mm-4539 === Subject: Re: compact space > Is it possible to define a topology on N (the set of natual numbers) > topology does not work. > Is it possible to give N a compact Hausdorff topology? Consider {0,1} with the discrete topology: it is compact > Hausdorff. Consider the product {0,1}x{0,1}x{0,1}x... (countable > product) equipped with the product topology. It is compact Hausdorff > by Tychonoff's theorem. Any bijection between > {0,1}x{0,1}x{0,1}x... and N yields a compact Hausdorff topology on N. pg. Easier yet: the range of a convergent sequence with distinct points, together with its limit, is a compact countable subset of the real line. Make up your own bijection. === Subject: help with proof DLM/4.0_1.0.0.1,gzip(gfe),gzip(gfe) I am having a hard time writing a probability proof of the following result: If the sequence Yn, n=1,2.... is bounded in probability and if {Cn} is a sequence of random variables tending to 0 in probability, then Cn Yn converges in probability to 0. I think I should start with, For every epsilon, there exists n* such that, for n>n*, |Cn| n0, P(|Yn| =< K) > 1 - delta This leads to There exists n0 such that, P(|Yn| > K) < delta Now I am thinking I should write P(|Cn Yn| > K|Cn| ) < delta That is where I am stuck. I would appreciate some help. === Subject: Re: help with proof > I am having a hard time writing a probability proof of the following > result: > If the sequence Yn, n=1,2.... is bounded in probability and if {Cn} is > a sequence of random variables tending to 0 in probability, then > Cn Yn converges in probability to 0. I think I should start with, > For every epsilon, there exists n* such that, for n>n*, |Cn| For any delta, there exists K, and there exists n0 such that, for all > n > n0, P(|Yn| =< K) > 1 - delta This leads to > There exists n0 such that, P(|Yn| > K) < delta Now I am thinking I should write > P(|Cn Yn| > K|Cn| ) < delta That is where I am stuck. I would appreciate some help. Hint: |Cn Yn| > K epsilon implies |Cn| > epsilon or |Yn| > K. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Solution Manual (MODERN CONTROL SYSTEM 4th Edition by OGATA ) If you need a Solution manual immediately . Please Email me at MANUALSELLER (at) GMAIL(dot)COM . donÍt give up Email Me. All the manual solutions author names are capitalize for you to find easily . ** donÍt post your message here ** I wonÍt be here again . ** email me at manualseller at gmail (dot)com ** Mechanical Vibration 3th Edition by RAO !!! including project ( 95% same as 4th ed ) Fluid Mechanics 6th edition WHITE !! Thermodynamic 6th Edition by CENGEL ! ! ! Fundamentals of Engineering Thermodynamics (Solutions Manual) (M. J. Moran & H. N. Shapiro Solution DIFFERENTIAL EQUATIONS -classic 5th edition DENNIS G. ZILL! Engineering Mechanics Dynamics 11th Edition by HIBBELER !!! MicroComputer Engineering Third Edition by GENE.H.MILLER 68HC11 Semiconductor devices--physics and technology [SZE, s m ] !! Control Systems Engineering -4th edition NISE !! Introduction To Fluid Mechanics 5th edition FOX ! ! ! Adaptive Control 2nd Edition by KARL.J.ASTROM !!! Antenna for all application 3rd Edition by JOHN . D . KRAUS !!! Applied Numerical Analysis 7Ed CURTIS F.GERALD PATRICK O WHEATLEY ! Communication Systems Engineering 2nd Edition by PROAKIS J !! Design of Analog CMOS Integrated Circuits solutions McGraw RAZAVI !! Digital Communications 4th Edition Solution manual by PROAKIS !! Digital Image Processing 2nd Edition by GONZALEZ !!! Digital Integrated Circuits by 2nd Edition by RABAEY ! ! ! Digital Signal Processing 2nd Edition by MITRA ! ! ! Digital Signal Processing by THOMAS J CAVICCHI ! ! Introduction to electric circuits_6th edition DORF SVABODA !! Engineering Circuit Analysis 6TH EDITION by HAYT !! Engineering Mathematics 4th EDITION by JOHN BIRD !! Fundamentals of Electric Circuits 2nd.ed.by C.K.ALEXANDER M.N.O.SADIKU Introduction to electrodynamics 3RD EdITION by DAVID GRIFFITHS !!! Linear circuit analysis 2nd Edition By R. A.DECARLO AND P. LIN !! Microwave and RF design of wireless systems by POZAR !! Microwave Engineering 2e David M POZAR !! Modern Control Engineering by K OGATA !!! Modern Digital and Analog Communications Systems B P LATHI !!! Principles and Applications of Electrical Engineering by GIORGIO RIZZON I !! Probability by SHELDON M. ROSE ! Probability Random Variables And Stochastic Processes edition 4th by PAPOULIS ! RF circuit Design Theory and Application by LUDWIG BRETCHKO !! Semiconductor Device Fundamentals by ROBERT . F.PIERRET !! Solid State Electronic Device by BEN STREETMAN !! Automatic Control Systems 8Ed KUO AND GOLNARAGHI ! Electronic Physics STRABMAN !!! Communication Systems 4th Edition by SIMON HAYKIN !! Chemical And Engineering Thermodynamics 3Edition WILEY !! Calculus of Variations solution manual [CapitalEth] RUSSAK Communication Systems 4TH Edition by CARLSON !! Control Systems Engineering -4th edition NISE !! Digital Signal Processing by SANJIT K MITRA COMPUTER BASED APPROCHED Engineering Fluid Mechanics [CapitalEth] 7TH edition by CLANTON T CROWE ! ! Engineering Electromagnetics 6th Edition William H HAYT JR, JOHN A BUCK !! Fluids 8th edition DONALD ELGER and CLAYTON CROWE ! ! Engineering Mechanics Dynamics 11th Edition by HIBBELER !!! Mechanics Of Materials 4TH edition , HIBBELER ! ! ! Mechanics of Materials by BEER, JOHSTON & DEWOLF ! ! Microelectronic Circuits SEDRA 5thed SEDAT !! Microwave and Rf Design of Wireless Systems [CapitalEth] D.M. POZAR ! ! Organic Chemistry - Carey 5th ed Prentice hall - c++ algorithms for realtime DSP !! Probability & Statistic WALPOLE !! Quantum Mechanics [CapitalEth] YUNG KUO LIM !! Fundamentals of electric circuits 2nd ed SADIKU !! System Dynamics 3rd edition [CapitalEth] KATSUHIKO OGATA !! Thomas' Calculus 11th ed - G Thomas, M WEIR, et al !! VECTOR mechanic STATICS 7ed BY BEER !! Applied Statistics and Probabilty For Engineer 3e DOUGLAS C. MONTGEMORY AND G.C RUNGER Electromagnetic Ulaby Electric_machinery_fundamentals by STEPHEN J. CHAPMAN !! Elementary Differential Equations 8E by Boyce !! Vector Mechanics 7 ED BY BEER !! Solution Manual Linear Systems And Signals B P LATHI !! Mechanics and Fluid MASSEY !! Still if you could not find out the solution above list . please email me at ** MANUALSELLER@GMAIL.COM ** OR **check this out ** http://www.manualseller.com** click on solution manuals for updated solutions !! === Subject: Solution Manual List posting-account=EMSFowoAAAAWVX65y-y79hF-VVqOrx2m Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ),gzip(gfe),gzip(gfe) If you need a Solution manual immediately . Please Email me at MANUALSELLER (at) GMAIL(dot)COM . don't give up Email Me. All the manual solutions author names are capitalize for you to find easily . ** don't post your message here ** I won't be here again . ** email me at manualseller at gmail (dot)com ** Mechanical Vibration 3th Edition by RAO !!! including project ( 95% same as 4th ed ) Fluid Mechanics 6th edition WHITE !! Thermodynamic 6th Edition by CENGEL ! ! ! Fundamentals of Engineering Thermodynamics (Solutions Manual) (M. J. Moran & H. N. Shapiro Solution DIFFERENTIAL EQUATIONS -classic 5th edition DENNIS G. ZILL! Engineering Mechanics Dynamics 11th Edition by HIBBELER !!! MicroComputer Engineering Third Edition by GENE.H.MILLER 68HC11 Semiconductor devices--physics and technology [SZE, s m ] !! Control Systems Engineering -4th edition NISE !! Introduction To Fluid Mechanics 5th edition FOX ! ! ! Adaptive Control 2nd Edition by KARL.J.ASTROM !!! Advanced Modern Engineering Mathematics 3rd Edition GLYN Antenna for all application 3rd Edition by JOHN . D . KRAUS !!! Applied Numerical Analysis 7Ed CURTIS F.GERALD PATRICK O WHEATLEY ! Communication Systems Engineering 2nd Edition by PROAKIS J !! Design of Analog CMOS Integrated Circuits solutions McGraw RAZAVI !! Digital Communications 4th Edition Solution manual by PROAKIS !! Digital Image Processing 2nd Edition by GONZALEZ !!! Digital Integrated Circuits by 2nd Edition by RABAEY ! ! ! Digital Signal Processing 2nd Edition by MITRA ! ! ! Digital Signal Processing by THOMAS J CAVICCHI ! ! Introduction to electric circuits_6th edition DORF SVABODA !! Electronic Circuit Analysis and Design 2nd Edition by DONALD .A. Engineering Circuit Analysis 6TH EDITION by HAYT !! Engineering Mathematics 4th EDITION by JOHN BIRD !! Fundamentals of Electric Circuits 2nd.ed.by C.K.ALEXANDER M.N.O.SADIKU Introduction to electrodynamics 3RD EdITION by DAVID GRIFFITHS !!! Linear circuit analysis 2nd Edition By R. A.DECARLO AND P. LIN !! Microwave and RF design of wireless systems by POZAR !! Microwave Engineering 2e David M POZAR !! Modern Control Engineering by K OGATA !!! Modern Digital and Analog Communications Systems B P LATHI !!! Principles and Applications of Electrical Engineering by GIORGIO RIZZON I !! Probability by SHELDON M. ROSE ! Probability Random Variables And Stochastic Processes edition 4th by PAPOULIS ! RF circuit Design Theory and Application by LUDWIG BRETCHKO !! Semiconductor Device Fundamentals by ROBERT . F.PIERRET !! Solid State Electronic Device by BEN STREETMAN !! Automatic Control Systems 8Ed KUO AND GOLNARAGHI ! Manual Device Electronics for Integrated Circuits 3Edition MULLER Electronic Physics STRABMAN !!! Communication Systems 4th Edition by SIMON HAYKIN !! Chemical And Engineering Thermodynamics 3Edition WILEY !! Calculus of Variations solution manual - RUSSAK Communication Systems 4TH Edition by CARLSON !! Control Systems Engineering -4th edition NISE !! Digital Signal Processing by SANJIT K MITRA COMPUTER BASED APPROCHED Engineering Fluid Mechanics - 7TH edition by CLANTON T CROWE ! ! Engineering Electromagnetics 6th Edition William H HAYT JR, JOHN A BUCK !! Fluids 8th edition DONALD ELGER and CLAYTON CROWE ! ! Engineering Mechanics Dynamics 11th Edition by HIBBELER !!! Mechanics Of Materials 4TH edition , HIBBELER ! ! ! Mechanics of Materials by BEER, JOHSTON & DEWOLF ! ! Microelectronic Circuits SEDRA 5thed SEDAT !! Microwave and Rf Design of Wireless Systems - D.M. POZAR ! ! Organic Chemistry - Carey 5th ed Prentice hall - c++ algorithms for realtime DSP !! Probability & Statistic WALPOLE !! Quantum Mechanics - YUNG KUO LIM !! Fundamentals of electric circuits 2nd ed SADIKU !! System Dynamics 3rd edition - KATSUHIKO OGATA !! Thomas' Calculus 11th ed - G Thomas, M WEIR, et al !! VECTOR mechanic STATICS 7ed BY BEER !! Applied Statistics and Probabilty For Engineer 3e DOUGLAS C. MONTGEMORY AND G.C RUNGER Electromagnetic Ulaby Electric_machinery_fundamentals by STEPHEN J. CHAPMAN !! Elementary Differential Equations 8E by Boyce !! Vector Mechanics 7 ED BY BEER !! Solution Manual Linear Systems And Signals B P LATHI !! Mechanics and Fluid MASSEY !! Still if you could not find out the solution above list . please email me at ** MANUALSELLER @GMAIL.COM ** OR **check this out ** http://www.manualseller.com** click on solution manuals for updated solutions !! === Subject: Re: hypercomplex numbers such that j^2=i or -i Cc: rokirovka@gmail.com posting-account=s4dLjAkAAAA9r-dpsX5X1Hsh9_J24ktp Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) (My apologies if this is posted twice -- browser froze while posting.) Here is another motivation for using the j-number algebra. In the j-number algebra of numbers a + bi +- cj +- dij, where +-j^2 = +-i, the square roots of i, and of -i, are {+-j, +-ij}. But this expression by itself is hardly a motivation for using the j- number algebra, because the square roots of i and of -i have a perfectly elegant expression within the complex numbers themselves: (+-z)^2 = i, (+-z-conjugate)^2 = -i = i-conjugate, where z=e^(i(pi)/4), which is quite elegant when written in a font that can mark the bar over z to express the conjugate. The conjugation of the square roots of i matches the conjugation of i. But the extension of the j-number algebra to an 8-dimensional k- algebra does provide a motivation for using this system of algebras to express the roots of i and -i. We define the elements of the k-algebra to have the form x1 + x2i +- x3j +- x4ij +- x5k +- x6ik +- x7jk +- x8ijk, where +-j^2 = +-i and +-k^2 = +-j. Like the j-number algebra, the k-number algebra is commutative and all elements have reciprocals. Also like the j-number algebra, the k- number algebra is not a field because the binary operations of k- number addition and multiplication take two elements and return multiple elements, not one element. In the k-number algebra we can express the 4th roots of i and -i in an elegant form that is lacking within the complex numbers themselves. First we note that in the k-number algebra, the square roots of +-j are {+-k, +-ik}. Also, the square roots of +-ij are {+-jk, +-ijk}. Thus in the k-number algebra the 4th roots of i and of -i are {+-k, +-ik, +-jk, +-ijk}. This is much more elegant and concise than any expression of the complex 4th roots of i and of -i, because negation and conjugation of a complex number do not suffice to express the 4th roots of i or of -i the way they suffice to express the square roots. The point is that the real and imaginary coefficients of different complex 4th roots of i and of -i are transposed. To express that fact concisely, we have to write out z as (a,b), which is less concise. Further, setting z=e^(i(pi)/8), we have (+-(a,b))^4 = i and (+-(a,b)-conjugate)^4 = -i = i-conjugate, but for the other 4th roots of i and of -i we have (+-(b,a))^4 = -i = i-conjugate and (+-(b,a)-conjugate)^4 = i. Thus the conjugation of the 4th roots of i no longer matches the conjugation of i. The 8-dimensional k-number algebra, on the other hand, allows us to express the 4th roots of i and of -i in the much more concise and elegant form {+-k, +-ik, +-jk, +-ijk}. === Subject: Re: hypercomplex numbers such that j^2=i or -i Cc: rokirovka@gmail.com posting-account=s4dLjAkAAAA9r-dpsX5X1Hsh9_J24ktp Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) There are many systems of hypercomplex numbers with various > properties, but all the hypercomplex number systems that I have seen > have the common feature that the square of each imaginary unit > (usually labelled i, j, k, or e) is a real number: -1 or 1 or 0. Has any hypercomplex number system been created in which the square of > some imaginary unit, let us call it j, equals i or -i (j^2=i, j^2=-i)? Of course such a construct immediately runs into the problem that i > and -i each have two perfectly good square roots within the complex > number system itself. But +1 has two perfectly good square roots > within the real number system, and that didn't prevent the creation of > algebras such as the split-complex numbers x+yj, where j^2=1, or the > hyperbolic quaternions, the forerunner of the Minkowski space of > special relativity, where i^2=j^2=k^2=1. The natural idea for an extension of complex numbers with an > additional imaginary base j such that j^2=i or j^2=-i, would be a > system of numbers of the form w+zj, where w and z are complex numbers. > That is, w+zj = a + bi + cj + dij. The problem is that setting j^2=i > or j^2=-i leads to ambiguities in the +- signs of some values when we > start multiplying these numbers by each other. But if it is not possible to resolve this problem by eliminating all > the +- sign ambiguities, perhaps it is possible to resolve the problem > by making the +- sign ambiguities a general feature of the system. > That is, define the unit j such that +j = -j, and +-j^2 = +-i. Then > every power of j will have ambiguous +- sign, and every product of j > and a real or complex number will also have ambiguous +- sign. The > equality j = -j may appear absurd at first glance, but 1/i = -i also > appears absurd at first glance, yet it holds for the standard > imaginary unit i. Thus a + bi + cj + dij = a + bi - cj - dij = a + bi + cj - dij = > a + bi - cj + dij, > where a,b,c,d are real numbers, i is the imaginary unit with the > standard definition, and j is the imaginary unit as defined above. > Note that, if we say w=a+bi and z=c+di, then +z, -z, and +-conjugate > of z are all equivalent in this system after multiplication by j. Once we accept these sign ambiguities throughout the system, we have > multiplication that is associative, distributive, and commutative. The > complex numbers, with all their properties and without any sign > ambiguities, are still contained within this system as the special > case where +-c = +-d = +-0. (Note that +-0 is the one case of +- sign > ambiguity within the real and complex number system.) While the arbitrary hypercomplex number a + bi +- cj +- dij in this > system is equivalent to four different sets of real numbers {a,b,c,d}, > the product of two such hypercomplex numbers > (a1 + b1i +- c1j +- d1ij)*(a2 + b2i +- c2j +- d2ij) > is equivalent to 2^12=4096 different sets of real numbers, > corresponding to the 12 terms in the product with coefficients with > ambiguous sign: +-c1d2, +-d1c2, +-c1c2i, +-d1d2i, +-a1c2j, +-b1d2j, +- > c1a2j, +-d1b2j, +-a1d2ij, +-b1c2ij, +-c1b2ij, +-d1a2ij. (To determine > which coefficients belong to the real part, imaginary part, j-part, > and ij-part of the product, note that since i^2=-1 and j^2=+-i, > therefore ij*i=+-j, ij*j=i*(+-i)=+-1, and ij*ij=(-1)*(+-i)=+-i.) Of course there is no well-defined norm on this algebra, but we can > define the multi-valued |a+bi+-cj+-dij| = sqrt(a^2+b^2+-c^2+-d^2). > Note that the mean value of the 4 values of this pseudo-norm equals > the norm of the complex number a+bi. When one takes the multi-valued > pseudo-norm of a product of two such hypercomplex numbers, the number > of values becomes astronomical. Some terms are 2-valued because they > occur once with ambiguous sign. Other terms occur twice with ambiguous > sign, so they are 3-valued: their sum may be two positives, two > terms occur four times with ambiguous sign, so they are 5-valued: > their sum may be four positive, four negatives, three positives and a > negative, three negatives and a positive, or two positives and two > pseudo-norm of a product of two such hypercomplex numbers = > (2^12)*(3^16)*(5^3) = ~2.2*10^13, over 22 trillion values. The mean > value of all these values still equals the norm of the product of the > complex parts of the two numbers, a1 + b1i and a2 + b2i. It is > apparent that the distribution of all the possible values is a > Gaussian distribution with the peak at the mean value. Jeff Caveney I will show an example of how these hypercomplex numbers, which I will > call j-numbers for brevity and to distinguish them from other systems > of hypercomplex numbers, can be used to express properties of > functions of complex numbers in an elegant way. The fact that a j-number (a + bi +- cj +- dij) has multi-valued j-part > and ij-part and single-valued real part and imaginary part, means that > we can use purely algebraic operations on the j-numbers to produce > expressions that represent multi-valued functions of complex numbers. > In particular, I will show that the multi-valued logarithm of a > complex number can be represented by a linear combination of squares > of j-numbers. I will begin with the most important and basic examples, > the complex logarithms of the complex units 1, -1, i and -i. Note first that the real part of the logarithm of a complex number is > always single-valued, while the imaginary part is multi-valued: log z > = log z +- 2n(pi)i. Thus to represent the multi-valued logarithm of a > complex number, we select a j-number such that its square will have a > single-valued real part and a multi-valued imaginary part. Such a j- > number cannot have both a nonzero j-part and a nonzero ij-part, > because their product would introduce a multi-valued real part into > the square, since (+-ij)*(+-j) = +-i*i = +-1. But if it has nonzero j- > part and no ij-part, then whatever its real and imaginary parts, its > square must have a single-valued real part and a multi-valued > imaginary part. This is precisely the form of the logarithm of a > complex number. We begin by showing the linear combinations of squares of j-numbers > that generate the logarithms of 1 and -1. Since log(1) = 0 +- 2n(pi)i > and log(-1) = (pi)i +- 2n(pi)i, it might appear natural at first > glance to begin by constructing a j-number whose square has the values > +-2(pi)i. But in fact the correct approach is to begin with a j-number > whose square has the values +-(pi)i. Then we can take 2 times the > square, (+-(pi)i) + (+-(pi)i), which has 3 values: (pi)i + (pi)i = 2(pi)i, > -(pi)i - (pi)i = -2(pi)i, and > (pi)i - (pi)i or -(pi)i + (pi)i = 0. In this case, the square has the two values of log(-1) with the > smallest norms, +-(pi)i, and 2 times the square has the three values > of log(1) with the smallest norms, 0 and +-2(pi)i. Thus we begin by constructing a j-number whose square has the values +- > (pi)i. That j-number is not hard to find: +-sqrt(pi)j. As noted above, (+-sqrt(pi)j)^2 = +-(pi)i = two values of log(-1), and > 2(+-sqrt(pi)j)^2 = +-2(pi)i or 0 = three values of log(1). We see further that 3(+-sqrt(pi)j)^2 = (+-(pi)i) + (+-(pi)i) + (+- > (pi)i) = +-(pi)i or +-3(pi)i, that is, the four values of log(-1) with > the smallest norms. > Likewise 4(+-sqrt(pi)j)^2 = 0 or +-2(pi)i or +-4(pi)i, the five values > of log(1) with the smallest norms. And in general, 2n(+-sqrt(pi)j)^2 = the 2n+1 values of log(1) with the > smallest norms, and > (2n-1)(+-sqrt(pi)j)^2 = the 2n values of log(-1) with the smallest > norms. > That is, the even multiples of (+-sqrt(pi)j)^2 are values of log(1), > and the odd multiples of (+-sqrt(pi)j)^2 are values of log(-1). Expressing the relation in the most complete form, we have the limit as n approaches infinity of 2n(+-sqrt(pi)j)^2 = log(1), and > the limit as n approaches infinity of (2n-1)(+-sqrt(pi)j)^2 = log(-1). It may be worthwhile here to note that the appearance of sqrt(pi) in > such an elementary equation is interesting, since the value of the > gamma function at 1/2 is sqrt(pi), and the value of the gamma function > at each half-integer is sqrt(pi) times a rational coefficient. To obtain the algebraic j-number expressions for log(i) and log(-i), > we will need to add multiples of (+-sqrt(pi)j)^2 to cover successively > greater numbers of values of these logarithms, as we did for log(1) > and log(-1). But we cannot begin with (+-sqrt(pi)j)^2, because we need > to have a single-valued part of the imaginary part, (pi)i/2 or -(pi)i/ > 2, to which multiples of +-(pi)i are added, to obtain the values of > log(i) and log(-i). Therefore we begin with a complex square root of -(pi)i/2, sqrt(pi)/2 > - sqrt(pi)i/2, as the complex part of our initial j-number, and to > this we add the same j-part we used above, +-sqrt(pi)j. The square of > this j-number gives the desired result: (sqrt(pi)/2 - sqrt(pi)i/2 +- sqrt(pi)j)^2 = > -(pi)i/2 +-(pi)i +-(pi)j/2 +-(pi)j/2 +-(pi)ij/2 +-(pi)ij/2. The complex part of this expression has 2 values, (pi)i/2 and -3(pi)i/ > 2, which are the two values of log(i) with the smallest norms. It is > true that the expression has a j-part and ij-part as well. Note that > the two multi-valued j-parts may or may not sum to zero, and the two > multi-valued ij-parts also may or may not sum to zero. Thus 2 values > of this j-number expression equal exactly two values of log(i), and > other values of this expression equal values of log(i) plus or minus a > j-part and/or plus or minus an ij-part. When we add (+-sqrt(pi)j)^2 to (sqrt(pi)/2 - sqrt(pi)i/2 +- > sqrt(pi)j)^2, we get -(pi)i/2, 3(pi)i/2, -5(pi)i/2, the three values > of log(-i) with the smallest norms, again plus or minus a j-part and/ > or ij-part. Now we add multiples of (+-sqrt(pi)j)^2 to (sqrt(pi)/2 - sqrt(pi)i/2 > +- sqrt(pi)j)^2, obtaining successively greater numbers of values of > log(i) when we add even multiples of (+-sqrt(pi)j)^2 and values of > log(-i) when we add odd multiples of (+-sqrt(pi)j)^2. Note that the > plus or minus j-parts or ij-parts do not change or increase as we add > multiples of (+-sqrt(pi)j)^2; they were only produced by the initial j- > number square of this sequence, (sqrt(pi)/2 - sqrt(pi)i/2 +- > sqrt(pi)j)^2. Thus we have the limit as n approaches infinity of > (sqrt(pi)/2 - sqrt(pi)i/2 +- sqrt(pi)j)^2 + 2n(+-sqrt(pi)j)^2 = > log(i) +-(pi)j +-(pi)ij, and > the limit as n approaches infinity of > (sqrt(pi)/2 - sqrt(pi)i/2 +- sqrt(pi)j)^2 + (2n-1)(+-sqrt(pi)j)^2 = > log(-i) +-(pi)j +-(pi)ij. Again, it may be worth noting the appearance of sqrt(pi)/2 as the real > part and -sqrt(pi)i/2 as the imaginary part of the j-number in the > first term of these expressions. sqrt(pi)/2 is the value of the gamma > function at 3/2, or the value of the product (pi) function at 1/2. In conclusion, we see that in the j-number system -- where the > imaginary unit +-j has ambiguous sign, +-j^2 = +-i, and the standard > complex numbers with unambiguous signs are a proper subset of the j- > numbers, just as the real numbers are a proper subset of the complex > numbers -- we can express the multi-valued logarithms of complex > numbers as linear combinations of squares of j-numbers. Here is another motivation for using the j-number algebra. In the j-number algebra, the square roots of i, and of -i, are {+-j, +- ij}. But this expression by itself is hardly a motivation for using the j-number algebra, because the square roots of i and of -i have a perfectly elegant expression within the complex numbers themselves: (+-z)^2 = i, (+-z-conjugate)^2 = -i = i-conjugate, where z=e^(i(pi)/4), which is quite elegant when written in a font that can mark the bar over z to express the conjugate. The conjugation of the square roots of i matches the conjugation of i. But the extension of the j-number algebra to an 8-dimensional k- algebra does provide a motivation for using this system of algebras to express the roots of i and -i. We define the elements of the k-algebra to have the form x1 + x2i +- x3j +- x4ij +- x5k +- x6ik +- x7jk +- x8ijk, where +-j^2 = +-i and +-k^2 = +-j. Like the j-number algebra, the k-number algebra is commutative and all elements have reciprocals. Also like the j-number algebra, the k- number algebra is not a field because the binary operations of k- number addition and multiplication take two elements and return multiple elements, not one element. In the k-number algebra we can express the 4th roots of i and -i in an elegant form that is lacking within the complex numbers themselves. First we note that in the k-number algebra, the square roots of +-j are {+-k, +-ik}. Also, the square roots of +-ij are {+-jk, +-ijk}. Thus in the k-number algebra the 4th roots of i and of -i are {+-k, +-ik, +-jk, +-ijk}. This is much more elegant and concise than any expression of the complex 4th roots of i and of -i, because negation and conjugation of a complex number do not suffice to express the 4th roots of i or of -i the way they suffice to express the square roots. The point is that the real and imaginary coefficients of different complex 4th roots of i and of -i are transposed. To express that fact concisely, we have to write out z as (a,b), which is less concise. Further, setting z=e^(i(pi)/8), we have (+-(a,b))^4 = i and (+-(a,b)-conjugate)^4 = -i = i-conjugate, but for the other 4th roots of i and of -i we have (+-(b,a))^4 = -i = i-conjugate and (+-(b,a)-conjugate)^4 = i. Thus the conjugation of the 4th roots of i no longer matches the conjugation of i. The 8-dimensional k-number algebra, on the other hand, allows us to express the 4th roots of i and of -i in the much more concise and elegant form {+-k, +-ik, +-jk, +-ijk}. === Subject: Re: Factors mod p <4790c6a1$0$47103$892e7fe2@authen.yellow.readfreenews.net> posting-account=504E-QkAAAA2v90r8nGnJKpfySa_yBSU 5.1),gzip(gfe),gzip(gfe) > Possibly the form I've given has made this result difficult for many > of you to understand, so I'm shifting to directly considering > composite factors modulo p. Given a target composite T and integer factors f 1 and f 2, such that > f 1*f 2 = nT, and any prime p, the following relations must be true: f 1 = ak mod p and f 2 = a^{-1}(1 + a^2)k mod p letsa pick a f, > æ æ ænow pick a n, > æ æ æ æ æ æanda picka a, > æ æ æ æ æ æ æ æ æanda guessa k Whats that spell ? ....... Whats that Spell ? æ...... æ æfnak æ! æ yea, æ æFNAK it. doesn't the mod ætake more cpu time than straight divide? æSeems like this > approach is doomed from the start. James, let's get down to basics. CPU execution time for any operation is precisely equal to the number of computer clock cycles required to perform the operation. I really don't believe that you have any comprehension of how computers perform arithmetic operations, including staight divides which are never simple. So, let's begin with a simple binary addition. What the computer chip for an addition is to XOR' each binary digit with the other, and determine if the result is a 1, in which case there will be no ripple down, or if the result is 0 resulting from two 1s. the result will of course be Zero with a ripple carry. I assume that by posting that you know the difference between an XOR, an OR and a AND. If you don't then it becomes obvious that you know nothing about computers, hence your post is total noise in this newsgroup. Harry C, === Subject: Re: Factors mod p > Possibly the form I've given has made this result difficult for many > of you to understand, so I'm shifting to directly considering > composite factors modulo p. Given a target composite T and integer factors f_1 and f_2, such that > f_1*f_2 = nT, and any prime p, the following relations must be true: f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p letsa pick a f, > now pick a n, > anda picka a, > anda guessa k Whats that spell ? ....... Whats that Spell ? ...... fnak ! yea, FNAK it. doesn't the mod take more cpu time than straight divide? Seems like this > approach is doomed from the start. [James, let's get down to basics. CPU execution time for any operation [is precisely equal to the number of computer clock cycles required to [perform the operation. I really don't believe that you have any [comprehension of how computers perform arithmetic operations, [including staight divides which are never simple. [So, let's begin with a simple binary addition. What the computer chip [for an addition is to XOR' each binary digit with the other, and [determine if the result is a 1, in which case there will be no [ripple down, or if the result is 0 resulting from two 1s. the [result will of course be Zero with a ripple carry. I assume that by [posting that you know the difference between an XOR, an OR and a [AND. If you don't then it becomes obvious that you know nothing [about computers, hence your post is total noise in this newsgroup. [Harry C, you are at too low a level, Harry. remember it is shift and add none of your operations above have carry bits, you are just doing dirt simple logic. (remember NAND and NOR too they are used more, most all memory is NAND) break out your old FORTRAN manual and look it up number of cycles per function in increasing order + - (also basic logic fns like your above) multiply divide mod exp series expansion for more complex fns SO, using standard lib functions like MOD, will never be as fast as divide and with the rest of poor old JSH's non-optomized equations with powers and roots and all, it will always be one of the slowest bicycles with a flat on the Expressway. === Subject: Re: Factors mod p <4790c6a1$0$47103$892e7fe2@authen.yellow.readfreenews.net> <4791361c$0$47115$892e7fe2@authen.yellow.readfreenews.net> posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) >Possibly the form I've given has made this result difficult for many >of you to understand, so I'm shifting to directly considering >composite factors modulo p. Given a target composite T and integer factors f_1 and f_2, such that >f_1*f_2 = nT, and any prime p, the following relations must be true: f_1 = ak mod p and f_2 = a^{-1}(1 + a^2)k mod p letsa pick a f, > now pick a n, > anda picka a, > anda guessa k Whats that spell ? ....... Whats that Spell ? ...... fnak ! yea, FNAK it. doesn't the mod take more cpu time than straight divide? Seems like this > approach is doomed from the start. [James, let's get down to basics. CPU execution time for any operation > [is precisely equal to the number of computer clock cycles required to > [perform the operation. I really don't believe that you have any > [comprehension of how computers perform arithmetic operations, > [including staight divides which are never simple. [So, let's begin with a simple binary addition. What the computer chip > [for an addition is to XOR' each binary digit with the other, and > [determine if the result is a 1, in which case there will be no > [ripple down, or if the result is 0 resulting from two 1s. the > [result will of course be Zero with a ripple carry. I assume that by > [posting that you know the difference between an XOR, an OR and a > [AND. If you don't then it becomes obvious that you know nothing > [about computers, hence your post is total noise in this newsgroup. [Harry C, you are at too low a level, Harry. > remember it is shift and add > none of your operations above have carry bits, you are just doing dirt > simple logic. > (remember NAND and NOR too they are used more, most all memory is NAND) > break out your old FORTRAN manual and look it up number of cycles per function in increasing order + - (also basic logic fns like your above) > multiply > divide > mod > exp > series expansion for more complex fns SO, using standard lib functions like MOD, will never be as fast as divide > and with the rest of poor old JSH's non-optomized equations with powers and > roots and all, it will always be one of the slowest bicycles with a flat on > the Expressway. The congruence relations are fundamental mathematics at core level. They define how all composite factorizations occur, over infinity. There is no factoring done, even now, that isn't encompassed by them. So whatever people already know about factoring can be quickly re-done in context, as these were decision equations all along, affecting everything, but they were in the shadows until now. James Harris === Subject: Re: Factors mod p > SO, using standard lib functions like MOD, will never be as fast as > divide > and with the rest of poor old JSH's non-optomized equations with powers > and > roots and all, it will always be one of the slowest bicycles with a flat > on > the Expressway. The congruence relations are fundamental mathematics at core level. They define how all composite factorizations occur, over