficients yet x^2 -2x -1 is a monic polynomial with integer coefficients and 1 - sqrt(2) and 1 +sqrt(2) are both units. (3x-1)(x^2 -2x -1) is not a monic polynomial but it is not irreducible over Q and so over Z. If a non-monic polynomial is the product of several polynomials with integer coefficients, one of these polynomial factors could be monic and so its roots would be algebraic integers. If (x + u2)is to be a monic polynomial with integer coefficients then u2 must be a rational integer. If u2 is not a rational integer say sqrt(2) then (x + sqrt(2)) is not a monic polynomial with integer coefficients but this does not mean that no monic polynomial with integral coefficients with -sqrt(2) as a root cannot be found. i.e x^2 -2 =0 === Subject: Re: Understanding the mistake with algebraic integers > It turns out that the math showing that the current teaching on the > ring of algebraic integers is wrong, with horrible implications for > much of what was thought to be mathematics in algebraic number theory, > is harder to show than explaining the problem and how it entered the > field. > > It goes back to irreducibility over rationals typically shortened to > the math slang, irreducible over Q, or just irreducible. > > Look at x^2 + 3x + 2, and use the quadratic formula (though I know, > itÕs not necessary with something that easy) and you get > > x = (-3 +/- sqrt(9 - 8))/2, > > and with the convention that you take the positive square root you can > use that recipe to get an answer, as > > x = (-3 +/- 1)/2 = -1 or -2. > > Now thatÕs trivial but what happens with x^2 + 5x + 2? > > Well you use the recipe again and get > > x = (-5 +/- sqrt(25 - 8))/2 > > and youÕre stopped because now you have > > x = (-5 +/- sqrt(17))/2 > > and the recipe says to take the square root of 17, but you donÕt have > an natural number for it like before, so you stop. > > Now, what if *one* of the xÕs STILL has only 2 as a factor while the > other is really like a unit? > > Can you tell? Well, people playing around with algebra for quite a > while COULD NOT tell. Try as they might, no matter what they did, > there didnÕt seem to be any way you could answer that question. > Both roots belong to the field Q(sqrt(17)), that is, the field of > numbers of the form r + s*sqrt(17) where r and s are rational. > The members of that field are algebraic numbers; the algebraic integers > in the field are of the form a + b(1 + sqrt(17))/2 where a and b are > integers. > The set of all these algebraic integers is a ring, for which the > notation R(sqrt(17)) is found in at least one book[1]. I forgot this: [1] Harvey Cohn, _Advanced_ _Number_ _Theory_, pb Dover Press. === Subject: Re: Calculus Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAJDJw304943; >this is a problem I got in my Calculus I class, and IÕm stumped. I >was wondering if anyone could help me here: >Use NewtonÕs Method to approximate the solution: >You are in a boat 2 miles from the nearest point on the coast. You >are to go to a point Q, which is 3 miles down the coast and 1 mile >inland. You can row at 3 miles per hour and walk at 4 miles per hour. >Toward what point X on the coast should you row in order to reach >point Q in the least time? Do you remember the Descartes-Snell law of reßexion? 2 media with diifferent speeds good luck,Alain. === Subject: Re: Calculus Problem >this is a problem I got in my Calculus I class, and IÕm stumped. I >was wondering if anyone could help me here: >Use NewtonÕs Method to approximate the solution: >You are in a boat 2 miles from the nearest point on the coast. You >are to go to a point Q, which is 3 miles down the coast and 1 mile >inland. You can row at 3 miles per hour and walk at 4 miles per > hour. >Toward what point X on the coast should you row in order to reach >point Q in the least time? > Do you remember the Descartes-Snell law of reßexion? > 2 media with diifferent speeds That should be refraction. RonL -- Ignorance is the most delightful of the sciences .. === Subject: Semi Perfect Tessellations by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAJIqIi03959; IÕm looking for help in the math community. A number of years ago, I worked with a Math PhD who had a poster on the wall. I believe it was titled the 13 semi-perfect tessellations ItÕs been a while, so I may not have the number or title exactly right. Has anyone seen this poster or know where I can get a copy? ItÕs for an elementary school. Rich === Subject: Re: 2x2 determinant and area of triangle-please HELP by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAJJ0oe04856; I will give you an example Area of negativelly oriented triangle A(-5,-1),B(x,4) and C(4,-3) is 29.5.What is x? if A(x1,y1) then |x2-x1 y2-y1| 1/2| |=-29.5 ---> x=2 ; correct |x3-x1 y3-y1| if B(x1,y1) and C(x2,y2) |x2-x1 y2-y1| 1/2| |=-29.5 ---> x=2 ; correct |x3-x1 y3-y1| BUT if B(x1,y1) and A(x2,y2) |x2-x1 y2-y1| 1/2| |=29.5 ---> this is false since result should be -29.5 |x3-x1 y3-y1| So IÕm asking how to know what points(A,B,c) can be used for x1,y1 or x2,y2 or x3,y3,since as shown above it DOES matter? === Subject: Re: 2x2 determinant and area of triangle-please HELP > I will give you an example > Area of negativelly oriented triangle A(-5,-1),B(x,4) and C(4,-3) is > 29.5.What is x? > if A(x1,y1) then > |x2-x1 y2-y1| > 1/2| |=-29.5 ---> x=2 ; correct > |x3-x1 y3-y1| > > if B(x1,y1) and C(x2,y2) > |x2-x1 y2-y1| > 1/2| |=-29.5 ---> x=2 ; correct > |x3-x1 y3-y1| > BUT if B(x1,y1) and A(x2,y2) > |x2-x1 y2-y1| > 1/2| |=29.5 ---> this is false since result should be -29.5 > |x3-x1 y3-y1| > So IÕm asking how to know what points(A,B,c) can be used for x1,y1 or > x2,y2 or x3,y3,since > as shown above it DOES matter? If you write the area formula as a 3x3 determinant 1/2 X |x1 y1 1| |x2 y2 1| |x3 y3 1| you know that swapping two rows changes the sign of the determinant (assumming the 3 points are not colinear) so you can easily see which orientations are positive and which negative === Subject: Re: 2x2 determinant and area of triangle-please HELP by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iALEU4D08503; IÕd like to add a little explanation to the reply of I.M.Davidson (see below) about computing the area of a triangle in the plane by computing a 3X3 determinant. Given 3 points A,B,C in the plane R^2, lift them up to points E,F,G in 3-space R^3 just by putting 1 in as the third coordinate: so if A=(A1,A2) or (A1,A2,0) then E=(A1,A2,1), etc. Then Area(A,B,C)=Area(E,F,G) and the same for the oriented areas. The tetrahedron T with vertices O,E,F,G has 1/6 the volume of the box B which has one vertex at O and edges going out from O to E,F,G. (That is true for any maximal tetrahedral corner lopped off of any parallelopiped.) The volume of the tetrahedron T is (1/3) area of the base Area(E,F,G) multiplied by the altitude. The altitude is 1 here since the plane of EFG lies 1 unit above the parallel plane which contains the origin O and A,B,C. Thus: det(E,F,G)=OrientedVol(B)=6*(OrientedVol(T))= 6*(1/3)*OrientedArea(E,F,G)=2*(OrientedArea(A,B,C)). Finally, note that this same idea should work similarly in any dimension. The n-volume of a maximal n-simplex cut off of a corner of an n-box is (1/n!)*(n-volume of the box). An n-simplex is just a pyramid over an [n-1]-simplex as base. And so it follows easily that the n-volume of an n-simplex is (1/n)*altitude*([n-1]-volume of the [n-1]-simplex that is itÕs base). You might try this for n=2, i.e. points A,B on a Cartesian line. Then orienteddistance(A,B)=B-A =(-1)*det(E,F). ----------------------------------------------- >If you write the area formula as a 3x3 determinant >1/2 X >|x1 y1 1| >|x2 y2 1| >|x3 y3 1| >you know that swapping two rows >changes the sign of the determinant >(assumming the 3 points are not colinear) >so you can easily see which orientations >are positive and which negative === Subject: Re: 2x2 determinant and area of triangle-please HELP by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAJNY0Z30274; Let me try to summarize the method that I learned from the following place by using your example: http://www.ScieceOxygen.com/mathnote/vector206.html A(-5,-1), B(x,4) , C(4,-3) The positions of A, B, and C should be look like as follows in the coordinate system: | | | | B | ------------------------ | | | C A | | To list the three points counterclockwise, it can be A-C-B or C-B-A or B-A-C To list them clockwise, it can be A-B-C B-C-A C-A-B Anyway, point B will be above x-axis. Since the area of this triangle is 29.5, we just list the coordinates of the three points according to the counterclockwise order in the following way: | -5 4 x -5 | 1/2 | | = 29.5 | -1 -3 4 -1 | (-5)(-3)+ 16 -x - (-4-3x-20) = 59 Thus, x=2 But if you following the clockwise order, you get -29.5: | -5 2 4 -5 | 1/2 | | = -29.5 | -1 4 -3 -1 | ( It is not determinant; it is just a way to evaluate the area.) If we choose a new coordinate system by using A as the orgin for the new coordinate system, the question turns out to be to find the area enclosed by the three points: (0,0) , (7,5) , (9,-2) And again, we arrange them in counterclockwise sequence to find the area: | 0 9 7 0 | 1/2 | | = 29.5 | 0 -2 5 0 | but if you put in the clockwise way, you will get | 0 7 9 0 | 1/2 | | = -29.5 | 0 5 -2 0 | The result is the same as the determinant method used by you. But it can be applied ina a more ßexible way. The rest of the problem is: why counterclockwise leads to positive value, and clockwise leads to negative ? Please recall how this area foluma is derived: http://www.scienceoxygen.com/mathnote/vector206.html When you use the value of sine between two vectors, the angle is postive when it moves in counterclockwise way that leads to the postive value of sine; the angle is negative when it moves in clockwise way that leads to negative value of sine. Hopefully, it would help to clarify something. >I will give you an example >Area of negativelly oriented triangle A(-5,-1),B(x,4) and C(4,-3) is >29.5.What is x? >if A(x1,y1) then > |x2-x1 y2-y1| >1/2| |=-29.5 ---> x=2 ; correct > |x3-x1 y3-y1| > if B(x1,y1) and C(x2,y2) > |x2-x1 y2-y1| >1/2| |=-29.5 ---> x=2 ; correct > |x3-x1 y3-y1| >BUT if B(x1,y1) and A(x2,y2) > |x2-x1 y2-y1| >1/2| |=29.5 ---> this is false since result should be -29.5 > |x3-x1 y3-y1| >So IÕm asking how to know what points(A,B,c) can be used for x1,y1 or >x2,y2 or x3,y3,since >as shown above it DOES matter? === Subject: Re: 2x2 determinant and area of triangle-please HELP > I will give you an example > Area of negativelly oriented triangle A(-5,-1),B(x,4) and C(4,-3) is > 29.5.What is x? > if A(x1,y1) then > |x2-x1 y2-y1| > 1/2| |=-29.5 ---> x=2 ; correct > |x3-x1 y3-y1| > if B(x1,y1) and C(x2,y2) > |x2-x1 y2-y1| > 1/2| |=-29.5 ---> x=2 ; correct > |x3-x1 y3-y1| > BUT if B(x1,y1) and A(x2,y2) > |x2-x1 y2-y1| > 1/2| |=29.5 ---> this is false since result should be -29.5 > |x3-x1 y3-y1| > So IÕm asking how to know what points(A,B,c) can be used for x1,y1 or > x2,y2 or x3,y3,since > as shown above it DOES matter? You can choose the correspondence as you wish, provided that you account properly for the orientation. YouÕre told ABC is negatively oriented. Note the ordering A-->B-->C(-->A) is going clock-wise around the triangle If you assign (x1,y1), (x2,y2), (x3,y3) maintaining this orientation then you will get a negative area, because the orientation is negative. You can start the assignment where you like, but must keep going round clock-wise: (all of the above visit the vertexes in a clockwise fashion (negative orientation)). If you reverse the order in which you assign the vertexes, then you will be going round in an anti-clockwise fashion (positive orientation), and so the area will be positive: Hope this helps... Mike. === Subject: Re: 2x2 determinant and area of triangle-please HELP by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAJNY0530284; >Hope this helps... >Mike. It shure did. Are you like me and you just use the following formula |x2-x1 y2-y1| 1/2 | | |x3-x1 y3-y1|,or can you actually reason as to why no matter what kind of triangle there is this formula works for it ? And why for Ō-oriented triangle you have to go clock-wise around the triangle when assigning (as you know I couldnÕt figure that out by myself by just looking at that formula ? === Subject: Re: 2x2 determinant and area of triangle-please HELP >Hope this helps... >Mike. > It shure did. > Are you like me and you just use the following formula > |x2-x1 y2-y1| > 1/2 | | > |x3-x1 y3-y1|,or can you actually reason as to why no matter > what kind of triangle there is this formula works for it ? > And why for Ō-oriented triangle you have to go clock-wise around the > triangle when assigning (as you know I couldnÕt figure that out by > myself by just looking at that formula ? IÕm not sure how much vector algebra youÕve done, so maybe what follows wonÕt help you, but you asked how I remember the formula, so IÕll do my best... (if you havenÕt done cross-products yet, nothing that follows will make much sense!) I remember the result by thinking about the vectors involved, not the formula with all the coordinates etc. - I know that if we have two vectors U and V, then their vector cross product W has magnitude |U| * |V| * sin(theta) where theta is the angle between the vectors. This is the same as the area of the parallelogram made by the two vectors. The direction of the cross product W is perpendicular to both U and V, and the three vectors [U, V, W] make a right-handed set. C ^--------------------- /.................... / V /.................... / /theta .............. / A--------------------->B U (sorry for the crappy diagram, I was never any good at art!) :-) With the vectors U,V as in the diagram, W will point straight out of your screen towards you. (If your screen is the x,y plane, then W is pointing along the positive z-axis.) When you have a triangle in the x,y plane, the vectors for the sides of the triangle all lie in the x,y plane, so the cross product is purely in the z-direction. The formula you are using is just (1/2) * z-component of the cross product. (Sorry, this will make absolutely no sense if you havenÕt learned about cross-products yet!) The (1/2) comes in because the area of the triangle is 1/2 the area of the parallelogram made by the two vectors. If we have a triangle ABC with a positive orientation, then vectors AB, AC, ABxAC (their cross-product) will form a right handed set with ABxAC pointing along the positive z-axis, and so itÕs z-component will be positive - thatÕs why the formula depends on the orientation of the triangle. In fact for any vectors U,V, the cross-product is anti-symmetric i.e. U x V = -(V x U) so if we look at a triangle with a negative orientation, this is just reversing the order of the vectors used in the cross-product, so the sign of the z-component changes, and