mm-569 === Subject: Re: A multi-variable integration challenge >This is purely a recreational puzzle; I wanted to see if I could come >up with a 3-space surface, symmetrical with respect to each dimension, >which would asymptotically extend to infinity around each axis, but >enclosing a finite volume. The first thing I came up with was the >following: >(x^2 * y^2) + (y^2 * z^2) + (x^2 * z^2) = 1 / (x^4 + y^4 + z^4) >This should fit my first two criteria; it is symmetrical, and as any >variable becomes very large, the right hand side becomes small, and >the left side approximates to the equation for a circle. For example, >as z becomes large the equation approximates to y^2 + x^2 = 1/z^2, > No: x^2 + y^2 = 1/((x^4 + y^4 + z^4) z^2) - x^2 y^2/z^2 < 1/z^6 What an embarassing mistake! I thought I needed the higher exponents on the right to get a decreasing radius, but of course I get that for free when dividing through by z^2, starting with a constant on the Scott === Subject: Re: A multi-variable integration challenge Surface of this sort with a 4by4 matrix determinant: Det[M]=Det[{{x,y,z,0},{y,z,0,-x},{z,0,-x,-y},{0,-x,-y,-z}}]=x ^4+y^4+z^4-4*x* y*y+z-2*x*x*z*z Det[M]-1=0 Surfaces of this type should also have limited surface area as well as limited volume? > This is purely a recreational puzzle; I wanted to see if I could come > up with a 3-space surface, symmetrical with respect to each dimension, > which would asymptotically extend to infinity around each axis, but > enclosing a finite volume. The first thing I came up with was the > following: > (x^2 * y^2) + (y^2 * z^2) + (x^2 * z^2) = 1 / (x^4 + y^4 + z^4) > This should fit my first two criteria; it is symmetrical, and as any > variable becomes very large, the right hand side becomes small, and > the left side approximates to the equation for a circle. For example, > as z becomes large the equation approximates to y^2 + x^2 = 1/z^2, > terms not involving z becoming insignificant. Then my question > becomes, does the asymptotic surface around each axis converge to the > axis rapidly enough to make the volume finite? Unfortunately when I > try to solve this equation for one variable I get some very complex > terms. The best I can do to separate the x's with algebra is > x^6(y^2 + z^2) + > x^4(y^2 * z^2) + > x^2(y^6 + y^4*z^2 + y^2*z^4) + > y^6*z^2 + y^2*z^6 - 1 = 0 > But how to turn this into x = ... is beyond me; after that, > integrating on the remaining two variables is something I could > conceivably do, but it's been a while since I took multivariable calc, > so anyone who finds this an interesting challenge is welcome to help. > And of course if the surface turns out to enclose infinite volume, I'd > appreciate suggestions for other equations that fit all my criteria. > Scott Forschler Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : URL : http://home.earthlink.net/~tftn URL : http://victorian.fortunecity.com/carmelita/435/ === Subject: Re: Laplace transform of u(t)v(t) > Is/are there any useful ways to express the Laplace transform of the product > of two functions? i.e. L(v(t).u(t)). > Ideally I'd like to express this in terms of the individual transforms > L(v(t)) and L(u(t)). I've been wondering if integrating by parts might help, > but have lost my way... The Laplace transform of the product is the convolution of the Laplace transforms. One would think this should be in any mathematics text that deals with laplace transforms. http://mathworld.wolfram.com/Convolution.html -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Laplace transform of u(t)v(t) > Is/are there any useful ways to express the Laplace transform of the product > of two functions? i.e. L(v(t).u(t)). > Ideally I'd like to express this in terms of the individual transforms > L(v(t)) and L(u(t)). I've been wondering if integrating by parts might help, > but have lost my way... >The Laplace transform of the product is the convolution of the Laplace >transforms. Not an ordinary convolution, but a kind of convolution in the complex plane. Specifically, if u and v are exponentially bounded, for any real numbers a and b such that all singularities of U(s) = L(u) and V(s) = L(v) are to the left of Re(s) = a and Re(s) = b respectively, L(u v)(s) = (2 pi i)^(-1) int_C U(p) V(s-p) dp for Re(s) > a+b where C is the straight line {a + it: t=-infinity .. infinity}. > One would think this should be in any mathematics text >that deals with laplace transforms. I don't think this is in most texts dealing with Laplace transforms. A typical DE text that talks about Laplace transforms, e.g. Boyce and DiPrima, doesn't use complex analysis at all. Even complex analysis texts that talk about Laplace