infinity. There is no factoring done, even now, that isn't encompassed by them. So whatever people already know about factoring can be quickly re-done > in context, as these were decision equations all along, affecting > everything, but they were in the shadows until now. > James Harris But you already have shown this in your post on 1-16-08; >In the previously posted derivation I show how you get the factoring >congruences so I'll just give them here versus deriving them in detail >again: >z = (2a)^{-1} (1 + 2a^2)k mod p >k^2 = (a^2+1)^{-1}(nT) mod p >and >y = (1+2a^2)^{-1} z mod p, or y = -(1+2a^2)^{-1} z mod p. sorry, but you have mod, power and divide as your core functions implemented, so it will be slower than straightforward attack, especially if you have loops with in loops going on like you do. === Subject: Re: Factors mod p > SO, using standard lib functions like MOD, will never be as fast as > divide > and with the rest of poor old JSH's non-optomized equations with powers > and > roots and all, it will always be one of the slowest bicycles with a flat > on > the Expressway. > The congruence relations are fundamental mathematics at core level. > They define how all composite factorizations occur, over infinity. > There is no factoring done, even now, that isn't encompassed by them. > So whatever people already know about factoring can be quickly re-done > in context, as these were decision equations all along, affecting > everything, but they were in the shadows until now. > James Harris > But you already have shown this in your post on 1-16-08; >In the previously posted derivation I show how you get the factoring >congruences so I'll just give them here versus deriving them in detail >again: >z = (2a)^{-1} (1 + 2a^2)k mod p >k^2 = (a^2+1)^{-1}(nT) mod p >and >y = (1+2a^2)^{-1} z mod p, or y = -(1+2a^2)^{-1} z mod p. sorry, but you have mod, power and divide as your core functions > implemented, so it will be slower than straightforward attack, especially > if you have loops with in loops going on like you do. Another side of the story JSH is unaware of, is the optimization of the code for the computer for speed. Just a direct attack using simple divides, is far faster than the complexity of JSH. === Subject: manual soluation of digital design third edition posting-account=Qc0kZwoAAACrQbcheQ68crpjWmx4XKUI CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.2),gzip(gfe),gzip(gfe) hello i need the manual soluation of digital design third edition for M.MORRIS MANO plz i have a exam after three days and i will be happy if you send it to my e-mial === Subject: Re: What is this shape called? posting-account=G4HhgwkAAAB0h1X9lFcN2_L63S7FUgWk 1.0.3705; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > Okay ... if you start with a cube, and you cut the eight corners off, > so that the eight original corners become eight regular triangular > faces, all touching corner-to-corner, then you have a fourteen-sided > solid with eight new triangular faces and six (reduced-size) square > faces. If you then remove more material from the original corners, so that > the eight triangles become larger and start to overlap and intersect, > they eventually become perfect hexagons ... then you have a different > fourteen-sided solid, that now has six smaller square faces and eight > larger hexagonal faces. This second semi-regular solid (which probably also counts as a > truncated octahedron) is special, because multiple copies of it will > stack to produce a perfect space-filling lattice. > That's a very nice (and unusual) property. SO: > Q1: æDoes anyone know the name of this second solid? Q2: æDoes anyone know what the resulting lattice is called? Q3: æAre there any common examples of crystals that use this lattice? > (if I knew the answers to Q1 or Q2, I could probably look this up > myself, but I don't) Q4: æSince each intersection point in the lattice connects to four > others, at equal distances, the obvious candidate material for this > sort of crystalline lattice structure would /seem/ to be carbon, with > its four covalent bonds. > Is there any particular reason why one couldn't build the thing out of > carbon atoms, and end up with something like a more spongiform, > lower-density version of diamond? > PS: I've looked up a few pages on the allotropes of carbon, and they > deal with graphite, diamond, and buckyballs and the more exotic > graphenes, but not this shape. Perhaps the big voids mean that it's > not a default formation for carbon unless you mix in some other big > molecule to pack out the voids and get the shape started, dunno. PPS: I dimly remember being taught in organic chem that there were no > regular allotropes of carbon other than graphite and diamond, because > these were (supposedly) the only two repeating four-bond structures > that could exist in nature. We were told that this had been proved > geometrically, and that it was, in fact, the most certain fact in all > of organic chemistry. Setting aside the more recent scale-specific > allotropes (buckyballs and buckytubes) this would /seem/ to be a third > (entirely legal) form of possible four-bond lattice structure. æ æ PPPS: Yes, it really does repeat. I made one of those fun lattice > models from wire twists and drinking straws to check. It works. If anyone can tell me what the correct name is for this polyhedron and > the resulting lattice (and ideally some background detail on the > carbon issue), I'd be very interested. Eric You might enjoy POLY, from http://www.peda.com/download/ === Subject: Re: What is this shape called? posting-account=qKxGxgkAAADAPfYVCc-ZQkIzl0senr2M Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Okay ... if you start with a cube, and you cut the eight corners off, > so that the eight original corners become eight regular triangular > faces, all touching corner-to-corner, then you have a fourteen-sided > solid with eight new triangular faces and six (reduced-size) square > faces. If you then remove more material from the original corners, so that > the eight triangles become larger and start to overlap and intersect, > they eventually become perfect hexagons ... then you have a different > fourteen-sided solid, that now has six smaller square faces and eight > larger hexagonal faces. This second semi-regular solid (which probably also counts as a > truncated octahedron) is special, because multiple copies of it will > stack to produce a perfect space-filling lattice. > That's a very nice (and unusual) property. SO: > Q1: Does anyone know the name of this second solid? A truncated octahedron, specifically, a cubically truncated octahedron. A Special case of the tetrakaidecahedron (from the Greek for 'fourteen'). Q2: Does anyone know what the resulting lattice is called? A body-centered cubic lattice. See your shape embedded here: http://ciks.cbt.nist.gov/~garbocz/closedcell/node1.html It is one of the classic Bravais lattices in crystallography: http://ece-www.colorado.edu/~bart/book/bravais.htm > Q3: Are there any common examples of crystals that use this lattice? > (if I knew the answers to Q1 or Q2, I could probably look this up > myself, but I don't) Tungsten (see the link above) - also iron, chromium, sodium - it's a very common lattice. > Q4: Since each intersection point in the lattice connects to four > others, at equal distances, the obvious candidate material for this > sort of crystalline lattice structure would /seem/ to be carbon, with > its four covalent bonds. Count again. Each atom in the lattice had 8 nearest neighbors (touching the octahedral faces). > Is there any particular reason why one couldn't build the thing out of > carbon atoms, and end up with something like a more spongiform, > lower-density version of diamond? Build it out of carbon atoms and you will have a higher density version of diamond. The body-centered cubic structure has a packing fraction of 0.6802 - less than that of the hexagonal close-packed lattice of londsdaleite (0.7405) http://www.answers.com/topic/lonsdaleite?cat=technology It is possible that carbon could form a body-centered cubic lattice (packing fraction - 0.6802) under high pressures (it is a very high density lattice). The real diamond lattice is already far more open (packing fraction = 0.3401) because each atom can have only 4 nearest neighbors - at the corners of a tetrahedron: http://www-ee.ccny.cuny.edu/www/web/crouse/I3600/Homework/Homework%201%20Sol utions%20Fall03.htm PS: I've looked up a few pages on the allotropes of carbon, and they > deal with graphite, diamond, and buckyballs and the more exotic > graphenes, but not this shape. Perhaps the big voids mean that it's > not a default formation for carbon unless you mix in some other big > molecule to pack out the voids and get the shape started, dunno. The big voids in the diamond lattice are driven by quantum mechanics. Carbon in its highest symmetry has only 4 outer orbitals (sp3 hybrids), which are arranged at the corners of a tetrahedron: http://www.answers.com/topic/orbital-hybridisation > PPS: I dimly remember being taught in organic chem that there were no > regular allotropes of carbon other than graphite and diamond, because > these were (supposedly) the only two repeating four-bond structures > that could exist in nature. We were told that this had been proved > geometrically, and that it was, in fact, the most certain fact in all > of organic chemistry. Setting aside the more recent scale-specific > allotropes (buckyballs and buckytubes) this would /seem/ to be a third > (entirely legal) form of possible four-bond lattice structure. Once believed to be true, but as Einstein said, one great fact can spoil a good theory. See the above link on londsdaleite. > PPPS: Yes, it really does repeat. I made one of those fun lattice > models from wire twists and drinking straws to check. It works. If anyone can tell me what the correct name is for this polyhedron and > the resulting lattice (and ideally some background detail on the > carbon issue), I'd be very interested. Eric HTH Tom Davidson Richmond, VA === Subject: Re: What is this shape called? > Okay ... if you start with a cube, and you cut the eight corners off, > so that the eight original corners become eight regular triangular > faces, all touching corner-to-corner, then you have a fourteen-sided > solid with eight new triangular faces and six (reduced-size) square > faces. > If you then remove more material from the original corners, so that > the eight triangles become larger and start to overlap and intersect, > they eventually become perfect hexagons ... then you have a different > fourteen-sided solid, that now has six smaller square faces and eight > larger hexagonal faces. > This second semi-regular solid (which probably also counts as a > truncated octahedron) is special, because multiple copies of it will > stack to produce a perfect space-filling lattice. > That's a very nice (and unusual) property. > SO: > Q1: Does anyone know the name of this second solid? A truncated octahedron, specifically, a cubically truncated >octahedron. A Special case of the tetrakaidecahedron (from the Greek >for 'fourteen'). > Q2: Does anyone know what the resulting lattice is called? A body-centered cubic lattice. See your shape embedded here: >http://ciks.cbt.nist.gov/~garbocz/closedcell/node1.html >It is one of the classic Bravais lattices in crystallography: >http://ece-www.colorado.edu/~bart/book/bravais.htm Sorry. Bad wording on my part. That page /does/ seem to give the shape, and the stacking method ( someone else has kindly pointed me towards http://en.wikipedia.org/wiki/Bitruncated_cubic_honeycomb ), but the lattice shape that I was interested in was actually the skeleton /between/ the polyhedra, rather than the shape made by mapping the polygon centres themselves, and their touching connections. I should have made that clearer. :( > Q3: Are there any common examples of crystals that use this lattice? > (if I knew the answers to Q1 or Q2, I could probably look this up > myself, but I don't) Tungsten (see the link above) - also iron, chromium, sodium - it's a >very common lattice. Yep, those'd presumably be BCC lattices, rather than the lattice formed by the outlines of the stacked polygons. Again, my bad -- I should have been more specific about what I meant. > Q4: Since each intersection point in the lattice connects to four > others, at equal distances, the obvious candidate material for this > sort of crystalline lattice structure would /seem/ to be carbon, with > its four covalent bonds. Count again. Each atom in the lattice had 8 nearest neighbors >(touching the octahedral faces). Yep, each solid has 8 nearest neighbours. I was instead thinking of the connections that you'd get if you removed the solid polyhedra and used the wireframe skeleton to describe a lattice, with a carbon atom at each vertex, and with the edges linking the vertices representing bonds. That gives four bonds per atom, like diamond ... but it's not the standard diamond lattice. > Is there any particular reason why one couldn't build the thing out of > carbon atoms, and end up with something like a more spongiform, > lower-density version of diamond? Build it out of carbon atoms and you will have a higher density >version of diamond. The body-centered cubic structure has a packing >fraction of 0.6802 - less than that of the hexagonal close-packed >lattice of londsdaleite (0.7405) > http://www.answers.com/topic/lonsdaleite?cat=technology Ooo! Lonsdaleite is new to me! That's cool! 1967? Looks like my ancient chemistry 101 classes were probably already obsolete before the textbooks were published! I read the buckyball papers in Nature when the story broke, and I don't /remember/ any of them mentioning a name like lonsdaleite ... but maybe it was mentioned and didn't register on my noggin. Or maybe it was just really obscure! I wonder how many of these different allotrope thingies there actually are? The thing described by the polygon mesh isn't lonsdaleite, because as far as I can make out by squinting at the diagram, lonsdaleite seems to be made of hexagons, whereas this mesh has a mix of hexagons and squares. >It is possible that carbon could form a body-centered cubic lattice >(packing fraction - 0.6802) under high pressures (it is a very high >density lattice). The real diamond lattice is already far more open >(packing fraction = 0.3401) because each atom can have only 4 nearest >neighbors - at the corners of a tetrahedron: >http://www-ee.ccny.cuny.edu/www/web/crouse/I3600/Homework/Homework%201%20So lutions%20Fall03.htm > PS: I've looked up a few pages on the allotropes of carbon, and they > deal with graphite, diamond, and buckyballs and the more exotic > graphenes, but not this shape. Perhaps the big voids mean that it's > not a default formation for carbon unless you mix in some other big > molecule to pack out the voids and get the shape started, dunno. The big voids in the diamond lattice are driven by quantum >mechanics. Carbon in its highest symmetry has only 4 outer orbitals >(sp3 hybrids), which are arranged at the corners of a tetrahedron: >http://www.answers.com/topic/orbital-hybridisation Confusion ... my fault again. The mesh created by the /edges and vertices/ of the bitruncated cubic honeycomb creates voids that correspond to the centres of the original polygons. Those are what I meant. Apologies again. > PPS: I dimly remember being taught in organic chem that there were no > regular allotropes of carbon other than graphite and diamond, because > these were (supposedly) the only two repeating four-bond structures > that could exist in nature. We were told that this had been proved > geometrically, and that it was, in fact, the most certain fact in all > of organic chemistry. Setting aside the more recent scale-specific > allotropes (buckyballs and buckytubes) this would /seem/ to be a third > (entirely legal) form of possible four-bond lattice structure. Once believed to be true, but as Einstein said, one great fact can >spoil a good theory. See the above link on londsdaleite. Well, I'm definitely bookmarking londsdaleite, along with all the other goodies that people have kindly linked to! > PPPS: Yes, it really does repeat. I made one of those fun lattice > models from wire twists and drinking straws to check. It works. > If anyone can tell me what the correct name is for this polyhedron and > the resulting lattice (and ideally some background detail on the > carbon issue), I'd be very interested. > Eric HTH Tom Davidson >Richmond, VA >Eric === Subject: Re: What is this shape called? | | > | >| | >| Okay ... if you start with a cube, and you cut the eight corners off, | >| so that the eight original corners become eight regular triangular | >| faces, all touching corner-to-corner, then you have a fourteen-sided | >| solid with eight new triangular faces and six (reduced-size) square | >| faces. | > | >Ok... | > | >| | >| If you then remove more material from the original corners, so that | >| the eight triangles become larger and start to overlap and intersect, | >| they eventually become perfect hexagons... | > | >HAHAHAHA! | > | > | > | >| then you have a different | >| fourteen-sided solid, that now has six smaller square faces and eight | >| larger hexagonal faces. | > | >Nope. Nothing left, you cut too much off. It's a POLY - GONE. | > | > http://www.androcles01.pwp.blueyonder.co.uk/Polygon.jpg | > | | Hi Androcles! | What I hadn't realised was that my attempt at describing how to | create the shape turns out to correspond to the naming system. | | http://en.wikipedia.org/wiki/Truncation_(geometry) | | (reads wiki pages) So-o ... you can take an octahedron and cut the | corners off to give six new square faces, and to crop the original | triangular faces down into hexagons, and that first regularish shape | that you get by lopping off the corners is called a truncated X | | If you lop off more, so that you get to the next regularish shape, | it's a bitruncated X | | So in this case, the same solid could presumably either be referred to | as a form of multiply-truncated cube (I'm not sure about the official | level of truncation here, the naming scheme seems slightly | arbitrary) or as a truncated octahedron (less ambiguous). | | | Whew! I had no idea that Wikipedia's coverage of this stuff was so | thorough! | | Hm. I wonder what you call the fractal solids that you get by | repeatedly truncating the curved remnant of a truncated sphere (if | truncation is considered a legal term for shaving bits off spheres)? | That gives a family of fractal solids with an infinite number of | circular faces ... ... :) | You'll have much the same problem in astronomy. Perigee - point of closest approach to the Earth. Perihelion - point of closest approach to the Sun. Perijove - point of closest approach to Jupiter. Periastron - point of closest approach to a star. Periapsis - point of closest approach.<--(period) The actual number of polygons and polyhedrons is infinite, one cannot name them all so only the simplest are named. | Anyhow, more seriously, back to the quest for interesting carbon | allotropes and variants... | ... the skeleton of a bitruncated cubic honeycomb (it's so nice to | know the name!) essentially consists of hexagons and squares, but I | don't know enough about carbon chemistry to know how happy and stable | carbon covalent bonds might be at those angles. If they are okay-ish, | then I don't see why this shouldn't be a legitimate carbon allotrope. Do you know about buckey balls (Buckminster-Fullerine)? http://users.omskreg.ru/~kolosov/atlas/3D-crystals/bonding.html What's happened is a follow-on from the traditions started long ago in the Natural Sciences where species are name for their discoverer or location - we have units of force called the newton, current in amperes, Volta gave us the volt, weber is a unit of magnetism and homo neanderthalensis was found in the Neander Valley. A woman went to a psychiatrist because she was terrified of birds. The psychiatrist told her she was suffering from ornithophobia, so she paid his fee and went away happy, knowing that her condition was recognised with a Latin name. She's still terrified of birds but at least now she knows she's suffering from fear of birds (aka ornithophobia). A fool and his money are soon parted. What I'd like to know is how the fool got the money in the first place. | Plus, it has all those fun big spaces bounded by stressed hexagonal | portals where you could hide heavy-metal ions or other fun goodies to | change the bulk properties. | | Ooo! (sudden thought) | I wonder if you could use this stuff as a hydrogen sponge? | | Eric Genius is 1% inspiration and 99% perspiration. Follow that thought, you may find soot soaks up excess water next time the dominant half of your family asks you to wash the kitchen floor. BTW, I use Google Sketchup to draw.. it's free, it's fast, it's intuitive, in short ... EXCELLENT. Not much in life will give you so much value for so little money. http://www.sketchup.com/ === Subject: Re: What is this shape called? > ... >Do you know about buckey balls (Buckminster-Fullerine)? > http://users.omskreg.ru/~kolosov/atlas/3D-crystals/bonding.html A little bit. But C60 is a closed, self-contained shell consisting of hexagons and pentagons: the thing that I was interested in is a continuous lattice consisting of hexagons and squares. It seems to be equivalent to a network of 24-atom cages, and the individual cages did remind me a little bit of smaller versions of a buckyball. >What's happened is a follow-on from the traditions started long >ago in the Natural Sciences where species are name for their >discoverer or location - we have units of force called the newton, >current in amperes, Volta gave us the volt, weber is a unit of >magnetism and homo neanderthalensis was found in the Neander Valley. A woman went to a psychiatrist because she was terrified of birds. >The psychiatrist told her she was suffering from ornithophobia, >so she paid his fee and went away happy, knowing that her condition >was recognised with a Latin name. She's still terrified of birds but >at least now she knows she's suffering from fear of birds (aka >ornithophobia). Yes, official technical names can be very reassuring, can't they? I think there's supposed to be a handy latin phrase that doctors use when someone has died during treatment and they have no idea of the cause, which simply translates as unknown complications. So the doctor can say, Mrs Smith, the port-mortem showed that your husband died of [fancy latin term]. It sounds like a nasty medical condition and placates the relative, but all it means is The patient died during the procedure and we'll probably never really know why, it's just something that happens from time to time. People are funny like that.. >| Plus, it has all those fun big spaces bounded by stressed hexagonal >| portals where you could hide heavy-metal ions or other fun goodies to >| change the bulk properties. >| >| Ooo! (sudden thought) >| I wonder if you could use this stuff as a hydrogen sponge? >| >| Eric Genius is 1% inspiration and 99% perspiration. Follow >that thought, you may find soot soaks up excess water >next time the dominant half of your family asks you to >wash the kitchen floor. Well, I know that people have already suggested using buckyballs and buckytubes for hydrogen storage. The sudden thought was that since this was a continuous structure (with some of the same cagelike attibutes), it might be a more efficient way of building cages than buckyballs or buckytubes, because in this case the cage components are shared. If we're dealing with 24-atom cages, and each of the 24 atoms is a corner, and gets used and shared by four adjacent cages, then the average number of carbon-atoms-required-per-cage in the bulk material would only seem to be be ... what ... 24/4 ... six? Have I got that right? it seems awfully low! Eric === Subject: Re: What is this shape called? | | | > ... | >Do you know about buckey balls (Buckminster-Fullerine)? | > http://users.omskreg.ru/~kolosov/atlas/3D-crystals/bonding.html | | | A little bit. | | But C60 is a closed, self-contained shell consisting of hexagons and | pentagons: the thing that I was interested in is a continuous lattice | consisting of hexagons and squares. | | It seems to be equivalent to a network of 24-atom cages, and the | individual cages did remind me a little bit of smaller versions of a | buckyball. | | | >What's happened is a follow-on from the traditions started long | >ago in the Natural Sciences where species are name for their | >discoverer or location - we have units of force called the newton, | >current in amperes, Volta gave us the volt, weber is a unit of | >magnetism and homo neanderthalensis was found in the Neander Valley. | > | >A woman went to a psychiatrist because she was terrified of birds. | >The psychiatrist told her she was suffering from ornithophobia, | >so she paid his fee and went away happy, knowing that her condition | >was recognised with a Latin name. She's still terrified of birds but | >at least now she knows she's suffering from fear of birds (aka | >ornithophobia). | | Yes, official technical names can be very reassuring, can't they? | | I think there's supposed to be a handy latin phrase that doctors use | when someone has died during treatment and they have no idea of the | cause, which simply translates as unknown complications. | | So the doctor can say, Mrs Smith, the port-mortem showed that your | husband died of [fancy latin term]. It sounds like a nasty medical | condition and placates the relative, but all it means is The patient | died during the procedure and we'll probably never really know why, | it's just something that happens from time to time. People are funny | like that.. | Yes indeed, the legal profession also likes to use Latin. In Newton's day it was the lingua franca of the European universities, there being so many European languages and anyone of learning would be familiar with Latin through the Church of Rome's teachings. Today it is a form of snobbery and also polite, plain Anglo-Saxon being considered vulgar. I was amused to see that the computer game Harry Potter has both British commentary and American commentary versions, not only for accent but for dialect as well. One cannot say toilet to American children, it has to be bathroom; rest room is even better, even if it does conjure up cartoon images of someone with their pants around their ankles, newspaper in their lap and head against the side of the cubicle, snoring. American - Post mortem - after death (Latin) British - Autopsy - an examination of a body after death (Greek) The examination itself is not to be mentioned but is understood. | >| Plus, it has all those fun big spaces bounded by stressed hexagonal | >| portals where you could hide heavy-metal ions or other fun goodies to | >| change the bulk properties. | >| | >| Ooo! (sudden thought) | >| I wonder if you could use this stuff as a hydrogen sponge? | >| | >| Eric | > | >Genius is 1% inspiration and 99% perspiration. Follow | >that thought, you may find soot soaks up excess water | >next time the dominant half of your family asks you to | >wash the kitchen floor. | | Well, I know that people have already suggested using buckyballs and | buckytubes for hydrogen storage. The sudden thought was that since | this was a continuous structure (with some of the same cagelike | attibutes), it might be a more efficient way of building cages than | buckyballs or buckytubes, because in this case the cage components are | shared. | | If we're dealing with 24-atom cages, and each of the 24 atoms is a | corner, and gets used and shared by four adjacent cages, then the | average number of carbon-atoms-required-per-cage in the bulk material | would only seem to be be ... what ... 24/4 ... six? | Have I got that right? it seems awfully low! | | Eric You seem to be headed toward catalysts (inorganic/industrial) and enzymes (organic/biological). Cages are only useful if you can open and close the door. I gave up chemistry a long time ago in preference to engineering but I'll agree it is an interesting subject. Unfortunately that's about as far as I can go, I simply don't have the knowledge or desire to gain it for any further analysis. For example, the tetrahedron of diamond and the cubic graphite are both carbon but nobody is suggesting they can be cages like the buckey ball. One presumes the cage has to have a minimum volume to qualify as a cage, which is where your 24 atom cage may fail and the buckey ball succeed, but I don't have enough background to say what will or will not work. === Subject: Re: -- minimal elements for certain classes of subrings of a field [...] >The above questions have yes answers for the following fields K: > any finite field > > the rationals > k(x) where k is any field >Right, because in each of those cases, the poset r(K) has a minimum >which is an element of both ic(K) and ufd(K), so you can always pick >the minimum. >For a finite field and for the rationals -- yes, but not for k(x). >For K = k(x), the posets ic(K) and ufd(K) are equal, and every element >is greater or equal to a minimal element, but there is no minimum >element. Right; no minimum in general. But if there is a minimum for k, say R, >then R[x] will be a minimum for k(x), if I'm not mistaken. So for >Q(x), for example, Z[x] will be a minimum. No, I don't think that's right. Z[x] is minimal, but not a minimum. For example, consider Z[1/x]. Or for that matter, Z[1/(x-c)], where c is an arbitrary rational. quasi === Subject: Re: -- minimal elements for certain classes of subrings of a field days. My association with the Department is that of an alumnus. > [...] >For K = k(x), the posets ic(K) and ufd(K) are equal, and every element >is greater or equal to a minimal element, but there is no minimum >element. >Right; no minimum in general. But if there is a minimum for k, say R, >then R[x] will be a minimum for k(x), if I'm not mistaken. So for >Q(x), for example, Z[x] will be a minimum. No, I don't think that's right. Z[x] is minimal, but not a minimum. For example, consider Z[1/x]. Right you are. >Or for that matter, Z[1/(x-c)], where c is an arbitrary rational. Or Z[x+a]... etc. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: -- minimal elements for certain classes of subrings of a field > [...] >For K = k(x), the posets ic(K) and ufd(K) are equal, and every element >is greater or equal to a minimal element, but there is no minimum >element. >Right; no minimum in general. But if there is a minimum for k, say R, >then R[x] will be a minimum for k(x), if I'm not mistaken. So for >Q(x), for example, Z[x] will be a minimum. >No, I don't think that's right. >Z[x] is minimal, but not a minimum. >For example, consider Z[1/x]. Right you are. >Or for that matter, Z[1/(x-c)], where c is an arbitrary rational. Or Z[x+a]... etc. Well, Z[x+a] is the same ring as Z[x], whereas the rings Z[1/(x-a)] are all distinct (although isomorphic). One can ask whether such minimal integrally closed rings and UFDs, when they exist for a given field K, are necessarily isomorphic. Ok, I'll ask that! Note that for K = Q, the answer is clearly yes. For K = k(x), the answer is also yes, but that proof, though elementary, is harder. quasi === Subject: Re: -- minimal elements for certain classes of subrings of a field > [...] >For K = k(x), the posets ic(K) and ufd(K) are equal, and every element >is greater or equal to a minimal element, but there is no minimum >element. >Right; no minimum in general. But if there is a minimum for k, say R, >then R[x] will be a minimum for k(x), if I'm not mistaken. So for >Q(x), for example, Z[x] will be a minimum. >No, I don't think that's right. >Z[x] is minimal, but not a minimum. >For example, consider Z[1/x]. >Right you are. >Or for that matter, Z[1/(x-c)], where c is an arbitrary rational. >Or Z[x+a]... etc. Well, Z[x+a] is the same ring as Z[x], Whoops, sorry -- they are not always the same. I was thinking, when I read it, that a was intended to be an integer. In that case, Z[x+a] and Z[x] would be the same. In general, if a,b are arbitrary rationals, the rings Z[x+a] and Z[x+b] are the same iff b-a is an integer. >whereas the rings Z[1/(x-a)] are all distinct (although isomorphic). One can ask whether such minimal integrally closed rings and UFDs, >when they exist for a given field K, are necessarily isomorphic. Ok, >I'll ask that! Note that for K = Q, the answer is clearly yes. For K = >k(x), the answer is also yes, but that proof, though elementary, is >harder. quasi === Subject: Re: -- minimal elements for certain classes of subrings of a field posting-account=ZuqAEgoAAABijw9pOcO0pFSQPzQkFppl MathPlayer 2.10b; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > æ [...] >For K = k(x), the posets ic(K) and ufd(K) are equal, and every element >is greater or equal to a minimal element, but there is no minimum >element. >Right; no minimum in general. But if there is a minimum for k, say R, >then R[x] will be a minimum for k(x), if I'm not mistaken. So for >Q(x), for example, Z[x] will be a minimum. >No, I don't think that's right. >Z[x] is minimal, but not a minimum. >For example, consider Z[1/x]. >Right you are. >Or for that matter, Z[1/(x-c)], where c is an arbitrary rational. >Or Z[x+a]... etc. Well, Z[x+a] is the same ring as Z[x], Whoops, sorry -- they are not always the same. I was thinking, when I read it, that a was intended to be an integer. > In that case, Z[x+a] and Z[x] would be the same. In general, if a,b are arbitrary rationals, the rings Z[x+a] and > Z[x+b] are the same iff b-a is an integer. whereas the rings Z[1/(x-a)] are all distinct (although isomorphic). One can ask whether such minimal integrally closed rings and UFDs, >when they exist for a given field K, are necessarily isomorphic. Ok, >I'll ask that! Note that for K = Q, the answer is clearly yes. For K = >k(x), the answer is also yes, but that proof, though elementary, is >harder. quasi- Hide quoted text - - Show quoted text - It is hard to see where you are going with it. Maybe because I am not too deeply interested in it. But it is always good to consider extreme situations and simple cases first. For example looking at D = cap R where R in ic(K) and asking: Does D have K for a field of fractions? Same for E = cap R where R in uf(K). It may be useful to read a paper by Alan Loper and F. Tartarone. The title is something like: A classification of integrally closed rings of polynomial containing Z[X]. It was a pre-print when I saw. Hopefully it has appeared or may be in the process of being published. Muhammad Muhammad === Subject: Re: -- minimal elements for certain classes of subrings of a field > æ [...] >For K = k(x), the posets ic(K) and ufd(K) are equal, and every element >is greater or equal to a minimal element, but there is no minimum >element. >Right; no minimum in general. But if there is a minimum for k, say R, >then R[x] will be a minimum for k(x), if I'm not mistaken. So for >Q(x), for example, Z[x] will be a minimum. >No, I don't think that's right. >Z[x] is minimal, but not a minimum. >For example, consider Z[1/x]. >Right you are. >Or for that matter, Z[1/(x-c)], where c is an arbitrary rational. >Or Z[x+a]... etc. >Well, Z[x+a] is the same ring as Z[x], > Whoops, sorry -- they are not always the same. > I was thinking, when I read it, that a was intended to be an integer. > In that case, Z[x+a] and Z[x] would be the same. > In general, if a,b are arbitrary rationals, the rings Z[x+a] and > Z[x+b] are the same iff b-a is an integer. >whereas the rings Z[1/(x-a)] are all distinct (although isomorphic). >One can ask whether such minimal integrally closed rings and UFDs, >when they exist for a given field K, are necessarily isomorphic. Ok, >I'll ask that! Note that for K = Q, the answer is clearly yes. For K = >k(x), the answer is also yes, but that proof, though elementary, is >harder. It is hard to see where you are going with it. Maybe because I am not >too deeply interested in it. The idea is, given a field, to find a natural subring in which to settle before we start factoring things. We want to make sure the ring is big enough to generate the field, but other than that, we want as small a ring as possible, subject to maintaining unique factorization, or at least staying integrally closed. We certainly don't want the whole field, since that destroys all the would be irreducibles. Looked at another way, we want as few units as possible. The two prototypes are Q and Q(x). In each case, there is a natural subring. For Q, the natural subring is Z, which is clearly minimum among all UFDs in Q with quotient field Q. For Q(x), the natural subring is Z[x], which is minimal, but not minimum, since there are other minimal UFD subrings with quotient field Q(x) -- for example, Z[1/x] or, more generally, Z[1/(x-a)] for an arbitrary a in Q. Thus, instead of starting with a ring and asking about factorization, I am starting with a field, and trying to squeeze it down to an appropriate ring. >But it is always good to consider extreme situations and simple cases >first. For example looking at D = cap R where R in ic(K) and asking: >Does D have K for a field of fractions? In general no -- consider K = Q(x). The intersection would just be Z, which has fraction field Q, not Q(x). >Same for E = cap R where R in uf(K). Nope -- same counterexample. >It may be useful to read a paper by Alan Loper and F. >Tartarone. The title is something like: A classification of integrally >closed rings of polynomial containing Z[X]. It was a pre-print when I >saw. Hopefully it has appeared or may be in the process of being >published. Ok, I'll look at it, but the field Q(x) is not interesting for my questions -- I already know the answers for that field. Also, there is a difference in point of view. I'm starting with a field K and going down to some minimal subring satisfying my conditions, whereas the paper is starting from a particular subring and going up. I'm very much interested in whether the minimal subrings generally exist, and if not, for which fields they do exist. Thus, my questions relate to a property (as yet unclear) of the field itself. For a given field K, when the minimal subrings exist, are they unique? When they exist but are not unique, are they at least isomorphic? My guess is that many of these questions can be settled in sci.math by simple proofs or (more likely) counterexamples, depending on the question. quasi === Subject: Re: -- minimal elements for certain classes of subrings of a field posting-account=ZuqAEgoAAABijw9pOcO0pFSQPzQkFppl MathPlayer 2.10b; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > æ [...] >For K = k(x), the posets ic(K) and ufd(K) are equal, and every element >is greater or equal to a minimal element, but there is no minimum >element. >Right; no minimum in general. But if there is a minimum for k, say R, >then R[x] will be a minimum for k(x), if I'm not mistaken. So for >Q(x), for example, Z[x] will be a minimum. >No, I don't think that's right. >Z[x] is minimal, but not a minimum. >For example, consider Z[1/x]. >Right you are. >Or for that matter, Z[1/(x-c)], where c is an arbitrary rational. >Or Z[x+a]... etc. >Well, Z[x+a] is the same ring as Z[x], > Whoops, sorry -- they are not always the same. > I was thinking, when I read it, that a was intended to be an integer. > In that case, Z[x+a] and Z[x] would be the same. > In general, if a,b are arbitrary rationals, the rings Z[x+a] and > Z[x+b] are the same iff b-a is an integer. >whereas the rings Z[1/(x-a)] are all distinct (although isomorphic). >One can ask whether such minimal integrally closed rings and UFDs, >when they exist for a given field K, are necessarily isomorphic. Ok, >I'll ask that! Note that for K = Q, the answer is clearly yes. For K = >k(x), the answer is also yes, but that proof, though elementary, is >harder. It is hard to see where you are going with it. Maybe because I am not >too deeply interested in it. The idea is, given a field, to find a natural subring in which to > settle before we start factoring things. We want to make sure the ring > is big enough to generate the field, but other than that, we want as > small a ring as possible, subject to maintaining unique factorization, > or at least staying integrally closed. We certainly don't want the > whole field, since that destroys all the would be irreducibles. Looked > at another way, we want as few units as possible. The two prototypes are Q and Q(x). In each case, there is a natural subring. For Q, the natural subring is Z, which is clearly minimum among all > UFDs in Q with quotient field Q. For Q(x), the natural subring is Z[x], which is minimal, but not > minimum, since there are other minimal UFD subrings with quotient > field Q(x) -- for example, Z[1/x] or, more generally, Z[1/(x-a)] for > an arbitrary a in Q. Thus, instead of starting with a ring and asking about factorization, > I am starting with a field, and trying to squeeze it down to an > appropriate ring. But it is always good to consider extreme situations and simple cases >first. For example looking at D = cap R where R in ic(K) and asking: >Does D have K for a field of fractions? In general no -- consider K = Q(x). The intersection would just be Z, > which has fraction field Q, not Q(x). Same for E = cap R where R in uf(K). Nope -- same counterexample. Good. In general no. You have two cases for D . (1) K = qf(D), in this case you will only have to study only the integrally closed rings between D and K, inclusive. (2) If D is not equal to qf(D), D is a subring that is integrally closed in K (check). Can you talk about some other property? So the members of ic(K) are some kind of D algebras. Now you would have to look into what a D-subalgebra of K has to do to be just in ic(K). (That may possibly be your minimal member.) Ask the question: Is the set S = {X: X is a D-subalgebra of K not in ic(K)} inductive with inclusion as a partial order? I wish I had the mental presence or interest to know the answer right away. But if the answer is yes you will have some maximal elements in S. (In this case you will have to see how you can use them.) If the answer is no, you can think of something else. We can look into the case of E, later. It may be useful to read a paper by Alan Loper and F. >Tartarone. The title is something like: A classification of integrally >closed rings of polynomial containing Z[X]. It was a pre-print when I >saw. Hopefully it has appeared or may be in the process of being >published. Ok, I'll look at it, but the field Q(x) is not interesting for my > questions -- I already know the answers for that field. Also, there is a difference in point of view. I'm starting with a > field K and going down to some minimal subring satisfying my > conditions, whereas the paper is starting from a particular subring > and going up. I'm very much interested in whether the minimal subrings generally > exist, and if not, for which fields they do exist. Thus, my questions > relate to a property (as yet unclear) of the field itself. For a given field K, when the minimal subrings exist, are they unique? > When they exist but are not unique, are they at least isomorphic? My guess is that many of these questions can be settled in sci.math by > simple proofs or (more likely) counterexamples, depending on the > question. quasi- Hide quoted text - - Show quoted text - Producing new results takes interest, (valid reason for interest) patience, techniques, a lot of hard work and at times willingness to indulge in wild goose chase. Muhammad === Subject: Re: -- minimal elements for certain classes of subrings of a field > æ [...] >For K = k(x), the posets ic(K) and ufd(K) are equal, and every element >is greater or equal to a minimal element, but there is no minimum >element. >Right; no minimum in general. But if there is a minimum for k, say R, >then R[x] will be a minimum for k(x), if I'm not mistaken. So for >Q(x), for example, Z[x] will be a minimum. >No, I don't think that's right. >Z[x] is minimal, but not a minimum. >For example, consider Z[1/x]. >Right you are. >Or for that matter, Z[1/(x-c)], where c is an arbitrary rational. >Or Z[x+a]... etc. >Well, Z[x+a] is the same ring as Z[x], > Whoops, sorry -- they are not always the same. > I was thinking, when I read it, that a was intended to be an integer. > In that case, Z[x+a] and Z[x] would be the same. > In general, if a,b are arbitrary rationals, the rings Z[x+a] and > Z[x+b] are the same iff b-a is an integer. >whereas the rings Z[1/(x-a)] are all distinct (although isomorphic). >One can ask whether such minimal integrally closed rings and UFDs, >when they exist for a given field K, are necessarily isomorphic. Ok, >I'll ask that! Note that for K = Q, the answer is clearly yes. For K = >k(x), the answer is also yes, but that proof, though elementary, is >harder. >It is hard to see where you are going with it. Maybe because I am not >too deeply interested in it. > The idea is, given a field, to find a natural subring in which to > settle before we start factoring things. We want to make sure the ring > is big enough to generate the field, but other than that, we want as > small a ring as possible, subject to maintaining unique factorization, > or at least staying integrally closed. We certainly don't want the > whole field, since that destroys all the would be irreducibles. Looked > at another way, we want as few units as possible. > The two prototypes are Q and Q(x). > In each case, there is a natural subring. > For Q, the natural subring is Z, which is clearly minimum among all > UFDs in Q with quotient field Q. > For Q(x), the natural subring is Z[x], which is minimal, but not > minimum, since there are other minimal UFD subrings with quotient > field Q(x) -- for example, Z[1/x] or, more generally, Z[1/(x-a)] for > an arbitrary a in Q. > Thus, instead of starting with a ring and asking about factorization, > I am starting with a field, and trying to squeeze it down to an > appropriate ring. >But it is always good to consider extreme situations and simple cases >first. For example looking at D = cap R where R in ic(K) and asking: >Does D have K for a field of fractions? > In general no -- consider K = Q(x). The intersection would just be Z, > which has fraction field Q, not Q(x). >Same for E = cap R where R in uf(K). > Nope -- same counterexample. Good. In general no. You have two cases for D . (1) K = qf(D), in this case you will only have to study only >the integrally closed rings between D and K, inclusive. That's going the wrong way -- I want to go down, not up. If K = qf(D), we are done -- D is the minimum. >(2) If D is not equal to qf(D), D is a subring that is integrally closed in K (check). Can you talk about some other property? So the members of >ic(K) are some kind of D algebras. Now you would have to look >into what a D-subalgebra of K has to do to be just in ic(K). >(That may possibly be your minimal member.) Ask the question: Is the set S = {X: X is a D-subalgebra of K >not in ic(K)} inductive with inclusion as a partial order? I wish I >had the mental presence or interest to know the answer right away. But if the answer is yes you will have some maximal elements in S. >(In this case you will have to see how you can use them.) If the answer is >no, you can think of something else. We can look into the case of E, later. >It may be useful to read a paper by Alan Loper and F. >Tartarone. The title is something like: A classification of integrally >closed rings of polynomial containing Z[X]. It was a pre-print when I >saw. Hopefully it has appeared or may be in the process of being >published. > Ok, I'll look at it, but the field Q(x) is not interesting for my > questions -- I already know the answers for that field. > Also, there is a difference in point of view. I'm starting with a > field K and going down to some minimal subring satisfying my > conditions, whereas the paper is starting from a particular subring > and going up. > I'm very much interested in whether the minimal subrings generally > exist, and if not, for which fields they do exist. Thus, my questions > relate to a property (as yet unclear) of the field itself. > For a given field K, when the minimal subrings exist, are they unique? > When they exist but are not unique, are they at least isomorphic? > My guess is that many of these questions can be settled in sci.math by > simple proofs or (more likely) counterexamples, depending on the > question. Producing new results takes interest, (valid reason for interest) I think I've motivated it. The question is not just a random question. >patience, techniques, a lot of hard work Well, I may not be _that_ serious about this problem, but I thought it would be a good one to bat around in sci.math. It offers lots of potential for examples, proofs, counterexamples, as well as further questions. >and at times willingness to indulge in wild goose chase. Hard work? Maybe, maybe not, it depends. But wild goose chase? Sure, why not -- as long as it appears that it _might_ reveal some interesting insights. quasi === Subject: Re: -- minimal elements for certain classes of subrings of a field posting-account=ZuqAEgoAAABijw9pOcO0pFSQPzQkFppl MathPlayer 2.10b; .NET CLR 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) > æ [...] >For K = k(x), the posets ic(K) and ufd(K) are equal, and every element >is greater or equal to a minimal element, but there is no minimum >element. >Right; no minimum in general. But if there is a minimum for k, say R, >then R[x] will be a minimum for k(x), if I'm not mistaken. So for >Q(x), for example, Z[x] will be a minimum. >No, I don't think that's right. >Z[x] is minimal, but not a minimum. >For example, consider Z[1/x]. >Right you are. >Or for that matter, Z[1/(x-c)], where c is an arbitrary rational. >Or Z[x+a]... etc. >Well, Z[x+a] is the same ring as Z[x], > Whoops, sorry -- they are not always the same. > I was thinking, when I read it, that a was intended to be an integer. > In that case, Z[x+a] and Z[x] would be the same. > In general, if a,b are arbitrary rationals, the rings Z[x+a] and > Z[x+b] are the same iff b-a is an integer. >whereas the rings Z[1/(x-a)] are all distinct (although isomorphic). >One can ask whether such minimal integrally closed rings and UFDs, >when they exist for a given field K, are necessarily isomorphic. Ok, >I'll ask that! Note that for K = Q, the answer is clearly yes. For K = >k(x), the answer is also yes, but that proof, though elementary, is >harder. It is hard to see where you are going with it. Maybe because I am not >too deeply interested in it. The idea is, given a field, to find a natural subring in which to > settle before we start factoring things. We want to make sure the ring > is big enough to generate the field, but other than that, we want as > small a ring as possible, subject to maintaining unique factorization, > or at least staying integrally closed. We certainly don't want the > whole field, since that destroys all the would be irreducibles. Looked > at another way, we want as few units as possible. The two prototypes are Q and Q(x). In each case, there is a natural subring. For Q, the natural subring is Z, which is clearly minimum among all > UFDs in Q with quotient field Q. For Q(x), the natural subring is Z[x], which is minimal, but not > minimum, since there are other minimal UFD subrings with quotient > field Q(x) -- for example, Z[1/x] or, more generally, Z[1/(x-a)] for > an arbitrary a in Q. Thus, instead of starting with a ring and asking about factorization, > I am starting with a field, and trying to squeeze it down to an > appropriate ring. But it is always good to consider extreme situations and simple cases >first. For example looking at D = cap R where R in ic(K) and asking: >Does D have K for a field of fractions? In general no -- consider K = Q(x). The intersection would just be Z, > which has fraction field Q, not Q(x). Same for E = cap R where R in uf(K). Nope -- same counterexample. Good. In general no. You have two cases for D . (1) K = qf(D), in this > case you will only have to study only the integrally closed rings > between D and K, inclusive. (2) If D is not equal to qf(D), D is a > subring that is integrally closed in K (check). Can you talk about > some other property? So the members of æ ic(K) are some kind of D > algebras. Now you would have to look into what a D-subalgebra of K has > to do to be just in ic(K). (That may possibly be your minimal > member.) Ask the question: Is the set S = {X: X is a D-subalgebra of K > not in ic(K)} inductive with inclusion as a partial order? I wish I > had the mental presence or interest to know the answer right away. But > if the answer is yes you will have some maximal elements in S. (In > this case you will have to see how you can use them.) If the answer is > no, you can think of something else. > We can look into the case of E, later. It may be useful to read a paper by Alan Loper and F. >Tartarone. The title is something like: A classification of integrally >closed rings of polynomial containing Z[X]. It was a pre-print when I >saw. Hopefully it has appeared or may be in the process of being >published. Ok, I'll look at it, but the field Q(x) is not interesting for my > questions -- I already know the answers for that field. Also, there is a difference in point of view. I'm starting with a > field K and going down to some minimal subring satisfying my > conditions, whereas the paper is starting from a particular subring > and going up. I'm very much interested in whether the minimal subrings generally > exist, and if not, for which fields they do exist. Thus, my questions > relate to a property (as yet unclear) of the field itself. For a given field K, when the minimal subrings exist, are they unique? > When they exist but are not unique, are they at least isomorphic? My guess is that many of these questions can be settled in sci.math by > simple proofs or (more likely) counterexamples, depending on the > question. quasi- Hide quoted text - - Show quoted text - Producing new results takes interest, (valid reason for interest) > patience, techniques, a lot of hard work and at times willingness to > indulge in wild goose chase. > Muhammad- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - (1) My D-subalgebras of K are supposed to contain (a copy of) D. (2) I do not see a proof of S being inductive. Sorry. (3) It would be helpful if you could give a clear description of minimal. Muhammad === Subject: Re: -- minimal elements for certain classes of subrings of a field >(1) My D-subalgebras of K are supposed to contain (a copy of) D Sure, that's automatic from the terminology D-subalgebra. >(2) I do not see a proof of S being inductive. Sorry. Alternatively, and more directly tied to the problem, one can try to show ic(K) or ufd(K) is inductive (going down), but that's not so clear either. In fact, my guess is it can't be proved. In other words, I don't believe that all fields have minimal such rings. The question then becomes -- which ones do? >(3) It would be helpful if you could give a clear description of >minimal. Minimal subring in the specified set of subrings of K, and where the partial order on the specified set of subrings is just inclusion. quasi === Subject: Solutions to Mechanical Vibration (William J. Palm, III) posting-account=xvhWegoAAAAPtYypD1BouH-4sUGwfLkd Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) I am not really sure how this group works, but I have been searching for a way to get a copy of the solutions to Mechanical Vibration (William J. Palm, III) or Book ISBN:0-471-34555-5. I was hoping that I might be able to get some more information on how all of this works. calvin.anders@gmail.com === Subject: Re: Axiom of choice posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU 5.1),gzip(gfe),gzip(gfe) > there, you write that the infinite sum 1 + 1 + 1.... equals aleph > zero...*sigh*. > You need to study some maths. I don't see anything wrong with the statement that 1 + 1 + 1 ... equals aleph 0. -- If Tonio's objection is that aleph 0 is not a real number, then note that since the OP is discussing the Axiom of Choice, we are clearly dealing with set theory, not real analysis. So we may properly deal with cardinals/ordinals in a set theoreic setting. -- If Tonio's objection is that 1 + 1 + 1 ... containsa an ellipsis and is thus not really an infinite sum, then keep in mind that set theorists use this notation regularly, such as: epsilon 0 = 1 + omega + omega^omega + ... (See, for examplt the mathworld page: http://mathworld.wolfram.com/OrdinalNumber.html and scroll to the bottom for epsilon 0.) where one uses the + symbol to denote the limit, or union, of infinitely many ordinals. -- If Tonio's objection is that since these are ordinals, we should be using the ordinal omega, not the cardinal aleph 0, then note that set theorists identify the wellordered cardinals with ordinals. In ZFC, a cardinal is by definition, an ordinal that is not equinumerous with any smaller ordinal. Even in ZF (since the OP rejects AC), the class of all wellordered cardinals may still be taken as a subclass of the class of all ordinals. Thus we have that aleph 0 = omega is a theorem of ZF. In any case, 1 + 1 + 1 ... = aleph 0 is a theorem of ZF, when rewritten first using limit ordinal notation and then reduced to the primitives, is a theorem of ZF. Therefore, at least on this particular matter, the OP is correct, and Tonio's objection is not relevant here. === Subject: Re: Axiom of choice > In any case, 1 + 1 + 1 ... = aleph_0 is a > theorem of ZF, when rewritten first using limit > ordinal notation and then reduced to the > primitives, is a theorem of ZF. Written that way, I would say it is a cardinal statement. Addition of a family of cardinals is a known operation. And in this case, a true result. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Axiom of choice posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e SIMBAR Enabled; SIMBAR={70306B22-CB8C-4d52-BFF4-18424E217075}; MathPlayer 2.10b; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) there, you write that the infinite sum 1 + 1 + 1.... equals aleph > zero...*sigh*. > You need to study some maths. I don't see anything wrong with the statement that > 1 + 1 + 1 ... equals aleph 0. -- If Tonio's objection is that aleph 0 is not a > real number, then note that since the OP is > discussing the Axiom of Choice, we are clearly > dealing with set theory, not real analysis. So > we may properly deal with cardinals/ordinals in > a set theoreic setting. -- If Tonio's objection is that 1 + 1 + 1 ... > containsa an ellipsis and is thus not really an > infinite sum, then keep in mind that set > theorists use this notation regularly, such as: epsilon 0 = 1 + omega + omega^omega + ... (See, for examplt the mathworld page: http://mathworld.wolfram.com/OrdinalNumber.html and scroll to the bottom for epsilon 0.) where one uses the + symbol to denote the limit, > or union, of infinitely many ordinals. -- If Tonio's objection is that since these are > ordinals, we should be using the ordinal omega, > not the cardinal aleph 0, then note that set > theorists identify the wellordered cardinals > with ordinals. In ZFC, a cardinal is by > definition, an ordinal that is not equinumerous > with any smaller ordinal. Even in ZF (since the > OP rejects AC), the class of all wellordered > cardinals may still be taken as a subclass of > the class of all ordinals. Thus we have that > aleph 0 = omega is a theorem of ZF. In any case, 1 + 1 + 1 ... = aleph 0 is a > theorem of ZF, when rewritten first using limit > ordinal notation and then reduced to the > primitives, is a theorem of ZF. Therefore, at least on this particular matter, > the OP is correct, and Tonio's objection is > not relevant here. ******************************************************** Correction: it may not be relevant for you since you didn't even understand what the OP meant, most probably because you didn't read I quote from the OP's link that he himself posted in his first post: Negation of the axiom of choice and Evil Negation of the axiom of choice and Evil Beside the particular case of the axiom of choice CC(2 through m), countable choice for sets of n elements n=2 through m, there is the particular case where the whole axiom is negated, no choice at all. In All things are numbers in Logic Colloquium 2001, and in About numbers of attributes of Evil are infinite sums and products of integers. If there is no choice at all, which is a standing valid case, as I am changing my mind to include this cas beside CC(2 through m). The only infinite sums or products well defined (existing) is the infinite sum 1+1+1+.... which is equal to aleph zero. So, Evil is really restricted.... Now read the above,and if you still think that my criticism of the OP's post, which was linked to his work in his site, is not relevant, then go ahead. Oh, and beware of The Evil. Tonio Pd. It was funny seeing Tommy1729, who has posted so ashtonishing ammounts of nonsenses and cranky posts in the few last months (just take a look at his last Riemann Series Theorem and divergent series....hilarious!), trying to defend the OP... === Subject: Re: Axiom of choice posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU 5.1),gzip(gfe),gzip(gfe) > Therefore, at least on this particular matter, > the OP is correct, and Tonio's objection is > not relevant here. > Correction: it may not be relevant for you since you didn't even > understand what the OP meant, most probably because you didn't read > Now read the above,and if you still think that my criticism of the > OP's post, which was linked to his work in his site, is not > relevant, then go ahead. At first I thought the OP was just arguing why ZF+~AC may be superior to ZFC, but all this about Choice being evil is a little over the top. The only evil number, of course, is 666... === Subject: Re: Axiom of choice > there, you write that the infinite sum 1 + 1 + > 1.... equals aleph > zero...*sigh*. > You need to study some maths. I don't see anything wrong with the statement that > 1 + 1 + 1 ... equals aleph_0. neither do i. Omega would have been good to, but thats an ordinal. however tonico and dennis feldmann said feldmann ( -1/2 ) tonico ( forgot what he said , not aleph or omega at least) notice that these 2 are big enemies of my set theorems and of me in general. but even by the standard set theory they accept , they are wrong !! they are just critics !!! claiming to defend standard theory , yet apparantly dont even understand that. its a typical reaction for critics of me. however i do need to give dennis 1 point since he probably is aware of zeta-regularized sums , but thats not set theory !! -- If Tonio's objection is that aleph_0 is not a > real number, then note that since the OP is > discussing the Axiom of Choice, we are clearly > dealing with set theory, not real analysis. So > we may properly deal with cardinals/ordinals in > a set theoreic setting. agreed. -- If Tonio's objection is that 1 + 1 + 1 ... > containsa an ellipsis and is thus not really an > infinite sum, then keep in mind that set > theorists use this notation regularly, such as: epsilon_0 = 1 + omega + omega^omega + ... agreed. though i hate epsilon and omega. (See, for examplt the mathworld page: http://mathworld.wolfram.com/OrdinalNumber.html and scroll to the bottom for epsilon_0.) where one uses the + symbol to denote the limit, > or union, of infinitely many ordinals. -- If Tonio's objection is that since these are > ordinals, we should be using the ordinal omega, > not the cardinal aleph_0, then note that set > theorists identify the wellordered cardinals > with ordinals. In ZFC, a cardinal is by > definition, an ordinal that is not equinumerous > with any smaller ordinal. Even in ZF (since the > OP rejects AC), the class of all wellordered > cardinals may still be taken as a subclass of > the class of all ordinals. Thus we have that > aleph_0 = omega is a theorem of ZF. agreed. in ZF and my set theories. one wonders why omega is still written since it equals aleph_0 !! yes ordinal , but infinite ( *rolleyes* ) like i said ; i hate omega. (and epsilon ) just another cantor blunder or whoever invented it. ( up to date i only support Riemann and Hilbert's Hotel ideas of infinity ) ( and timothy golden :p ) In any case, 1 + 1 + 1 ... = aleph_0 is a > theorem of ZF, when rewritten first using limit > ordinal notation and then reduced to the > primitives, is a theorem of ZF. Therefore, at least on this particular matter, > the OP is correct, and Tonio's objection is > not relevant here. indeed. Tonio is not a very good critic. his bad reply almost strenghtens the OP. same for dennis feldmann (confusing calculus with set theory ) oh well. tommy1729 === Subject: Re: Axiom of choice <1720646.1200700300575.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > aleph 0 = omega is a theorem of ZF. > agreed. in ZF and my set theories. > one wonders why omega is still written since it equals aleph 0 !! > yes ordinal , but infinite ( *rolleyes* ) > like i said ; i hate omega. (and epsilon ) > just another cantor blunder or whoever invented it. Even though aleph 0 = omega in ZF, the reason we distinguish between the two is mainly because cardinal arithmetic differs greatly from ordinal arithmetic. With cardinals, aleph 0 + 1 = aleph 0. This is simply telling us that it is possible to add one person to a full Hilbert Hotel -- we see below that tommy1729 himself supports the Hilbert Hotel model: > ( up to date i only support Riemann and Hilbert's Hotel ideas of infinity ) But with ordinals, omega + 1 is distinct from omega. The reason is that, by definition order makes a difference with ordinals. With a set of order type omega, there exists only one element with no predecessor (the first one), but with a set of order type omega + 1, there exist two elements with no predecessor (the first and the last). But then, we might ask, why can't we simply add 1 to both sides of the equation aleph 0 = omega to get aleph 0 + 1 = omega + 1, and thus conclude that aleph 0 + 1 = aleph 0 if and only if omega + 1 = omega? The reason is that the symbol + is actually representing two distinct operations -- one being cardinal addition and the other being ordinal addition. On the Metamath website, one distinguishes between the two by writing + CR for the former and + o for the latter. But most set theorists do not write + CR and + o -- they simply write + for both operations. Yet it is no more valid to assume that cardinal addition and ordinal addition are the same operation, any more than it is to assume that addition and multiplication are the same operation. In computer science, we refer to the use of one symbol to represent two distinct operations operator overloading. And yes, believe it or not, mathematicians overload operators all of the time! We use + for cardinal addition, ordinal addition, and analytic (real or complex) addition, and ditto for multiplication and exponentiation. Notice that mathematicians justify the use of + for both cardinal and ordinal addition by pointing out that they agree for finite cardinals. (Then again, addition and multiplication actually agree for wellordered infinite cardinals -- which are all the infinite cardinals in ZFC -- yet no one uses that as a reason to use the same symbol for both cardinal addition and multiplication!) Metamath, on the other hand, never overloads operators. Even when that site is constructing the real numbers from Dedekind cuts, different symbols are used for ordinal addition (+ o), rational addition (+ Q), Dedekind cut addition (+ P), and finally complex addition (+). The same is true of the numbers themselves, where the site distinguishes the ordinal 1 (1 o) from the rational 1 (1 Q), the Dedekind cut 1 (1 P), and the complex number 1. (Notice that, for example, although mathematicians identity the set of finite ordinals with a subset of the rationals, reals, etc., they are not the same sets as far as the Axiom of Extensionality is concerned. The ordinal 1 is just the singleton of the empty set, while the rational 1 contains infinitely many pairs of ordinals, etc.) Only complex numbers and operators are written without some sort of subscript (like CR, o, Q, etc.). Since mathematicians write + for both cardinal and ordinal addition, how do they distinguish between the two? They do so by writing the operands as cardinals or ordinals. So aleph 0 + 1 denotes cardinal addition, while omega + 1 denotes ordinal addition. But strictly speaking, aleph 0 and omega are exactly the same set -- the two instances of + are what differ. In Metamath, we would write the cardinal addition as omega +CR 1 and the ordinal addition as omega +o 1, thereby emphasizing that the true difference between the two is the operator , not the operands. === Subject: Re: Axiom of choice posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG 2.0.50727; .NET CLR 3.0.04506.30; InfoPath.2),gzip(gfe),gzip(gfe) > The axiom of choice is false because its negation is applied > successfully in > http://jebara.topcities.com > You need to study some maths. I don't see anything wrong with the statement that > 1 + 1 + 1 ... equals aleph_0. Why not say it is equal to w (omega) instead? Or perhaps just is undefined? At any rate, Adib Ben Jebara's theorems are meaningless in the realm of mathematics. Perhaps he'd care to define exactly what an evil number is? I'd also like to know how an axiom can be proved false. Successfully applying the negation of an axiom only shows that the negated axiom can be used to form a consistent axiomatic system. But it says nothing about the original axiom. === Subject: Re: Axiom of choice > adib.jebara@topnet.tn a .8ecrit : > Nonsense, as expected. In one of your rather > there, you write that the infinite sum 1 + 1 + > 1.... equals aleph > zero...