you get a negative area. So I guess the short answer is that I donÕt memorise your formula, I just remember that I can use vector cross-products to get the required area, and that the cross-product follows the right-hand rule which in turn relates to the orientation of the triangle... Mike. === Subject: Re: 2x2 determinant and area of triangle-please HELP by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAJHDGw27979; > If you arrange the three points in counterclockwise order, > you will get postive value for the area of the triangle; if > you put it in clockwise ordre, you get negative. hi I computed an area of triangle(negativelly oriented) each time using different vertices for x1,y1 etc and sometimes I got positive and sometimes negative result.So if I was given a task of figuring out if it is + or - oriented triangle,what vertices go with x1,y1 etc could matter,since getting positive result for area would suggest it was + oriented triangle,even if not true?! === Subject: Continuous function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAJNXwh30236; Let g: D->Real How do you prove that if a is in D, g is continuous at a, and g(a)> 0, then g is positive near a? === Subject: Re: Continuous function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAK471820431; It can be proved directly from the definition of continuity: ( http://www.scienceoxygen.com/mathnote/calculus201.html ) Since g(a)> 0, letÕs say g(a)=c where c> 0. And g is continuous at a, then for any e>0, there exists a corresponding d such that | g(a)-g(b)| < e when |a-b| < d So, we can properly choose e such that g(b)-e > 0. Then -e < g(a)-g(b) < e 0< g(b)-e < g(a) < g(b)+e That proves g is positive in the small neighborhood around a. Hopefully, this would help. >Let g: D->Real >How do you prove that if a is in D, g is continuous at a, and g(a)> 0, >then g is positive near a? === Subject: Re: Continuous function > It can be proved directly from the definition of continuity: > ( http://www.scienceoxygen.com/mathnote/calculus201.html ) > Since g(a)> 0, letÕs say g(a)=c where c> 0. > And g is continuous at a, then for any e>0, there exists > a corresponding d such that > | g(a)-g(b)| < e when |a-b| < d > So, we can properly choose e such that g(b)-e > 0. > Then > -e < g(a)-g(b) < e > 0< g(b)-e < g(a) < g(b)+e > That proves g is positive in the small neighborhood around a. > Hopefully, this would help. Of course it helps. But is it only me or does anyone else find this handing out of solutions to problems a bit unethical? Students need to be challenged in college. I see no problem with handing out hints (as the previous poster did), but to lay out the details of an argument defeats the purpose of a college education. Maybe I am a bit annoyed after watching the ABC special on cheating last night but this is getting ridiculous. DM === Subject: Re: Continuous function > Let g: D->Real > How do you prove that if a is in D, g is continuous at a, and g(a)> 0, > then g is positive near a? Consider the point (a, g(a)). ItÕs above the x-axis. Can you draw a you can derive the formal proof. === Subject: I need help with a riddle by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iAK471b20399; Riddle: Three little ladies, all dressed in white, no use in the morning, but all in use at night. Who/what are they? === Subject: Re: I need help with a riddle >Riddle: >Three little ladies, >all dressed in white, >no use in the morning, >but all in use at night. >Who/what are they? This IS NOT a riddle group. It is a math group! === Subject: Re: I need help with a riddle >Riddle: .... > This IS NOT a riddle group. It is a math group! To be a little more positive, I suggest you send riddles to the rec.puzzles news group rather than here. Ken Pledger. === Subject: Re: I need help with a riddle > Riddle: > Three little ladies, > all dressed in white, > no use in the morning, > but all in use at night. > Who/what are they? The north pole star and the two stars on the big dipper that point to the pole star. === Subject: Optomisation Problem We have a rectangle inscribed in a semi-circle, and weÕre looking to maximize the area of the rectangle, assuming that one side of the rectangle lies on the diameter of the semi-circle. D= 97. Now, I understand that the appropriate way to attack this problem is to a) find a formual for the rectangleÕs area, and b) to write this formula in one variable. Now, the issue IÕm having is, IÕm having no luck trying to write out the rectangleÕs area in terms of one variable. The biggest trouble IÕm having is that since a rectangle inscribed in a semi-circle must have a side smaller than the diameter. I donÕt even know where to begin. I looked up a similar problem in my textbookÕs answer guide, and it used a^2 + b^2 = c^2 to define y in terms of X and R, but I donÕt see how that worked out as valid. IÕd appreciate any tips. === Subject: Re: Optomisation Problem Never mind. I got the right answer, assuming that as y approaches 0, x approaches d, and ... whatever. Harder problem, and one that, trying to attack it in the same fashion, gets me nowhere: Given a sphere of radius 85 find the radius R and the altitude 2H of the right circular cylinder with largest volume, that can be inscribed in the sphere. The volume of the cylinder is V= pi(r^2)(2h) where R and H relate to the cylinder (not the sphere). Find R, and H. Vsphere = 2572440.785 Area of a Circle with radius 85 (or, rather, the area of the cylinder where h approaches 0 and radius approaches 85), is 22698.00692. I donÕt know where to begin to relate the sphere to the cylinder, algebraically. === Subject: Re: Optomisation Problem > Never mind. I got the right answer, assuming that as y approaches 0, x > approaches d, and ... whatever. > Harder problem, and one that, trying to attack it in the same fashion, gets > me nowhere: > Given a sphere of radius 85 find the radius R and the altitude 2H of the > right circular cylinder with largest volume, that can be inscribed in the > sphere. The volume of the cylinder is V= pi(r^2)(2h) where R and H relate to > the cylinder (not the sphere). Find R, and H. > Vsphere = 2572440.785 > Area of a Circle with radius 85 (or, rather, the area of the cylinder where > h approaches 0 and radius approaches 85), is 22698.00692. > I donÕt know where to begin to relate the sphere to the cylinder, > algebraically. A good way of attacking many problems in 3D geometry is to draw one or more plane projections which reduce the problem to 2D geometry. You choose the orientation of the projection(s) so as to make the resulting 2D geometry as simple as possible. In this case, draw the the view you would see if you were looking straight at the centre of the sphere in a direction perpendicular to the axis of the cylinder. You will find that the (outline of the) sphere is a circle of the same radius, and the (outline of the) cylinder is a rectangle of height 2H and width 2R inscribed inside the circle. The relationship between R, H and the radius of the sphere follows from the equation of the circle. Since you solved the first problem you should now be able to solve this one in a similar way. === Subject: Re: Optomisation Problem if u reduce the problem, its the same as the one b4, the height will be length of rectange of the example b4, the width will be double the width of the example b4. > Never mind. I got the right answer, assuming that as y approaches 0, x > approaches d, and ... whatever. > Harder problem, and one that, trying to attack it in the same fashion, > gets me nowhere: > Given a sphere of radius 85 find the radius R and the altitude 2H of the > right circular cylinder with largest volume, that can be inscribed in the > sphere. The volume of the cylinder is V= pi(r^2)(2h) where R and H relate > to the cylinder (not the sphere). Find R, and H. > Vsphere = 2572440.785 > Area of a Circle with radius 85 (or, rather, the area of the cylinder > where h approaches 0 and radius approaches 85), is 22698.00692. > I donÕt know where to begin to relate the sphere to the cylinder, > algebraically. === Subject: Re: Optomisation Problem i meant the radius of the cylinder is double the width of the problem b4, but at first you have to replace the radius of the example b4 with the the radius of the sphere in this example. > if u reduce the problem, its the same as the one b4, the height will be > length of rectange of the example b4, the width will be double the width > of the example b4. >> Never mind. I got the right answer, assuming that as y approaches 0, x >> approaches d, and ... whatever. >> Harder problem, and one that, trying to attack it in the same fashion, >> gets me nowhere: >> Given a sphere of radius 85 find the radius R and the altitude 2H of the >> right circular cylinder with largest volume, that can be inscribed in the >> sphere. The volume of the cylinder is V= pi(r^2)(2h) where R and H relate >> to the cylinder (not the sphere). Find R, and H. >> Vsphere = 2572440.785 >> Area of a Circle with radius 85 (or, rather, the area of the cylinder >> where h approaches 0 and radius approaches 85), is 22698.00692. >> I donÕt know where to begin to relate the sphere to the cylinder, >> algebraically. === Subject: Re: Optimisation Problem > Given a sphere of radius 85 find the radius R and the altitude 2H of the > right circular cylinder with largest volume, that can be inscribed in the > sphere. The volume of the cylinder is V= pi(r^2)(2h) where R and H relate to > the cylinder (not the sphere). Find R, and H. > Vsphere = 2572440.785 > Area of a Circle with radius 85 (or, rather, the area of the cylinder where > h approaches 0 and radius approaches 85), is 22698.00692. > I donÕt know where to begin to relate the sphere to the cylinder, > algebraically. Draw a sphere. Draw a cylinder inscribed within the sphere. Put point A at the center of the sphere. Put point B at the center of one end of the cylinder. Put point C anywhere at the edge of one end of the cylinder. Note that by the definition of inscribing, C is also on the surface of the sphere. Consider the right triangle ABC. How do the sides of that triangle relate to H, R, and 85? Once you have that figured out, you should be able to solve the problem. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Optimisation Problem >> Given a sphere of radius 85 find the radius R and the altitude 2H of the >> right circular cylinder with largest volume, that can be inscribed in the >> sphere. The volume of the cylinder is V= pi(r^2)(2h) where R and H relate >> to >> the cylinder (not the sphere). Find R, and H. >> Vsphere = 2572440.785 >> Area of a Circle with radius 85 (or, rather, the area of the cylinder >> where >> h approaches 0 and radius approaches 85), is 22698.00692. >> I donÕt know where to begin to relate the sphere to the cylinder, >> algebraically. > Draw a sphere. Draw a cylinder inscribed within the sphere. Put > point A at the center of the sphere. Put point B at the center of one > end of the cylinder. Put point C anywhere at the edge of one end of > the cylinder. Note that by the definition of inscribing, C is also on > the surface of the sphere. > Consider the right triangle ABC. How do the sides of that triangle > relate to H, R, and 85? Well, H is the length of the line connecting A and B. C, the hypotenuse, will equal 85. So 85 = r^2 + h^2. So, h^2 = 7225 - r ^ 2 h = sqrt (85^2 - r^2). And from there... === Subject: Help w/ Transitive Property Can someone help me understand the proper use of the transitive property for equality? My understanding is that if x=y and y=z then x=z. This seems to work except when y is a constant. For example, if x=10 and z=10 then we conclude that x=z. The problem is that the statement x=z would appear to be true only when x and z are both equal to 10. However, the statement x=z does not explictly impose that limitation and would imply that x=z is true for all values (e.g., x=z=9). Steven === Subject: Re: Help w/ Transitive Property >Can someone help me understand the proper use of the transitive property >for equality? >My understanding is that if x=y and y=z then x=z. This seems to work >except when y is a constant. For example, if x=10 and z=10 then we >conclude that x=z. The problem is that the statement x=z would appear to >be true only when x and z are both equal to 10. However, the statement x=z >does not explictly impose that limitation and would imply that x=z is true >for all values (e.g., x=z=9). >Steven I presume you are studying equivalence relations, of which ordinary = between numbers is an example. Surely you understand that if x, y, and z are numbers with the property that x = y and y = z, then x = z. Here the letters represent specific numbers, and the equal sign does not represent an identity in the sense that (x - 1)^2 == x^2 -2x + 1 represents an equation that is true for any value of the variable x. It might help you to consider other examples of equivalence relations. Let S be the set of living humans. Lets say P and Q in S are related by the statement: P ~ Q means P and Q are the same sex. Do you see that this is transitive? (Also, of course, reßexive and symmetric). Or, considering the same S, say P ~ Q if P is shorter than Q. Is this relation transitive? In other words if P ~ Q and Q ~ R can you conclude P ~ R? You might also consider whether this relation is symmetric and/or reßexive. Hope this helps. --Lynn === Subject: Re: Help w/ Transitive Property >>Can someone help me understand the proper use of the transitive >>property for equality? >><>Steven > I presume you are studying equivalence relations, of which ordinary > = between numbers is an example. Surely you understand that if x, y, > and z are numbers with the property that x = y and y = z, then x = z. > Here the letters represent specific numbers, and the equal sign does > not represent an identity in the sense that (x - 1)^2 == x^2 -2x + 1 > represents an equation that is true for any value of the variable x. > < Hope this helps. > --Lynn Lynn, Steven === Subject: Optimization, last problem A poster is to contain 113 square cm of printed matter, with margins of 5.5 cm each top and bottom and 3.5 on each of the sides. Find the minimal cost of the poster if it is to be made of material costing 10 cents per square cm. Minimal cost in dollars is: Okay, well... I did about five pages of work on this problem, hand-written, so forgive me for not putting out all my work here. Now, area is xy =113. However, weÕre only considering printed area, so I Solving for y in terms of x, IÕm left with [{113/(x-3.5)} + 5.5]. Now, creating a cost function (multiplying by ten, and dividing by a hundred -ultimately, just dividing by ten), I get: C: (1/10)(x-3.5) [{113/(x-3.5)} + 5.5] To minimize cost, we find CÕ. I found C Ō to be: (x-3.5)(-113) + 113 + (5.5)(x-3.5) / LCD. LCD = 10[(x-3.5)^2] I set C Ō to 0, and solve for X. I solve X to be 4.551162791. Now, plugging it back into the cost function to get the minimal cost, I get 11.87813954. It is, however, incorrect. I also tried doing this all with (x-7), where y is defined as [(113/x-7)+11]. Still wrong. Also tried with (x+7)(y+11), still wrong. DonÕt know where IÕm going wrong. === Subject: Re: Optimization, last problem alt.math.undergrad: >A poster is to contain 113 square cm of printed matter, with margins of 5.5 >cm each top and bottom and 3.5 on each of the sides. Find the minimal cost >of the poster if it is to be made of material costing 10 cents per square >cm. Minimal cost in dollars is: >Okay, well... I did about five pages of work on this problem, hand-written, >so forgive me for not putting out all my work here. >Now, area is xy =113. However, weÕre only considering printed area, so I You seem uncertain whether x and y should be the dimensions of the paper or the dimensions of the print. You seem to have settled n x and y as the dimensions of the paper, and as youÕve discovered that leads to grief. (Your equation is not correct even so: if the top and bottom margins are each 5.5 cm and the height of the paper is y cm then the printed area is y-11 cm high, not y-5.5; and similarly for width. Try putting x and y as the dimensions of printed area; then xy=113 and itÕs easy to isolate either of the variables. n this case you want to minimize the quantity (x-7)(y-11) = xy - 11x - 7y + 77 = 113 - 11x - 7y + 77 = 190 - 11x - 7y I think youÕll find thatÕs not too horrendous after you eliminate one variable. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youÕre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: Optimization, last problem Cc: the_stan_brown@fastmail.fm > Try putting x and y as the dimensions of printed area; then xy=113 > and itÕs easy to isolate either of the variables. n this case you > want to minimize the quantity > (x-7)(y-11) If x and y are the dimensions of the printed area, you want to minimize (x + 7)(y + 11) since you need to add the margins to the size of the printed area to get the size of the poster, not subtract them. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Optimization, last problem alt.math.undergrad: >> Try putting x and y as the dimensions of printed area; then xy=113 >> ... you want to minimize the quantity >> (x-7)(y-11) >If x and y are the dimensions of the printed area, >you want to minimize > (x + 7)(y + 11) >since you need to add the margins to the size of the >printed area to get the size of the poster, not subtract them. Rich is 100% correct. I apologize for my clumsy mistake. (Also IÕm grateful to Rich because he identified his courtesy e-mail copy as such, so I knew to reply in this forum.) -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youÕre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: Optimization, last problem > A poster is to contain 113 square cm of printed matter, with margins of 5.5 > cm each top and bottom and 3.5 on each of the sides. Find the minimal cost > of the poster if it is to be made of material costing 10 cents per square > cm. Minimal cost in dollars is: Minimize C = 0.10*x*y subject to (x-2*3.5)*(y-2*5.5) - 113 = 0. I think you forgot that there are four edges so that there are two margins in each direction, one along each edge. === Subject: Re: Optimization, last problem > A poster is to contain 113 square cm of printed matter, with margins of 5.5 > cm each top and bottom and 3.5 on each of the sides. Find the minimal cost > of the poster if it is to be made of material costing 10 cents per square > cm. Minimal cost in dollars is: You made a fundamental mistake setting up the problem, I believe. Let: x = width of *printed matter* y = height of *printed matter* then width of *poster* = x + 3.5 + 3.5 = x + 7 height of *poster* = y + 5.5 + 5.5 = y + 11 then cost = (1/10)(x + 7)(y + 11) But we know that xy = 113 and therefore y = 113/x then cost = C(x) = (1/10)(x + 7)(113/x + 11) = (1/10)(113 + 11x + 791/x + 77) = (1/10)(11x + 791/x + 190) -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Optimization, last problem >> A poster is to contain 113 square cm of printed matter, with margins of >> 5.5 >> cm each top and bottom and 3.5 on each of the sides. Find the minimal >> cost >> of the poster if it is to be made of material costing 10 cents per square >> cm. Minimal cost in dollars is: > You made a fundamental mistake setting up the problem, I believe. > Let: > x = width of *printed matter* > y = height of *printed matter* > then > width of *poster* = x + 3.5 + 3.5 = x + 7 > height of *poster* = y + 5.5 + 5.5 = y + 11 > then > cost = (1/10)(x + 7)(y + 11) > But we know that > xy = 113 > and therefore > y = 113/x > then > cost = C(x) = (1/10)(x + 7)(113/x + 11) > = (1/10)(113 + 11x + 791/x + 77) > = (1/10)(11x + 791/x + 190) Ah, I see where I went wrong. I had tried (x+7)(y+11), but then IÕd used (x+7(y+11)=113 to solve y in terms of x, and since xy=113, IÕd gotten the incorrect solution. (And yes, I was confused as to just what they were referring to as having the area of 113. I wasnÕt *Too* confused, given I ended up on settling on it. I used represent the 113/printed material, and then xy to equal to the full area of the poster.) === Subject: Re: Optimization, last problem > (And yes, I was confused as to just what they were referring to as having > the area of 113. I wasnÕt *Too* confused, given I ended up on settling on > x-7/y-11, but apparently that was the wrong way to go about it. I used > x-7/y-11 on the assumption that I could use x-7/y-11 to represent the > 113/printed material, and then xy to equal to the full area of the poster.) That would have worked as well, as long as you were consistent. You could have done: Width of poster = x Height of poster = y Width of printed matter = x - 7 Height of printer matter = y - 11 Then said: cost = (1/10)xy and (x - 7)(y - 11) = 113 and solved that for y in terms of x and plugged the answer into cost. But that would be messier. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Optimization, last problem >> (And yes, I was confused as to just what they were referring to as having >> the area of 113. I wasnÕt *Too* confused, given I ended up on settling >> on >> x-7/y-11, but apparently that was the wrong way to go about it. I used >> x-7/y-11 on the assumption that I could use x-7/y-11 to represent the >> 113/printed material, and then xy to equal to the full area of the >> poster.) > That would have worked as well, as long as you were consistent. > You could have done: > Width of poster = x > Height of poster = y > Width of printed matter = x - 7 > Height of printer matter = y - 11 > Then said: > cost = (1/10)xy > and > (x - 7)(y - 11) = 113 > and solved that for y in terms of x and plugged > the answer into cost. > But that would be messier. That was what I did, but apparently got wrong. Perhaps somewhere in the arithmetic. === Subject: Statistics problem, help would be appreciated... Yeah, I am an undergrad and am taking an online course in business statistics. I wanted to see if anybody could give me an answer on this problem. Its been bugging me and the professor is unavaliable. Process 1 Process 2 Process 3 Total Process Totals ($ 100Õs) 137 108 107 352 Sample Size 10 10 10 30 Sum of Squares 1893 1188 1175 4256 Calculate F. Choices are given as .086, 1.168, 11.56, and 13.50. Any ideas? === Subject: Re: Statistics problem, help would be appreciated... > Yeah, I am an undergrad and am taking an online course in business > statistics. I wanted to see if anybody could give me an answer on > this problem. Its been bugging me and the professor is unavaliable. > Process 1 Process 2 Process 3 Total > Process Totals ($ 100Õs) 137 108 107 352 > Sample Size 10 10 10 30 > Sum of Squares 1893 1188 1175 4256 > Calculate F. > Choices are given as .086, 1.168, 11.56, and 13.50. Any ideas? No, I have no idea. I donÕt know what F is supposed to be. Sum of squares of what? Perhaps the best lesson here is to present clearly to others what you are trying to communicate. Bill === Subject: Trig Help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iALEU2u08348; can anyone help me solve this equation? 2sin2x=1 === Subject: Re: Trig Help > can anyone help me solve this equation? > 2sin2x=1 If 2 sin(2x) = 1 then sin(2x) = 0.5. The solutions of sin(2x) = 0.5 are 2x = 2n*pi + pi/6 and 2x = 2n*pi + 5*pi/6 (n an integer). So x = (n + 1/12)pi and x = (n + 5/12)pi (n an integer). Those who offer a finite number of solution are short changing you! === Subject: Re: Trig Help >> can anyone help me solve this equation? >> 2sin2x=1 > If 2 sin(2x) = 1 then sin(2x) = 0.5. > The solutions of sin(2x) = 0.5 are > 2x = 2n*pi + pi/6 and 2x = 2n*pi + 5*pi/6 (n an integer). > So x = (n + 1/12)pi and x = (n + 5/12)pi (n an integer). > Those who offer a finite number of solution are short changing you! Those who offer you =complete= solutions are short changing you. It seems to me that it is detrimental to a studentÕs learning to just hand them a complete answer. Rather, they should get a hint pointing them in the right direction. DM === Subject: Re: Trig Help >Those who offer you =complete= solutions are short changing you. >It seems to me that it is detrimental to a studentÕs learning >to just hand them a complete answer. Rather, they should get >a hint pointing them in the right direction. Galdor of the Havens, who sat near by, overheard him. ŌYou speak for me also,he cried. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ And if youÕre afraid of butter, which many people are nowa- days, (long pause) you just put in cream. --Julia Child === Subject: Re: Trig Help >can anyone help me solve this equation? >2sin2x=1 Inscribe a hexagon in the unit circle so that the longest diagonal extends from value -1 on horizonal axis to +1 on the horizonatal axis. G C === Subject: Re: Trig Help > can anyone help me solve this equation? > 2sin2x=1 sin(2x) = 1/2 Since sine is positive, the angle is in the First or Second Quadrant. 2x = 30 degrees or 2x = (180-30) degrees x = 15 degrees or x = 75 degrees. === Subject: Re: Trig Help > can anyone help me solve this equation? > 2sin2x=1 What have you tried to do? I would suggest isolating sin(2x) and then using your knowledge of the values of the sine function at special angles like 30, 45, 60, 90, etc. degrees. DM === Subject: What am I by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iALEU1308293; I am sometimes strong and sometimes weak, but I am nobodyÕs fool. For there is no language that I cannot speak, though I never went to school. What am I? === Subject: Re: What am I > I am sometimes strong and sometimes weak, but I am nobodyÕs fool. For > there is no language that I cannot speak, though I never went to > school. What am I? The dollar? === Subject: Re: What am I > I am sometimes strong and sometimes weak, but I am nobodyÕs fool. For > there is no language that I cannot speak, though I never went to > school. What am I? Please post riddles in the rec.puzzles news group rather than a mathematical group. Ken Pledger. === Subject: Re: What am I > I am sometimes strong and sometimes weak, but I am nobodyÕs fool. For > there is no language that I cannot speak, though I never went to > school. What am I? A tunafish sandwich. === Subject: Re: What am I >I am sometimes strong and sometimes weak, > but I am nobodyÕs fool. For there is no language > that I cannot speak, though I never went to > school. What am I? The signal from a cable TV satellite. === Subject: Re: What am I >>I am sometimes strong and sometimes weak, >>but I am nobodyÕs fool. For there is no language >>that I cannot speak, though I never went to >>school. What am I? > The signal from a cable TV satellite. ThatÕs a moon with a really long cord. -- john === Subject: Re: What am I > I am sometimes strong and sometimes weak, but I am nobodyÕs fool. For > there is no language that I cannot speak, though I never went to > school. What am I? Another question: what has it got to do with undergraduate mathematics? === Subject: Definition of the imaginary unit Hello again Recently, while contemplating the meaning of ... the meaning of ... ... erm ... oh yes ... the meaning of exponentiation in the domain of complex numbers, a thought suddenly struck me. books I have propping up my wobbly table, so maybe someone can enlighten me. ItÕs usually said that i is defined as the square root of -1. But we know that the equation x^2 = -1 has TWO solutions, which we would write as i and -i. My question therefore is: 1. How do we know which value to designate i, and which to designate -i? 2. If we DONÕT know, then almost any expression involving i would seem to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is there any easy way to understand why this never matters? === Subject: Re: Definition of the imaginary unit > ItÕs usually said that i is defined as the square root of -1. But we > know that the equation > x^2 = -1 > has TWO solutions, which we would write as i and -i. > My question therefore is: > 1. How do we know which value to designate i, and which to designate > -i? > 2. If we DONÕT know, then almost any expression involving i would seem > to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is > there any easy way to understand why this never matters? The symmetry between i and -i is essentially the symmetry between clockwise and anti-clockwise. You have this symmetry in any plane. Here is a way to see what this symmetry means. Suppose I turn a bike upside down, rotate one of its wheels, and ask if the wheel is rotating clockwise or anti-clockwise. The answer depends on which side we choose to look at the bike from, so the wheel in itself is rotating neither clockwise nor anti-clockwise, it is simply rotating. Suppose now that I rotate both wheels of the bike and ask if they are rotating in the same direction. The answer is either yes or no. Suppose finally that in addition to rotating a wheel on this bike, I rotate a wheel on another bike. I then ask if the wheels are rotating in the same direction. Unless the bikes are aligned somehow, this question will be meaningless. In my opinion, one should stop speaking about /the/ complex plane and /the/ imaginary unit and instead speak about complex planes (in plural), each having two imaginary units. Mathematical objects are only determined up to isomorphism, and one should not pretend otherwise. Some things to note: * Although the square roots of -1 have identical properties in C, they have different properties in Z_(4k+1), where k is any positive integer. For example, in Z_5 we have (2)(2)=4=-1 as well as (-2)(-2)=4=-1, but 2 and -2 have different properties in Z_5. * Although sqrt(2) and -sqrt(2) have different properties (sqrt(2)>0 while -sqrt(2)<0), if we forget about the ordering properties of the real numbers