*sigh*. > You need to study some maths. > Tonio To what is equal 1+1+1+....then ? > To -1/2 (see > http://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_% > C2%B7_%C2%B7_%C2%B7 ) Besides this, *plonk* until you have studied some > maths... > Adib Ben Jebara. to -1/2 dennis ? well , i cant say your totally wrong ; you used regularized zeta sum for that i assume. ( did not look at your link ) but ive never seen 1+1+1+1+1+... to be -1/2 in any set theory context. many types of zero's and infinities depending on the axioms accepted or denied, but never -1/2 ... what are talking about seems like regularized zeta sums. this is not what the OP was referring to. however i dont agree with the OP for clarity. tommy1729 === Subject: Re: -- nonunique factorization (clarified) a finite field is a galois field. >so i am intrested in non-unique factorization in > galois fields. If you read some of the replies in the other thread, > you should now >realize that fields lack irreducible elements, so > are trivially (and >vacuously) always UFDs (unique factorization > domains). For finite fields, you get can another kind of > unique factorization >as follows ... Let K be a finite field. Then the multiplicative > group of nonzero >elements is always cyclic. Choose a group generator, > x say. If K has m >elements, then the group has order m-1, so every > nonzero element other >than 1 has a unique representation in the form x^k > where k is a >positive integer <= m-1. Correction: where k is a positive integer _less_ than > m-1. (since I chose to exclude 1) > >So finite fields can't exhibit non-unique > factorization directly. However infinite extensions of finite fields can > have subrings which >lack unique factorization. quasi are those infinite extensions of the finite fields then isomorphic to rings , such that my two intrests are actually one and the same ? i think so if i understood correctly. tommy1729 ps : so what remains are non-unique factorizations of rings and rings with mod ( what the name for it ? ) srr im not an expert at this. === Subject: Re: -- nonunique factorization (clarified) > > a finite field is a galois field. >so i am intrested in non-unique factorization in > galois fields. >If you read some of the replies in the other thread, > you should now >realize that fields lack irreducible elements, so > are trivially (and >vacuously) always UFDs (unique factorization > domains). >For finite fields, you get can another kind of > unique factorization >as follows ... >Let K be a finite field. Then the multiplicative > group of nonzero >elements is always cyclic. Choose a group generator, > x say. If K has m >elements, then the group has order m-1, so every > nonzero element other >than 1 has a unique representation in the form x^k > where k is a >positive integer <= m-1. > > Correction: where k is a positive integer _less_ than > m-1. > > (since I chose to exclude 1) > >So finite fields can't exhibit non-unique > factorization directly. >However infinite extensions of finite fields can > have subrings which >lack unique factorization. > > quasi are those infinite extensions of the finite fields then isomorphic >o rings , such that my two intrests are actually one and the same ? i think so if i understood correctly. No, the extensions are fields, but they may have subrings which are not UFDs. >ps : so what remains are non-unique factorizations of rings Yes, essentially, but for another interpretation, see my the thread -- factorization in fields (inspired by your question) >rings with mod ( what the name for it ? ) If you mean rings such that m*1 = 0 for some positive integer m -- for example, Z_m (i.e., Z/mZ), then these are called rings of characteristic m. Now suppose R is a commutative ring R (with 1) of characteristic m. If m is prime and R is finite, then R must be a field, so, as already discussed, there are no irreducible elements. If m is composite, then R has zero divisors, so R can't be a subring of any field. Even though there are zero divisors, one can discuss factorization, but the concept of irreducible elements can be problematic. For example in the ring Z_6, the nonunits are 2,3,4. Are any of those irreducible? For example, note that 2*2*2 = 2. In any case, for fields, there are no irreducibles, so when discussing factorization in a field, in order to have some irreducibles to factor things with, one typically restricts the domain of factorization to an appropriate subring of the field. >srr im not an expert at this. That's ok. Rings are strange beasts, hard to get to know. Whenever you think you know some general fact about them, unless it's trivially true, it's probably false. Worse, even if it is true, they will typically refuse to admit it. quasi === <20080118151233.676$Iq@newsreader.com> posting-account=a6woBRAAAADpNFZJBA7ZBx35zXaKmaP4 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) I am about to lose a bet with a co-worker. I need help with this one. Determine all ordered pairs (m, n) of positive integers >for which (n^3 + 1)/(mn - 1) is an integer. This is equivalent to asking for a divisor of n^3 + 1 >congruent to -1 mod n. Such divisors occur in pairs >since k = -1 mod n implies (n^3 + 1)/k = -1 mod n. There are trivial cases (2,1) and (2,2). For n > 2, neither n+1 nor n^2 - n + 1 can have a divisor >congruent to -1 mod n. But it is possible to have such a >divisor of n^3 + 1 whose factors split across these two. For example, with n = 5, n^3 + 1 = 126 = 9 * 14 with both >factors congruent to -1 mod 5. What exactly is the bet or conjecture (if there is one)? I suspect the conjecture is that the only solutions are > (n,m) = (2,1), (2,2), (2,5), (3,1), (3,5), (5,2) and (5,3). > There are no others for n up to 10000, if my Maple program is > correct. You seem to have overlooked solutions having n = 1. David When these are included, the symmetry of the solution set is surprising: (n,m) <--> (m,n) (1,2),(1,3),(2,1),(2,2),(2,5),(3,1),(3,5),(5,2),(5,3) --c === > I am about to lose a bet with a co-worker. I need help with this > one. > Determine all ordered pairs (m, n) of positive integers > for which (n^3 + 1)/(mn - 1) is an integer. This is equivalent to asking for a divisor of n^3 + 1 >congruent to -1 mod n. Such divisors occur in pairs >since k = -1 mod n implies (n^3 + 1)/k = -1 mod n. There are trivial cases (2,1) and (2,2). For n > 2, neither n+1 nor n^2 - n + 1 can have a divisor >congruent to -1 mod n. But it is possible to have such a >divisor of n^3 + 1 whose factors split across these two. For example, with n = 5, n^3 + 1 = 126 = 9 * 14 with both >factors congruent to -1 mod 5. What exactly is the bet or conjecture (if there is one)? I suspect the conjecture is that the only solutions are >(n,m) = (2,1), (2,2), (2,5), (3,1), (3,5), (5,2) and (5,3). >There are no others for n up to 10000, if my Maple program is >correct. You seem to have overlooked solutions having n = 1. David When these are included, the symmetry of the solution > set is surprising: (n,m) <--> (m,n) (1,2),(1,3),(2,1),(2,2),(2,5),(3,1),(3,5),(5,2),(5,3) Hmm. I have a nagging feeling that the symmetry shouldn't surprise us. Indeed, it might be the case that, if the power 3 in the original expression is changed to any nonnegative integer p, -- so that we are asked to determine all ordered pairs (m, n) of positive integers for which (n^p + 1)/(mn - 1) is an integer -- we still find the same symmetry. David === > I am about to lose a bet with a co-worker. I need help with this > one. > Determine all ordered pairs (m, n) of positive integers > for which (n^3 + 1)/(mn - 1) is an integer. >This is equivalent to asking for a divisor of n^3 + 1 >congruent to -1 mod n. Such divisors occur in pairs >since k = -1 mod n implies (n^3 + 1)/k = -1 mod n. >There are trivial cases (2,1) and (2,2). >For n > 2, neither n+1 nor n^2 - n + 1 can have a divisor >congruent to -1 mod n. But it is possible to have such a >divisor of n^3 + 1 whose factors split across these two. >For example, with n = 5, n^3 + 1 = 126 = 9 * 14 with both >factors congruent to -1 mod 5. >What exactly is the bet or conjecture (if there is one)? >I suspect the conjecture is that the only solutions are >(n,m) = (2,1), (2,2), (2,5), (3,1), (3,5), (5,2) and (5,3). >There are no others for n up to 10000, if my Maple program is >correct. > You seem to have overlooked solutions having n = 1. > David > When these are included, the symmetry of the solution > set is surprising: > (n,m) <--> (m,n) > (1,2),(1,3),(2,1),(2,2),(2,5),(3,1),(3,5),(5,2),(5,3) Hmm. I have a nagging feeling that the symmetry shouldn't surprise us. >Indeed, it might be the case that, if the power 3 in the original >expression is changed to any nonnegative integer p, -- so that we are asked to determine all ordered pairs (m, n) of positive >integers for which (n^p + 1)/(mn - 1) is an integer -- we still find the same symmetry. Not so surprising ... If n is a positive integer such that, for some positive integer a, n^3 + 1 is divisible by (an - 1) then a^3 + 1 is also divisible by (an - 1) proof: Suppose n is a positive integer such that, for some positive integer a, n^3 + 1 is divisible by (an - 1) Let x = an - 1. Then an = 1 (mod x) and n^3 = -1 (mod x). But n^3 = -1 (mod x) => a^3*n^3 = -a^3 (mod x) => (an)^3 = -a^3 (mod x) => 1 = -a^3 (mod x) => a^3 + 1 = 0 (mod x) which proves the claim. Thus, any pair (n,a) induces the pair (a,n). This explains the apparent symmetry. The above proof would work just as well for n^k + 1, where k is an arbitrary positive integer, confirming David W. Cantrell's intuition in this regard. quasi === <20080118151233.676$Iq@newsreader.com> <20080118191117.508$B0@newsreader.com> posting-account=VR0DOgoAAADggPTteFeA2AkmHNhjcrDV Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > I am about to lose a bet with a co-worker. I need help with this > one. > Determine all ordered pairs (m, n) of positive integers > for which (n^3 + 1)/(mn - 1) is an integer. > This is equivalent to asking for a divisor of n^3 + 1 > congruent to -1 mod n. Such divisors occur in pairs > since k = -1 mod n implies (n^3 + 1)/k = -1 mod n. > There are trivial cases (2,1) and (2,2). > For n > 2, neither n+1 nor n^2 - n + 1 can have a divisor > congruent to -1 mod n. But it is possible to have such a > divisor of n^3 + 1 whose factors split across these two. > For example, with n = 5, n^3 + 1 = 126 = 9 * 14 with both > factors congruent to -1 mod 5. > What exactly is the bet or conjecture (if there is one)? I suspect the conjecture is that the only solutions are >(n,m) = (2,1), (2,2), (2,5), (3,1), (3,5), (5,2) and (5,3). >There are no others for n up to 10000, if my Maple program is >correct. You seem to have overlooked solutions having n = 1. David When these are included, the symmetry of the solution > set is surprising: (n,m) <--> (m,n) (1,2),(1,3),(2,1),(2,2),(2,5),(3,1),(3,5),(5,2),(5,3) Hmm. I have a nagging feeling that the symmetry shouldn't surprise us. > Indeed, it might be the case that, if the power 3 in the original > expression is changed to any nonnegative integer p, -- so that we are asked to determine all ordered pairs (m, n) of positive > integers for which (n^p + 1)/(mn - 1) is an integer -- we still find the same symmetry. David The symmetry is not surprising. If mn - i divides n^3 + 1, then mn - 1 must divide m^3n^3 + m^3. Use polynomial division treating n as the variable. You will find that the remainder is m^3 + 1. It follows that if mn - 1 divides n^3 + 1, then it must divide m^3 + 1. So by symmetry mn - 1 divides n^3 + 1 iff mn - 1 divides m^3 - 1. Thus if the pair (a, b) is good then the pair (b, a) must be good. === > I am about to lose a bet with a co-worker. I need help with this > one. > Determine all ordered pairs (m, n) of positive integers > for which (n^3 + 1)/(mn - 1) is an integer. > This is equivalent to asking for a divisor of n^3 + 1 > congruent to -1 mod n. Such divisors occur in pairs > since k = -1 mod n implies (n^3 + 1)/k = -1 mod n. > There are trivial cases (2,1) and (2,2). > For n > 2, neither n+1 nor n^2 - n + 1 can have a divisor > congruent to -1 mod n. But it is possible to have such a > divisor of n^3 + 1 whose factors split across these two. > For example, with n = 5, n^3 + 1 = 126 = 9 * 14 with both > factors congruent to -1 mod 5. > What exactly is the bet or conjecture (if there is one)? >I suspect the conjecture is that the only solutions are >(n,m) = (2,1), (2,2), (2,5), (3,1), (3,5), (5,2) and (5,3). >There are no others for n up to 10000, if my Maple program is >correct. >You seem to have overlooked solutions having n = 1. >David > When these are included, the symmetry of the solution > set is surprising: > (n,m) <--> (m,n) > (1,2),(1,3),(2,1),(2,2),(2,5),(3,1),(3,5),(5,2),(5,3) > Hmm. I have a nagging feeling that the symmetry shouldn't surprise us. > Indeed, it might be the case that, if the power 3 in the original > expression is changed to any nonnegative integer p, > -- so that we are asked to determine all ordered pairs (m, n) of positive > integers for which (n^p + 1)/(mn - 1) is an integer -- > we still find the same symmetry. > David >The symmetry is not surprising. If mn - i divides n^3 + 1, then mn - >1 must divide >m^3n^3 + m^3. Use polynomial division treating n as the variable. >You will find that the remainder is m^3 + 1. It follows that if mn - >1 divides n^3 + 1, then it must divide m^3 + 1. So by symmetry mn - 1 divides n^3 + 1 iff mn - 1 divides m^3 - 1. >Thus if the pair (a, b) is good then the pair (b, a) must be >good. Ha! I just (independently) posted a similar comment, and with almost the same starting words. quasi === >I am about to lose a bet with a co-worker. I need help with this one. >Determine all ordered pairs (m, n) of positive integers for which (n^3 + 1)/(mn - 1) is an integer. Suppose n^3 + 1 is divisible by an - 1. Let z = n^3 + 1 and x = an - 1. Then z = xy for some positive integer y. Since z = 1 (mod n) and x = (-1) mod n, it follows that y = (-1) mod (n) > >Hence y = bn - 1 for some positive integer b. The equation z = xy is then n^3 + 1 = (an - 1)*(bn -1) > >Without loss of generality, assume a <= b. > >Expand, subtract 1, then divide by n. This yields n^2 - (ab)n + (a + b) = 0 > >Then the discriminant D = (ab)^2 - 4(a + b) > >must be a perfect square. Since D is a perfect square and (ab)^2 > D, we must have (ab - 1)^2 >= D, > >so we have the inequality (ab - 1)^2 >= (ab)^2 - 4(a + b) > >which simplifies to (a - 2)*(b - 2) <= 9/2 > >Hence (a - 2)*(b - 2) <= 4 > > Since a <= b we must have a <= 4. If a = 4, then solving (a - 2)*(b - 2) <= 4 for b yields b <= 4, >hence, a = b = 4, but then D = 224 which is not a perfect square, >contradiction. If a = 3, then we get 3 <= b <= 6, which yields 4 possible pairs >(a,b). but in each case, the discriminant is not a perfect square. If a = 2, then D = 4b^2 - 4b - 8 = (2b - 1)^2 - 9 Hence, since D is a perfect square, we must have (2b - 2)^2 >= D. Then (2b - 2)^2 >= 4b^2 - 4b - 8 gives b <= 3. If a = 2 and b = 2, then n = 2. If a = 2 and b = 3, then n = 5. Finally, if a = 1, then D = b^2 - 4b - 4 = (b - 2)^2 - 8. Hence, since D is a perfect square, we must have (b - 3)^2 >= D. Then (b - 3)^2 >= b^2 - 4b - 4 gives b <= 6. For b = 1,2,3,4,5,6 and a = 1, D is only a perfect square if b = 5, >which gives the solution a = 1, b = 5, n = 3. In summary, the only solutions (a,b,n) are (2,2,2) > (2,3,5) > (1,5,3). > >In particular, for the original problem, the only pairs (m,n) are (1,3) > (2,2) > (2,5) > (3,5) > >quasi Hmmm ... It appears that I missed a few solutions. I'm pretty sure my basic solution strategy is correct. Probably an elementary algebra error somewhere. I'll check my work. quasi === >I am about to lose a bet with a co-worker. I need help with this one. >Determine all ordered pairs (m, n) of positive integers for which (n^3 + 1)/(mn - 1) is an integer. >Suppose n^3 + 1 is divisible by an - 1. >Let z = n^3 + 1 and x = an - 1. >Then z = xy for some positive integer y. >Since z = 1 (mod n) and x = (-1) mod n, it follows that > y = (-1) mod (n) > >Hence y = bn - 1 for some positive integer b. >The equation z = xy is then > n^3 + 1 = (an - 1)*(bn -1) > >Without loss of generality, assume a <= b. > >Expand, subtract 1, then divide by n. This yields > n^2 - (ab)n + (a + b) = 0 > >Then the discriminant > D = (ab)^2 - 4(a + b) > >must be a perfect square. >Since D is a perfect square and (ab)^2 > D, we must have > (ab - 1)^2 >= D, > >so we have the inequality > (ab - 1)^2 >= (ab)^2 - 4(a + b) > >which simplifies to > (a - 2)*(b - 2) <= 9/2 > >Hence > (a - 2)*(b - 2) <= 4 > > Since a <= b we must have a <= 4. >If a = 4, then solving (a - 2)*(b - 2) <= 4 for b yields b <= 4, >hence, a = b = 4, but then D = 224 which is not a perfect square, >contradiction. >If a = 3, then we get 3 <= b <= 6, which yields 4 possible pairs >(a,b). but in each case, the discriminant is not a perfect square. >If a = 2, then D = 4b^2 - 4b - 8 = (2b - 1)^2 - 9 >Hence, since D is a perfect square, we must have (2b - 2)^2 >= D. >Then (2b - 2)^2 >= 4b^2 - 4b - 8 gives b <= 3. >If a = 2 and b = 2, then n = 2. >If a = 2 and b = 3, then n = 5. >Finally, if a = 1, then D = b^2 - 4b - 4 = (b - 2)^2 - 8. >Hence, since D is a perfect square, we must have (b - 3)^2 >= D. >Then (b - 3)^2 >= b^2 - 4b - 4 gives b <= 6. >For b = 1,2,3,4,5,6 and a = 1, D is only a perfect square if b = 5, >which gives the solution a = 1, b = 5, n = 3. >In summary, the only solutions (a,b,n) are > (2,2,2) > (2,3,5) > (1,5,3). > >In particular, for the original problem, the only pairs (m,n) are > (1,3) > (2,2) > (2,5) > (3,5) > >quasi Hmmm ... It appears that I missed a few solutions. I'm pretty sure my basic solution strategy is correct. Probably an elementary algebra error somewhere. I'll check my work. Ok, I found the error. I neglected I smaller solution of the quadratic equation when I solved for n. With that correction get 2 more triples. Writing the triples this time as (n,a,b), the only triples are (1,2,3) (2,1,5) (2,2,2) (3,1,5) (5,2,3) and for the original problem, the only pairs (n,m) are (1,2) (1,3) (2,1) (2,2) (2,5) (3,1) (3,5) (5,2) (5,3) quasi === Subject: Re: #17 recent news of antimatter in Milky Way galactic center and black holes posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG 2.0.50727; .NET CLR 3.0.04506.30; InfoPath.2),gzip(gfe),gzip(gfe) > There is some news recently of discovered Antimatter clouds near the > Milky Way galactic center. Of course this is more evidence that black-holes are science fiction > and science fakery. Okay, I'll bite. How does the existence of antimatter disprove the existence of black holes? What do your equations of gravity tell you about an object with a mass greater than 10 solar masses? === Subject: Re: #17 recent news of antimatter in Milky Way galactic center and black holes posting-account=fsC03QkAAAAwkSNcSEKmlcR-W_HNitEd Gecko/20021120 Netscape/7.01,gzip(gfe),gzip(gfe) > There is some news recently of discovered Antimatter clouds near the > Milky Way galactic center. Of course this is more evidence that black-holes are science fiction > and science fakery. Okay, I'll bite. How does the existence of antimatter disprove the existence > of black holes? What do your equations of gravity tell you about an object with > a mass greater than 10 solar masses? No, you do not need to bite. What you need is to learn a basic principle of discussion. That you engage in discussion only with people who you have mutual respect, whether you disagree or agree with their ideas, there is at least a parity of respect. Somehow your immaturity never learned this. And so you should not engage in any of my threads for your mind is very much imbalanced, and your mind is not focused on science but some of your pitiful psychological chips on your shoulder. You should have studied psychology but not any of the hard sciences === Subject: Re: a^b = b^a ... commutative exponentiation and R ^^ R : set tetration And even > Andrew Robbins, the inventor of the extension of > tetration to > the reals, Andrew Robbins is the inventor of ONE particular > extension to the reals. He is > not the inventor of any THE extension to the reals, > as the criteria for what > constitutes THE tetration extension are variable. Robbins' extension is no more valid than any other > attempt, particularly since > he hasn't published it in a peer-reviewed journal, > showing its supposed > superiority over other tetration extensions. > -- > I.N. Galidakis > i am aware you worked in this field too sir. and i respect you for that. an inventer of an extension. happy now ? ;-) Andrew Robbins is a nice example : he did work on tetration + is critical about cantor = => not accepted in peer reviewed journals, despite his good math. as JSH would say : it's a conspiracy !!! and Robbins isnt the only example its however not that much the mathematicians but rather the media around them like mags that sell lies. (however there is still a cantor conspiracy ) ironicly JSH is a good example to it. he made it to a math magazine and Robbins dealing with tetration did not !!! and many good pdf's on the internet of talented pro and amateur mathematicians deserve a place in a math mag ... but it will never happen... dont be mad on lwalke or l.Renfro , there nice people and good mathematicians and they did not try to defend you. tommy1729 === Subject: Help posting-account=tg4q8goAAABTkYMicvXSqOY9lBrlUs6z Dealio Toolbar 3.1.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Hi I need the solution manual for Digital Communications: Fundamentals and Application === Subject: -- Birational? If E is a submanifold of R^m and F is a submanifold of R^n, and if a given homeomorphism f: E --> F maps all points of E having rational coordinates to points of F having rational coordinates, and if f^{-1} maps all points of F having rational coordinates to points of E having rational coordinates, must the coordinate (component) functions of f and f^{-1} be quotients of polynomial functions with integer coefficients? (I'm not sure of the correct terminology to use. On the Web, I looked up birational, which I'd vaguely heard of, and was led to a short MathWorld entry about Weber's Theorem, using the term rational correspondence - which however is not defined there. Is a rational correspondence in fact what I'm talking about?) (The example leading to my question was Exercise 5.6.2 in John Stillwell's Mathematics and Its History (2nd ed. 2002). This involves a mapping between R^2 and the cone x^2 + y^2 = z^2 in R^3. One direction is given by x = (u^2 - v^2)/v, y = 2u, and z = (u^2 + v^2)/v, and the other by u = y/2 and v = (z - x)/2. To make this work, you have to delete the lines given by (x = z and y = 0) in R^3, and v = 0 in R^2. In the exercise, (x, y, z) are the lengths of the sides of a right-angled triangle, so you can assume they are all strictly positive. The main interest in the parameterisation of (x, y, z) by (u, v) is that it can be used to describe all triangles with rational side lengths: this was done by Brahmagupta in 628 CE.) (Totally irrelevant niggle: rather than writing R^3 and R^2, it would surely be clearer to write something like R^{'x','y','z'} and R^{'u','v'}. I probably shouldn't mention it, but this kind of thing repeatedly annoys me, so I'm just letting off a little steam.) -- Angus Rodgers Contains mild peril === Subject: Re: -- Birational? >[...] The main interest >in the parameterisation of (x, y, z) by (u, v) is that it can >be used to describe all triangles with rational side lengths >[...] Oops! I left out the phrase ... and rational area. -- Angus Rodgers Contains mild peril === Subject: Re: -- Birational? > If E is a submanifold of R^m and F is a submanifold of R^n, and > if a given homeomorphism f: E --> F maps all points of E having > rational coordinates to points of F having rational coordinates, > and if f^{-1} maps all points of F having rational coordinates > to points of E having rational coordinates, must the coordinate > (component) functions of f and f^{-1} be quotients of polynomial > functions with integer coefficients? No, take m = n = 1, E = F = R. Define f : R -> R to be x on (-oo, 0], 2x on [0, oo). > (I'm not sure of the correct terminology to use. On the Web, I > looked up birational, which I'd vaguely heard of, and was led > to a short MathWorld entry about Weber's Theorem, using the term > rational correspondence - which however is not defined there. > Is a rational correspondence in fact what I'm talking about?) (The example leading to my question was Exercise 5.6.2 in John > Stillwell's Mathematics and Its History (2nd ed. 2002). This > involves a mapping between R^2 and the cone x^2 + y^2 = z^2 in > R^3. One direction is given by x = (u^2 - v^2)/v, y = 2u, and > z = (u^2 + v^2)/v, and the other by u = y/2 and v = (z - x)/2. > To make this work, you have to delete the lines given by (x = z > and y = 0) in R^3, and v = 0 in R^2. In the exercise, (x, y, z) > are the lengths of the sides of a right-angled triangle, so you > can assume they are all strictly positive. The main interest > in the parameterisation of (x, y, z) by (u, v) is that it can > be used to describe all triangles with rational side lengths: > this was done by Brahmagupta in 628 CE.) (Totally irrelevant niggle: rather than writing R^3 and R^2, it > would surely be clearer to write something like R^{'x','y','z'} > and R^{'u','v'}. I probably shouldn't mention it, but this kind > of thing repeatedly annoys me, so I'm just letting off a little > steam.) === Subject: Re: -- Birational? > If E is a submanifold of R^m and F is a submanifold of R^n, and > if a given homeomorphism f: E --> F maps all points of E having > rational coordinates to points of F having rational coordinates, > and if f^{-1} maps all points of F having rational coordinates > to points of E having rational coordinates, must the coordinate > (component) functions of f and f^{-1} be quotients of polynomial > functions with integer coefficients? No, take m = n = 1, E = F = R. Define f : R -> R to be x on (-oo, 0], >2x on [0, oo). So much for mere homeomorphisms, then! But what if f and f^{-1} are supposed to be [continuously] differentiable, or even smooth? -- Angus Rodgers Contains mild peril === Subject: Re: -- Birational? > If E is a submanifold of R^m and F is a submanifold of R^n, and > if a given homeomorphism f: E --> F maps all points of E having > rational coordinates to points of F having rational coordinates, > and if f^{-1} maps all points of F having rational coordinates > to points of E having rational coordinates, must the coordinate > (component) functions of f and f^{-1} be quotients of polynomial > functions with integer coefficients? >No, take m = n = 1, E = F = R. Define f : R -> R to be x on (-oo, 0], >2x on [0, oo). So much for mere homeomorphisms, then! But what if f and f^{-1} >are supposed to be [continuously] differentiable m = n = 1 E = R F = (-oo, 1) f(x) = x (x <= 0) f(x) = x/(1 + x) (x >= 0) f^{-1}(y) = y (y <= 0) f^{-1}(y) = y/(1 - y) (0 <= y < 1) >or even smooth? Or C^2? -- Angus Rodgers Contains mild peril === Subject: Re: -- Birational? <0gg3p3t91e7a55ggqfavhpqfl87nmugvpm@4ax.com> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20071213 Fedora/2.0.0.10-3.fc8 Firefox/2.0.0.10,gzip(gfe),gzip(gfe) > If E is a submanifold of R^m and F is a submanifold of R^n, and > if a given homeomorphism f: E --> F maps all points of E having > rational coordinates to points of F having rational coordinates, > and if f^{-1} maps all points of F having rational coordinates > to points of E having rational coordinates, must the coordinate > (component) functions of f and f^{-1} be quotients of polynomial > functions with integer coefficients? >No, take m = n = 1, E = F = R. Define f : R -> R to be x on (-oo, 0], >2x on [0, oo). So much for mere homeomorphisms, then! But what if f and f^{-1} >are supposed to be [continuously] differentiable m = n = 1 > E = R > F = (-oo, 1) > f(x) = x (x <= 0) > f(x) = x/(1 + x) (x >= 0) > f^{-1}(y) = y (y <= 0) > f^{-1}(y) = y/(1 - y) (0 <= y < 1) or even smooth? Or C^2? You can play that game all the way up to C^infty. -- m === Subject: Re: -- Birational? >So much for mere homeomorphisms, then! But what if f and f^{-1} >are supposed to be [continuously] differentiable > m = n = 1 > E = R > F = (-oo, 1) > f(x) = x (x <= 0) > f(x) = x/(1 + x) (x >= 0) > f^{-1}(y) = y (y <= 0) > f^{-1}(y) = y/(1 - y) (0 <= y < 1) >or even smooth? > Or C^2? You can play that game all the way up to C^infty. I can't even see what the next move should be, let alone what might happen all the way up to (and including?) C^oo. I can only see that for r >= 0, one might have tried defining f(x) = x + x^{r+1} (x >= 0), so that f is in C^r, but that's no good, because f^{-1} also has to preserve the set of rationals. Am I missing some trick that can be carried on for all r? -- Angus Rodgers Contains mild peril === Subject: Re: -- Birational? >So much for mere homeomorphisms, then! But what if f and f^{-1} >are supposed to be [continuously] differentiable > m = n = 1 > E = R > F = (-oo, 1) > f(x) = x (x <= 0) > f(x) = x/(1 + x) (x >= 0) > f^{-1}(y) = y (y <= 0) > f^{-1}(y) = y/(1 - y) (0 <= y < 1) >or even smooth? > Or C^2? >You can play that game all the way up to C^infty. I can't even see what the next move should be, let alone what might >happen all the way up to (and including?) C^oo. I can only see that >for r >= 0, one might have tried defining f(x) = x + x^{r+1} (x >= 0), >so that f is in C^r, but that's no good, because f^{-1} also has to >preserve the set of rationals. Am I missing some trick that can be >carried on for all r? No, I think you are right. Certainly there is no function of the form f(x) = x if x <= 0 g(x) if x > 0 such that (1) g is a rational function (2) f is injective (3) f(x) is rational iff x is rational (4) f is twice differentiable at x = 0. quasi === Subject: Re: -- Birational? >So much for mere homeomorphisms, then! But what if f and f^{-1} >are supposed to be [continuously] differentiable > m = n = 1 > E = R > F = (-oo, 1) > f(x) = x (x <= 0) > f(x) = x/(1 + x) (x >= 0) > f^{-1}(y) = y (y <= 0) > f^{-1}(y) = y/(1 - y) (0 <= y < 1) >or even smooth? > Or C^2? >You can play that game all the way up to C^infty. >I can't even see what the next move should be, let alone what might >happen all the way up to (and including?) C^oo. I can only see that >for r >= 0, one might have tried defining f(x) = x + x^{r+1} (x >= 0), >so that f is in C^r, but that's no good, because f^{-1} also has to >preserve the set of rationals. Am I missing some trick that can be >carried on for all r? No, I think you are right. Certainly there is no function of the form f(x) = > x if x <= 0 > g(x) if x > 0 such that (1) g is a rational function (2) f is injective (3) f(x) is rational iff x is rational (4) f is twice differentiable at x = 0. This calls for some conjectures ... Conjectures: (1) If f is a continuous function from R to R which such that f(Q) is a subset Q, then f is a piecewise rational function (with possibly infinitely many pieces). (2) If f is a twice differentiable, injective function from R to R such that f(x) is rational iff x is rational, then f is a linear polynomial. Remarks: For conjecture (1), injectivity of f is not specified, however, if the conjecture fails as stated, then revise it to require that f is also injective. quasi === Subject: Re: -- Birational? > >So much for mere homeomorphisms, then! But what if f and f^{-1} >are supposed to be [continuously] differentiable > m = n = 1 > E = R > F = (-oo, 1) > f(x) = x (x <= 0) > f(x) = x/(1 + x) (x >= 0) > f^{-1}(y) = y (y <= 0) > f^{-1}(y) = y/(1 - y) (0 <= y < 1) >or even smooth? > Or C^2? >You can play that game all the way up to C^infty. >I can't even see what the next move should be, let alone what might >happen all the way up to (and including?) C^oo. I can only see that >for r >= 0, one might have tried defining f(x) = x + x^{r+1} (x >= 0), >so that f is in C^r, but that's no good, because f^{-1} also has to >preserve the set of rationals. Am I missing some trick that can be >carried on for all r? No, I think you are right. Certainly there is no function of the form f(x) = > x if x <= 0 > g(x) if x > 0 such that (1) g is a rational function (2) f is injective (3) f(x) is rational iff x is rational (4) f is twice differentiable at x = 0. This calls for some conjectures ... Conjectures: (1) If f is a continuous function from R to R which such that f(Q) is > a subset Q, then f is a piecewise rational function (with possibly > infinitely many pieces). To get an injective counterxample, let C(x) denote the Cantor function on [0,1]. Define f(x) = x on (-oo, 0], f(x) = x + C(x) on [0, 1], f(x) = 1 + x on [1,oo). Then f is strictly increasing and continuous on R. Suppose x is rational and not in the Cantor set. Then it's easy to see f(x) is rational. If x is rational and in the Cantor set, then the ternary expansion of x eventually repeats, and therefore so does the binary expansion of C(x), which implies C(x) is rational, hence f(x) is rational. > (2) If f is a twice differentiable, injective function from R to R > such that f(x) is rational iff x is rational, then f is a linear > polynomial. Remarks: For conjecture (1), injectivity of f is not specified, > however, if the conjecture fails as stated, then revise it to require > that f is also injective. === Subject: Re: -- Birational? > >So much for mere homeomorphisms, then! But what if f and f^{-1} >are supposed to be [continuously] differentiable > m = n = 1 > E = R > F = (-oo, 1) > f(x) = x (x <= 0) > f(x) = x/(1 + x) (x >= 0) > f^{-1}(y) = y (y <= 0) > f^{-1}(y) = y/(1 - y) (0 <= y < 1) >or even smooth? > Or C^2? >You can play that game all the way up to C^infty. >I can't even see what the next move should be, let alone what might >happen all the way up to (and including?) C^oo. I can only see that >for r >= 0, one might have tried defining f(x) = x + x^{r+1} (x >= 0), >so that f is in C^r, but that's no good, because f^{-1} also has to >preserve the set of rationals. Am I missing some trick that can be >carried on for all r? No, I think you are right. Certainly there is no function of the form f(x) = > x if x <= 0 > g(x) if x > 0 such that (1) g is a rational function (2) f is injective (3) f(x) is rational iff x is rational (4) f is twice differentiable at x = 0. This calls for some conjectures ... Conjectures: (1) If f is a continuous function from R to R which such that f(Q) is > a subset Q, then f is a piecewise rational function (with possibly > infinitely many pieces). No, in fact there are non-polynomial entire functions that do this. I think Robert Israel was the first on sci.math to give such an example. One way to do this is let {r_1, r_2, ...} be the rationals. For each n, let f_n(z) = (1/2^n)*product(k=1,n) [(z-r_k)/(n+|r_k|)]^2. Then each f_n takes Q into Q. We have f_n = 0 on {r_1, ..., r_n}, and |f_n| < 1/2^n on {|z| < n}. So if we set f(z) = sum(n=1,oo) f_n(z), then f is entire, and f(Q) is contained in Q (note only finitely many f_n's are nonzero at each fixed rational). This f is non-polynomial because on R, all summands are positive, which implies f grows faster than any polynomial at oo. > (2) If f is a twice differentiable, injective function from R to R > such that f(x) is rational iff x is rational, then f is a linear > polynomial. Remarks: For conjecture (1), injectivity of f is not specified, > however, if the conjecture fails as stated, then revise it to require > that f is also injective. quasi === Subject: Re: complex number set > possible ??? > Again I'm here to ask you a question. Generally we > express set of all > real numbers by R. similarly N for all natural > number and so > on.......But is there any notation by which we > can express all > complex number. Blackboard C is a common choice, as is boldface C. -- > Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man > schweigen > - Ludwig Wittgenstein, Tractatus > s Logico-Philosophicus yes C in calculus , algebra and many other fields of math. in set theory this would be R^2. (isomorphic to it at least) btw complex numbers can also be given by real matrices , but i guess you already knew that. tommy1729 === Subject: Re: proofs by contradiction > >Hi! I've been trying to prove a theorem by trying to > show that the >negation of the theorem leads to the following > contadiction: >there exists a set whose cardinality is infinite > and finite. I'm >wondering if this is enough. Since the mathematics > I'm using contains arithmetic isn't it true that I > can never be sure whether or not >any particular contradiction can be derived from > first principles so >don't I first have to prove that my contradiction is > imposible? No. Say A is the arithmetic system you're using, and > T is > what you're trying to prove. Now, if you're using A, then of course you can't be > proving > T in any absolute sense, all you can be proving is > that T > follows from A, in symbols A |- T. That is in fact a > limitation, > but it has nothing to do with proof by contradiction > per se. If you show that A plue not lead to a contradition, > that is, A + ~T |- P + ~P then A |- T follows, whether A is consistent or not. > >Respectfully, >Robert Kaufman > ************************ David C. Ullrich This might come as a shock ; but i agree with David. === Subject: Re: proofs by contradiction > >Hi! >I've been trying to prove a theorem by trying to > show that the >negation of the theorem leads to the following > contadiction: >there exists a set whose cardinality is infinite > and finite. I'm >wondering if this is enough. Since the mathematics > I'm using contains arithmetic isn't it true that I > can never be sure whether or not >any particular contradiction can be derived from > first principles so >don't I first have to prove that my contradiction is > imposible? > > No. Say A is the arithmetic system you're using, and > T is > what you're trying to prove. > > Now, if you're using A, then of course you can't be > proving > T in any absolute sense, all you can be proving is > that T > follows from A, in symbols A |- T. That is in fact a > limitation, > but it has nothing to do with proof by contradiction > per se. > > If you show that A plue not lead to a contradition, > that is, > > A + ~T |- P + ~P > > then A |- T follows, whether A is consistent or not. > >Respectfully, >Robert Kaufman > > > ************************ > > David C. Ullrich This might come as a shock ; but i agree with David. Hehe. Perhaps there's hope for you after all! quasi === Subject: -- factorization in fields This is an attempt to explore the concept of factorization in fields. Of course, every field is trivially a UFD, so the ring-theoretic concept of factorization is not useful if the goal is to classify fields with respect to their factorization properties. Instead, let's focus on the group structure of the multiplicative group of nonzero elements of the field. For a given field K, let the subset ru(K) be the set of elements of K which are roots of unity. Explicitly: ru(K) = {x in K | x^n = 1 for some positive integer n} Call K a UFF (unique factorization field) if there is a free abelian subgroup G of K^* such that K^* = ru(K) x G, an internal direct product. Once again, let me emphasize -- the terminology UFF is for fields only, and is not be confused with the class of rings call UFDs. Every field, when regarded as a ring, is a UFD, but not every field is a UFF. Also don't confuse UFFs with the class of objects called UFOs. UFF's don't fly (but the elements of a UFF do factor nicely). Example: Consider the field Q of rational numbers. Then ru(Q) = {1, -1}. Let S = set of positive integer primes and let G be the subgroup of Q^* generated by S. Then G is a free abelian subgroup of Q^* and Q^* = ru(Q) x G. Hence Q is a UFF. To show how the concept of UFF relates to factorization in Q, observe that every element x of Q has a unique representation of the form x = u*(p_1^e_1)*(p_2^e_2)*(p_3^e_3)*... where u = 1 or -1, p_n is the n'th prime number (in N), each e_n is a nonnegative integer e_n = 0 for all but finitely many n. This interpretation of factorization gives a nontrivial meaning to tommy1729's question about unique factorization in fields. Question: Let K be a UFF. Must there exist a subring R (with 1) of K such that (1) The quotient field of R is K. (2) R is a UFD. (3) The group of units of R, modulo the subgroup ru(K), is a free abelian group. Note: Since R is a UFD (and hence integrally closed) with quotient field K, it's automatic ru(K) is a subset of R, and hence ru(K) is a subgroup of the group of units of R. Remark: I posed the above problem as a question, rather than a conjecture, because I feel that an answer of yes would be too good to be true. quasi === Subject: Re: -- factorization in fields >This is an attempt to explore the concept of factorization in fields. Of course, every field is trivially a UFD, so the ring-theoretic >concept of factorization is not useful if the goal is to classify >fields with respect to their factorization properties. Instead, let's focus on the group structure of the multiplicative >group of nonzero elements of the field. For a given field K, let the subset ru(K) be the set of elements of K >which are roots of unity. Explicitly: ru(K) = {x in K | x^n = 1 for some positive integer n} > >Call K a UFF (unique factorization field) if there is a free abelian >subgroup G of K^* such that K^* = ru(K) x G, an internal direct >product. Once again, let me emphasize -- the terminology UFF is for fields >only, and is not be confused with the class of rings call UFDs. Every >field, when regarded as a ring, is a UFD, but not every field is a >UFF. Also don't confuse UFFs with the class of objects called UFOs. UFF's >don't fly (but the elements of a UFF do factor nicely). Example: Consider the field Q of rational numbers. Then ru(Q) = {1, -1}. Let S = set of positive integer primes and let G be the subgroup of >Q^* generated by S. Then G is a free abelian subgroup of Q^* and Q^* = ru(Q) x G. Hence Q is a UFF. To show how the concept of UFF relates to factorization in Q, observe >that every element x of Q has a unique representation of the form x = u*(p_1^e_1)*(p_2^e_2)*(p_3^e_3)*... > > where > > u = 1 or -1, > > p_n is the n'th prime number (in N), > > each e_n is a nonnegative integer Whoops: the above line should be each e_n is an integer (possibly negative or 0) > e_n = 0 for all but finitely many n. > >This interpretation of factorization gives a nontrivial meaning to >tommy1729's question about unique factorization in fields. Question: Let K be a UFF. Must there exist a subring R (with 1) of K such that (1) The quotient field of R is K. > (2) R is a UFD. > (3) The group of units of R, modulo the subgroup ru(K), > is a free abelian group. > >Note: Since R is a UFD (and hence integrally closed) with quotient >field K, it's automatic ru(K) is a subset of R, and hence ru(K) is a >subgroup of the group of units of R. Remark: I posed the above problem as a question, rather than a >conjecture, because I feel that an answer of yes would be too good >to be true. quasi === Subject: -- Boyce & Brannan, a view after the first week of class FWIW, I just finished my first week of Differential Equations at Austin Community College. The book they use is a custom-published book from Wiley that pulls together parts of Boyce & DiPrima's book along with Boyce & Brannan. The professor told us that the first half of the book was different than the DiPrima text and that the last half was pulled directly from DiPrima. But it looks to me that Chapter 1 is the Boyce & DiPrima book. At any rate, I' like to say that if the book does not improve, which I doubt, it will be a long semester. When I went back to night school to study math a couple of years ago, I did not know what people meant by cookbook math. All I'd have to do to show someone what the phrase means is to let them look at the first chapter of this dreadful, poorly-written tome. Do this, do that, do something else and you will get a result. I am concurrently reviewing web sites, other books (Tenenbaum & Pollard's text from Dover arrived in today's mail), and the MIT videos to try and get some understanding of what the hell is going on. Perhaps that is just the way DE books are written, but hopefully not. I do not need some dense theorem-proof book, but something with clarity would be nice. I read an excerpt from Tenenbaum and am hoping that he does the trick. Alan === Subject: Need test data for Nelson Rules I'm writing some code to evaluate the Nelson Rules (http:// en.wikipedia.org/wiki/Nelson_rules). æI'm looking for some test data sets that I can use for testing my code. æDo such data sets exist? My apologies if this isn't the right group for a request like this. === Subject: Puzzle with functions posting-account=oIJ4EQoAAACxoPMGTzrZOJmogGjoMdQ4 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) Find infinite sequence of real polynomial functions fn such that exp = f1 o f2 o f3 o ... or e^x = lim_{n -> inf} f1(f2(f3(...fn(x)...))) Marc === Subject: Re: Puzzle with functions >Find infinite sequence of real polynomial functions fn such that exp = f1 o f2 o f3 o ... or e^x = lim_{n -> inf} f1(f2(f3(...fn(x)...))) I'm not sure if such a sequence of polynomials exists, but I suspect that if there is one such sequence, then there are infinitely many distinct such sequences. As a possible strategy, suppose f_n is of degree d_n, but with undetermined coefficients. One can try to impose the conditions (1) f_1(f_2(f_3...f_n(x)...))) matches the first n terms of the taylor polynomial for e^x (2) All other coefficients are nonnegative and less than or equal to the corresponding taylor polynomial coefficient. Since you can choose the degree d_n on the fly, you have a lot of flexibility to achieve condition (1). However, simultaneously satisfying condition (2) becomes problematic. Still, since condition (2) only requires matching within a range of values, it suggests that the solutions, if they exist at all, are probably not uniquely determined. quasi === Subject: Re: Puzzle with functions >Find infinite sequence of real polynomial functions fn such that >exp = f1 o f2 o f3 o ... >or e^x = lim_{n -> inf} f1(f2(f3(...fn(x)...))) I'm not sure if such a sequence of polynomials exists, but I suspect >that if there is one such sequence, then there are infinitely many >distinct such sequences. As a possible strategy, suppose f_n is of degree d_n, but with >undetermined coefficients. One can try to impose the conditions (1) f_1(f_2(f_3...f_n(x)...))) matches the first n terms of the taylor >polynomial for e^x (2) All other coefficients are nonnegative and less than or equal to >the corresponding taylor polynomial coefficient. Since you can choose the degree d_n on the fly, you have a lot of >flexibility to achieve condition (1). However, simultaneously >satisfying condition (2) becomes problematic. Still, since condition >(2) only requires matching within a range of values, it suggests that >the solutions, if they exist at all, are probably not uniquely >determined. Actually, now that I look at your question more carefully, it's clear there is no such sequence. You have the order wrong. There can't be a fixed _outer_ polynomial, otherwise the graph of your limit function is a subset of the graph of f_1, hence can't equal the graph of e^x. My suggested strategy above was based on the _opposite_ order. In other words, here's the problem I thought you posed, and for which the strategy I suggested in my original reply _might_ work. I'll pose it as an existence question ... Problem: Does there exist an infinite sequence f_1, f_2, f_3, ... of polynomials in the variable x, with real coefficients, such that the sequence of polynomials f_1, f_2 o f_1, f_3 o f_2 o f_1, ... converges pointwise to e^x? quasi === Subject: Re: Puzzle with functions <6pr2p3tf70pipo1k4fo4aslccrj8l0ljk4@4ax.com> <7a03p39vvsuaotppjsulgpsd6vcfogcl6c@4ax.com> posting-account=oIJ4EQoAAACxoPMGTzrZOJmogGjoMdQ4 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) >Find infinite sequence of real polynomial functions fn such that >exp = f1 o f2 o f3 o ... >or e^x = lim_{n -> inf} f1(f2(f3(...fn(x)...))) Actually, now that I look at your question more carefully, it's clear > there is no such sequence. You have the order wrong. I don't think so. > There can't be a fixed _outer_ polynomial, otherwise the graph of your > limit function is a subset of the graph of f_1, hence can't equal the > graph of e^x. Let consider the following example: fn(x) = (x^2 + 1)/2^n + x Numerically it seems that f = f1 o f2 o f3 ... make sense and certainly f is not a subset of f1. Marc === Subject: Re: Puzzle with functions >Find infinite sequence of real polynomial functions fn such that >exp = f1 o f2 o f3 o ... >or e^x = lim_{n -> inf} f1(f2(f3(...fn(x)...))) > Actually, now that I look at your question more carefully, it's clear > there is no such sequence. > You have the order wrong. I don't think so. > There can't be a fixed _outer_ polynomial, otherwise the graph of your > limit function is a subset of the graph of f_1, hence can't equal the > graph of e^x. You're right. My objection above is not correct. The input to f_1 is not x. >Let consider the following example: fn(x) = (x^2 + 1)/2^n + x Numerically it seems that f = f1 o f2 o f3 ... make sense and >certainly f is not a subset of f1. Yes, I agree. But I do like the other order better -- see my thread iterated composition of polynomials where I explore that version of the idea. quasi === Subject: Re: Puzzle with functions > Find infinite sequence of real polynomial functions fn such that exp = f1 o f2 o f3 o ... or e^x = lim_{n -> inf} f1(f2(f3(...fn(x)...))) f_1(x)=f_2(x)=f_3(x)=...=f_{n-1}(x)=x, f_n(x)=1+x+x^2/2+x^3/3!+...+x^n/n! > Marc -- I.N. Galidakis === Subject: Re: Puzzle with functions > Find infinite sequence of real polynomial functions fn such that > > exp = f1 o f2 o f3 o ... > > or e^x = lim_{n -> inf} f1(f2(f3(...fn(x)...))) f_1(x)=f_2(x)=f_3(x)=...=f_{n-1}(x)=x, f_n(x)=1+x+x^2/2+x^3/3!+...+x^n/n! That doesn't look right to me. That's not a well defined infinite sequence of functions. quasi === Subject: Re: Puzzle with functions > Find infinite sequence of real polynomial functions fn such that > exp = f1 o f2 o f3 o ... > or e^x = lim_{n -> inf} f1(f2(f3(...fn(x)...))) > f_1(x)=f_2(x)=f_3(x)=...=f_{n-1}(x)=x, > f_n(x)=1+x+x^2/2+x^3/3!+...+x^n/n! That doesn't look right to me. It doesn't look right, but not for the reason you mention, rather because f_n(x) is not a polynomial. Sorry, didn't see that. > That's not a well defined infinite sequence of functions. Nonsense. The sequence is perfectly well defined. Give me a n and I give you back a sequence: n=1: f_1(x)=1+x n=2: f_1(x)=x, f_2(x)=1+x+x^2/2! n=3: f_1(x)=x, f_2(x)=x, f_3(x)=1+x+x^2/2!+x^3/3!, etc. Indeed, lim_{n->oo}f_1(f_2(...f_n(x)...))=exp The problem is that f_n(x) is not a poly. > quasi -- I.N. Galidakis === Subject: Re: Puzzle with functions > Find infinite sequence of real polynomial functions fn such that > exp = f1 o f2 o f3 o ... > or e^x = lim_{n -> inf} f1(f2(f3(...fn(x)...))) > f_1(x)=f_2(x)=f_3(x)=...=f_{n-1}(x)=x, > f_n(x)=1+x+x^2/2+x^3/3!+...+x^n/n! > That doesn't look right to me. It doesn't look right, but not for the reason you mention, rather because f_n(x) >is not a polynomial. Sorry, didn't see that. > That's not a well defined infinite sequence of functions. Nonsense. The sequence is perfectly well defined. Give me a n and I give you >back a sequence: I'm not sure what question that would be answering, but it's not what is normally meant by defining an infinite sequence of functions. You don't get to be told a specific value of n. For each n, You have to define f_n(x), in terms of n (or possibly recursively in terms of the previous functions f_1, ..., f_(n-1)). >n=1: f_1(x)=1+x >n=2: f_1(x)=x, f_2(x)=1+x+x^2/2! >n=3: f_1(x)=x, f_2(x)=x, f_3(x)=1+x+x^2/2!+x^3/3!, etc. It looks like you are specifying an infinite sequence of finite sequences of functions, which is not what is desired here. >Indeed, lim_{n->oo}f_1(f_2(...f_n(x)...))=exp The problem is that f_n(x) is not a poly. It's not even a function. For example, you have more than one specification for f_1. f_1 can't depend on n. The variable n is just an index variable to identify which of the functions f_1, f_2, f_3, ... is being defined. But once defined, that's it -- you can't change it -- touch move! quasi === Subject: Re: Puzzle with functions [cut for brevity] > It looks like you are specifying an infinite sequence of finite > sequences of functions, which is not what is desired here. I am specifying a WHAT? And why is it not desired? > Indeed, > lim_{n->oo}f_1(f_2(...f_n(x)...))=exp > The problem is that f_n(x) is not a poly. It's not even a function. f_n(x)=1+x+x^2/2!+...+x^n/n! is a perfectly defined function. > For example, you have more than one specification for f_1. So? > f_1 can't depend on n. It doesn't. As n grows without bound, f_k(x)=x, 1<=k The variable n is just an index variable to identify which of the > functions f_1, f_2, f_3, ... is being defined. But once defined, > that's it -- you can't change it -- touch move! Nonsense again. Here's the formal proof: Defined as above, (excluding the fact that f_n(x) is not a poly), we have: For each nin N, with f_n defined as above: f_1(f_2(...f_n(x)...)=sum(k=0..n,x^k/k!). hence, |f_1(f_2(...f_n(x)...)-exp|=sum(k=n+1..oo,x^k/k!). Now give me an epsilon>0. Because the tail of the series goes to zero, I can find a n_0(epsilon), such that for all n>n_0: |f_1(f_2(...f_n(x)...)-exp| lim_{n->oo} f_1(f_2(...f_n(x)...)=exp In fact, defining them in the opposite order works as well: f_1(x)=sum(k=0..n,x^k/k!) f_2(x)=f_3(x)=...=f_n(x)=x As I said, the problem is that one of them (in this case f_1(x)) is not a poly. FWIW, I also think that a sequence of POLY functions satisfying the problem does not exist, but I can't see it right now. > quasi -- I.N. Galidakis === Subject: Re: Puzzle with functions [snip for brevity] > Defined as above, (excluding the fact that f_n(x) is not a poly), we > have: For each nin N, with f_n defined as above: f_1(f_2(...f_n(x)...)=sum(k=0..n,x^k/k!). hence, > > f_1(f_2(...f_n(x)...)-exp|=sum(k=n+1..oo,x^k/k!). Now give me an epsilon>0. Because the tail of the series goes to > zero, I can find a n_0(epsilon), such that for all n>n_0: > > f_1(f_2(...f_n(x)...)-exp| In fact, n_0 does not even depend on epsilon, because sum(k=0..n,x^n/n!) -> exp uniformly. [snip for brevity] -- I.N. Galidakis === Subject: Re: Puzzle with functions >[cut for brevity] > It looks like you are specifying an infinite sequence of finite > sequences of functions, which is not what is desired here. I am specifying a WHAT? Well, you cut the relevant definitions (of yours), so no context. >And why is it not desired? As far as what's desired, I'm making a judgement as to what the OP has in mind. In any case, I don't follow your construction at all. Perhaps it's a blind spot. quasi === Subject: ---minimum value Cc: deepkdeb@yahoo.com posting-account=iJfBPwgAAACaBAH7DreA6VfAOYvL5VDz 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; IEMB3; IEMB3),gzip(gfe),gzip(gfe) Consider the following equation under the given conditions. x = a/b + c/d (1) Conditions: x, a, b, c, d all are integers each > 1; a, b, c, d all are relatively prime integers taken two at a time. Odd x > 5. Question: Is there any minimum value of x? If so what is the value? Any helpful comment will be highly appreciated. === Subject: Re: ---minimum value >Consider the following equation under the given conditions. x = a/b + c/d (1) Conditions: x, a, b, c, d all are integers each > 1; a, b, c, d all >are relatively prime integers taken two at a time. Odd x > 5. Question: Is there any minimum value of x? If so what is the value? There's no such value of x, odd or even. The condition x > 5 is also irrelevant. quasi === Subject: Re: ---minimum value Cc: deepkdeb@yahoo.com posting-account=iJfBPwgAAACaBAH7DreA6VfAOYvL5VDz 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; IEMB3; IEMB3),gzip(gfe),gzip(gfe) Consider the following equation under the given conditions. x = a/b + c/d æ æ æ æ æ æ æ æ(1) Conditions: x, a, b, c, d all are integers each > 1; a, b, c, d all >are relatively prime integers taken two at a time. Odd x > 5. Question: Is there any minimum value of x? If so what is the value? There's no such value of x, odd or even. The condition x > 5 is also > irrelevant. quasi -------*** I understand you agree with the assertion that if a, b, c, d are relatively prime integers each > 0 then x cannot be an integer where x = a/b + c/d Kindly give your reason for agreeing with this assertion. A helpful and simple reply will be appreciated. -------*** === Subject: Re: ---minimum value > -------*** > I understand you agree with the assertion that if a, b, c, d are > relatively prime integers each > 0 > then x cannot be an integer where x = a/b + c/d > Kindly give your reason for agreeing with this assertion. A helpful > and simple reply will be appreciated. > -------*** If b and d are relatively prime, then the denominator (in lowest terms) of a/b + c/d is bd . So if x is an integer, then bd=1 . === Subject: Re: ---minimum value > -------*** > I understand you agree with the assertion that if a, b, c, d are > relatively prime integers each > 0 > then x cannot be an integer where x = a/b + c/d > Kindly give your reason for agreeing with this assertion. A helpful > and simple reply will be appreciated. > -------*** If b and d are relatively prime, then the denominator (in lowest terms) >of a/b + c/d is bd . So if x is an integer, then bd=1 . Right. To provide more details ... a/b + c/d = (ad + bc)/bd The claim is that the fraction on the RHS is already in lowest terms. Suppose p is a common prime factor of (ad + bc) and bd. Since p|bd, either p|b or p|d. Assume p|b (the argument for p|d is analogous). Since b,d are relatively prime, p does not divide d. Then p|b => p|bc => p | (ad + bc) - bc => p|ad => p|a contrary to the specification that a,b are relatively prime. Notice that the proof has nothing to do with even/odd. Similarly, the specification x > 5 is superfluous. I don't know why many of your problems include such extraneous conditions, but they distract from the main point. quasi === Subject: Re: ---minimum value posting-account=VR0DOgoAAADggPTteFeA2AkmHNhjcrDV Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Consider the following equation under the given conditions. x = a/b + c/d (1) Conditions: x, a, b, c, d all are integers each > 1; a, b, c, d all > are relatively prime integers taken two at a time. Odd x > 5. Question: Is there any minimum value of x? If so what is the value? Any helpful comment will be highly appreciated. I think I do not understand the question. The equation (1) is, since equation it follows that d divides bc, which, since d > 1, means that at least one of b or c is not relatively prime with d. === Subject: Re: ---minimum value posting-account=iJfBPwgAAACaBAH7DreA6VfAOYvL5VDz 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; IEMB3; IEMB3),gzip(gfe),gzip(gfe) Consider the following equation under the given conditions. x = a/b + c/d æ æ æ æ æ æ æ æ(1) Conditions: x, a, b, c, d all are integers each > 1; a, b, c, d all > are relatively prime integers taken two at a time. Odd x > 5. Question: Is there any minimum value of x? If so what is the value? Any helpful comment will be highly appreciated. > I think I do not understand the question. æThe equation (1) is, since > equation it follows that d divides bc, which, since d > 1, means that > at least one of b or c is not relatively prime with d. ***** So you agree (1) cannot exist under the given conditions *** === Subject: Re: Analysis with Cantor function... > Hello sir~ > Let F be Cantor ternary set. > Define that f(sum{n=1 to 00} 2.(z_n) / (3^n)) = sum{n=1 to 00} (z_n) / > (2^n) > (z_n = 0 or 1) > Show that f : F -> R is continuous. > ------------------------------------------------- > If F is isolated points, trivial. > But F is not isolated points.(Maybe, dense...) > Because, I can make a sequence (x_n) such that x_n -> c for each c in F. > (Of course, x_n in F). > Anyway, let's go... > Below we define a sequence of functions f_n on the interval that > converges to the Cantor function. > Let f_0(x) = x. > Then f_(n+1)(x) will be defined in terms of f_n(x). > Let f_(n+1)(x) = (0.5).f_(n)(3x) when 0<= x <= 1/3. > Let f_(n+1)(x) = 0.5 when 1/3 <= x <= 2/3. > Let f_(n+1)(x) = 0.5 + (0.5).f_(n)[3{x-(2/3)}] when 2/3 <= x <= 1. > so, f_n converges to the Cantor function f : [0,1] -> [0,1]. > For reference, http://en.wikipedia.org/wiki/Cantor_function > Lemma) > The uniform limit of continuous functions is continuous. > so, Cantor function f : [0,1] -> [0,1] is continous. > so, my original function f : F -> R is continuous. > How do you think about it ? I like to think about this a different way. Suppose 1/3^(n+1) < |y - > x| <= 1/3^n. Write x = sum_m x_m/3^m, where x_m = 0 or 2; same with y. > Exercise: x_m = y_m, m = 1, 2, ..., n. This implies |f(y) - f(x)| <= > 1/2^(n+1) + 1/2^(n+2) + ... = 1/2^n. We have 3^a = 2 for some a, 0 < a > < 1. So |f(y) - f(x)|/|y - x|^a <= (1/2^n)/(1/3^(n+1))^a = 2. In other > words, |f(y) - f(x)| <= 2|y - x|^a for all x, y in [0,1]. This of > course gives continuity immediately, but it also shows f is Lip_a. Wow...flamboyant technique. Good idea. === Subject: Re: A simple problem of analysis. posting-account=cEKB4goAAABJ5eifbZgDmj5s5T5PQpjx BESAGENT; .NET CLR 1.1.4322; BESAGENT),gzip(gfe),gzip(gfe) f : [a, b] -> R , differentiable. Show that f'(x) is continuous at least at one point. The problem is really simple to understand, but I have no idea to > solve it. I just think that it may need Mean Value Theorem, but I'm not sure. Any idea you've got? > I have a answer: > f is continuous, so is f n(x) = n ( f(x + 1/n) - f(x) ) . > f'(x) = lim (n->infinity) f n(x) > Continuous functions' limit function has dense continuous points. > So we get an answer to your question. This question belongs to REAL ANALYSIS. > You can refer to some advanced real analysis book. Hope these help you. > I am learning real analysis too. === Subject: Complex analysis posting-account=cEKB4goAAABJ5eifbZgDmj5s5T5PQpjx BESAGENT; .NET CLR 1.1.4322; BESAGENT),gzip(gfe),gzip(gfe) D : open connected subset in C(complex number set) f : continuous on D Show that if f^2 is analytic(or holomorphic) on D, then f is also anaytic on D. === Subject: Re: Complex analysis > D : open connected subset in C(complex number set) f : continuous on D Show that if f^2 is analytic(or holomorphic) on D, > then f is also > anaytic on D. > As just a quick thought, wouldn't that follow from de Branges' Theorem (Bieberbach Conjecture)? Tom === Subject: Re: Complex analysis <17439561.1200761985519.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20071213 Fedora/2.0.0.10-3.fc8 Firefox/2.0.0.10,gzip(gfe),gzip(gfe) > D : open connected subset in C(complex number set) f : continuous on D Show that if f^2 is analytic(or holomorphic) on D, > then f is also > anaytic on D. As just a quick thought, wouldn't that follow from > de Branges' Theorem (Bieberbach Conjecture)? In what way? -- m === Subject: Re: Complex analysis >D : open connected subset in C(complex number > set) f : continuous on D Show that if f^2 is analytic(or holomorphic) on > D, >then f is also >anaytic on D. As just a quick thought, wouldn't that follow from > de Branges' Theorem (Bieberbach Conjecture)? In what way? -- m I should know better than to say A if I don't plan to say B. :-) Honestly, I was only thinking quickly of the implication that geometric extremality implies metric extremality. Tom === Subject: Re: Complex analysis <15791141.1200763527427.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20071213 Fedora/2.0.0.10-3.fc8 Firefox/2.0.0.10,gzip(gfe),gzip(gfe) >D : open connected subset in C(complex number > set) f : continuous on D Show that if f^2 is analytic(or holomorphic) on > D, >then f is also >anaytic on D. As just a quick thought, wouldn't that follow from >de Branges' Theorem (Bieberbach Conjecture)? In what way? -- m I should know better than to say A if I don't plan to > say B. :-) Honestly, I was only thinking quickly > of the implication that geometric extremality implies > metric extremality. Honestly, I fail to see in what possible way either of those two are related to the original question. -- m === Subject: Re: Complex analysis posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > D : open connected subset in C(complex number set) f : continuous on D Show that if f^2 is analytic(or holomorphic) on D, then f is also > anaytic on D. I expect that someone who knows something about analysis knows a clever way to do this. But I shall attempt a brute-force skeleton argument: Consider first a point z_0 such that f(z_0) != 0, so f^2(z_0) != 0. Let U be an open ball containing f^2(z_0) which does not contain zero. Then we may define the square root on U to be single-valued and analytic - denote this square root function by sqrt(). Let V be the image of U under sqrt(). Then V and -V are disjoint open sets, and for each z in (f^2)^-1(U), we have f(z) = +sqrt(f^2(z)) in V or f(z) = - sqrt(f^2(z)) in -V. Since f^2 is continuous there exists an open ball B containing z_0 such that f^2(B) is contained in U, and since f is continuous the image f(B) is connected and therefore contained in either V or -V. Therefore we either have that f(z) = +sqrt(f^2(z)) on the whole of B or f(z) = -sqrt(f^2(z)) on the whole of B, and so f is analytic on B. Now consider a point z_0 such that f(z_0) = 0. Let w = z - z_0 and, by Taylor expanding, write f^2(z) = w^n * g(z), where g(z) is analytic and non-zero at z_0. As before, choose an open set U containing g(z_0) but not the origin, and an open ball B containing z_0 such that g(B) is contained in U. Also as before define sqrt() to be single-valued and analytic on U, so that sqrt(g(z)) is analytic and non-zero on B. Then we may show that n is even. For suppose otherwise: then f(z)/ sqrt(g(z)) = +/-w^(n/2), but no choice of + or - sign will make this function continuous at w = 0 if n is odd. Therefore we have f(z) = +/- w^m * sqrt(g(z)) on B, and since f is continuous the choice of + or - must be independent of z, so f is analytic on B. === Subject: Saturday Challenging Problem !!! posting-account=h94dGgoAAAALp7NtJSAjbsSEO3J9lAcK BESAGENT; .NET CLR 1.1.4322; BESAGENT),gzip(gfe),gzip(gfe) Let d_n be derangement number for n>=1, that is, the number of permutations of S_n which has no fixed point, and d_0 be 1. Prove that for n>=2, sum{k=1 to n} [(k^2)*(_n C_k)*(d_(n-k))] = 2 (n !) ( _n C_k = n! / (k! * (n-k)!) ) === Subject: Re: -- nonunique factorization posting-account=euF15goAAACbw3KIqEWxZHCIPUc2KPmU .