and just consider the algebraic properties (i. e., properties that can be expressed in terms of addition and multiplication), the two square roots of 2 have identical properties. For example, both are roots of the equation x^2-2=0, and neither is a root of the equation x^2+1=0. * Some expressions, like (exp(ix)-exp(-ix))/(2i) at first appear to contain a reference to an imaginary unit i of a complex plane, but can be rewritten into a form where no imaginary unit appears (the expression above can be simplified into sin(x)). The imaginary unit works like a bound (a.k.a. apparent or dummy) variable in this case, and if we take into account that in (Integral from x=a1 to x=a2 of f(x))+(Integral from x=b1 to x=b2 of g(x)), x in the first integral has nothing to do with x in the second integral, it seems we should say that in (i/i)+(i/i), i in the first parenthesis has nothing to do with i in the second parenthesis. Mattias === Subject: Re: Definition of the imaginary unit > ItÕs usually said that i is defined as the square root of -1. That is never said in any well grounded introduction to complex numbers; furthermore it doesnÕt make mathematical sense. As mathedman said, the easy way out (which goes back to Gauss I believe) is to define a multiplication on pairs of real numbers (x,y), identify the reals with the pairs (x,0), and then define i = (0,1). It is then a theorem that i^2 = -1. === Subject: Re: Definition of the imaginary unit > ItÕs usually said that i is defined as the square root of -1. > That is never said in any well grounded introduction to complex numbers; > furthermore it doesnÕt make mathematical sense. As mathedman said, the easy > way out (which goes back to Gauss I believe) is to define a multiplication > on pairs of real numbers (x,y), identify the reals with the pairs (x,0), > and then define i = (0,1). It is then a theorem that i^2 = -1. ThatÕs interesting. Would defining i = (0,-1) make any difference to anything, or is it just an arbitrary choice? === Subject: Re: Definition of the imaginary unit >Hello again >Recently, while contemplating the meaning of ... the meaning of ... >... erm ... oh yes ... the meaning of exponentiation in the domain of >complex numbers, a thought suddenly struck me. >books I have propping up my wobbly table, so maybe someone can >enlighten me. >ItÕs usually said that i is defined as the square root of -1. But we >know that the equation > x^2 = -1 >has TWO solutions, which we would write as i and -i. HUH??? The complex plane is the ordinary x-y plane with a definition of a multiplication and addition operations which make it an algebraic field. The Number i (Europeans largely use j so as to mimic vectors in the x-y plane) is the complex number (0,1). -i is the complex number (0, -1) . When we casually say the square root of -1, as we do for folks who have never known about the complex plane, we mean the square root of the complex number (-1, 0). It has those two complex-number roots named above. And surely one can tell which is which! Once logarithms are defined in this plane [and ln(z) is NOT a function but an infinitely many valued relation], exponentiation is defined by: z^w = e^(w*ln(z)) . This MAY have infinitely many values or a finite number or a single value depending upon the values of z and w. >My question therefore is: >1. How do we know which value to designate i, and which to designate >-i? >2. If we DONÕT know, then almost any expression involving i would seem >to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is >there any easy way to understand why this never matters? === Subject: Re: Definition of the imaginary unit >Hello again >Recently, while contemplating the meaning of ... the meaning of ... >... erm ... oh yes ... the meaning of exponentiation in the domain of >complex numbers, a thought suddenly struck me. >books I have propping up my wobbly table, so maybe someone can >enlighten me. >ItÕs usually said that i is defined as the square root of -1. But we >know that the equation > x^2 = -1 >has TWO solutions, which we would write as i and -i. > HUH??? > The complex plane is the ordinary x-y plane with a definition of a > multiplication and addition operations which make it an algebraic > field. > The Number i (Europeans largely use j so as to mimic vectors > in the x-y plane) is the complex number (0,1). In the UK at least I think that i is pretty much universal in maths. I think that j might be used in physics possibly, but I know even less about physics than I do about maths so thatÕs not saying much... > -i is the complex > number (0, -1) . When we casually say the square root of -1, as we > do for folks who have never known about the complex plane, > we mean the square root of the complex number (-1, 0). It has those > two complex-number roots named above. And surely one can tell which is > which! How, exactly? === Subject: Re: Definition of the imaginary unit > >Hello again Recently, while contemplating the meaning of ... the meaning of ... >... erm ... oh yes ... the meaning of exponentiation in the domain of >complex numbers, a thought suddenly struck me. books I have propping up my wobbly table, so maybe someone can >enlighten me. ItÕs usually said that i is defined as the square root of -1. But we >know that the equation x^2 = -1 has TWO solutions, which we would write as i and -i. > > > HUH??? > The complex plane is the ordinary x-y plane with a definition of a > multiplication and addition operations which make it an algebraic > field. > The Number i (Europeans largely use j so as to mimic vectors > in the x-y plane) is the complex number (0,1). > In the UK at least I think that i is pretty much universal in maths. > I think that j might be used in physics possibly, but I know even > less about physics than I do about maths so thatÕs not saying much... > -i is the complex > number (0, -1) . When we casually say the square root of -1, as we > do for folks who have never known about the complex plane, > we mean the square root of the complex number (-1, 0). It has those > two complex-number roots named above. And surely one can tell which is > which! > How, exactly? Ignore that question. Having read some of the further postings I now understand what youÕre saying. === Subject: Re: Definition of the imaginary unit > ... > The Number i (Europeans largely use j so as to mimic vectors > in the x-y plane) ... I donÕt think so. Electrical engineers use j so that it doesnÕt get confused with a common name for current. === Subject: Re: Definition of the imaginary unit days. My association with the Department is that of an alumnus. >>Hello again >>Recently, while contemplating the meaning of ... the meaning of ... >>... erm ... oh yes ... the meaning of exponentiation in the domain of >>complex numbers, a thought suddenly struck me. >>books I have propping up my wobbly table, so maybe someone can >>enlighten me. >>ItÕs usually said that i is defined as the square root of -1. But we >>know that the equation >> x^2 = -1 >>has TWO solutions, which we would write as i and -i. >HUH??? >The complex plane is the ordinary x-y plane with a definition of a >multiplication and addition operations which make it an algebraic >field. This is not the only way to define the complex numbers. One can also define them to be equal to R[x]/(x^2+1), for example. In that case, the question not only makes perfect sense, but understanding the answer is the first step along the way to Galois Theory. > The Number i (Europeans largely use j so as to mimic vectors >in the x-y plane) I didnÕt know this was a European vs. non-European thing. Are you sure itÕs not just some field vs another? I think some engineering schools prefer to use j (since the use i for indices)... > is the complex number (0,1). -i is the complex >number (0, -1) . When we casually say the square root of -1, as we >do for folks who have never known about the complex plane, >we mean the square root of the complex number (-1, 0). It has those >two complex-number roots named above. And surely one can tell which is >which! And, if you start from R[x]/(x^2+1), can you tell me please whether x(x^2+1) corresponds to i or to -i? And why? -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Definition of the imaginary unit >>Hello again >>Recently, while contemplating the meaning of ... the meaning of ... >>... erm ... oh yes ... the meaning of exponentiation in the domain of >>complex numbers, a thought suddenly struck me. >>books I have propping up my wobbly table, so maybe someone can >>enlighten me. >>ItÕs usually said that i is defined as the square root of -1. But we >>know that the equation >> x^2 = -1 >>has TWO solutions, which we would write as i and -i. >HUH??? >The complex plane is the ordinary x-y plane with a definition of a >multiplication and addition operations which make it an algebraic >field. > This is not the only way to define the complex numbers. One can also > define them to be equal to R[x]/(x^2+1), for example. IÕm trying hard to keep up with this, but much of the maths is over my head. This R[x]/(x^2+1) thing also appeared in an earlier response. Could someone explain what this means? What is R? (R stands for Real?) What is x? What is R[x]? How does R[x]/(x^2+1) define the complex numbers? === Subject: Re: Definition of the imaginary unit <41a209d1.92621240@netnews.worldnet.att.net> >ItÕs usually said that i is defined as the square root of -1. But we >>know that the equation >> x^2 = -1 >>has TWO solutions, which we would write as i and -i. The complex plane is the ordinary x-y plane with a definition of a >multiplication and addition operations which make it an algebraic >field. > This is not the only way to define the complex numbers. One can also > define them to be equal to R[x]/(x^2+1), for example. > IÕm trying hard to keep up with this, but much of the maths is over my > head. This R[x]/(x^2+1) thing also appeared in an earlier response. > Could someone explain what this means? What is R? (R stands for > Real?) What is x? What is R[x]? How does R[x]/(x^2+1) define the > complex numbers? R is the real numbers. R[x] is the ring of polynomials with real coefficients, ax^2 + bc + c for example in R[x]. (x^2 + 1) is all multiplies of the polynomial x^2 + 1 R[x]/(x^2 + 1) is the polynomials modulus x^2 + 1. Just like integers modulus n, you divide away factors of n to keep the remainder. For example, x^3 + x^2 + 2x + 1 = x(x^2 + 1) + x^2 + x + 1 = (x + 1)(x^2 + 1) + x would have remainder x. Auturo means not x(x^2 + 1), but instead x + (x^2 + 1) or the polynomial x modulus x^2 + 1 Thus x^2 = x^2 + 1 - 1 = -1 modulus x^2 + 1 Also (-x)^2 = x^2 + 1 - 1 = -1 modulus x^2 + 1 Tho _algebraically_ it makes no difference which is i, the natural way to define i would be i = x + (x^2 + 1) in R/(x^2 + 1) or the polynomial x modulus x^2 + 1 -i = -x + (x^2 + 1) or the polynomial -x modulus x^2 + 1 Otherwise youÕll be crossed iÕd with i = -x in R/(x^2 + 1) -i = x in R/(x^2 + 1) === Subject: Re: Definition of the imaginary unit > .... > The Number i (Europeans largely use j so as to mimic vectors >in the x-y plane) > I didnÕt know this was a European vs non-European thing. Are you > sure itÕs not just some field vs another? I think some engineering > schools prefer to use j (since they use i for indices)... For instantaneous current, I think. ThatÕs why j is often preferred when complex numbers are used in A.C. circuit calculations. Ken Pledger. === Subject: Re: Definition of the imaginary unit > .... > ItÕs usually said that i is defined as the square root of -1. But we > know that the equation > x^2 = -1 > has TWO solutions, which we would write as i and -i. > My question therefore is: > 1. How do we know which value to designate i, and which to designate > -i? > 2. If we DONÕT know, then almost any expression involving i would seem > to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is > there any easy way to understand why this never matters? ThatÕs a good question, which has come up before. HereÕs what I > There is an oddity in the way complex numbers are introduced.... > -1 has two square roots, which are conjugates of each other, but > only one of them is given its own symbol. Why pretend there is asymmetry > when there is none? > > If we accept that there is complete symmetry, then we are faced with a > strange consequence: we cannot name any single non-real complex number.... > with complex numbers we can only name pairs of numbers, such as the pair > that satisfies the equation x^2+1=0.... > perhaps evaded the real issue. I agree with your main concern, and I > think the answer lies in the connection between mathematics and our > physical diagrams. > As you suggest, complex conjugation is an automorphism of the complex > field, leaving invariant every real number, but swapping i with -i. > That means that most algebraic statements we make about complex numbers > remain true if i and -i are transposed, so how do we know which is > which? The mathematics assures us that the equation z^2 + 1 = 0 has two > solutions, but it treats them even-handedly, giving us no way to stick a > special label on one of them rather than the other. > The down-to-earth everyday answer is that our diagrams represent the > real line horizontally with positive numbers to the right, and the > imaginary axis perpendicular to it with i above and -i below. That > depends upon our notions of right and left (or clockwise and > anti-clockwise), which we acquired as small children by being shown > something *physical*. You canÕt define right or left mathematically. A > mathematical argument (perhaps about real vector spaces) can tell you that > there are two sides, but it treats them even-handedly, giving us no way to > stick a special label on one of them rather than the other. IÕve repeated > those words, to suggest that your question about i versus -i is pretty > close to the more basic question of right versus left. > If youÕre doing really pure mathematics, then calling certain complex > numbers i and -i isnÕt going to cause any trouble. You never *need* > to decide which is which. But if youÕre a normal human being who uses a > physical plane diagram to help in thinking about complex numbers, then you > put 1 on the right, -1 on the left, i above and -i below. ItÕs > as humdrum as that. > Ken Pledger. === Subject: Re: Definition of the imaginary unit days. My association with the Department is that of an alumnus. >Hello again >Recently, while contemplating the meaning of ... the meaning of ... >... erm ... oh yes ... the meaning of exponentiation in the domain of >complex numbers, a thought suddenly struck me. >books I have propping up my wobbly table, so maybe someone can >enlighten me. >ItÕs usually said that i is defined as the square root of -1. But we >know that the equation > x^2 = -1 >has TWO solutions, which we would write as i and -i. >My question therefore is: >1. How do we know which value to designate i, and which to designate >-i? We donÕt. But in a very real sense it doesnÕt matter. Suppose martians are all left handed, and so when they came up with complex numbers, what they designated as i is what ->we<- would designate as -i. There is then a perfect translation between what the martians will do and what we do. It is called complex conjugation. Namely, complex conjugation is a bijection function from C to C which has the following properties: (i) f(x+y) = f(x) + f(y) for all complex numbers x and y; (ii) f(xy) = f(x)f(y) for all complex numbers x and y; (iii) f(r) = r for all real numbers r; (iv) f(i) = -i. This automorphism of the complex numbers tells you that, regardless of which of the two roots of x^2+1 you choose to designate by i and which you choose to designate by -i, everything you can do with one choice you can also do with the other choice. The two choices yield algebraically indistinguishable constructions. >2. If we DONÕT know, then almost any expression involving i would seem >to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is >there any easy way to understand why this never matters? No, it is not ambiguous. Choose your favorite one to be i; the only other choice will be algebraically indistinguishable, and therefore there is no problem with the arbitrariness of this choice. Just reßect everything about the X axis if you made the other choice. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Definition of the imaginary unit >ItÕs usually said that i is defined as the square root of -1. But we >know that the equation > x^2 = -1 >has TWO solutions, which we would write as i and -i. >My question therefore is: >1. How do we know which value to designate i, and which to designate >-i? > We donÕt. But in a very real sense it doesnÕt matter. > Suppose martians are all left handed, and so when they came up with > complex numbers, what they designated as i is what ->we<- would > designate as -i. There is then a perfect translation between what > the martians will do and what we do. It is called complex > conjugation. > Namely, complex conjugation is a bijection function from C to C which > has the following properties: > (i) f(x+y) = f(x) + f(y) for all complex numbers x and y; > (ii) f(xy) = f(x)f(y) for all complex numbers x and y; > (iii) f(r) = r for all real numbers r; > (iv) f(i) = -i. > This automorphism of the complex numbers tells you that, regardless of > which of the two roots of x^2+1 you choose to designate by i and > which you choose to designate by -i, everything you can do with one > choice you can also do with the other choice. The two choices yield > algebraically indistinguishable constructions. >2. If we DONÕT know, then almost any expression involving i would seem >to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is >there any easy way to understand why this never matters? > No, it is not ambiguous. Choose your favorite one to be i; the only > other choice will be algebraically indistinguishable, and therefore > there is no problem with the arbitrariness of this choice. Just > reßect everything about the X axis if you made the other choice. I think I get what youÕre saying. Let me just explain it back in a slightly different way to make sure: Any true (false) mathematical statement involving i will be equally true (false) if i and -i are interchanged. Is that correct? The only potential counterexample I can think of so far is arg(i) = pi/2 (1) Swapping i and -i, you get arg(-i) = pi/2 (2) which is wrong. Although this looks to be a counterexample, I THINK there must be a subtle reason why it isnÕt. Maybe the DEFINITION of the arg function depends on the choice of i vs -i, so the arg function in (1) is not the same as the arg function in (2)? [After writing this, have just read Ken PledgerÕs response, which includes: ... most algebraic statements we make about complex numbers remain true if i and -i are transposed... Almost the same as mine, except that I have all and KP has most!!] === Subject: Re: Definition of the imaginary unit days. My association with the Department is that of an alumnus. >>ItÕs usually said that i is defined as the square root of -1. But we >>know that the equation >> x^2 = -1 >>has TWO solutions, which we would write as i and -i. >>My question therefore is: >>1. How do we know which value to designate i, and which to designate >>-i? >> We donÕt. But in a very real sense it doesnÕt matter. >> Suppose martians are all left handed, and so when they came up with >> complex numbers, what they designated as i is what ->we<- would >> designate as -i. There is then a perfect translation between what >> the martians will do and what we do. It is called complex >> conjugation. >> Namely, complex conjugation is a bijection function from C to C which >> has the following properties: >> (i) f(x+y) = f(x) + f(y) for all complex numbers x and y; >> (ii) f(xy) = f(x)f(y) for all complex numbers x and y; >> (iii) f(r) = r for all real numbers r; >> (iv) f(i) = -i. >> This automorphism of the complex numbers tells you that, regardless of >> which of the two roots of x^2+1 you choose to designate by i and >> which you choose to designate by -i, everything you can do with one >> choice you can also do with the other choice. The two choices yield >> algebraically indistinguishable constructions. >>2. If we DONÕT know, then almost any expression involving i would seem >>to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is >>there any easy way to understand why this never matters? >> No, it is not ambiguous. Choose your favorite one to be i; the only >> other choice will be algebraically indistinguishable, and therefore >> there is no problem with the arbitrariness of this choice. Just >> reßect everything about the X axis if you made the other choice. >I think I get what youÕre saying. Let me just explain it back in a >slightly different way to make sure: >Any true (false) mathematical statement involving i will be equally >true (false) if i and -i are interchanged. >Is that correct? No. Any ->algebraic<- statement. The complex numbers and the complex numbers after conjugation are ALGEBRAICALLY indistinguishable. I said that many times. What made you think that the word algebraically was superßuous? >The only potential counterexample I can think of so far is > arg(i) = pi/2 (1) >Swapping i and -i, you get > arg(-i) = pi/2 (2) >which is wrong. Because arg is a function which is not defined algebraically. >Although this looks to be a counterexample, I THINK there must be a >subtle reason why it isnÕt. Maybe the DEFINITION of the arg function >depends on the choice of i vs -i, so the arg function in (1) is not >the same as the arg function in (2)? >[After writing this, have just read Ken PledgerÕs response, which >includes: >... most algebraic statements we make about complex numbers remain >true if i and -i are transposed... >Almost the same as mine, except that I have all and KP has most!!] You are doing the same thing you did with the exponentials thread: you read a response, drop all sort of assumptions, and then wonder why your conclusions are false after you dropped the bunch of conditions. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Definition of the imaginary unit ... >Any true (false) mathematical statement involving i will be equally >true (false) if i and -i are interchanged. >Is that correct? > No. Any ->algebraic<- statement. The complex numbers and the complex > numbers after conjugation are ALGEBRAICALLY indistinguishable. I said > that many times. What made you think that the word algebraically was > superßuous? The answer is simple, my friend. To you, the word algebraically obviously has a precise technical meaning which is essential to the point you are making. To me, the word algebraically merely conveyed the idea of some collection of symbols with a mathematical meaning and did not register as being particularly important. === Subject: Re: Definition of the imaginary unit days. My association with the Department is that of an alumnus. >... >>Any true (false) mathematical statement involving i will be equally >>true (false) if i and -i are interchanged. >>Is that correct? >> No. Any ->algebraic<- statement. The complex numbers and the complex >> numbers after conjugation are ALGEBRAICALLY indistinguishable. I said >> that many times. What made you think that the word algebraically was >> superßuous? >The answer is simple, my friend. To you, the word algebraically >obviously has a precise technical meaning which is essential to the >point you are making. To me, the word algebraically merely conveyed >the idea of some collection of symbols with a mathematical meaning >and did not register as being particularly important. Look at calculus textbooks. Some of them have the words Early Transcendentals on them. This refers to the early introduction of functions such as the logarithms, the exponential functions, and most importantly, the trigonometric and inverse trigonometric functions. They are transcendealt functions. Algebraic functions are those that involve only the use of addition, multiplication, division, exponentiation, and fractional powers. The arg function in complex analysis is defined, usually, via a branch of the inverse tangent, a transcendental function. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Definition of the imaginary unit > Look at calculus textbooks. Some of them have the words Early > Transcendentals on them. This refers to the early introduction of > functions such as the logarithms, the exponential functions, and most > importantly, the trigonometric and inverse trigonometric > functions. They are transcendealt functions. Algebraic functions are > those that involve only the use of addition, multiplication, division, > exponentiation, and fractional powers. The arg function in complex > analysis is defined, usually, via a branch of the inverse tangent, a > transcendental function. Copy that. IÕm actually pretty familiar with all the ... erm .... familiar transcendental functions, and the basic calculus associated with them. Just didnÕt make the contrast transcendental vs algebraic here. I think of everything like arctan(x), sin(x), exp(x) etc. as just algebra. (Comes from having no formal training you know!) Let me coin the term parity violation for the situation when interchanging i and -i turns a true statement into a false one, or vice versa. (I donÕt know the correct term, or even if there is one.) Most familiar transcendental functions can be (or, if you prefer, ARE) defined as the sum of an infinite series of algebraic terms. I canÕt currently see how an infinite sum can violate parity when none of its individual (algebraic) terms do. Though I DO know that some very weird and counterintuitive things can happen with infinite sums. Maybe this one of them? Right now, I actually think that the arg function causes a parity violation NOT because itÕs transcendental, but because its definition depends on the handedness of the complex plane (which, debatably, is arbitrary). For example, I could equally well define the very simple quadrant function, which takes the values 1, 2, 3, 4 (ignoring boundary cases here). Then, quadrant(1+i) = 1 and yet quadrant(1-i) =/= 1 Thus violating parity. In fact, IÕm currently thinking that IF a statement (algebraic, transcendental, anything) violates parity, then some part of its DEFINITION must depend on the (debatably arbitrary) assignation of i (aka handedness of the complex plane) - like arg and quadrant. None of that was terribly precisely explained, but hopefully you will get the gist of what I am trying to say. Can you think of any parity-violating statements involving plain old arctan (NOT the handed version used by arg), or plain old sine, cosine, exp, log etc.? IÕd be very interested to see an example, as that would blow my current theory on this matter out of the water. === Subject: Re: Definition of the imaginary unit > Let me coin the term parity violation for the situation when > interchanging i and -i turns a true statement into a false one, or > vice versa. (I donÕt know the correct term, or even if there is one.) To make this precise: You have a statement S(z) which asserts something about the complex number z=a+bi. S is then a parity violator if there exist a, b such that S(a+bi) is true but S(a-bi) is false. > Most familiar transcendental functions can be (or, if you prefer, ARE) > defined as the sum of an infinite series of algebraic terms. I canÕt > currently see how an infinite sum can violate parity when none of its > individual (algebraic) terms do. Though I DO know that some very weird > and counterintuitive things can happen with infinite sums. Maybe this > one of them? No, it is not. The algebraic VS. transcendental distinction has nothing, or almost nothing, to do with the symmetry between i and -i. > Right now, I actually think that the arg function causes a parity > violation NOT because itÕs transcendental, but because its definition > depends on the handedness of the complex plane (which, debatably, is > arbitrary). You are absolutely right. The function arg causes parity violation, as does the function Im. However, if we define f(z)=i*Im(z) and g(z)=i*arg(z), then neither f nor g will cause parity violation. Let me use zto denote the conjugate of z. Then the functions Re, Im, abs, and arg can be defined by the following symmetrical-looking expressions: (z+zÕ)*(1/2)=Re z (z-zÕ)*(1/2)=i*Im z (z*zÕ)^(1/2)=abs(z) (z/zÕ)^(1/2)=exp(i*arg z) (To use the last expression as a definition of arg, you must of course first define the exponential function for complex numbers without using the arg function, but this can be done using power series.) Mattias === Subject: Re: Definition of the imaginary unit > Let me coin the term parity violation for the situation when > interchanging i and -i turns a true statement into a false one, or > vice versa. (I donÕt know the correct term, or even if there is one.) > To make this precise: You have a statement S(z) which asserts > something about the complex number z=a+bi. S is then a parity violator > if there exist a, b such that S(a+bi) is true but S(a-bi) is false. Absolutely, except that the statement neednÕt be concerned with just one complex number. It could be, for example, z_1 * z_2 = z_3, or arctan(z_1)/z_2 + z_3 = 0, or whatever. In this case i and -i must be interchanged wherever they occur. > > Most familiar transcendental functions can be (or, if you prefer, ARE) > defined as the sum of an infinite series of algebraic terms. I canÕt > currently see how an infinite sum can violate parity when none of its > individual (algebraic) terms do. Though I DO know that some very weird > and counterintuitive things can happen with infinite sums. Maybe this > one of them? > No, it is not. bonkers or (ii) making a meaningless statement. > The algebraic VS. transcendental distinction has > nothing, or almost nothing, to do with the symmetry between i and -i. Right. ThatÕs interesting. WeÕre getting to the heart of something IÕm pretty confused about. HereÕs three statements that I currently think are true, but people can agree or disagree with: 1. No algebraic parity-violating statements can be constructed (In other words, algebraic operations cannot break the symmetry.) All 2. No partity-violating statements consisting of algebraic operations and transcendental functions definable as the limit of an infinite sequence of algebraic operations can be constructed. In other words, contrasting with (1), the algebraic vs transcendental distinction IS irrelevant when the transcendental function(s) in question are of this nature. 