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Sorry to bother you but could you check Wikipedia again? It seems that > you have copied only one part of the definition. I think I can find > tons of domains that fit your definition but are not UFD's. D'oh! if q 1,...,q m are irreducible elements of R such that x = q 1 q 2 ... q m, then m = n and there exists a bijective map phi : {1,...,n} -> {1,...,n} such that p i is associated to q phi(i) for i = 1, ..., n. The uniqueness part is sometimes hard to verify, which is why the following equivalent definition is useful: a unique factorization domain is an integral domain R in which every non-zero non-unit can be written as a product of prime elements of R. I apologize for the confusion. === Subject: var bounded in probability posting-account=HwWEpQoAAADVZ2R1X9i1OJsBpZb3TGwM Gecko/20071127 Firefox/2.0.0.11 eMusic DLM/4.0_1.0.0.1,gzip(gfe),gzip(gfe) I am stalling on the following question: If Yn converges in probability to c (c finite), then Yn is bounded in probability, meaning P(|Yn| =< K) > 1 -delta for all n > n0. I know that I should start with something that resembles For any small delta and eps, there exists n*, such that n>n*, P(|Yn-c| > eps) < delta and that I can then write -P(|Yn-c| > eps) > -delta 1-P(|Yn-c| > eps) > 1-delta P(|Yn-c| =< eps) > 1-delta But I am not too sure how to go from here to P(|Yn| =< K) > 1 -delta === Subject: Math Trek: Checking It Twice [MATHTREK] Checking It Twice Election officials have had no practical way to guarantee a correct ballot count--until now. === Subject: -- iterated composition of polynomials These questions are based on the ideas from the thread Puzzle with functions by marcjhg however I define the partial composites in the opposite order from the way it was specified in that thread. Let's start with some definitions, then ask some general questions ... In principle, we can start with any base set of generating functions, but for now, let's stay with polynomial generators. So let's try this ... Given a sequence of polynomials f_1, f_2, f_3, ... in the variable x, with real coefficients, define a sequence of partial composites, g_1, g_2, g_3, ... by g_1 = f_1 g_2 = f_2 o f_1 g_3 = f_3 o f_2 o f_1 ... Then, for each real x, define g(x) by g(x) = lim g_n(x), as n approaches infinity provided that the limit exists. Let S be the set of all functions g which can be realized as specified above, and such that g(x) is defined for all real x. Thus, we are mainly interested in full functions from R to R, not partial functions. Remarks: (1) If any one of the f_1, f_2, f_3, ... is constant, then g may or may not be defined, but if g is defined anywhere (i.e., g(x) is defined for some x), then g is constant (and hence defined for all x).. (2) If none of the f_n is constant, and if all but finitely many of the f_n are linear, then g(x) exists for all x iff, for the linear polynomials, the product of the leading coefficients converges and the sum of the constant terms converges. (3) For an arbitrary polynomial f, let f_1 = f, and f_n = x for all n > 1. Then g = f. Thus S contains all polynomials. (4) I don't immediately see an example of a non-polynomial element of S. The general question is to characterize the set of functions which are elements of S. Here are some specific questions ... (1) Does S have any elements other than polynomials? (2) If yes to (1), is e^x in S? [Note -- the above question is essentially marcjhg's question, but with the order of composition reversed.] (3) If yes to (1), does S contain all continuous functions from R to R? (4) If yes to (3), does S contain all functions from R to R? quasi === Subject: Re: -- iterated composition of polynomials posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20071213 Fedora/2.0.0.10-3.fc8 Firefox/2.0.0.10,gzip(gfe),gzip(gfe) > These questions are based on the ideas from the thread Puzzle with functions by marcjhg however I define the partial composites in the opposite order from the > way it was specified in that thread. Let's start with some definitions, then ask some general questions ... In principle, we can start with any base set of generating functions, > but for now, let's stay with polynomial generators. So let's try this ... Given a sequence of polynomials f_1, f_2, f_3, ... in the variable x, > with real coefficients, define a sequence of partial composites, > g_1, g_2, g_3, ... by g_1 = f_1 > g_2 = f_2 o f_1 > g_3 = f_3 o f_2 o f_1 > ... Then, for each real x, define g(x) by g(x) = lim g_n(x), as n approaches infinity provided that the limit exists. Let S be the set of all functions g which can be realized as specified > above, and such that g(x) is defined for all real x. Thus, we are mainly interested in full functions from R to R, not > partial functions. Remarks: (1) If any one of the f_1, f_2, f_3, ... is constant, then g may or > may not be defined, but if g is defined anywhere (i.e., g(x) is > defined for some x), then g is constant (and hence defined for all > x).. (2) If none of the f_n is constant, and if all but finitely many of > the f_n are linear, then g(x) exists for all x iff, for the linear > polynomials, the product of the leading coefficients converges and the > sum of the constant terms converges. (3) For an arbitrary polynomial f, let f_1 = f, > and f_n = x for all n > 1. Then g = f. Thus S contains all > polynomials. (4) I don't immediately see an example of a non-polynomial element of > S. The general question is to characterize the set of functions which are > elements of S. Here are some specific questions ... (1) Does S have any elements other than polynomials? Pick any increasing sequence (a_n : n >= 1) of positive integers such that for all n we have a_1 a_2 ... a_n < a_{n+1} For example, one can take a_n = 2^(2^n). Now consider a sequence of numbers (b_n) which goes to zero very fast, and put f_n(x) = x + x^(a_n) / b_n Then the limit f(x) = lim_{n --> infty} (f_n o ... o f_1)(x) exists for all (complex) x if one makes sure that the sequence b_n grows sufficiently fast, and it is clear, when you prove this, that the limit (which converges uniformly on compact subsets) is a non-polynomial entire function. This provides lots of examples, of course. One can do examples in which the composition factors are of bounded degree. For example, let f_1 = x + x^2 Compute the maximum modulus of f_1(x)^2 for |x| <= 1, name it a_1, and put f_2 = x + x^2 / (2 a_1) Now compute the maximum modulus of (f_2 o f_1)^2 on |x| <= 2, name it a_2, and set f_3 = x + x^2 / (2^2 a_2) and continue in this way: the next step is, compute the maximum modulus of (f_3 o f_2 o f_1)^2 on |x| <= 3, name it a_3, and set f_4 = x + x^2 / (2^3 a_3). Then lim (f_n o ... o f_1) (x) exists for all complex x. One can check that the limit is not a polynomial in this case, too. -- m === Subject: Re: -- iterated composition of polynomials > These questions are based on the ideas from the thread > Puzzle with functions by marcjhg > however I define the partial composites in the opposite order from the > way it was specified in that thread. > Let's start with some definitions, then ask some general questions ... > In principle, we can start with any base set of generating functions, > but for now, let's stay with polynomial generators. > So let's try this ... > Given a sequence of polynomials f_1, f_2, f_3, ... in the variable x, > with real coefficients, define a sequence of partial composites, > g_1, g_2, g_3, ... by > g_1 = f_1 > g_2 = f_2 o f_1 > g_3 = f_3 o f_2 o f_1 > ... > Then, for each real x, define g(x) by > g(x) = lim g_n(x), as n approaches infinity > provided that the limit exists. > Let S be the set of all functions g which can be realized as specified > above, and such that g(x) is defined for all real x. > Thus, we are mainly interested in full functions from R to R, not > partial functions. > Remarks: > (1) If any one of the f_1, f_2, f_3, ... is constant, then g may or > may not be defined, but if g is defined anywhere (i.e., g(x) is > defined for some x), then g is constant (and hence defined for all > x).. > (2) If none of the f_n is constant, and if all but finitely many of > the f_n are linear, then g(x) exists for all x iff, for the linear > polynomials, the product of the leading coefficients converges and the > sum of the constant terms converges. > (3) For an arbitrary polynomial f, let f_1 = f, > and f_n = x for all n > 1. Then g = f. Thus S contains all > polynomials. > (4) I don't immediately see an example of a non-polynomial element of > S. > The general question is to characterize the set of functions which are > elements of S. > Here are some specific questions ... > (1) Does S have any elements other than polynomials? Pick any increasing sequence (a_n : n >= 1) of positive integers such >that for all n we have a_1 a_2 ... a_n < a_{n+1} For example, one can take a_n = 2^(2^n). Now consider a sequence of numbers (b_n) which goes to zero >very fast, It's clear that you actually meant (1/b_n) goes to zero very fast >and put f_n(x) = x + x^(a_n) / b_n I see ... The structure of f together with the inequalities on the a's insures that the initial part of f_n is just f_(n-1), and the fast growth of the b's is designed to insure all convergence. >Then the limit f(x) = lim_{n --> infty} (f_n o ... o f_1)(x) exists for all (complex) x if one makes sure that >the sequence b_n grows sufficiently fast, and it is clear, >when you prove this, that the limit (which converges >uniformly on compact subsets) is a non-polynomial >entire function. Right, because, if I follow the idea fully, the composition effectively yields a convergent power series, convergent for all x, with infinitely many nonzero terms. >This provides lots of examples, of course. Yes -- a cool technique. >One can do examples in which the composition factors are >of bounded degree. For example, let f_1 = x + x^2 Compute the maximum modulus of f_1(x)^2 for |x| <= 1, >name it a_1, and put f_2 = x + x^2 / (2 a_1) Now compute the maximum modulus of (f_2 o f_1)^2 >on |x| <= 2, name it a_2, and set f_3 = x + x^2 / (2^2 a_2) and continue in this way: the next step is, >compute the maximum modulus of (f_3 o f_2 o f_1)^2 >on |x| <= 3, name it a_3, and set f_4 = x + x^2 / (2^3 a_3). Then lim (f_n o ... o f_1) (x) exists for all complex x. One can check that the limit is >not a polynomial in this case, too. Very nice constructions. quasi === Subject: Re: -- iterated composition of polynomials posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > Given a sequence of polynomials f_1, f_2, f_3, ... in the > variable x, with real coefficients, define a sequence of > partial composites, g_1, g_2, g_3, ... by g_1 = f_1 > g_2 = f_2 o f_1 > g_3 = f_3 o f_2 o f_1 > ... Then, for each real x, define g(x) by g(x) = lim g_n(x), as n approaches infinity provided that the limit exists. Let S be the set of all functions g which can be realized > as specified above, and such that g(x) is defined for > all real x. > The general question is to characterize the set of > functions which are elements of S. Here are some specific questions ... (1) Does S have any elements other than polynomials? (2) If yes to (1), is e^x in S? [Note -- the above question is essentially marcjhg's > question, but with the order of composition reversed.] (3) If yes to (1), does S contain all continuous functions > from R to R? (4) If yes to (3), does S contain all functions from R to R? I have some papers at home that might deal with this topic from a Baire category standpoint in C[0,1], but I don't remember enough about them to know if they have any direct relevance to your specific questions. However, (4) is easy. Your set only includes pointwise limits of continuous functions, and so every element of S must be a Baire one function, which is a long way away from being all function (or even all Lebesgue measurable functions, or even all Borel measurable functions). Dave L. Renfro === Subject: Re: -- iterated composition of polynomials > Given a sequence of polynomials f_1, f_2, f_3, ... in the > variable x, with real coefficients, define a sequence of > partial composites, g_1, g_2, g_3, ... by > g_1 = f_1 > g_2 = f_2 o f_1 > g_3 = f_3 o f_2 o f_1 > ... > Then, for each real x, define g(x) by > g(x) = lim g_n(x), as n approaches infinity > provided that the limit exists. > Let S be the set of all functions g which can be realized > as specified above, and such that g(x) is defined for > all real x. > The general question is to characterize the set of > functions which are elements of S. > Here are some specific questions ... > (1) Does S have any elements other than polynomials? > (2) If yes to (1), is e^x in S? > [Note -- the above question is essentially marcjhg's > question, but with the order of composition reversed.] > (3) If yes to (1), does S contain all continuous functions > from R to R? > (4) If yes to (3), does S contain all functions from R to R? I have some papers at home that might deal with this topic >from a Baire category standpoint in C[0,1], but I don't >remember enough about them to know if they have any direct >relevance to your specific questions. However, (4) is easy. Your set only includes pointwise >limits of continuous functions, and so every element of >S must be a Baire one function, which is a long way away >from being all function (or even all Lebesgue measurable >functions, or even all Borel measurable functions). Cool. I didn't believe (4) would hold. quasi === Subject: convergence in probability of cn*Yn when Yn bounded in probability rv posting-account=r5Mu_AoAAAA4Or1qD_TiZuru2F3PYCEL Gecko/20071127 Firefox/2.0.0.11 eMusic DLM/4.0_1.0.0.1,gzip(gfe),gzip(gfe) I am having a hard time writing a probability proof of the following result: If the sequence Yn, n=1,2.... is bounded in probability and if {Cn} is a sequence of random variables tending to 0 in probability, then Cn Yn converges in probability to 0. I think I should start with, For any small delta and epsilon, there exists n* such that, for n>n*, P(|Cn|> eps) < delta For any delta1, there exists K, and there exists n0 such that, for all n > n0, P(|Yn| =< K) > 1 - delta1 This leads to There exists n0 such that, for n>n0, P(|Yn| > K) < delta1 Now I was given the following hint |Cn Yn| > K epsilon implies |Cn| > epsilon or |Yn| > K which lead me to write P{|Cn Yn| > K epsilon} < P{|Cn| > epsilon or |Yn| > K }=P{|Cn| > epsilon} +P{ |Yn| > K } meaning P{|Cn Yn| > K epsilon} max(n*,n0) Now take n to infty delta/2? I greatly appreciate your help. === Subject: Re: Comprehensive Solution Manual for Textbooks posting-account=MtqluwoAAABjHv0Q6j-IGFrq_Hj2wMHc Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) I wanted to check if you have the following solutions manuals: 1- TITLE: Statistics for Engineers And Scientists by William C. Navidi ISBN: 0073-30949-4 2- TITLE: Biomechanics and Biomaterials in Orthopedics by K.-G. Thorngren, R. Kotz, Dominique Poitout, ISBN: 1852-33481-9 if you have it please let me know, you can email me on gmail or yassorah@yahoo.com === Subject: Re: Comprehensive Solution Manual for Textbooks posting-account=ajjnwgoAAABec-rtvO49tR-Ova2Hz3Xj Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > I wanted to check if you have the following solutions manuals: 1- TITLE: Statistics for Engineers And Scientists > by William C. Navidi > ISBN: 0073-30949-4 2- TITLE: Biomechanics and Biomaterials in Orthopedics > by K.-G. Thorngren, R. Kotz, Dominique Poitout, > ISBN: 1852-33481-9 if you have it please let me know, you can email me on gmail or > yasso...@yahoo.com > I would like to purchase the solutions to: Applied Linear Algebra - Chehrzad Shakiban, Peter J. Olver (1st edition) (ISBN: 0131473824) I === Subject: Re: Comprehensive Solution Manual for Textbooks posting-account=ajjnwgoAAABec-rtvO49tR-Ova2Hz3Xj Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > I would like to purchase the solutions to: Applied Linear Algebra - > Chehrzad Shakiban, Peter J. Olver (1st edition) (ISBN: 0131473824) === Subject: Number with sqrt... Hello sir~ I showed a method that finding sqrt(9). Namely, 1) Divide 9 by 2. so, 4.5 2) Subtract 1. so, 3.5 3) Subtract 2. so, 1.5 4) Subtract 3. so, impossible in positive. anwer is 1.5*2 = 3 = sqrt(9). Can you explain this logically ? (with principle...) Apply this with sqrt(169). 1) Divide 169 by 2. so, 84.5 2) Substract 1. so, 83.5 3) Substract 2. so, 81.5 4) Substract 3. so, 78.5 5) Substract 4. so, 74.5 6) Substract 5. so, 69.5 7) Substract 6. so, 63.5 8) Substract 7. so, 56.5 9) Substract 8. so, 48.5 10) Substract 9. so, 39.5 11) Substract 10. so, 29.5 12) Substract 11. so, 18.5 13) Substract 12. so, 6.5 14) Substract 13. so, impossible. answer is 6.5*2 = 13 = sqrt(169) === Subject: Re: Number with sqrt... > I showed a method that finding sqrt(9). > Namely, > 1) Divide 9 by 2. so, 4.5 > 2) Subtract 1. so, 3.5 > 3) Subtract 2. so, 1.5 > 4) Subtract 3. so, impossible in positive. > anwer is 1.5*2 = 3 = sqrt(9). Can you explain this logically ? (with principle...) Apply this with sqrt(169). > 1) Divide 169 by 2. so, 84.5 > 2) Substract 1. so, 83.5 > 3) Substract 2. so, 81.5 > 4) Substract 3. so, 78.5 > 5) Substract 4. so, 74.5 > 6) Substract 5. so, 69.5 > 7) Substract 6. so, 63.5 > 8) Substract 7. so, 56.5 > 9) Substract 8. so, 48.5 > 10) Substract 9. so, 39.5 > 11) Substract 10. so, 29.5 > 12) Substract 11. so, 18.5 > 13) Substract 12. so, 6.5 > 14) Substract 13. so, impossible. > answer is 6.5*2 = 13 = sqrt(169) Let _k_ be a natural number. Your method of finding _k_ starting from k^2 is this: k = 2*(k^2/2 - (1 + 2 + 3 + ... + n)), (*) where _n_ is the largest natural such that 1 + 2 + 3 + ... + n < k^2/2. But 1 + 2 + 3 + ... + n = n(n + 1)/2, and so (*) = k^2 - n(n + 1) = k^2 - n^2 - n. Well, the largest natural _n_ such n(n + 1) < k^2 is k - 1, and therefor k^2 - n^2 - n = k^2 - (k - 1)^2 - k + 1 = k. Jose Carlos Santos === Subject: Re: Number with sqrt... > I showed a method that finding sqrt(9). > Namely, > 1) Divide 9 by 2. so, 4.5 > 2) Subtract 1. so, 3.5 > 3) Subtract 2. so, 1.5 > 4) Subtract 3. so, impossible in positive. > anwer is 1.5*2 = 3 = sqrt(9). > Can you explain this logically ? (with principle...) > Apply this with sqrt(169). > 1) Divide 169 by 2. so, 84.5 > 2) Substract 1. so, 83.5 > 3) Substract 2. so, 81.5 > 4) Substract 3. so, 78.5 > 5) Substract 4. so, 74.5 > 6) Substract 5. so, 69.5 > 7) Substract 6. so, 63.5 > 8) Substract 7. so, 56.5 > 9) Substract 8. so, 48.5 > 10) Substract 9. so, 39.5 > 11) Substract 10. so, 29.5 > 12) Substract 11. so, 18.5 > 13) Substract 12. so, 6.5 > 14) Substract 13. so, impossible. > answer is 6.5*2 = 13 = sqrt(169) Let _k_ be a natural number. Your method of finding _k_ starting > from k^2 is this: k = 2*(k^2/2 - (1 + 2 + 3 + ... + n)), (*) where _n_ is the largest natural such that 1 + 2 + 3 + ... + n < k^2/2. But 1 + 2 + 3 + ... + n = n(n + 1)/2, and so (*) = k^2 - n(n + 1) = k^2 - n^2 - n. Well, the largest natural _n_ such n(n + 1) < k^2 is k - 1, and > therefor k^2 - n^2 - n = k^2 - (k - 1)^2 - k + 1 = k. Yes, wise and bright... === Subject: Re: Number with sqrt... >Hello sir~ I showed a method that finding sqrt(9). >Namely, >1) Divide 9 by 2. so, 4.5 >2) Subtract 1. so, 3.5 >3) Subtract 2. so, 1.5 >4) Subtract 3. so, impossible in positive. >anwer is 1.5*2 = 3 = sqrt(9). Can you explain this logically ? (with principle...) Apply this with sqrt(169). >1) Divide 169 by 2. so, 84.5 >2) Substract 1. so, 83.5 >3) Substract 2. so, 81.5 >4) Substract 3. so, 78.5 >5) Substract 4. so, 74.5 >6) Substract 5. so, 69.5 >7) Substract 6. so, 63.5 >8) Substract 7. so, 56.5 >9) Substract 8. so, 48.5 >10) Substract 9. so, 39.5 >11) Substract 10. so, 29.5 >12) Substract 11. so, 18.5 >13) Substract 12. so, 6.5 >14) Substract 13. so, impossible. >answer is 6.5*2 = 13 = sqrt(169) Let f(n) = 1 + 2 + ... + n. Then f(n) = (n*(n+1))/2. Suppose x^2 is given, where x is a positive integer. The goal is to find x. Suppose n is a positive integer such that x^2/2 - f(n-1) > 0 >= x^2/2 - f(n) The claim is that x = 2*(x^/2 - f(n-1)). x^2/2 - f(n-1) > 0 implies n^2 - n < x^2 0 >= x^2 - f(n) implies x^2 <= n^2 + n Thus n^2 - n < x^2 <= n^2 + n. It follows that x^2 = n^2, and hence x = n. The 2*(x^2 - f(n-1)) = x^2 - (n^2 - n) = x^2 - (x^2 - x) = x, as claimed. quasi === Subject: Re: Number with sqrt... ... something related to ... 1 + 2 + 3 + ... + n = n(n+1)/2 Rainer === Subject: Re: Number with sqrt... ... something related to ... 1 + 2 + 3 + ... + n = n(n+1)/2 I almost gave the same hint, but then decided to just solve it. I should have just given a hint -- sorry mina. But if you (mina) read this one first, don't open my other reply. quasi === Subject: I can't understand a proof in a book. posting-account=d9DpDwoAAADlGCWGhrCHDkiN-0F85Exg MathPlayer 2.10b; Avant Browser; .NET CLR 1.1.4322; .NET CLR 2.0.50727; TheWorld),gzip(gfe),gzip(gfe) in R^n, E is compact iff E is bounded closed. R is real line .D subset R and is bound. E subset D. B_i is finite interval, i=1 to k boundary of E subset /_i B_i. bd E is bounded closed(compact). Then there existx d>0, for any parting of D: D_i,i=1 to k, if diam D_i < d,i=1 to k, then we have bd E subset ( / D_i | D_i / bd E notequal none) subset /_(i=1 to k)B_i. I can't understand the last sentens.Why does d exists ? === Subject: Re: I can't understand a proof in a book. > in R^n, E is compact iff E is bounded closed. R is real line .D subset R and is bound. D is bounded? > E subset D. > B_i is finite interval, i=1 to k > boundary of E subset /_i B_i. > bd E is bounded closed(compact). In language a space is suppose to proceed ( In math expressions, the rule is different, ie f(x). > Then there existx d>0, for any parting of D: D_i,i=1 to k, I think you mean 'partition'. > if diam D_i < d,i=1 to k, then we have > bd E subset ( / D_i | D_i / bd E notequal none) > subset /_(i=1 to k)B_i. > A Frequently used ascii notation for 'not equal' is /=. > I can't understand the last sentens.Why does d exists ? > Because D is bounded there is some d > 0 with diam D < d and because for j = 1 to k, D_j subset D, diam Dj <= diam D. === Subject: Re: I can't understand a proof in a book. posting-account=d9DpDwoAAADlGCWGhrCHDkiN-0F85Exg MathPlayer 2.10b; Avant Browser; .NET CLR 1.1.4322; .NET CLR 2.0.50727; TheWorld),gzip(gfe),gzip(gfe) D is bounded. > Because D is bounded there is some d > 0 with diam D < d > and because for j = 1 to k, D_j subset D, diam Dj <= diam D. No. I only mean every diam D_j <= d. Exists d>0, when every diam D_j <= d, below formular holds. bd E subset ( / D_i | D_i / (bd E) /= none ) subset /_(i=1 to k)B_i. why does d exist and formular hold? === Subject: Re: inconsistent fourier transforms of periodic signals? <3pb1p31nc7m5uha7nnps2u20ms02cs1pjt@4ax.com> posting-account=Qa-OoAoAAABkrOXD3qFsQ3h0Gy0hySxp 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0),gzip(gfe),gzip(gfe) I am getting inconsistent answers for the fourier transform of the >dirac delta impulse train: x(t) = summation over k from -inf to +inf > ædelta(t-4k). If I consider this signal as periodic, then the answer I get is X(jw) >= pi/2 summation over k from -inf to +inf ædelta(w - pi*k/2). However, if I consider the signal as a summation of shifted deltas, >then accordingly, the fourier transform of delta(t-4k) = e^(jw4k). >Then adding these shifted fourier transforms i get: X(jw) = summation >over k from -inf to +inf e^(jw4k) Why do I get such inconsistencies, or is it simply that I am going >wrong somewhere? Believe it or not the two answers are not inconsistent, there is > a flavor of convergence where the sum of those exponentials does > converge to exactly that sum of delta functions. Hard to know exactly what to say by way of explanation, without > knowing how much math you know. > ************************ David C. Ullrich k from -inf to +inf e^(jw4k) converges, since the exponential goes to infinity as k tends towards -inf. If you could direct me to a proof of the flavor of convergence that you are refering to, it would be === Subject: Reference on Stone Representation Theorem posting-account=4m6i3AoAAACkBWCtXNbpbnHMq5T0a0Fu I am looking for a good reference (aside from Johnstone, please) to cite the version of Stone representation theorem that says that the category of Boolean rings and that of Stone Spaces are contravariantly equivalent. This equivalence is given by taking a Boolean ring A to its prime spectrum Spec A (this is the map for the objects of these categories) and taking a morphism in Boolean ring to its corresponding spectral map. I find Johnstone to general and not particularly very easy to read when one just wants to know this (It goes on with Frames and Locales and whatnots and uses a more generalized Stones rep. theorem and then just states the above as a corollary without even giving how the equivalence is particularly done). So I want to see if I can cite another reference that might be easier for the reader and states what Jose Capco === Subject: Re: Reference on Stone Representation Theorem >I am looking for a good reference (aside from Johnstone, please) to >cite the version of Stone representation theorem that says that the >category of Boolean rings and that of Stone Spaces are contravariantly >equivalent. This >equivalence is given by taking a Boolean ring A to its prime spectrum >Spec A (this is the map for the objects of these categories) and >taking a morphism in Boolean ring to its corresponding spectral >map. I find Johnstone to general and not particularly very easy to >read when one just wants to know this (It goes on with Frames and >Locales and whatnots and uses a more generalized Stones rep. theorem >and >then just states the above as a corollary without even giving how the >equivalence is particularly done). So I want to see if I can cite >another reference that might be easier for the reader and states what Possibly (it's been ages since I studied any of this stuff, and even then I wasn't concentrating too well!): Halmos, Lectures on Boolean Algebras Koppelberg, Handbook of Boolean Algebras, vol. 1? Davey & Priestley, Introduction to Lattices and Order Of these, Halmos is probably the one most to the point, but, sadly, both it and Koppelberg seem to be out of print. Davey & Priestley is now in a second edition (I have the first): it (naturally!) treats the more general Priestley representation theorem, for bounded distributive lattices, so extracting the exact statement of the Stone theorem might require some work. (I'm afraid I don't feel up to thinking hard enough about it at the moment to make sure. Maybe later ...) There's a new book, Kaye, The Mathematics of Logic: A Guide to Completeness Theorems and Their Applications, which has something about the Stone representation theorem in it, but I haven't seen a copy, so I don't know if it gives the full statement that you require. There might perhaps be something in John L. Bell, Set Theory: Boolean-Valued Models and Independence Proofs (now in a third edition), or in Jech, Set Theory, or Kunen, Set Theory: An Introduction to Independence Proofs. (However, I may have got these listed as references for reasons only loosely related to the Stone representation theorem, so it's a bit of a long shot). -- Angus Rodgers Contains mild peril === Subject: Re: Reference on Stone Representation Theorem posting-account=4m6i3AoAAACkBWCtXNbpbnHMq5T0a0Fu æHalmos, Lectures on Boolean Algebras > æKoppelberg, Handbook of Boolean Algebras, vol. 1? > æDavey & Priestley, Introduction to Lattices and Order Of these, Halmos is probably the one most to the point, but, > sadly, both it and Koppelberg seem to be out of print. æDavey > & Priestley is now in a second edition (I have the first): it > (naturally!) treats the more general Priestley representation > theorem, for bounded distributive lattices, so extracting the > exact statement of the Stone theorem might require some work. > (I'm afraid I don't feel up to thinking hard enough about it > at the moment to make sure. æMaybe later ...) I've heard about Halmos, never got the chance to read it though. We have all three volumes of Koppelberg in the library, I think this must also have an easy formulation of Stones representation theorem. I will borrow the Halmos book from the library, I know they are out of print.. but fortunately our small library has them =) Jose Capco > Angus Rodgers > Contains mild peril === Subject: Re: Geometry with difficult .... posting-account=n4TzyQkAAADLWxrRHqyiUZ-1SZdOB4vv Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > [...] Back to the imaginary. I still do not grasp it. >Let me pose my question again. >And it is about, what You express here: > It is allways hard the first time, just because imaginary > points don't have any real existence. > You can't show where they are. > Better understand them as just a way to do calculations. I do not believe in mathematical magic. There is no magic here, just definitions and their > consequences. This is the result of my second try: >... >Now You get two points. Are these the imaginary points You >are talking of? Of course not. The imaginary points being imaginary are not (repeated quote) > imaginary points don't have any real existence. > You can't show where they are. You really _can't_ show (neither draw) imaginary points ! > It is an abstraction. You have to leave the ordinary Euclidean plane and use a > complexified plane, in which the coordinates of points are no > more real numbers but complex numbers (in C), that is C^2 .... All others just can't be shown on the Euclidean plane : > they are imaginary points ! > .............. > You have to leave the ordinary Euclidean plane and use a > complexified plane, in which the coordinates of points are no > more real numbers but complex numbers (in C), that is C^2 A geometric representation of ordered couples of ordered couples of real numbers, C x C, can be done in a lot of ways. It seems to me, that the first of such a representation was done from Caspar Wessel in the second part of On the Analytical Representation of Direction, ca.1797 The only complete edition of this text in the internet is a french translation http://gallica.bnf.fr/ark:/12148/bpt6k99681g With friendly greetings Hero PS Around 1968 one could read on a wall in Paris something like 'L' imagination a la puissance'. Can anyone help with more info about it? === Subject: Re: Geometry with difficult ....Butterflies and simple Owls posting-account=n4TzyQkAAADLWxrRHqyiUZ-1SZdOB4vv Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) some explanations to a theorem, mina proved. > This is very hard to me. > Maybe, this is expert solution. > pencil, polar, conjugate, harmonic range... > I'm unfamiliar... Let me try, let's start again: > Hello sir~ There is a circle with center 0. > There is a line L outside this circle. > There is a point P on this line such that line OP and line L are > perpendicular. > Draw any two secant line from P to circle. > Label point C, D and E, F respectively. > Draw the line DE. > Label the intersection point A on L. > Draw the line CF. > Label the intersection point B on L. Show that AP = BP. > This is a drawing. http://board-2.blueweb.co.kr/user/math565/data/math/appr001.jpg This is a answer. http://board-2.blueweb.co.kr/user/math565/data/math/APBP001.jpg You give a correct proof. Philippe recognized in Your description a butterfly, four points on a circle joint by two crossing chords ( in the point Cross or X) and two further chords with no common point inside, but with the crosspoint P outside. And he gave a link to Alexander Bogumolny's website, where lot's of butterflies are treated. http://www.cut-the-knot.org/pythagoras/Butterfly.shtml Here are some speculations of mine, which You might like to prove too, if You feel for this. Let's give these butterflies a body, connect the center O of the circle with radius r with the point Cross X and extend to the perimeter. Draw a line halving the angle between OP and OX. Reflect X through this line onto the line OP on a new point Y. Reflect Y through the perimeter and You will get P. So OX times OP = r * r or OX / r = r / OP. > I know that Given collinear points A, P, B, and Q, > P and Q are harmonic conjugates with respect to A and B > if |AP|/|PB| = |AQ|/|BQ|. Draw a perpendicular to the line OX through X. This line intersects the two chords CD and EF in two points Ainside and Binside. Reflect them through the linie, which divides the angle XOP, and reflect them through the circle and they will land on A and B.(Actually the endpoints Ainside Binside only considered, not the lines connecting with X). So XAinside times PA = r * r or XAinside / r = r / PA. Given Your assumptions one can have a lot of different butterflies, symmetric or not. They all have a body, divided in the ratio OX to (r - OX). And a common line in their wings extend from the body, from X perpendicular to both sides with the length XAinside. This together expresses the amount of symmetry and harmony in a butterfly. In a co-discussion in the german de.sci.mathematik Schmetterlinge praesentiert von Philippe und die Eule von Klaus Klaus Loeffler gave us a beautiful applet of this http://www.krloeffler.de/HT/HerosTheorem.html It looks like an owl with two big eyes and the butterfly is flapping around , but not leaving the nose of the owl, the point X of intersection ( Klaus notation P). In dragging the points Q and either A or B in Klaus applet, one can change the relation of OX to XAinside. The comparision with an owl is not quite correct mathematically, just an association, the eyes can extend partly outside the face. And philippe gave us another applet, where one can move the point of intersection X to the outside, inverting the points (not the lines) of a butterfly (and lot of more interesting material, relating the butterflies to conic sections and to projetive geometry): http://chephip.free.fr/pbg_en/sol150.html It was Philippes idea to look for a relation to complex numbers (a, b ), ordered pairs of real numbers, their addition componentwise and their multiplication ( a, b ) * (u, v ) = ( au - bv, av + bu) = (x, y ) = x + sqrt ( - 1 ) * y. The harmonic symmetry of an asymmetric or not butterfly is given by a complex number ( OX, X Ainside). ( OX, X Ainside) times ( OP, PA) = (1, 0 ). It is a time of cold and long dark nighths. One day-dreams about warmth and butterflies, like Dschuang Dsi, a chinese, about a ' southern land of flowers' and butterflies. Is this southern land Your home Korea, do You have flowers now and butterflies, or do You have to be content with mathematical flinders? With friendly greetings Hero PS What is the relation to the second point of intersection outside, of the intersection of the two chords CE and DF extended.? === Subject: Re: Geometry with difficult ....Butterflies and simple Owls > some explanations to a theorem, mina proved. > This is very hard to me. > Let me try, let's start again: > Hello sir~ > > There is a circle with center 0. > There is a line L outside this circle. > There is a point P on this line such that line OP and line L are > perpendicular. > Draw any two secant line from P to circle. > Label point C, D and E, F respectively. > Draw the line DE. > Label the intersection point A on L. > Draw the line CF. > Label the intersection point B on L. > > Show that AP = BP. > This is a drawing. > http://board-2.blueweb.co.kr/user/math565/data/math/appr001.jpg > > This is a answer. > http://board-2.blueweb.co.kr/user/math565/data/math/APBP001.jpg You give a correct proof. > Philippe recognized in Your description a butterfly, four points on a > circle joint by two crossing chords (in the point Cross or X) Here are some speculations of mine, which You might like to prove > too, if You feel for this. > Let's give these butterflies a body, connect the center O of the > circle with radius r with the point Cross X and extend to the > perimeter. > Draw a line halving the angle between OP and OX. > Reflect X through this line onto the line OP on a new point Y. > Reflect Y through the perimeter and You will get P. False, if by Reflect Y through the perimeter you mean Circle inversion, that is OY*OZ = r^2, you don't get P but some point Z allways nearer than P. (OZ <= OP) http://i2.tinypic.com/6l2torb.gif > So OX times OP = r * r or > OX / r = r / OP. That is OX * OP >= r*r > I know that Given collinear points A, P, B, and Q, > P and Q are harmonic conjugates with respect to A and B > if |AP|/|PB| = |AQ|/|BQ|. Draw a perpendicular to the line OX through X. This line intersects > the two chords CD and EF in two points Ainside and Binside. For simplicity, call them just a, b Oh, BTW, a,b might be ... outside the circle, even with X inside ! > Reflect > them through the linie, which divides the angle XOP, and reflect them > through the circle and they will land on A and B. This is also false. It was Philippes idea to look for a relation to complex numbers However, I didn't intend *these* complex numbers... (to represent a real point of the plane with coordinates (x, y) by the complex number (x+i*y)), but rather to have points with coordinates x, y which are themselves both complex numbers, that is one point have coordinates (x, y) with x = a + i*b and y = c + i*d. > (a, b ), ordered pairs of real numbers, their addition componentwise > and their multiplication > ( a, b ) * (u, v ) = ( au - bv, av + bu) = (x, y ) = > x + sqrt ( - 1 ) * y. What we write using i = sqrt(-1) : x + i*y > The harmonic symmetry of an asymmetric or not butterfly is given by > a complex number ( OX, X Ainside). > ( OX, X Ainside) times ( OP, PA) = (1, 0 ). In unit circle, that is with r = 1. Hum... (answer also on dsm, now I see which of (OX, Xa) times ( OP, PA) = r*r you meant there). It can't be true, because this implies a relation with just lengths OX, Xa, OP, PA, independant of r (the component 0 in (r*r, 0)), which can't exist as from given just OX, OP, PA lengths you can get any Xa you want (changing r). > PS What is the relation to the second point of intersection outside, > of the intersection of the two chords CE and DF extended.? what relation between Mina's Butterfly, and normal Alexander Bogumolny's butterfly. All the properties are exactly the same, P being inside or outside, and X being independantly inside or outside. -- Philippe C., mail : chephip+news@free.fr site : http://chephip.free.fr/ (recreational mathematics) === Subject: A question in real analysis! posting-account=zn2mMAoAAAB8GcQbJgFi7C8YMkm06Ath 5.1),gzip(gfe),gzip(gfe) Hi! I want to share some combersome problem with you. That is, if P is nonconstant real polynomial, show that lim_n->inf integral_0^ 1 {e^(i*n*P(x)) dx = 0. How to solve it? === Subject: Re: A question in real analysis! posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq InfoPath.1),gzip(gfe),gzip(gfe) > I want to share some combersome problem with you. That is, if P is nonconstant real polynomial, show that lim_n->inf integral_0^ 1 {e^(i*n*P(x)) dx = 0. How to solve it? I found it interesting that you used real analysis in the subject thread and then employed i = sqrt(-1) in your question! That said, there are a lot of books titled real analysis that do this, such as when a little Fourier series is covered (and even if not, then it still often comes up when the Riemann-Lebesgue theorem shows up, either in the text or the exercises). I'm not really sure about this, but I suppose the term real is used (exclusively) as long as major incursions into complex analysis aren't made (Cauchy integral theorem, residues, etc.). Dave L. Renfro === Subject: Re: A question in real analysis! posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20071213 Fedora/2.0.0.10-3.fc8 Firefox/2.0.0.10,gzip(gfe),gzip(gfe) > I want to share some combersome problem with you. That is, if P is nonconstant real polynomial, show that lim_n->inf integral_0^ 1 {e^(i*n*P(x)) dx = 0. How to solve it? I found it interesting that you used real analysis > in the subject thread and then employed i = sqrt(-1) > in your question! That said, there are a lot of books > titled real analysis that do this, such as when > a little Fourier series is covered (and even if not, > then it still often comes up when the Riemann-Lebesgue > theorem shows up, either in the text or the exercises). I'm not really sure about this, but I suppose the > term real is used (exclusively) as long as major > incursions into complex analysis aren't made (Cauchy > integral theorem, residues, etc.). Well, real analysis usually includes, for example, the theory of analytic interpolation of Banach spaces in order to do things such as showing boundedness of various very-`real-analysis' operators on Lp spaces, and that tends to be an orgy of complex analysis. Hadamard is usually quoted as saying that the shortest way between a question an an answer of real analysis goes through the complex domain. -- m === Subject: Re: A question in real analysis! >I want to share some combersome problem with you. That is, if P is nonconstant real polynomial, > show that lim_n->inf integral_0^ 1 {e^(i*n*P(x)) dx = 0. How to solve it? I found it interesting that you used real > analysis > in the subject thread and then employed i = > sqrt(-1) > in your question! That said, there are a lot of > books > titled real analysis that do this, such as when > a little Fourier series is covered (and even if > not, > then it still often comes up when the > Riemann-Lebesgue > theorem shows up, either in the text or the > exercises). I'm not really sure about this, but I suppose the > term real is used (exclusively) as long as major > incursions into complex analysis aren't made > (Cauchy > integral theorem, residues, etc.). Well, real analysis usually includes, for example, > the theory of analytic interpolation of Banach spaces > in order to do things such as showing boundedness of > various very-`real-analysis' operators on Lp spaces, > and that tends to be an orgy of complex analysis. Hadamard is usually quoted as saying that the > shortest way > between a question an an answer of real analysis goes > through > the complex domain. -- m Quite so. The result is what interests us. Analysis implies continuous functions. The 2 dimensional property of the complex plane allows us a conveniently shorter path by supplying more room to reach a destination while guaranteeing continuity. Tom === Subject: Re: A question in real analysis! >I want to share some combersome problem with you. That is, if P is nonconstant real polynomial, show that lim_n->inf integral_0^ 1 {e^(i*n*P(x)) dx = 0. How to solve it? This is a problem in stationary phase. Take a look at . The rate at which this tends to 0 depends on the greatest degree of a zero of P'(x) in [0,1]. The greater the degree of this zero, the slower the convergence to 0. For example, if P'(x) is not 0 in [0,1], the integral tends to 0 like 1/n. If P'(x) has a zero of degree k, the integral tends to 0 like n^{-1/(k+1)}. Rob Johnson take out the trash before replying === Subject: Re: JSH: Congradulations on your success and New Job <4790beb2$0$47130$892e7fe2@authen.yellow.readfreenews.net> posting-account=3WPJYgoAAAA55VjhzK9i07RN8h8u8eEs Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) By the time a > clearance at this level is granted, the government will know more > about you than either you or your parents know. Yes, a 10 second search of James on Google should pretty much do it (if his total lack of any mathematical experience doesn't). OTOH, if the govt wants to do research into narcissistic , paranoid, delusional types or they're looking for another unstable person to install in a dictatorship, James may be a shoo-in. M === Subject: Number with sqrt...2 Hello sir~ A method finding sqrt... Ok, let's go... sqrt(2) = (a1). (a2)(a3).... a1, a2, a3, .... are nonnegative integer. (a1). (a2)(a3).... is decimal notation as 1. 41... I will find (a1). [(a1). (a2)(a3)...]^2 = 2. so, [(a1) + 0. (a2)(a3)...]^2 = 2. so, (a1)^2 + 2.(a1).{0. (a2)(a3)...} + {0. (a2)(a3)...}^2 = 2. so, (a1)^2 + {0. (a2)(a3)...}*[{0. (a2)(a3)...} + 2.(a1)] = 2. Choose the largest (a1) such that (a1)^2 <= 2. I will find (a2). so, 2 - (a1)^2 = {0. (a2)(a3)...}*[{0. (a2)(a3)...} + 2.(a1)]. so, (100)[2 - (a1)^2] = {(a2). (a3)...}*[{a2). (a3)...} + 20.(a1)] = [(a2) + {0. (a3)...}]*[((a2) + {0. (a3)...}) + 20.(a1)] = [(a2).{(a2) + 20.(a1)}] + {0.(a3)...}*[~~~]. Choose the largest (a2) such that [(a2).{(a2) + 20.(a1)}] <= (100)[2 - (a1)^2]. I will find (a3). so, (100)(100)[2 - (a1)^2] - [(a2).{(a2) + 20.(a1)}] = ~~~~~~~~. I can make a diagram. Namely, http://board-2.blueweb.co.kr/user/math565/data/math/didiag.jpg === Subject: Re: Number with sqrt...2 > Hello sir~ A method finding sqrt... > Ok, let's go... sqrt(2) = (a1). (a2)(a3).... > a1, a2, a3, .... are nonnegative integer. > (a1). (a2)(a3).... is decimal notation as 1. 41... I will find (a1). > [(a1). (a2)(a3)...]^2 = 2. > so, [(a1) + 0. (a2)(a3)...]^2 = 2. > so, (a1)^2 + 2.(a1).{0. (a2)(a3)...} + {0. (a2)(a3)...}^2 = 2. > so, (a1)^2 + {0. (a2)(a3)...}*[{0. (a2)(a3)...} + 2.(a1)] = 2. > Choose the largest (a1) such that (a1)^2 <= 2. I will find (a2). > so, 2 - (a1)^2 = {0. (a2)(a3)...}*[{0. (a2)(a3)...} + 2.(a1)]. > so, (100)[2 - (a1)^2] = {(a2). (a3)...}*[{a2). (a3)...} + 20.(a1)] > = [(a2) + {0. (a3)...}]*[((a2) + {0. (a3)...}) + 20.(a1)] > = [(a2).{(a2) + 20.(a1)}] + {0.(a3)...}*[~~~]. > Choose the largest (a2) such that [(a2).{(a2) + 20.(a1)}] <= (100)[2 - > (a1)^2]. I will find (a3). > so, (100)(100)[2 - (a1)^2] - [(a2).{(a2) + 20.(a1)}] = ~~~~~~~~. I can make a diagram. > Namely, > http://board-2.blueweb.co.kr/user/math565/data/math/didiag.jpg Apply this cubic root. cubic root(2) = (a1). (a2)(a3).... will find (a1). [(a1). (a2)(a3)...]^3 = 2. so, [(a1) + 0. (a2)(a3)...]^3 = 2. so, (a1)^3 + 3.(a1)^2.{0. (a2)(a3)...} + 3.(a1).{0. (a2)(a3)...}^2 + {0. (a2)(a3)...}^3 = 2. so, (a1)^3 + {0. (a2)(a3)...}*[3.(a1)^2 + 3.(a1).{0. (a2)(a3)...} + {0. (a2)(a3)...}^2] = 2. Choose the largest (a1) such that (a1)^3 <= 2. I will find (a2). so, 2 - (a1)^3 = {0. (a2)(a3)...}*[3.(a1)^2 + 3.(a1).{0. (a2)(a3)...} + {0. (a2)(a3)...}^2]. so, (1000)[2 - (a1)^3] = {(a2). (a3)...}* (100).[3.(a1)^2 + 3.(a1).{0. (a2)(a3)...} + {0. (a2)(a3)...}^2]. = [(a2) + {0. (a3)...}]*[300.(a1)^2 + 30.(a1).{(a2). (a3)...} + {(a2). (a3)...}^2]. = [(a2).{300.(a1)^2 + 30.(a1).(a2) + 1.(a2)^2}] + {0.(a3)...}*[~~~]. Choose the largest (a2) such that [(a2).{300.(a1)^2 + 30.(a1).(a2) + 1.(a2)^2}] <= (100)[2 - (a1)^3]. so, Key with sqrt = {(a2) + 20.(a1)} Key with cubic root = {300.(a1)^2 + 30.(a1).(a2) + 1.(a2)^2} I can make a diagram. Namely, http://board-2.blueweb.co.kr/user/math565/data/math/cubdiag.jpg === Subject: Re: Number with sqrt...2 > Hello sir~ > A method finding sqrt... > Ok, let's go... > sqrt(2) = (a1). (a2)(a3).... > a1, a2, a3, .... are nonnegative integer. > (a1). (a2)(a3).... is decimal notation as 1. 41... > I will find (a1). > Choose the largest (a1) such that (a1)^2 <= 2. > I will find (a2). > Choose the largest (a2) such that > [(a2).{(a2) + 20.(a1)}] <= (100)[2 - (a1)^2]. etc... Apply this cubic root. > cubic root(2) = (a1). (a2)(a3).... will find (a1). > [(a1). (a2)(a3)...]^3 = 2. > Choose the largest (a1) such that (a1)^3 <= 2. I will find (a2). > Choose the largest (a2) such that > [(a2).{300.(a1)^2 + 30.(a1).(a2) + 1.(a2)^2}] <= (100)[2 - (a1)^3]. > This was the method I learnt in school (at least for the square root) to get digits of sqrt's one at a time. Ok, it works (and your calculations as a proof of the method seem to be correct), but it is never so efficient than the Newton's method, which roughly doubles the number of digits at each step ! for instance sqrt(2) step 1 : 1 step 2 : 1 - (1^2 - 2)/(2*1) = 1.5 step 3 : 1.5 - (1.5^2 - 2)/(2*1.5) = 1.416 step 4 : 1.416 - (1.416^2 - 2)/(2*1.416) = 1.414215 step 5 : 1.414215 - (1.414215^2 - 2)/(2*1.414215) = 1.414213562374 step 6 : 1.414213562374 - (1.414213562374^2 - 2)/(2*1.414213562374) = 1.4142135623730951 and we get allready 15 exact digits (limited by accuracy of the javascript used for printing that). There are refinements to that method to avoid dividing by successive values. -- Philippe C., mail : chephip+news@free.fr site : http://chephip.free.fr/ (recreational mathematics) === Subject: Re: Meaning of religion > I am truly an admirer of the West, but a Democratic, Atheistic, > Scientific and Artistic West. > We cannot survive otherwise! > Survive as exactly what? > As a Civilization! > How well have atheistic civilisations survived? > Start a new thread! > Okay, let us ask, what do we mean by the term civilisation. > Sigge > In addition we might want to know what > we mean by the term atheistic. > It means not having any sort of supernatural consciousness to impact any > aspect of personal, social and political life. > Religion seems to be the stepchild of > civilization, or at least religion as we > currently know it. > Civilisation as we knew it was the direct product of religion. For in > all old civilisations they had kings who were the representatives of gods > on earth. Without the concept of the Divine, they could not have the > concept of the king, and without the king they could not have any sort of > state with fixed geographical boundaries. Of course, here the implicit > meaning of civilisation is that it has to be related to human settlement > in a fixed area (not any nomadic existence) and that there should be a > concept of ownership of land and other property. > The Roman Empire seems to have been very > tolerant of people's religions and in the > later days, seemed to not have much to > do with religion until it became Christian > in the 4th century. > It would seem that our own, post enlightenment, > liberal civilization could be considered > atheistic, but no one can claim that the > American people are atheistic as a group. > discuss... > the whole arrogant parasites feeding upon the creativity and productivity > of the humble theists. > Yet Western civilization collpased when it became Christian. Yes, they had the Dark Ages when there was no safety. However, Christian or not, Europe today is a very safe and admirable place! In 2006 my daughter travelled all over Europe by herself! Isn't that wonderful? -- > Corporate society looks after everything. All it asks of anyone, all it > has ever asked of anyone, is that they do not interfere with management === Subject: Re: Meaning of religion Oh, i see, we can discuss religion in Mathforum as well. It is a very nice social forum? What about discussing discussing politics behind religions? So far I see, there are some political reasons behind the spread of religions? These divides people in the name of god and religion. The very question is, if the purpose of every religion is the preaching of peace and civilization, then why are so much killing on the earth in the name of religion? === Subject: Re: Meaning of religion > Oh, i see, we can discuss religion in Mathforum as well. on any subject matter at all. Write about your bizarrest sexual fantasies, anything at all. Whether you _ought_ to is another matter entirely. Please dont. Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: Meaning of religion > I am truly an admirer of the West, but a Democratic, Atheistic, > Scientific and Artistic West. > We cannot survive otherwise! > Survive as exactly what? > As a Civilization! > How well have atheistic civilisations survived? > Start a new thread! > Okay, let us ask, what do we mean by the term civilisation. > Sigge > In addition we might want to know what > we mean by the term atheistic. > It means not having any sort of supernatural consciousness to impact any > aspect of personal, social and political life. > Religion seems to be the stepchild of > civilization, or at least religion as we > currently know it. > Civilisation as we knew it was the direct product of religion. For in > all old civilisations they had kings who were the representatives of gods > on earth. Without the concept of the Divine, they could not have the > concept of the king, and without the king they could not have any sort of > state with fixed geographical boundaries. Of course, here the implicit > meaning of civilisation is that it has to be related to human settlement > in a fixed area (not any nomadic existence) and that there should be a > concept of ownership of land and other property. Civilization is the direct product of organized > agriculture. Organization seemed to become necessary > when harvests were excessive in good years and > inadequate in bad years. Agriculture also resulted > directly in increases in population which created > a need for organization to reduce violence between > people. And exactly how do organised agriculturists protect themselves from raiders? Organised agriculture or for that matter anything organised implicitly results from the notion of some protector. Without some sort of a king to begin with, there can be no organisation and no protection. > The preceding statement is horribly simplistic, but > seems to tally with what we know from the > archaeological record about the development of > early civilizations. It is not only simplistic, it is also unconvincing. > I might suggest Jared Diamond as good reading on > the subject. I would suggest that a study of modern socities undergoing transition from hunter-gatherer to agriculture state could be more rewarding and revealing. My father is trying to do his bit here, by trying to bring the neglected tribals of Jharkhand from their hunter-gatherer+minimal agriculture state to more modern styles. In this, the role of the Govt. as a protector and guide is most important. So long as there was no Govt. to take care, these people were neglected or oppressed and could never make a break-through. Modern democracy has given them some power, and they are using it to their own benefit. In short, you need a protector FIRST to create prosperity and its results - proper civilisation. Organized agriculture/civilization creates wealth. > It also creates it's opposite, poverty. Poverty > generates resentment in those who are impovrished. If you imply that poor people are normally and naturally resentful, that may be more true of Western societies than the Indian ones I am more familiar with. The failure of Communism (so far) even in the poorest parts of India should bear me out. Of course, people can be misguided and made to behave violently as a result of propaganda and theories. > Religion (again, very simplisticly) arose as an > adjunct to the organizers and became partners with > the organizers as a means to educate the poor to > the fact that even though they are poor, they will > have their reward in the next life. This is classic Communist theory and neglects that fact that personal wealth has little or nothing to do with one's relationship with the Divine. How curious, or how methodical one is, about the supernatural, is an individual's business and his choice. Only simple minds will swallow the communist logic, but there was a time when their numbers were legion. Religion is man's greatest invention or discovery or blessing, as it allows him to conjecture entities necessarily far superior to himself in the abstract, and model his behaviour along such superior lines as a consequence. All human gains have ultimately resulted from this great blessing, which is as far as known denied to other life forms. > The Roman Empire seems to have been very > tolerant of people's religions and in the > later days, seemed to not have much to > do with religion until it became Christian > in the 4th century. > It would seem that our own, post enlightenment, > liberal civilization could be considered > atheistic, but no one can claim that the > American people are atheistic as a group. > discuss... > the whole arrogant parasites feeding upon the creativity and productivity > of the humble theists. We (the imperial we, meaning I) don't agree with > Religion developed as a need arose. The organizers > (kings, chiefs, etc.) found it easier to control > populations with fear instilled by religious leaders > than with whips, though whips are very effective > and continued to be use even after religion was > developed. There is true religion, but there are also so many cults. Where men take on the role of gods, religion gets debased into personality cults. > Theists, the rank and file, can indeed be humble, > but they are because their leaders tell them they > must be to enter the promised land. The leaders, > on the other hand, are far from humble and more > closely resemble your definition of atheists, which, > of course doesn't agree with my definition of > atheists at all. No doubt, for you are talking from the consequences of the wretched post-Socratic theistic positions now dominant in the Western world. === Subject: Re: Meaning of religion > I am truly an admirer of the West, but a Democratic, Atheistic, > Scientific and Artistic West. > We cannot survive otherwise! > Survive as exactly what? > As a Civilization! > How well have atheistic civilisations survived? > Start a new thread! > Okay, let us ask, what do we mean by the term civilisation. > Sigge > In addition we might want to know what > we mean by the term atheistic. > It means not having any sort of supernatural consciousness to impact any > aspect of personal, social and political life. > Religion seems to be the stepchild of > civilization, or at least religion as we > currently know it. > Civilisation as we knew it was the direct product of religion. For in > all old civilisations they had kings who were the representatives of gods > on earth. Without the concept of the Divine, they could not have the > concept of the king, and without the king they could not have any sort of > state with fixed geographical boundaries. Of course, here the implicit > meaning of civilisation is that it has to be related to human settlement > in a fixed area (not any nomadic existence) and that there should be a > concept of ownership of land and other property. > Civilization is the direct product of organized > agriculture. Organization seemed to become necessary > when harvests were excessive in good years and > inadequate in bad years. Agriculture also resulted > directly in increases in population which created > a need for organization to reduce violence between > people. And exactly how do organised agriculturists protect themselves from raiders? > Organised agriculture or for that matter anything organised implicitly > results from the notion of some protector. Without some sort of a king to > begin with, there can be no organisation and no protection. > > The preceding statement is horribly simplistic, but > seems to tally with what we know from the > archaeological record about the development of > early civilizations. It is not only simplistic, it is also unconvincing. > > I might suggest Jared Diamond as good reading on > the subject. I would suggest that a study of modern socities undergoing transition from > hunter-gatherer to agriculture state could be more rewarding and revealing. > My father is trying to do his bit here, by trying to bring the neglected > tribals of Jharkhand from their hunter-gatherer+minimal agriculture state to > more modern styles. In this, the role of the Govt. as a protector and guide > is most important. So long as there was no Govt. to take care, these people > were neglected or oppressed and could never make a break-through. Modern > democracy has given them some power, and they are using it to their own > benefit. In short, you need a protector FIRST to create prosperity and its > results - proper civilisation. > Organized agriculture/civilization creates wealth. > It also creates it's opposite, poverty. Poverty > generates resentment in those who are impovrished. If you imply that poor people are normally and naturally resentful, that may > be more true of Western societies than the Indian ones I am more familiar > with. The failure of Communism (so far) even in the poorest parts of India > should bear me out. Of course, people can be misguided and made to behave > violently as a result of propaganda and theories. > > Religion (again, very simplisticly) arose as an > adjunct to the organizers and became partners with > the organizers as a means to educate the poor to > the fact that even though they are poor, they will > have their reward in the next life. This is classic Communist theory and neglects that fact that personal wealth > has little or nothing to do with one's relationship with the Divine. How > curious, or how methodical one is, about the supernatural, is an > individual's business and his choice. Only simple minds will swallow the > communist logic, but there was a time when their numbers were legion. Religion is man's greatest invention or discovery or blessing, as it allows > him to conjecture entities necessarily far superior to himself in the > abstract, and model his behaviour along such superior lines as a > consequence. All human gains have ultimately resulted from this great > blessing, which is as far as known denied to other life forms. > > The Roman Empire seems to have been very > tolerant of people's religions and in the > later days, seemed to not have much to > do with religion until it became Christian > in the 4th century. > It would seem that our own, post enlightenment, > liberal civilization could be considered > atheistic, but no one can claim that the > American people are atheistic as a group. > discuss... > the whole arrogant parasites feeding upon the creativity and productivity > of the humble theists. > We (the imperial we, meaning I) don't agree with > Religion developed as a need arose. The organizers > (kings, chiefs, etc.) found it easier to control > populations with fear instilled by religious leaders > than with whips, though whips are very effective > and continued to be use even after religion was > developed. There is true religion, but there are also so many cults. Where men take on > the role of gods, religion gets debased into personality cults. > > Theists, the rank and file, can indeed be humble, > but they are because their leaders tell them they > must be to enter the promised land. The leaders, > on the other hand, are far from humble and more > closely resemble your definition of atheists, which, > of course doesn't agree with my definition of > atheists at all. No doubt, for you are talking from the consequences of the wretched > post-Socratic theistic positions now dominant in the Western world. I will save this post for a time when I have enough time to respond in a thoughtful way. It seems to me that you are thinking and writing from the point of view of a civilized person who has bought into the NEED for civilization. That's not necessarily a bad thing overall, but it does cloud a proper view of the past. Consider that agriculture may be as much as 12,000 years old but human occupation of the planet is much, much older, maybe as much as a million years in more or less modern form. Try Jared Diamond, Guns, Germs and Steel. Don't reject an argument out of hand simply because the communists used it. That's as bad as communism rejecting any and all religion out of hand. Look at the argument only for it's value. Not for who used it. === Subject: (41)^n in terms of 2^k How can we express (41)^n in terms of 2^k? === Subject: Re: (41)^n in terms of 2^k posting-account=4n0P8QoAAACPj0DJnja1mCT1wUiU6txx Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > How can we express (41)^n in terms of 2^k? Given: 41^n = 2^k Solve for: k in terms of n [1] Solution: lg is the base 2 logarithm. 41^n = 2^k lg(41^n) = lg(2^k) lg(41^n) = k k = n * lg(41) Therefore: 41^n = 2^(n * lg(41)) [1] Ambiguous as your question is, I may have misunderstood it. If so, please clarify. === Subject: Re: (41)^n in terms of 2^k >How can we express (41)^n in terms of 2^k? if n and k are integers, there is no way. If you allow real values, then take logarithms of both sides: n log(41) = k log(2) So we can compute k = n log(41)/log(2). Rob Johnson take out the trash before replying === Subject: Re: Algebra with inverse... posting-account=z1ZA6AoAAACEgXDaRRTJFG5d4vJvYyOY SV1),gzip(gfe),gzip(gfe) > Hello sir~ f(x) = x^3 - 2x + 2 is a irreducible polynomial over Q. Let c be a root of f(x). Find multiplicative inverseof 3+c in Q(c). ------------------------------------------------------------- 1) c^3 - 2c + 2 = 0. so, c^3 - 2c + 3 = 1. so, (c^3 - 2c + 3) = (c+3)(c^2 - 3c + 7) - 18 so, 1 = (c+3)(c^2 - 3c + 7) - 18 so, (c+3)(c^2 - 3c + 7) = 19 so, (c+3)^{-1} = (1/19)(c^2 - 3c + 7) 2) For e + f.c + g.c^2 in Q(c), (3+c)(e + f.c + g.c^2) = 1. so, (3e-2g) + c.(3f+e+2g) + c^2(3g+f) = 1. so, 3e-2g = 1 , 3f+e+2g = 0 , 3g+f = 0. so, g = 1/19 , f = -3/19, e = 7/19. so, (c+3)^{-1} = (1/19)(c^2 - 3c + 7) === === Subject: Re: Unsolvable Math/Science Problem? posting-account=y55HvwoAAAA0ux66xt1rddbAcfDqzDxu 2.0.50727),gzip(gfe),gzip(gfe) > In theory, a single pulse should accumulate to infinite pressure > in the center. But since it is a waste of money to risk that a > big ball of gold turns into a black hole, let us assume a stable > situation... Try to find a standing wave ... > Argue that each vibrating shell (i.e. parts between node surfaces) > should contain the same energy. > Stick the same energy in the center... Excellent approach (the method of shells). If your assumptions are correct the energy density in the core is inversely related to the square of the wavelength; outer shell volume decreases by 1/2 when the frequency is doubled, compared to a 1/8 decrease in volume for the central sphere. This means that wave shape is very important for single pulse (non-standing) waves. > Question 2: If a bubble of helium is located in the exact center of the sphere > as the pressure wave passes, what temperature (or should I say atomic > speed distribution) will it acquire? Hm, needs more physics. Yes, for a first approximation we could use the laws governing elastic collisions. This requires a system-wide conservation of both total momentum (m*v) and total kinetic energy (1/2 * m * v^2) at the atomic scale. A more precise calculation might take other things into consideration, like electron distributions and atomic energy/distance curves. > I'd guess that the typical speed of helium atoms should correspond > to a pressure that is 30,000 times increased, i.e. the temperature > should be increased by the same factor to about 9,000,000 K. Unfortunately, without knowing the size of the compressed bubble, there is little justification for that conclusion. The bubble's volume is an integral component influencing (and influenced by) the internal pressure and temperature. The best we can do appears to be an extreme case analysis using the Ideal Gas Law (PV=nRT). As the pressure wave passes, we consider two extreme cases: <1>: bubble volume remains constant. This requires a temperature increase of 30,000 fold. <2>: bubble temperature remains constant. This requires a volume decrease of 30,000 fold. > However, this suggests that some (nwanted?) side effects occur making > all rules of thumb look inadequate. Yes, fusion. P.S. I'm moving this discussion to sci.physics so the physicists can supply the details you guys need in order to build a decent mathematical model. Hope to see you there. === Subject: Re: Kuratowski Ordered Pair...is a set posting-account=n4TzyQkAAADLWxrRHqyiUZ-1SZdOB4vv Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) >Calling one element first ist not meant as first in time. f.e. there >is something like first and second in logical order, Moe mentioned to >me. And with Kuratowski we do not have to presuppose any meaning to >first and second, as used by set theory with regard to the Kuratowski >set, the first is even defined as >the element which is element in both elements of the Kuratowski set > The (Kuratowski) definition is: > = {{x} {xy}}. > Then > =