3. Every parity-violating statement contains an operator, function, whatever you want to call it, whose definition depends on the (debatably arbitrary) assignation of the symbol i to one of the roots of x^2 + 1 = 0, or, if you prefer, the assignation of i to the point (0,1) or (0,-1) on the complex plane. Now, IF every non-algebraic statement is transcendental, then, from 1, all parity-violating statements are transcendental. BUT ... IÕm not clear if this is the standard definition of the term transcendental. Many sources say that transcendental functions are non-algebraic, but I havenÕt been able to find an explicit statement that ALL non-algebraic functions are transcendental. For example, is the parity-violating Im() function transcendental? Is my parity-violating quadrant function transcendental? Not sure how much sense that made to anyone, but any views are, as always, very welcome... === Subject: Re: Definition of the imaginary unit > To make this precise: You have a statement S(z) which asserts > something about the complex number z=a+bi. S is then a parity violator > if there exist a, b such that S(a+bi) is true but S(a-bi) is false. > > Absolutely, except that the statement neednÕt be concerned with just > one complex number. It could be, for example, z_1 * z_2 = z_3, or > arctan(z_1)/z_2 + z_3 = 0, or whatever. In this case i and -i must be > interchanged wherever they occur. That is, a statement S(z1, z2, ..., zn) about complex numbers z1, z2, ..., zn is a parity violator if there exist complex numbers z1, z2, ..., zn such that S(z1, z2, ..., zn) is true, but S(z1Õ, z2Õ, ..., znÕ) is false, where Ō denotes conjugation. > The algebraic VS. transcendental distinction has > nothing, or almost nothing, to do with the symmetry between i and -i. > > Right. ThatÕs interesting. WeÕre getting to the heart of something IÕm > pretty confused about. > HereÕs three statements that I currently think are true, but people > can agree or disagree with: > 1. No algebraic parity-violating statements can be constructed (In > other words, algebraic operations cannot break the symmetry.) All Well, the function sqrt (which satisfies (I) sqrt(z)*sqrt(z)=z for all z, and (II) sqrt(-1)=i) is algebraic in spirit but can be used to construct the parity-violating statement S(z) <-> z=sqrt(-1). I would need an exact definition of algebraic to tell whether statement 1. is correct. > 3. Every parity-violating statement contains an operator, function, > whatever you want to call it, whose definition depends on the > (debatably arbitrary) assignation of the symbol i to one of the roots > of x^2 + 1 = 0, or, if you prefer, the assignation of i to the point > (0,1) or (0,-1) on the complex plane. Your basic intuition here is correct: If you combine ingredients, none of which can lead to parity-violation, the result cannot either. [...] > BUT ... IÕm not clear if this is the standard definition of the term > transcendental. Many sources say that transcendental functions are > non-algebraic, but I havenÕt been able to find an explicit statement > that ALL non-algebraic functions are transcendental. For example, is > the parity-violating Im() function transcendental? Is my > parity-violating quadrant function transcendental? Transcendental is synonymous with non-algebraic, but the algebraic/non-algebraic distinction does not apply to anything you can think of. Just as it would be strange to call the number i non-negative, so it would be strange to call your quadrant function, which is not even continuous, non-algebraic. Mattias === Subject: Re: Definition of the imaginary unit > 1. No algebraic parity-violating statements can be constructed (In > other words, algebraic operations cannot break the symmetry.) All > Well, the function sqrt (which satisfies (I) sqrt(z)*sqrt(z)=z for all > z, and (II) sqrt(-1)=i) is algebraic in spirit but can be used to > construct the parity-violating statement S(z) <-> z=sqrt(-1). I would > need an exact definition of algebraic to tell whether statement 1. > is correct. The definition of algebraic that IÕm working with is Algebraic functions are those that involve only the use of addition, multiplication, division, exponentiation, and fractional powers. That would seem to include Sqrt. The statement Sqrt(-1) = i is an interesting one. HereÕs how I look at it. If one accepts that Sqrt has two values then the function is not parity-violating in this instance because Sqrt(-1) = {i, -i} is TRUE Sqrt(-1) = {-i, i} is TRUE If one considers Sqrt as a single-valued function then Sqrt(-1) has the single value i (presumably), and Sqrt(-1) = i is TRUE Sqrt(-1) = -i is FALSE which is parity-violating. This is because the assignation of i to one of the roots is then implicit in the DEFINITION of Sqrt (as per my earlier statement 3). Where this leaves us with the algebraic vs non-algebraic vs transcendental question I do not know. Any ideas? === Subject: Re: Definition of the imaginary unit > algebraic parity-violating statements can be constructed (In other > words, algebraic operations cannot break the symmetry.) All > Well, the function sqrt (which satisfies (I) sqrt(z)*sqrt(z)=z for all > z, and (II) sqrt(-1)=i) is algebraic in spirit but can be used to > construct the parity-violating statement S(z) <-> z=sqrt(-1). I would > need an exact definition of algebraic to tell whether statement 1. > is correct. > The definition of algebraic that IÕm working with is Algebraic > functions are > those that involve only the use of addition, multiplication, division, > exponentiation, and fractional powers. That would seem to include > Sqrt. > The statement Sqrt(-1) = i is an interesting one. HereÕs how I look at > it. > If one accepts that Sqrt has two values then the function is not > parity-violating in this instance because > Sqrt(-1) = {i, -i} is TRUE > Sqrt(-1) = {-i, i} is TRUE > If one considers Sqrt as a single-valued function then Sqrt(-1) has > the single value i (presumably), and > Sqrt(-1) = i is TRUE > Sqrt(-1) = -i is FALSE > which is parity-violating. This is because the assignation of i to one > of the roots is then implicit in the DEFINITION of Sqrt (as per my > earlier statement 3). > Where this leaves us with the algebraic vs non-algebraic vs > transcendental question I do not know. > Any ideas? I havenÕt been following this thread, but I may now be able to shed some light (or, failing that, to create more confusion). Algebaic can be used in two different ways. In the link below, I mention a text describing the ways. See The function Sqrt(x) is transcendental! at As you have already noted, symmetry between i and -i is broken by the principal-valued square root function Sqrt, while symmetry is preserved under the multivalued relation sqrt. Using the sense of the term algebraic which is appropriate to this thread, sqrt is algebraic, while Sqrt is not. Perhaps that resolves your problem. David Cantrell === Subject: Re: Definition of the imaginary unit > algebraic parity-violating statements can be constructed (In other > words, algebraic operations cannot break the symmetry.) All Well, the function sqrt (which satisfies (I) sqrt(z)*sqrt(z)=z for all > z, and (II) sqrt(-1)=i) is algebraic in spirit but can be used to > construct the parity-violating statement S(z) <-> z=sqrt(-1). I would > need an exact definition of algebraic to tell whether statement 1. > is correct. > The definition of algebraic that IÕm working with is Algebraic > functions are > those that involve only the use of addition, multiplication, division, > exponentiation, and fractional powers. That would seem to include > Sqrt. > The statement Sqrt(-1) = i is an interesting one. HereÕs how I look at > it. > If one accepts that Sqrt has two values then the function is not > parity-violating in this instance because > Sqrt(-1) = {i, -i} is TRUE > Sqrt(-1) = {-i, i} is TRUE > If one considers Sqrt as a single-valued function then Sqrt(-1) has > the single value i (presumably), and > Sqrt(-1) = i is TRUE > Sqrt(-1) = -i is FALSE > which is parity-violating. This is because the assignation of i to one > of the roots is then implicit in the DEFINITION of Sqrt (as per my > earlier statement 3). > Where this leaves us with the algebraic vs non-algebraic vs > transcendental question I do not know. > Any ideas? > I havenÕt been following this thread, but I may now be able to shed some > light (or, failing that, to create more confusion). > Algebaic can be used in two different ways. In the link below, I mention > a text describing the ways. See The function Sqrt(x) is transcendental! at > As you have already noted, symmetry between i and -i is broken by the > principal-valued square root function Sqrt, while symmetry is preserved > under the multivalued relation sqrt. Using the sense of the term > algebraic which is appropriate to this thread, sqrt is algebraic, while > Sqrt is not. Perhaps that resolves your problem. > David Cantrell Yes, that would certainly resolve it. I guess the only downside is that this solution makes the square root function (i.e. the single-valued function) non-algebraic, which contradicts quite a few sources IÕve found. algebraic vs transcendental, and it this is definitely a terminological minefield for the unwary! === Subject: Re: Definition of the imaginary unit days. My association with the Department is that of an alumnus. >> Look at calculus textbooks. Some of them have the words Early >> Transcendentals on them. This refers to the early introduction of >> functions such as the logarithms, the exponential functions, and most >> importantly, the trigonometric and inverse trigonometric >> functions. They are transcendealt functions. Algebraic functions are >> those that involve only the use of addition, multiplication, division, >> exponentiation, and fractional powers. The arg function in complex >> analysis is defined, usually, via a branch of the inverse tangent, a >> transcendental function. >Copy that. IÕm actually pretty familiar with all the ... erm .... >familiar transcendental functions, and the basic calculus associated >with them. Just didnÕt make the contrast transcendental vs >algebraic here. I think of everything like arctan(x), sin(x), exp(x) >etc. as just algebra. (Comes from having no formal training you >know!) >Let me coin the term parity violation for the situation when >interchanging i and -i turns a true statement into a false one, or >vice versa. (I donÕt know the correct term, or even if there is one.) Fair enough. >Most familiar transcendental functions can be (or, if you prefer, ARE) >defined as the sum of an infinite series of algebraic terms. I canÕt >currently see how an infinite sum can violate parity when none of its >individual (algebraic) terms do. Again, you seem to be confusing things. When I said that i and -i are algebraically indistinguishable, we are talking about certain kinds of statements; infinite sums involve limits, and limits are not algebraic statements in this sense. When you say an infinite sum vs the individual terms, the statement is really meaningless. If you are defining the value of, say, arctan, via an infinite series, then you are saying that arctan(x) is the limit of the values of the partial sums. You can certainly talk about the partial sums. But what statement are you making about the partial sum? The only statement you could make about the partial sums without involving the value of the limit would be, perhaps, that the partial sums form a Cauchy-sequence. Even that statement is already non-algebraic, since it involves the notion of distance and proximity in the complex plane, which is an analytic (not algebraic) concept. So what statement are you making about partial sums which is algebraic and does not violate parity? > Though I DO know that some very weird >and counterintuitive things can happen with infinite sums. Maybe this >one of them? Before we can talk about whether something is weird and counterintuitive, we must have a clear idea of just what it is that we are talking about. What is it that you are talking about? What algebraic statement are you making about the partial sums, and what statement about the infinite series? And how are you defining convergence of the infinite series without first fixing the complex plane in some way? >Right now, I actually think that the arg function causes a parity >violation NOT because itÕs transcendental, but because its definition >depends on the handedness of the complex plane (which, debatably, is >arbitrary). I did not say that the transcendence of arctan (or of arg) is the ->cause<- of the violation. The point was that the violation by arg does NOT invalidate the statement that i and -i are algebraically indistinguishable because arg is not an algebraic function. The transcendental nature of arg is not what ->causes<- the violation; it is what ->allows<- the violation to happen. Do you see the difference there? -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Definition of the imaginary unit I am having problems posting this. It may appear multiple times. Google groups is as buggy as ... well ... a thing that is very buggy. >> Look at calculus textbooks. Some of them have the words Early >> Transcendentals on them. This refers to the early introduction of >> functions such as the logarithms, the exponential functions, and most >> importantly, the trigonometric and inverse trigonometric >> functions. They are transcendealt functions. Algebraic functions are >> those that involve only the use of addition, multiplication, division, >> exponentiation, and fractional powers. The arg function in complex >> analysis is defined, usually, via a branch of the inverse tangent, a >> transcendental function. >> >Copy that. IÕm actually pretty familiar with all the ... erm .... >familiar transcendental functions, and the basic calculus associated >with them. Just didnÕt make the contrast transcendental vs >algebraic here. I think of everything like