<-> (x=p & y=q) > is a THEOREM that comes from the definition > Then '1st(p)' and '2nd(p)' are defined, then another theorem: > p is an ordered pair <-> p = <1st(p) 2nd(p)> so > p is an ordered pair <-> p = {{1st(p)} {1st(p) 2nd(p)}}. In wiki a definition of first and second is given: >The statement that x is the first element of an ordered pair p can >then be formulated as >For all Y element of p : x element of Y >http://en.wikipedia.org/wiki/Ordered_pair >In other words, when the Kuratowski set has two elements, that element >which is element of both is called first. >We do not have to attach or associate any other interpretation, >meaning and content of first, used by set theory here, than just this. Actually this use of 'first', 'second' runs contrary to the common >math knowledge, in which first and second relates to numbers, being >ordinal versus the cardinal one, two. (Consider the difference of the >second house in a street, when starting at a certain end as ordinal- >from two houses in a street as cardinal). So it is uniquely defined in >this way for the replacement of the Kuratowski set with < a, b > and >in further developement first and second are used in their common >ordinal form, and this special definition of first and second is >never applied again. A very telling point. it seems to me. I was just reviewing this > whole thread. > ...I would love to have commented more on your ideas, about time > and Hamilton pairs, for example, if I was more sure of having something > coherent to say. As will have been obvious, I am only a novice in > mathematics. I would like to move on to the advanced stuff, but I keep > getting stuck trying to make sense of the fundamentals. Just to say I have > appreciated your posts. I feel with You. I always want to proceed to the residuum of complex functions and further on, but again and again i'm prevented, as i have to understand something more fundamental before, so i've to return to start. It took me more than three decennia to find out, that there is no mathematical definition of an imaginary axis and no definition of a complex plane (it is just the ordinary cartesian y-axis, even Riemann in 1850 knew, but later some mathemagicians changed this, and that's the way it is still taught most often). It is very satisfying to have someone, who shares the same view on things, especially when it is not mainstream. You are very cautious, but i think, after this discussion You can try and run around on this new territory, applying, what You found out. You might get stuck in some mud? Never mind, start anew. It was very good for us, that there were defenders of the view, that all of math can be based on logic and axiomatic set-theory, including especially the Kuratowski set. So we could learn more, than just telling us, how much we like each others point of view. And not all of their points are refuted, as seen fromout my standpoint. They became a little rare in this thread, i already miss Moe a bit. > I think we are agreed that there is a fudge about order at the > heart of set theory. I must say I find amusing the offer, sometimes offered > as argument in this thread, to take the ordered pair as primitive. How one > could hope to give an account of the Naturals, for example, or any other > sequence/order notion, having taken the ordered pair as undefined, is > beyond me. But that is the reductionist frame of mind at work: give me the > atom of the ordered pair and I will build you any molecule you desire. The chemists do not start with the atom and construct molecules. They had centuries before they found the atoms, and now time and again they have to return to the basic atoms finding out the electrons, neutrons and protons, or recently they found out that a basic square form of some molecule or part of it is a dynamical form (Petra Schiebel et.al.), a bit like the spokes of a wheel are creating a firm disc, when rotated with high speed, but here a square from rotation. Same for math, now and again we have to deduct from the math at work the basic elements and basic movements, revise, what was found out with the last attempts, and then, after a kind of axiomation turn to synthesize new stuff. For Moe mathworld is starting with ZFC-axioms and first and second order logic formulation. This theory is far over the climax already, as from defenders ( not Moe may be) Aristoteles, Euclid, Archimedes, Archytas, Euler and Newton, just to name some hellenic and some northern european, are not to be considered any more. Yet this wakens my appetite, and as they can be read in the internet online (mostly), i do. > I sympathize with your intuition that the theory of sets somehow > lacks a dynamic. On the other hand, perhaps in a very similar way so does > the calculus, which doesn't prevent it from being an effective tool for the > description of change. That is, what was made out of it: the variables,change and movement is only permitted to the inner kring of top mathematicians in the most abstract manner, to Perelman and Co:they do heat-flow, deformations, cut and paste. Actually they are making use of pre-cauchy calculus and differential geometry of pre-non euclidian times. Some temporal basics in simple math-language could make these theories understandable to the common mathematician. > There is obviously a lot more to be said about > cardinality and ordinality, and for me that would be the immediate focus. > (Perhaps in another post, before too long.) So You might go back to the very basics and try to differ cardinal sets and ordered sequences in it. When You see these three letters B E D it can be a word, denoting a sleeping place, but still today it can be three notes in music a notion of poetry ( in one or the other system) a notion of steps in a dance three numbers 2 5 4 like in old greek, or in computers 2 + n, 5 + n , 4 + n, where n is an offset, three numbers, ciphers in hex, or one number in hex with place value, or the numbers on a telephone: 2 and 3 ( pushed one time) and 3 ( pushed twice) - so 2 3 33. These triples, vectors or sequences are temporal (written down, as a trace, they appear spatial). When time is measured or better counted with a regular cyclic movement, the sequences can be related to this temporal movement ( clock, metronom, tact, beat, ..) can be measured out, normalized, standarized - a metric or a meter. In music, measured with a clock, a note with the length of 1 /440 of one second is called A. (And i continued, as Robert Kaplan was cited, regarding zero as a seperation sign): There might be other separations too, but there are intervalls of no sound, of resting in place of stepping, of not talking in between two words or two sentences. So in notation there are places of no note in between the notes. And there are signs for periods of time with a pause in it too: a caesura, a stroke |, a point, a comma (end of citing myself). So a telephone number is an ordinal of ciphers, You have to dial in a certain sequence. Even if there will be a caesura in this thread, movement, change, variables,Hamilton's ordered couples will continue to give understanding to math in sci.math. With friendly greetings Hero I will not forget Your name, Noel and i want to ask You a favour. As soon, as You are allowed to talk outside the web, what You think, just like in sci.math - that is, when You do not need an alias any more, please send me an e-mail with Your name of birth, or just popp around on a visit in east-frisia. See You later, alligator. === Subject: Three Year Old Girl Solves Rubik's Cube In 114 Seconds This kid is just amazing! Check out the video of this three year old chinese girl who solves the Rubik's cube in 114 seconds http://silverdenial.com/youtube/ === Subject: Re: Three Year Old Girl Solves Rubik's Cube In 114 Seconds posting-account=iBgNeAoAAADRhzuSC4Ai7MUeMmxtwlM7 Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > This kid is just amazing! > Check out the video of this three year old chinese girl who solves the > Rubik's cube in 114 secondshttp://silverdenial.com/youtube/ So? Can she pick pecans? talk.politics.misc === Subject: Re: Obama's family tree, father. posting-account=rsfFvAoAAADqHCqmunbjtmaKjfuP-pwP .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) [ nip.] > As I write the investigation intensifies. Expect a major blow-up > within ten days. > Remus At the barracks ..... Major Blowup: ****&%$$£££^&*()*CC+***)))(*&^%***, u! Col. I. M. Plode: QWERTYUIOP? T? The journey down? Major Blowup: Yes, a cup would be nice. Awful: (**(*(*(&&^^%&^&$*******))! **************************************************************************** ****** 'foolsrushin.' === Subject: Physics problem posting-account=13KbngoAAABfTatbRX9gGDZXi3iESPtC AppleWebKit/523.12.2 (KHTML, like Gecko) Version/3.0.4 Safari/523.12.2,gzip(gfe),gzip(gfe) Hi there! Can you please help me with this problem: a man with a mass of 84.4kg is on a boat of 425kg moving at the speed of 4.16m/s northward. The man starts moving northward at the speed of 2.08m/s and he stops after 18.2m. How far did the boat go meanwhile? Here's my solution: Initially speed of the center of mass(MC)=425kg*4.16m/s / (425+84.4)kg Speed of MC while the man is moving= (V * 425kg + 2.08m/s*84.4kg) / (425+84.4)kg where V is the new speed of the boat. The man takes 18.2m/2.08m/s to complete 18.2m The speed of MC is constant, so I can find V (new speed of the boat) Then I multiply V*18.2/2.08 and I get the result The problem is: my result=32.8m ; book result=33.4m Is the difference due to a different approximation, or my result is wrong? === Subject: Re: Physics problem posting-account=Cbgh4AoAAAAr0dt1RqLOClWCyUWii2fU Gecko/20071204 Ubuntu/7.10 (gutsy) Firefox/2.0.0.11,gzip(gfe),gzip(gfe) > Hi there! Can you please help me with this problem: > a man with a mass of 84.4kg is on a boat of 425kg moving at the speed > of 4.16m/s northward. The man starts moving northward at the speed of > 2.08m/s and he stops after 18.2m. How far did the boat go meanwhile? Here's my solution: > Initially speed of the center of mass(MC)=425kg*4.16m/s / (425+84.4)kg > Speed of MC while the man is moving= (V * 425kg + 2.08m/s*84.4kg) / > (425+84.4)kg > where V is the new speed of the boat. > The man takes 18.2m/2.08m/s to complete 18.2m > The speed of MC is constant, so I can find V (new speed of the boat) > Then I multiply V*18.2/2.08 and I get the result The problem is: my result=32.8m ; book result=33.4m > Is the difference due to a different approximation, or my result is > wrong? The center of mass keeps moving with v = 4.16m/s northward. The walk takes 18.2m / (2.08m/s) = 8.75s, thus the CM moves by 8.75s * 4.16m/s = 36.4m. However, the common MC is 84.4kg/(425+84.4)kg * 18.2m =~ 3.02m north of the boat MC, hence the boat moves by only 36.4m-3.02m = 33.38m =~ 33.4m. All intermediate results were exact except the 3.02m near the end. hagman === Subject: Solutions for ISBN: 0131473824 posting-account=ajjnwgoAAABec-rtvO49tR-Ova2Hz3Xj Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) I would like to purchase the solutions to: Applied Linear Algebra - Chehrzad Shakiban, Peter J. Olver (1st edition) (ISBN: 0131473824) === Subject: Re: ? def of complex number- > The books I read all define a complex number as a+i*b, where > i = sqrt(-1), a and b are real numbers. Are there other definitions > for complex number, e.g., a-i*b? Although there are different constructions of the field of complex > numbers from the real numbers, they are isomorphic, which is to say, > operationally indistinguishable. The field of complex numbers is, in this > reasonable sense, unique. The map a+i*b => a-i*b is an automorphism > of the complex numbers. You could define complex numbers as > ordered pairs with certain properties, univariate real polynomials modulo > x^2-1, two by two matrices with certain properties > (http://en.wikipedia.org/wiki/Complex_number#Matrix_representation_of_comple x _numbers), > geometric points in R^2 with certain properties, but these choices > lead to structures which are isomorphic to whichever choice you > consider to be the official C. Your a-i*b proposal will only lead to the > same field C or a field isomorphic to C and that won't be any different > from C in any way that really counts as far as the essential properties > of C goes. > Quotation from this post ------------------------ > (...) You could define complex numbers as > ordered pairs with certain properties, univariate real polynomials modulo > x^2-1, (...) I guess you meant modulo (x^2 + 1) !?! Johan E. Mebius === Subject: Leibniz's Monadology actually in mathematics? Has anyone ever figured it out? posting-account=TbPjIQkAAACzP8U_C0Zcf8kprNJ34-sk Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) governing them, many of Leibniz's speculative ideas about aspects of nature not reducible to statics and dynamics made little sense. For instance, he anticipated Einstein by arguing, against Newton, that space, time and motion are relative, not absolute. Leibniz's rule in interacting theories plays a role in supersymmetry and in the lattices of quantum mechanics. His principle of sufficient reason has been invoked in recent cosmology, and his identity of indiscernibles in quantum mechanics, a field some even credit him with having anticipated in some sense. Those who advocate digital philosophy, a recent direction in cosmology, claim Leibniz as a precursor. Leibniz's ideas were called Monadology. To the best of my knowledge, to support Monadology he devised a binary system of math that is now at work in computers. Also to the best of my knowledge, despite the support of technology, Monadology is still waiting to be expressed in mathematics as it was originally intended. There are some hints that it may be more suitable for quantum gravity than our Newtonian traditions. http://www.cloudmusiccompany.com/paper.htm === === Subject: intervals posting-account=HGa-KwoAAACh0I68xblsVW5UyCAVbd5b 4.90),gzip(gfe),gzip(gfe) How many intervals are there in the reals? Define the term 'intervals' to mean closed, non-overlapping, continuous subsets of the reals, which share endpoints. Define the term 'point' to mean a real number. For example, the set of all points between 1 and 2, inclusive, is an interval, and so is the set of all points between 2 and 3, inclusive, and they share the endpoint 2. There are at least aleph0 intervals, as seen by the fact that the unit-length intervals between the integers can be paired one-to-one with the integers. Is there any set of intervals with cardinality greater than aleph0? Consider the fact that each of the unit-length intervals mentioned above can be divided into a set of aleph0 intervals by halving it and then halving each half, and then halving each quarter, etc. But the total number of intervals thus created is still aleph0, because aleph0xaleph0=aleph0. However, infinitely continue this process* of creating aleph0 intervals from every interval created, and the total number of intervals becomes the aleph0 power of aleph0, does it not? And the cardinality of that is greater than aleph0. Another way of looking at this, though, is to note the fact that no matter how many intervals are created, by any infinite process, each interval contains a unique rational number. Thus it would seem that there can be no more intervals than the number of rationals, which is aleph0. ------------------- * by infinitely continue this process is meant any collection of accelerated concurrent infinite processes that one cares to imagine. === Subject: Re: intervals posting-account=O9zR9AkAAACmp918j6u5m5plppeILcze 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Is there any set of intervals with cardinality greater than aleph0? There is a rational number between every pair of reals, and the rational numbers are countable. Therefore, you can't partition the real line into more than a countable number of non-overlapping closed intervals. Dave === Subject: Re: intervals posting-account=AdyLXQoAAABgRay99CKv1O8Y_7jjivwq MathPlayer 2.0; .NET CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > How many intervals are there in the reals? Define the term 'intervals' to mean closed, non-overlapping, > continuous subsets of the reals, which share endpoints. > Define the term 'point' to mean a real number. In what follows, c = 2^(aleph_0) is the cardinality of the set of real numbers. There are at most c intervals, since there can be no more intervals than 4c + 4c^2 = c, where I'm simply counting possibilities for one-endpoint intervals (left or right, include or exclude endpoint) and two-endpoint intervals (for each pair of real numbers r < s, use r as left endpoint and s as right endpoint, and there are 4 possibilities in each case for including or excluding the two endpoints for each choice of r < s). There are at least c intervals, since every there are c many intervals of the form (r, infinity) for each real number r. On the other hand, if you meant to ask what is the maximum cardinality of a set of non-overlapping intervals, this is aleph_0, being a consequence of the Borel covering theorem (in the strengthened form given by Lindelof, Young, and Lebesgue around 1902-1905). In the case of a topological space, this cardinality is called the cellularity of the space. [Note: For the reals, it suffices to restrict attention to open intervals, and it's also not difficult to see that the resulting maximal cardinality possible is the same for any collection of pairwise disjoint nonempty open sets. Dave L. Renfro === Subject: Re: intervals posting-account=HGa-KwoAAACh0I68xblsVW5UyCAVbd5b 4.90),gzip(gfe),gzip(gfe) > On the other hand, if you meant to ask what is the > maximum cardinality of a set of non-overlapping intervals, > this is aleph 0, being a consequence of the Borel covering > theorem (in the strengthened form given by Lindelof, > Young, and Lebesgue around 1902-1905). In the case of a > topological space, this cardinality is called the cellularity > of the space. [Note: For the reals, it suffices to restrict > attention to open intervals, and it's also not difficult > to see that the resulting maximal cardinality possible is > the same for any collection of pairwise disjoint nonempty > open sets. I did say non-overlapping. === Subject: Re: intervals > How many intervals are there in the reals? Define the term 'intervals' to mean closed, non-overlapping, > continuous subsets of the reals, which share endpoints. > Define the term 'point' to mean a real number. For example, the set of all points between 1 and 2, inclusive, > is an interval, and so is the set of all points between 2 and 3, > inclusive, and they share the endpoint 2. There are at least aleph0 intervals, as seen by the fact that > the unit-length intervals between the integers can be paired > one-to-one with the integers. Is there any set of intervals with cardinality greater than aleph0? Consider {(0,x) : x > 0} > Consider the fact that each of the unit-length intervals mentioned > above can be divided into a set of aleph0 intervals by halving it > and then halving each half, and then halving each quarter, etc. But the total number of intervals thus created is still aleph0, > because aleph0xaleph0=aleph0. However, infinitely continue this process* of creating aleph0 > intervals > from every interval created, and the total number of intervals becomes > the aleph0 power of aleph0, does it not? And the cardinality of > that is greater than aleph0. Another way of looking at this, though, is to note the fact that > no matter how many intervals are created, by any infinite process, > each interval contains a unique rational number. Thus it would > seem that there can be no more intervals than the number of > rationals, which is aleph0. ------------------- > * by infinitely continue this process is meant any collection > of accelerated concurrent infinite processes that one cares to > imagine. === Subject: Re: intervals posting-account=HGa-KwoAAACh0I68xblsVW5UyCAVbd5b 4.90),gzip(gfe),gzip(gfe) > Consider {(0,x) : x > 0} Those intervals are overlapping. === Subject: Re: intervals posting-account=9QOSvAoAAACEOWJVSDuswW7dB_0wApQO Gecko/20071213 Fedora/2.0.0.10-3.fc8 Firefox/2.0.0.10,gzip(gfe),gzip(gfe) > How many intervals are there in the reals? Define the term 'intervals' to mean closed, non-overlapping, > continuous subsets of the reals, which share endpoints. > Define the term 'point' to mean a real number. For example, the set of all points between 1 and 2, inclusive, > is an interval, and so is the set of all points between 2 and 3, > inclusive, and they share the endpoint 2. There are at least aleph0 intervals, as seen by the fact that > the unit-length intervals between the integers can be paired > one-to-one with the integers. Is there any set of intervals with cardinality greater than aleph0? Consider the fact that each of the unit-length intervals mentioned > above can be divided into a set of aleph0 intervals by halving it > and then halving each half, and then halving each quarter, etc. But the total number of intervals thus created is still aleph0, > because aleph0xaleph0=aleph0. However, infinitely continue this process* of creating aleph0 > intervals > from every interval created, and the total number of intervals becomes > the aleph0 power of aleph0, does it not? And the cardinality of > that is greater than aleph0. Another way of looking at this, though, is to note the fact that > no matter how many intervals are created, by any infinite process, > each interval contains a unique rational number. Thus it would > seem that there can be no more intervals than the number of > rationals, which is aleph0. This is not `another way of looking at this': it is the correct way to look at it. In particular, it correctly reaches the correct conclusion that a set of intervals with disjoint subsets is at most countable. -- m === Subject: Re: intervals posting-account=n4TzyQkAAADLWxrRHqyiUZ-1SZdOB4vv Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) How many intervals are there in the reals? Define the term 'intervals' to mean closed, non-overlapping, > continuous subsets of the reals, which share endpoints. > Define the term 'point' to mean a real number. For example, the set of all points between 1 and 2, inclusive, > is an interval, and so is the set of all points between 2 and 3, > inclusive, and they share the endpoint 2. There are at least aleph0 intervals, as seen by the fact that > the unit-length intervals between the integers can be paired > one-to-one with the integers. Is there any set of intervals with cardinality greater than aleph0? > ........ This is not `another way of looking at this': it is > the correct way to look at it. In particular, > it correctly reaches the correct conclusion that > a set of intervals with disjoint subsets is at > most countable. > This of course leads to a question. Suppose, You have ordered the open intervalls of R, let them intersect, so You will have a sequence of disjoint open intervalls, each one having a rational number inside. Taken together in a union we will have the whole R minus the endpoints, transzendent numbers. Take one of the intervalls and devide it into two, seperated by a rational number in it. These intervalls are not in the sequence presupposed. So every seperation of R into disjoint intervalls will be countable. But these are not all possible intervalls. Now the question: how many seperations, every two of these seperations with no common intervall, are there? With friendly greetings Hero === Subject: Riemann-Siegel computations For t>=0, we can write: Z(t) = (+/-) exp(-i*arg(zeta(1/2+i*t))) * zeta(1/2+i*t) That way, Z(t) is real-valued, and in practice + changes to - when, as t increases, we go through a zero of zeta(1/2+i*t), and vice versa (- changes to + when (etc.)). I don't know how Z(t) behaves when passing through a double zero of zeta(1/2+i*t); however, as far as I know, no multiple zeros of t |-> zeta(1/2+i*t) are known. There are formulas to approximate Z(t), for example (35) from ``Computational strategies for the Riemann zeta function by J.M. Borwein, D.M. Bradley and R.E. Crandall in Journal of Computational and Applied Mathematics 121 (2000). Cf.: < http://cr.yp.to/bib/2000/borwein.pdf > In my computations, I decided to use vartheta(t) as given by Ken Takusagawa in Section 2 of his document on Z(t) computations, available from: < http://www.mit.edu/~kenta/six/parallel/2-Final-Report.html >. Andrew Odlyzko has several tables of the non-trivial zeros of the zeta function, one of them starting at zero number 10^12 + 1 : < http://www.dtc.umn.edu/~odlyzko/zeta_tables/zeros3 > (35), tau := sqrt(t/(2pi)), m:= floor(tau) and z = 2*(tau - m) -1. The quantity z only appears in terms of order t^(-1/4) or less in (35) and furthermore, dz/dt is of the order t^(-1/2) for intervals of 't' with no discontinuity in the integer tau. To the extent that z changes only slowly throughout the zeros3 table above, we can put all the higher order terms together in a constant. on the value of the constant. Then the simplifies formula from the one (35) that I chose is: Z(t) ~= 2*sum(X=1,206393,cos(t*log(X)-t/2*log(t/(2*Pi))+t/2+Pi/8)/sqrt(X))+0.0010102 Note that vartheta(t) doesn't depend on X, or the integer 'n' in (35). To test the accuracy, I needed a value of t far from the zeros used to fix the constant 0.0010102 . So using the Z(t) approximation above, I tried to find the first zero t' of the approximation after t = 267,653,397,000. Using a Newton-like method, I found: t' ~= 267653397000.28482386 [ error of 7* 10^(-8) ] The Z(t) formula above gives: Z(t') ~= 0.0000000705 t = 267653397000.2848239314 (gamma - 267653395647 = 1353.2848239314 in Odlyzko's table.) I haven't tested the Z(t)approximation for t where Z(t) is far from zero. I'm not sure what has been computed on large values of |Z(t)| compared to t. RH implies the Lindelof Hypothesis. I don't know of a growth condition on Z(t) which is known to be equivalent to RH. David Bernier === Subject: YOU MUST KNOW THIS MAN posting-account=1g2ShwoAAADtW0q8OloXyTG2V8z9nomW Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ; MEGAUPLOAD 2.0; .NET CLR 2.0.50727; GreenBrowser),gzip(gfe),gzip(gfe) .8cëæÍ.84[NonBre akingSpace].82.8a.8c[Times].8f.90[C apitalEAcute] .90íé[NonBreakingSpace]Ì[ AE][NonBreakingSpace]æô.84.87 [NonBreakingSpace].8d[CapitalEth].82[NonBreakingSpa ce].82.8a.84í[ADoubleDo t] ææêì[CapitalEDo ubleDot][NonBreakingSpace].82.8a.8c[Times][E Grave].90.83[NonBreakingSpace].82[ADoubleDo t].8c.8fí.8d.83[NonBreakingSp ace]ÌÍ.8a.82[NonBreak ingSpace].81.8a.8e[NonBreakingSpace][CapitalCCe dilla].8a.8b.87.82.84[NonB reakingSpace]ô.be[NonBreakingSpace].8cê .8cì[NonBreakingSpace]Í.8a .8e[NonBreakingSpace].82.8a.8a[CCe dilla][NonBreakingSpace]ô.8a.90.8d[Non BreakingSpace].8fî.8a.8c .87ææ.84í[ARing ][NonBreakingSpace].8a.8d[NonBreakingSpace]æ .84í.8c.83[NonBreakingSpace][ ARing]ëæÍ.84[CapitalEA cute][NonBreakingSpace]è.8c[NonBreakingSpace][Capi talEHat][CapitalEth].8b.84[NonBreakingSpace].8c[AE ].82.89é.8d[NonBreakingSpace][EGr ave]ï.8d.82ì[CapitalE Acute][NonBreakingSpace]éô.85 [NonBreakingSpace].82.8a.8b.87[Capital CCedilla].84[NonBreakingSpace].8a.8d[NonBrea kingSpace].8f.82.8aÌ[CapitalID oubleDot].8a.83[NonBreakingSpace]ô[AD oubleDot].8e Íì.89.8d[NonBreakingSpace][E Grave].8b.90.87[NonBreakingSpace]é.89[EH at][NonBreakingSpace]ì.90.be[NonBreakingSpace][Ca pitalCCedilla].8a.81î.8a[Capi talCCedilla].8c[NonBreakingSpace]î.82[AD oubleDot].8c.82[NonBreakingSpace].8c.be[NonBreak ingSpace].82.8aæê[Capit alNTilde].90.87[NonBreakingSpace].8b.8c[CapitalCCedill a][NonBreakingSpace].8f.89ô[NonBreakingSpace][AAc ute].90[NonBreakingSpace]ò.90.84[CCedil la][NonBreakingSpace].8c.be[NonBreakingSpace].82[ADo ubleDot]Ìì.90.82.be .8f.8b.90.87[NonBreakingSpace]Í.be[ CapitalUGrave][NonBreakingSpace].82.8a[CapitalNTil de]î.8f.8a[NonBreakingSpace]Í .8a.8e[NonBreakingSpace].82.8a[ADo ubleDot].8d[NonBreakingSpace]ô.8a.90[C Cedilla][NonBreakingSpace].8fî.8a.8c[No nBreakingSpace]Ì.8c.83[NonBreakingSpace][C apitalCCedilla]îæ[Times].82[Capi talUGrave]æ[NonBreakingSpace]Ì.be[NonBreaking Space]æé.be.90[NonBreakingSpace].87 .90[NonBreakingSpace].8f.89æ[NonBreakingSpace][ AHat]Í.90.84 Ì.8bé.84[NonBreakingSpace] ì.8f.8a.83[NonBreakingSpace] .90ê.8b.8c.8d.82[NonBre akingSpace].82.8a.81î[AD oubleDot].82.8c In The Name Of Allaah, Most Gracious, Most Merciful YOU MUST KNOW THIS MAN (May peace and blessings of God Almighty be upon him) You may be an atheist or an agnostic; or you may belong to anyone of the religious denominations that exist in the world today. You may be a Communist or a believer in democracy and freedom. No matter what you are, and no matter what your religious and political beliefs, personal and social habits happen to be - YOU MUST STILL KNOW THIS MAN! He was by far the most remarkable man that ever set foot on this earth. He preached a religion, founded a state, built a nation, laid down a moral code, initiated numberless social and political reforms, established a dynamic and powerful society to practice and represent his teachings, and completely revolutionized the worlds of human thought and action for all times to come. and he accomplished all these wonders in the unbelievably short span of twenty-three years. Muhammad, peace and blessings of God Almighty be upon him was born in Arabia on the 20th of August, in the year 570 of the Christian era, and when he died after 63 years, the whole of the Arabian Peninsula had changed from paganism and idol-worship to the worship of One God; from tribal quarrels and wars to national solidarity and cohesion; from drunkenness and debauchery to sobriety and piety; from lawlessness and anarchy to disciplined living; from utter moral bankruptcy to the highest standards of moral excellence. Human history has never known such a complete transformation of a people or a place before or since! The Encyclopedia Britannica calls him the most successful of all religious personalities of the world. Bernard Shaw said about him that if Muhammad were alive today he would succeed in solving all those problems which threaten to destroy human civilization in our times. Thomas Carlysle was simply amazed as to how one man, single- handedly, could weld warring tribes and wandering Bedouins into a most powerful and civilized nation in less than two decades. Napoleon and Gandhi never tired of dreaming of a society along the lines established by this man in Arabia fourteen centuries ago. Indeed no other human being ever accomplished so much, in such diverse fields of human thought and behavior, in so limited a space of time, as did Muhammad, peace and blessings of God Almighty be upon him. He was a religious teacher, a social reformer, a moral guide, a political thinker, a military genius, an administrative colossus, a faithful friend, a wonderful companion, a devoted husband, a loving father - all in one. No other man in history ever excelled or equaled him in any of these difficult departments of life. The world has had its share of great personalities. But these were one sided figures who distinguished themselves in but one or two fields such as religious thought or military leadership. None of the other great leaders of the world ever combined in himself so many different qualities to such an amazing level of perfection as did Muhammad, peace and blessings of God Almighty be upon him. The lives and teachings of other great personalities of the world are shrouded in the mist of time. There is so much speculation about the time and the place of their birth, the mode and style of their life, the nature and detail of their teachings and the degree and measure of their success or failure that it is impossible for humanity today to reconstruct accurately and precisely the lives and teachings of those men. Not so this man Muhammad, peace and blessings of God Almighty be upon him. Not only was he born in the fullest blaze of recorded history, but every detail of his private and public life, of his actions and utterances, has been accurately documented and faithfully preserved to our day. The authenticity of the information so preserved is vouched for not only by faithful followers but also by unbiased critics and open-minded scholars. At the level of ideas there is no system of thought and belief-secular the system which Muhammad peace and blessings of God Almighty be upon him propounded. In a fast changing world, while other systems have undergone profound transformations, Islaam alone has remained above all change and mutation, and retained its original form for the past 1400 years. What is more, the positive changes that are taking place in the world of human thought and behavior, truly and consistently reflect the healthy influence of Islam in these areas. Further, it is not given to the best of thinkers to put their ideas completely into practice, and to see the seeds of their labors grow and bear fruit, in their own lifetime. Except of course, Muhammad, peace and blessings of God Almighty be upon him, who not only preached the most wonderful ideas but also successfully translated each one of them into practice in his own lifetime. At the time of his death his teachings were not mere precepts and ideas straining for fulfillment, but had become the very core of the life of tens of thousands of perfectly trained individuals, each one of whom was a marvelous personification of everything that Muhammad peace and blessings of God Almighty be upon him taught and stood for. At what other time or place and in relation to what other political, social, religious system, philosophy or ideology-did the world ever witness such a perfectly amazing phenomenon? Indeed no other system or ideology secular or religious, social or political, ancient or modern - could ever claim the distinction of having been put into practice in its fullness and entirety EVEN ONCE in this world, either before or after the death of its founder. Except blessings of God Almighty be upon him which was established as a complete way of life by the teacher himself, before he departed from this world. History bears testimony to this fact and the greatest skeptics have no option but to concede this point. In spite of these amazing achievements and in spite of the countless absolutely convincing and authentic miracles performed by him and the phenomenal success which crowned his efforts, he did not for a moment claim to be God or God's incarnation or Son - but only a human being who was chosen and ordained by God to be a teacher of truth to man kind and a complete model and pattern for their actions. He was nothing more or less than a human being. But he was a man with a noble and exalted mission-and his unique mission was to unite humanity on the worship of ONE AND ONLY GOD and to teach them the way to honest and upright living in accordance with the laws and commands of God. He always described himself as A MESSENGER AND SERVANT OF GOD, and so indeed every single action and movement of his proclaimed him to be. A world which has not hesitated to raise to Divinity individuals whose very lives and missions have been lost in legend and who historically speaking did not accomplish half as much-or even one tenth-as was accomplished by Muhammad, peace and blessings of God Almighty be upon him should stop to take serious note of this remarkable man's claim to be God's messenger to mankind. Today after the lapse of some 1400 years the life and teachings of Prophet Muhammad, peace and blessings of God Almighty be upon him, have survived without the slightest loss, alteration or interpolation. Today they offer the same undying hope for treating mankind's many ills which they did when Prophet Muhammad, peace and blessings of God Almighty be upon him, was alive. This is our honest claim and this is the inescapable conclusion forced upon us by a critical and unbiased study of history. The least YOU should do as a thinking, sensitive, concerned human being is to stop for one brief moment and ask yourself: Could it be that these statements, extraordinary and revolutionary as they sound, are really true? Supposing they really are true, and you did not know this man Muhammad, peace and blessings of God Almighty be upon him or hear about his teachings? Or did not know him well and intimately enough to be able to benefit from his guidance and example? Isn't it time you responded to this tremendous challenge and made some effort to know him? It will not cost you anything but it may well prove to be the beginning of a completely new era in your life. Come, let us make a new discovery of the life of this wonderful man Muhammad, peace and blessings of God Almighty be upon him the like of whom never walked on this earth, and whose example and teachings can change YOUR LIFE and OUR WORLD for the better. May God shower His choicest blessings upon him! Written by S.H. Pasha