arctan(x), sin(x), exp(x) >etc. as just algebra. (Comes from having no formal training you >know!) >Let me coin the term parity violation for the situation when >interchanging i and -i turns a true statement into a false one, or >vice versa. (I donÕt know the correct term, or even if there is one.) > Fair enough. >Most familiar transcendental functions can be (or, if you prefer, ARE) >defined as the sum of an infinite series of algebraic terms. I canÕt >currently see how an infinite sum can violate parity when none of its >individual (algebraic) terms do. > Again, you seem to be confusing things. When I said that i and -i are > algebraically indistinguishable, we are talking about certain kinds of > statements; infinite sums involve limits, and limits are not algebraic > statements in this sense. > When you say an infinite sum vs the individual terms, the > statement is really meaningless. If you are defining the value of, > say, arctan, via an infinite series, then you are saying that > arctan(x) is the limit of the values of the partial sums. > You can certainly talk about the partial sums. But what statement are > you making about the partial sum? The only statement you could make > about the partial sums without involving the value of the limit would > be, perhaps, that the partial sums form a Cauchy-sequence. Even that > statement is already non-algebraic, since it involves the notion of > distance and proximity in the complex plane, which is an analytic (not > algebraic) concept. So what statement are you making about partial > sums which is algebraic and does not violate parity? > Though I DO know that some very weird >and counterintuitive things can happen with infinite sums. Maybe this >one of them? > Before we can talk about whether something is weird and > counterintuitive, we must have a clear idea of just what it is that we > are talking about. What is it that you are talking about? What > algebraic statement are you making about the partial sums, and what > statement about the infinite series? And how are you defining > convergence of the infinite series without first fixing the complex > plane in some way? Let me try to clarify. Consider the general statement: Finite algebraic expression 1 = Finite algebraic expression 2 (1) (By finite I mean an expression consisting of a finite number of algebraic operations. Throughout the remainder of this I will also assume that EVERYWHERE we are considering complex-valued expressions and functions. Obvious from the context, but some other things IÕve thought were obvious from the context havenÕt been!) it involves only algebraic operations. Correct? Now introduce one or more transcendental functions into the mix - maybe the familiar ones, maybe any others you choose - PROVIDED that the function(s) can be defined as the limit of an infinite sequence of algebraic operations. (DonÕt know how big a provided that might be.) This gives, in general Infinite algebraic expression 1 = Infinite algebraic expression 2 (2) (By infinite I mean the limit of an infinite sequence of algebraic operations.) If (2) can be parity-violating but (1) canÕt, then this is the possible weird and wonderful thing about infinite sums I was talking about. My current view is that it is impossible for (2) to violate parity, but proving it is well beyond me. If you know otherwise then IÕd be interested to know! In particular, if you can show me a parity-violating statement of the form Transcendental expression #1 = Transcendental expression #2 using only transcendental functions defined (or definable) as the limit of an infinite sum of algebraic expressions, then my theory is demolished. >Right now, I actually think that the arg function causes a parity >violation NOT because itÕs transcendental, but because its definition >depends on the handedness of the complex plane (which, debatably, is >arbitrary). > I did not say that the transcendence of arctan (or of arg) is the > ->cause<- of the violation. The point was that the violation by arg > does NOT invalidate the statement that i and -i are algebraically > indistinguishable because arg is not an algebraic function. Agreed. > The > transcendental nature of arg is not what ->causes<- the violation; it > is what ->allows<- the violation to happen. Do you see the difference > there? That depends on the definition of transcendental. If transcendental literally just means non-algebraic then my quadrant function is transcendental (because quadrant(z) cannot be defined in terms of algebraic operations on z), and I agree. (I note that this definition would also make such functions as abs, ßoor etc. transcendental. IÕve never considered them to be so, but you may be able to correct me on this one. Could be just another terminological blunder on my part.) If, however, my quadrant function isnÕt transcendental then no, I donÕt understand what the property of transcendence has to do with it. Whatever the case, I think that the transcendental nature of the (vanilla) ARCTAN function is irrelevant to the parity-violation of arg. As I say, my quadrant function has nothing to do with arctan and yet exhibits an identical parity-violation. Let me introduce another example which may help illustrate my point further. Define the function f of some complex number by f(a + ib) = b (3) This statement is TRUE (by definition). Swap i with -i and you get f(a - ib) = b which is FALSE. Parity is violated. My contention is that this is because (like arg and quadrant) the choice between i and -i is implicit in the DEFINITION of f. (Question: is f transcendental?) To help understand this, IÕll extend your Martian metaphor. Suppose that somewhere out in space the complex numbers exist and are uniquely defined. That is, the complex numbers a + ib and a - ib are distinct and distinguishable. (Actually, I am not currently entirely convinced that they ARE distinguishable, either by us or the Martians. Ref KPÕs response: The mathematics assures us that the equation z^2 + 1 = 0 has two solutions, but it treats them even-handedly, giving us no way to stick a special label on one of them rather than the other.) But anyway, suppose that here on earth we DO know which of this pair we call a + ib and which we call a - ib. The left-handed Martians do the opposite to us. Now pull a unique complex number, z, out of space. We call it a + ib, and the left-handed Martians call it a - ib. To us, f(z) = b but to the Martians f(z) = -b depends on the choice of i vs -i. It depends on whether we call the complex number that we picked out of space a + ib or a - ib, and THAT is why (3) is parity-violating. I am ASSUMING that the definitions of the familiar transcendental functions (exp, log, trig, inverse trig, hyperbolic etc.) do NOT depend on the choice of i vs -i, and are therefore NOT parity-violating. === Subject: Re: Definition of the imaginary unit I am having problems posting this. It may appear multiple times. Google groups is as buggy as ... well ... a thing that is very buggy. >> Look at calculus textbooks. Some of them have the words Early >> Transcendentals on them. This refers to the early introduction of >> functions such as the logarithms, the exponential functions, and most >> importantly, the trigonometric and inverse trigonometric >> functions. They are transcendealt functions. Algebraic functions are >> those that involve only the use of addition, multiplication, division, >> exponentiation, and fractional powers. The arg function in complex >> analysis is defined, usually, via a branch of the inverse tangent, a >> transcendental function. >> >Copy that. IÕm actually pretty familiar with all the ... erm .... >familiar transcendental functions, and the basic calculus associated >with them. Just didnÕt make the contrast transcendental vs >algebraic here. I think of everything like arctan(x), sin(x), exp(x) >etc. as just algebra. (Comes from having no formal training you >know!) >Let me coin the term parity violation for the situation when >interchanging i and -i turns a true statement into a false one, or >vice versa. (I donÕt know the correct term, or even if there is one.) > Fair enough. >Most familiar transcendental functions can be (or, if you prefer, ARE) >defined as the sum of an infinite series of algebraic terms. I canÕt >currently see how an infinite sum can violate parity when none of its >individual (algebraic) terms do. > Again, you seem to be confusing things. When I said that i and -i are > algebraically indistinguishable, we are talking about certain kinds of > statements; infinite sums involve limits, and limits are not algebraic > statements in this sense. > When you say an infinite sum vs the individual terms, the > statement is really meaningless. If you are defining the value of, > say, arctan, via an infinite series, then you are saying that > arctan(x) is the limit of the values of the partial sums. > You can certainly talk about the partial sums. But what statement are > you making about the partial sum? The only statement you could make > about the partial sums without involving the value of the limit would > be, perhaps, that the partial sums form a Cauchy-sequence. Even that > statement is already non-algebraic, since it involves the notion of > distance and proximity in the complex plane, which is an analytic (not > algebraic) concept. So what statement are you making about partial > sums which is algebraic and does not violate parity? > Though I DO know that some very weird >and counterintuitive things can happen with infinite sums. Maybe this >one of them? > Before we can talk about whether something is weird and > counterintuitive, we must have a clear idea of just what it is that we > are talking about. What is it that you are talking about? What > algebraic statement are you making about the partial sums, and what > statement about the infinite series? And how are you defining > convergence of the infinite series without first fixing the complex > plane in some way? Let me try to clarify. Consider the general statement: Finite algebraic expression 1 = Finite algebraic expression 2 (1) (By finite I mean an expression consisting of a finite number of algebraic operations. Throughout the remainder of this I will also assume that EVERYWHERE we are considering complex-valued expressions and functions. Obvious from the context, but some other things IÕve thought were obvious from the context havenÕt been!) it involves only algebraic operations. Correct? Now introduce one or more transcendental functions into the mix - maybe the familiar ones, maybe any others you choose - PROVIDED that the function(s) can be defined as the limit of an infinite sequence of algebraic operations. (DonÕt know how big a provided that might be.) This gives, in general Infinite algebraic expression 1 = Infinite algebraic expression 2 (2) (By infinite I mean the limit of an infinite sequence of algebraic operations.) If (2) can be parity-violating but (1) canÕt, then this is the possible weird and wonderful thing about infinite sums I was talking about. My current view is that it is impossible for (2) to violate parity, but proving it is well beyond me. If you know otherwise then IÕd be interested to know! In particular, if you can show me a parity-violating statement of the form Transcendental expression #1 = Transcendental expression #2 using only transcendental functions defined (or definable) as the limit of an infinite sum of algebraic expressions, then my theory is demolished. >Right now, I actually think that the arg function causes a parity >violation NOT because itÕs transcendental, but because its definition >depends on the handedness of the complex plane (which, debatably, is >arbitrary). > I did not say that the transcendence of arctan (or of arg) is the > ->cause<- of the violation. The point was that the violation by arg > does NOT invalidate the statement that i and -i are algebraically > indistinguishable because arg is not an algebraic function. Agreed. > The > transcendental nature of arg is not what ->causes<- the violation; it > is what ->allows<- the violation to happen. Do you see the difference > there? That depends on the definition of transcendental. If transcendental literally just means non-algebraic then my quadrant function is transcendental (because quadrant(z) cannot be defined in terms of algebraic operations on z), and I agree. (I note that this definition would also make such functions as abs, ßoor etc. transcendental. IÕve never considered them to be so, but you may be able to correct me on this one. Could be just another terminological blunder on my part.) If, however, my quadrant function isnÕt transcendental then no, I donÕt understand what the property of transcendence has to do with it. Whatever the case, I think that the transcendental nature of the (vanilla) ARCTAN function is irrelevant to the parity-violation of arg. As I say, my quadrant function has nothing to do with arctan and yet exhibits an identical parity-violation. Let me introduce another example which may help illustrate my point further. Define the function f of some complex number by f(a + ib) = b (3) This statement is TRUE (by definition). Swap i with -i and you get f(a - ib) = b which is FALSE. Parity is violated. My contention is that this is because (like arg and quadrant) the choice between i and -i is implicit in the DEFINITION of f. (Question: is f transcendental?) To help understand this, IÕll extend your Martian metaphor. Suppose that somewhere out in space the complex numbers exist and are uniquely defined. That is, the complex numbers a + ib and a - ib are distinct and distinguishable. (Actually, I am not currently entirely convinced that they ARE distinguishable, either by us or the Martians. Ref KPÕs response: The mathematics assures us that the equation z^2 + 1 = 0 has two solutions, but it treats them even-handedly, giving us no way to stick a special label on one of them rather than the other.) But anyway, suppose that here on earth we DO know which of this pair we call a + ib and which we call a - ib. The left-handed Martians do the opposite to us. Now pull a unique complex number, z, out of space. We call it a + ib, and the left-handed Martians call it a - ib. To us, f(z) = b but to the Martians f(z) = -b depends on the choice of i vs -i. It depends on whether we call the complex number that we picked out of space a + ib or a - ib, and THAT is why (3) is parity-violating. I am ASSUMING that the definitions of the familiar transcendental functions (exp, log, trig, inverse trig, hyperbolic etc.) do NOT depend on the choice of i vs -i, and are therefore NOT parity-violating. === Subject: Re: Definition of the imaginary unit I am having problems posting this. It may appear multiple times. Google groups is as buggy as ... well ... a thing that is very buggy. >> Look at calculus textbooks. Some of them have the words Early >> Transcendentals on them. This refers to the early introduction of >> functions such as the logarithms, the exponential functions, and most >> importantly, the trigonometric and inverse trigonometric >> functions. They are transcendealt functions. Algebraic functions are >> those that involve only the use of addition, multiplication, division, >> exponentiation, and fractional powers. The arg function in complex >> analysis is defined, usually, via a branch of the inverse tangent, a >> transcendental function. >> >Copy that. IÕm actually pretty familiar with all the ... erm .... >familiar transcendental functions, and the basic calculus associated >with them. Just didnÕt make the contrast transcendental vs >algebraic here. I think of everything like arctan(x), sin(x), exp(x) >etc. as just algebra. (Comes from having no formal training you >know!) >Let me coin the term parity violation for the situation when >interchanging i and -i turns a true statement into a false one, or >vice versa. (I donÕt know the correct term, or even if there is one.) > Fair enough. >Most familiar transcendental functions can be (or, if you prefer, ARE) >defined as the sum of an infinite series of algebraic terms. I canÕt >currently see how an infinite sum can violate parity when none of its >individual (algebraic) terms do. > Again, you seem to be confusing things. When I said that i and -i are > algebraically indistinguishable, we are talking about certain kinds of > statements; infinite sums involve limits, and limits are not algebraic > statements in this sense. > When you say an infinite sum vs the individual terms, the > statement is really meaningless. If you are defining the value of, > say, arctan, via an infinite series, then you are saying that > arctan(x) is the limit of the values of the partial sums. > You can certainly talk about the partial sums. But what statement are > you making about the partial sum? The only statement you could make > about the partial sums without involving the value of the limit would > be, perhaps, that the partial sums form a Cauchy-sequence. Even that > statement is already non-algebraic, since it involves the notion of > distance and proximity in the complex plane, which is an analytic (not > algebraic) concept. So what statement are you making about partial > sums which is algebraic and does not violate parity? > Though I DO know that some very weird >and counterintuitive things can happen with infinite sums. Maybe this >one of them? > Before we can talk about whether something is weird and > counterintuitive, we must have a clear idea of just what it is that we > are talking about. What is it that you are talking about? What > algebraic statement are you making about the partial sums, and what > statement about the infinite series? And how are you defining > convergence of the infinite series without first fixing the complex > plane in some way? Let me try to clarify. Consider the general statement: Finite algebraic expression 1 = Finite algebraic expression 2 (1) (By finite I mean an expression consisting of a finite number of algebraic operations. Throughout the remainder of this I will also assume that EVERYWHERE we are considering complex-valued expressions and functions. Obvious from the context, but some other things IÕve thought were obvious from the context havenÕt been!) it involves only algebraic operations. Correct? Now introduce one or more transcendental functions into the mix - maybe the familiar ones, maybe any others you choose - PROVIDED that the function(s) can be defined as the limit of an infinite sequence of algebraic operations. (DonÕt know how big a provided that might be.) This gives, in general Infinite algebraic expression 1 = Infinite algebraic expression 2 (2) (By infinite I mean the limit of an infinite sequence of algebraic operations.) If (2) can be parity-violating but (1) canÕt, then this is the possible weird and wonderful thing about infinite sums I was talking about. My current view is that it is impossible for (2) to violate parity, but proving it is well beyond me. If you know otherwise then IÕd be interested to know! In particular, if you can show me a parity-violating statement of the form Transcendental expression #1 = Transcendental expression #2 using only transcendental functions defined (or definable) as the limit of an infinite sum of algebraic expressions, then my theory is demolished. >Right now, I actually think that the arg function causes a parity >violation NOT because itÕs transcendental, but because its definition >depends on the handedness of the complex plane (which, debatably, is >arbitrary). > I did not say that the transcendence of arctan (or of arg) is the > ->cause<- of the violation. The point was that the violation by arg > does NOT invalidate the statement that i and -i are algebraically > indistinguishable because arg is not an algebraic function. Agreed. > The > transcendental nature of arg is not what ->causes<- the violation; it > is what ->allows<- the violation to happen. Do you see the difference > there? That depends on the definition of transcendental. If transcendental literally just means non-algebraic then my quadrant function is transcendental (because quadrant(z) cannot be defined in terms of algebraic operations on z), and I agree. (I note that this definition would also make such functions as abs, ßoor etc. transcendental. IÕve never considered them to be so, but you may be able to correct me on this one. Could be just another terminological blunder on my part.) If, however, my quadrant function isnÕt transcendental then no, I donÕt understand what the property of transcendence has to do with it. Whatever the case, I think that the transcendental nature of the (vanilla) ARCTAN function is irrelevant to the parity-violation of arg. As I say, my quadrant function has nothing to do with arctan and yet exhibits an identical parity-violation. Let me introduce another example which may help illustrate my point further. Define the function f of some complex number by f(a + ib) = b (3) This statement is TRUE (by definition). Swap i with -i and you get f(a - ib) = b which is FALSE. Parity is violated. My contention is that this is because (like arg and quadrant) the choice between i and -i is implicit in the DEFINITION of f. (Question: is f transcendental?) To help understand this, IÕll extend your Martian metaphor. Suppose that somewhere out in space the complex numbers exist and are uniquely defined. That is, the complex numbers a + ib and a - ib are distinct and distinguishable. (Actually, I am not currently entirely convinced that they ARE distinguishable, either by us or the Martians. Ref KPÕs response: The mathematics assures us that the equation z^2 + 1 = 0 has two solutions, but it treats them even-handedly, giving us no way to stick a special label on one of them rather than the other.) But anyway, suppose that here on earth we DO know which of this pair we call a + ib and which we call a - ib. The left-handed Martians do the opposite to us. Now pull a unique complex number, z, out of space. We call it a + ib, and the left-handed Martians call it a - ib. To us, f(z) = b but to the Martians f(z) = -b depends on the choice of i vs -i. It depends on whether we call the complex number that we picked out of space a + ib or a - ib, and THAT is why (3) is parity-violating. I am ASSUMING that the definitions of the familiar transcendental functions (exp, log, trig, inverse trig, hyperbolic etc.) do NOT depend on the choice of i vs -i, and are therefore NOT parity-violating. === Subject: Re: Definition of the imaginary unit days. My association with the Department is that of an alumnus. [.snip.] >>Let me coin the term parity violation for the situation when >>interchanging i and -i turns a true statement into a false one, or >>vice versa. (I donÕt know the correct term, or even if there is one.) >> Fair enough. >>Most familiar transcendental functions can be (or, if you prefer, ARE) >>defined as the sum of an infinite series of algebraic terms. I canÕt >>currently see how an infinite sum can violate parity when none of its >>individual (algebraic) terms do. >> Again, you seem to be confusing things. When I said that i and -i are >> algebraically indistinguishable, we are talking about certain kinds of >> statements; infinite sums involve limits, and limits are not algebraic >> statements in this sense. >> When you say an infinite sum vs the individual terms, the >> statement is really meaningless. If you are defining the value of, >> say, arctan, via an infinite series, then you are saying that >> arctan(x) is the limit of the values of the partial sums. >> You can certainly talk about the partial sums. But what statement are >> you making about the partial sum? The only statement you could make >> about the partial sums without involving the value of the limit would >> be, perhaps, that the partial sums form a Cauchy-sequence. Even that >> statement is already non-algebraic, since it involves the notion of >> distance and proximity in the complex plane, which is an analytic (not >> algebraic) concept. So what statement are you making about partial >> sums which is algebraic and does not violate parity? >> Though I DO know that some very weird >>and counterintuitive things can happen with infinite sums. Maybe this >>one of them? >> Before we can talk about whether something is weird and >> counterintuitive, we must have a clear idea of just what it is that we >> are talking about. What is it that you are talking about? What >> algebraic statement are you making about the partial sums, and what >> statement about the infinite series? And how are you defining >> convergence of the infinite series without first fixing the complex >> plane in some way? >Let me try to clarify. Consider the general statement: > Finite algebraic expression 1 = Finite algebraic expression 2 >(1) >(By finite I mean an expression consisting of a finite number of >algebraic operations. Throughout the remainder of this I will also >assume that EVERYWHERE we are considering complex-valued expressions >and functions. Obvious from the context, but some other things IÕve >thought were obvious from the context havenÕt been!) >it involves only algebraic operations. Correct? >Now introduce one or more transcendental functions into the mix - >maybe the familiar ones, maybe any others you choose - PROVIDED that >the function(s) can be defined as the limit of an infinite sequence of >algebraic operations. (DonÕt know how big a provided that might be.) >This gives, in general > Infinite algebraic expression 1 = Infinite algebraic expression 2 > (2) >(By infinite I mean the limit of an infinite sequence of algebraic >operations.) You are already introducing horrible notation that is bound to cause confusion. A limit of an infinite sequence is ->not<- an algebraic expression; it is an ->analytic<- expression. The use of the adjective algebraic here is ->bound<- to cause all sorts of headaches when you try to square it with the statement that i and -i are algebraically indistinguishable. >If (2) can be parity-violating but (1) canÕt, then this is the >possible weird and wonderful thing about infinite sums I was talking >about. You are talking about ENTIRELY DIFFERENT ANIMALS. Just because you throw in the word algebraically, it doesnÕt mean that there is such a close relation between them. >My current view is that it is impossible for (2) to violate >parity, How can that possibly be your current view, if you already have that the arctan function gives you different values for i and -i, and is an instance of (2)? >In particular, if you can show me a parity-violating statement of the >form > Transcendental expression #1 = Transcendental expression #2 >using only transcendental functions defined (or definable) as the >limit of an infinite sum of algebraic expressions, then my theory is >demolished. You already did! >> The >> transcendental nature of arg is not what ->causes<- the violation; it >> is what ->allows<- the violation to happen. Do you see the difference >> there? >That depends on the definition of transcendental. No. It depends on the meaning of the word cause. Which you improperly used. Is English your native language? It isnÕt mine, and it is obvious we are having some serious problems communicating. I have the feeling that I say things clearly, and they wash off you like water off a duckÕs back. You pick a few drops, misinterpret them, misrepresent them, misstate them, and then wonder where things went wrong. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Definition of the imaginary unit > ... it is obvious we > are having some serious problems communicating. points in my last posting got across at all. It obviously just didnÕt make any sense to you. IÕm still interested in your views, so if you get time read it again and see if you can figure out what IÕm trying to say. If that proves impossible then no worries ... letÕs call it a day on this one! rgds matt === Subject: Re: Definition of the imaginary unit > Hello again > Recently, while contemplating the meaning of ... the meaning of ... > ... erm ... oh yes ... the meaning of exponentiation in the domain of > complex numbers, a thought suddenly struck me. > books I have propping up my wobbly table, so maybe someone can > enlighten me. > ItÕs usually said that i is defined as the square root of -1. But we > know that the equation > x^2 = -1 > has TWO solutions, which we would write as i and -i. > My question therefore is: > 1. How do we know which value to designate i, and which to designate > -i? > 2. If we DONÕT know, then almost any expression involving i would seem > to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is > there any easy way to understand why this never matters? The mapping x + i*y -> x - i*y is a field isomorphism, so that, as far as the field of complex numbers is concerned, there is no difference that makes any difference. === Subject: Re: Definition of the imaginary unit > ItÕs usually said that i is defined as the square root of -1. But we > know that the equation > x^2 = -1 > has TWO solutions, which we would write as i and -i. > My question therefore is: > 1. How do we know which value to designate i, and which to designate > -i? C = R[x]/(x^2 + 1) i = x + (x^2 + 1) -i = -x + (x^2 + 1) Naturally, thatÕs how. Otherwise youÕd get cross eyed looking at i = -x + (x^2 + 1) -i = x + (x^2 + 1) > 2. If we DONÕT know, then almost any expression involving i would seem > to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is > there any easy way to understand why this never matters? > The mapping x + i*y -> x - i*y is a field isomorphism, so that, as far > as the field of complex numbers is concerned, there is no difference > that makes any difference.