transforms, e.g. Levinson and Redheffer, tend not to include this. Perhaps you're thinking of the Laplace transform of a convolution, rather than the Laplace transform of a product. ' === Subject: Re: Laplace transform of u(t)v(t) > Is/are there any useful ways to express the Laplace transform of the product > of two functions? i.e. L(v(t).u(t)). > Ideally I'd like to express this in terms of the individual transforms > L(v(t)) and L(u(t)). I've been wondering if integrating by parts might help, > but have lost my way... >The Laplace transform of the product is the convolution of the Laplace >transforms. > Not an ordinary convolution, but a kind of convolution in the complex > plane. Specifically, if u and v are exponentially bounded, for any real > numbers a and b such that all singularities of U(s) = L(u) and V(s) = L(v) > are to the left of Re(s) = a and Re(s) = b respectively, > L(u v)(s) = (2 pi i)^(-1) int_C U(p) V(s-p) dp for Re(s) > a+b > where C is the straight line {a + it: t=-infinity .. infinity}. > One would think this should be in any mathematics text >that deals with laplace transforms. > I don't think this is in most texts dealing with Laplace transforms. > A typical DE text that talks about Laplace transforms, e.g. Boyce > and DiPrima, doesn't use complex analysis at all. Even complex > analysis texts that talk about Laplace transforms, e.g. Levinson > and Redheffer, tend not to include this. Perhaps you're thinking > of the Laplace transform of a convolution, rather than the Laplace > transform of a product. Yet, for example, I found this relation in almost all my textbooks dealing with Laplace transforms - control systems textbooks, complex analysis textbooks and signal and systems textbooks. I believed that this is commonly covered in any course related to Laplace transforms. Interestingly enough all the books having this explained are either Russian textbooks or textbooks written in my native Serbian. None of textbooks written in English I have doesn't mention this relation. Maybe it is because here (and probably in Russia) Laplace (and Fourier) transforms are usually taught after or as part of complex analysis course (at least in an engineering curricula)? In what order are those topics usually taught in Canada/US? === Subject: Re: Laplace transform of u(t)v(t) > Is/are there any useful ways to express the Laplace transform of the > product of two functions? i.e. L(v(t).u(t)). > The Laplace transform of the product is the convolution of the Laplace > transforms. One would think this should be in any mathematics text > that deals with laplace transforms. Are you certain ? True: L(f(t)*g(t)) = F(s).G(s) And: L-1(F(s).G(s)) = f(t)*g(t) (* is convolution here, . multiplication) I am not exactly sure what you're saying, but I have never seen: L(f(t).g(t)) = F(s)*G(s) stated, and taking the convolution of two functions of s seems rather odd to me. I do not think that the relationship between convolution and the laplace transform aids in calculating the transform of the product of two functions. -- Michael Ballbach, N0ZTQ ballbach@rten.net -- PGP KeyID: 0xA05D5555 http://www.rten.net/ === Subject: Re: Set Theory homework - please help > Hello! > I just got some homework from the professor dealing with set theory, > only our books are not due in the book store until monday. I need help > with a problem if possible. I just started the class this week and > this is due tomorrow. I have done a little of it, but i don't have > enough info in order to finish it. I don't know where to go next. My > teacher hasn't gone through it thuroughly yet. > the question is this > (A-B) - (B-C) = A-B > so far i have this for the proof > A, B, AND C are all sets > & = INTERSECT > + = UNION > Ô = COMPLEMENT > LHS=(A-B) - (B-C) > =(A&B') - (B&C') //difference of sets in each set of parenthesis > =(A&B') & (B'+C) //difference of sets in the whole set problem > =((A&B') + B') & (A&B')+C) //distributive properties of the sets > --------------------------------- > ANOTHER PROBLEM > (A-B) + (B-A) = (A+B) - A&B > my proof so far > LHS = (A-B+ + (B-A) > = (A&B') + (B&A') > =((A&B') & B) + ((A&B') &A') > that it for that one too! > My teacher suggested that i use the distributive set in the last step > of both problems, which i did, and then i simplify, but i can't figure > out how in the world, that equals out to A-B > Can anyone suggest a Venn Diagram to explan it to me, or just let me > know in words. I have never taken this course before and we just > started it up this week. You might find this template easier: to prove E = F, prove e in E imples e in F and f in F imples f in E === Subject: Re: sequences of touching spheres === e === Subject: Re: Field extensions that are not algebraic by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7RIqOt13575; >We then get a bijection between: > >{ fields K/k of trans deg=1, such { smooth projective >curves > that K is algebraic over some k(T) } <-> over k, up to >isomorphism } > >In fact, this correspondence is _categorical_, meaning that a >homomorphism of fields K_1 -> K_2 on the left, corresponds to a map >the curves X_2 -> X_1 on the right (note the reversing of arrows), >which respects composition. > Actually this is not correct: your statement only holds in this > simple form if the field k is algebraically closed. > Otherwise on the left side you must take the conservative alg. > function fields K|k of transcendence degree 1. >Conservative< > means that the genus of K|k does not change under arbitrary > base extensions l|k. >algebraically closed in K? If not, is there an algebraic way to express >it? > Moreover, if k is not perfect, the category equivalence can > be generalized by considering regular instead of smooth curves. > H >Ah. The distinction between regular and smooth again. I still can't >grasp it well - I just know that Ôsmooth' is stronger than Ôregular'. >Are there any nice properties which can help me understand their In the case of an algebraic curve C a closed point P is regular iff the local ring O_(C,P) is a discrete valuation ring. Therefore the curve C is regular iff all local rings are integrally closed. This property is not stable under base change. Smoothness is the same thing as saying that the curve remains regular under arbitrary base change l|k, where l is a field extension of k. Grothendieck therefore calls >smoothness< also >geometric regularity<. Let C be an algebraic curve over k (curve here includes being irreducible and reduced). Let F|k be the function field of C. Then the following properties are equivalent: F|k is separably generated and k is alg. closed in F C remains irreducible and reduced under arbitrary base change l|k. (alternatively: C is geometrically integral). H C remains === Subject: [Collatz] was : Re: Status of Waring-problem - Collatz - Sorry to intrude. Am 25.08.04 10:40 schrieb Karl Easterly: > I have made the following observation and have tried to get an opinion > on its value a couple of times to no avail. Basically, it defines a > static structure all the collatz branches adhere to and provides a > means to calculate all values that have a stopping value of N without > traversing the tree to this depth. Whether it is useful is > questionable as the computational time to peg Ôintegers' in the set is > larger than the time to traverse the tree; however, I find it > insightful and this idea has given me a solid understanding into the > complexities of the problem. > It is fairly simple to state. An example would suffice. > Take the following > Vi - - -> (Vj,Vi') > ^ ^ > | | > | | > | | > Vo - - -> Vj' > Vo = Initial integer value > Vi = Vo * 2^i > Vj = F(Vi)[j] (j = recursive count) with F(x) defined as (2x - 1)/3 > Vj' = F(Vo)[j'] > Vi' = Vj' * 2^i' > We end up with the points Vj, Vi' where Vj - Vi' = constant and > independent of Vo. > I know that is not standard math formulary, but I believe it can be > followed. > Here is a concrete example. > (2 = Vi) - (1 = Vj) > | > (1 = Vo) > (2/3 = Vi') > | > (1 = Vo) - (1/3 = Vj') > Vj - Vi' = (1 - 2/3) = 1/3 > (i,j) = (1,1) > Independent of Vo, Vj - Vi' always equals 1/3. > The same can be said for any (i,j), Vj - Vi' always = same constant > for any specific (i,j) independent of Vo. Hi karl, let me try, whether I got you right. The first question is, is in your notation i' the same as i, and j' the same as j? In the following I assume, it is. Using your example with the initial value Vo = a 2*A - 1 A A ----> a = --------- ! ! 3 ! ! ! ! ! ! ! 2* A*2^i - 1 ! 2*A - 1 B = A*2^i ----> b = ------------ b'= -------- * 2^i 3 3 Now A*(2*2^i - 2*2^i) - 1 + 2^i b - b' = ---------------------------------- 3 This result is independent of A, since 3(b - b') = A ( 0 ) + 2^i-1 = 2^i - 1 b - b' = (2^i - 1) / 3 (given i=i' and j=j' ) You may be interested in a formalism, that I use for computations like that. I note a' = C(a ; A) for a' = (2^A *a - 1 )/3 or the inverse a = T(a'; A) for a = (3*a' +1 )/2^A where the capital letters are used for the exponents. Combine several steps like a' = C(a; A, B,C,D...,Z) for a 26-step C-transformation or a = T(a';Z,Y,...C,B,A) for its inverse transformation. Note that with N exponents and their sum = S you can always write 2^S a' = a --- - C(0;A,B,C,....,Z) [with N exponents and S=A+B+C...+Z)] 3^N Now, one horizontal-step in your example equivalents a' = C(a;1) and j steps are just a' = C(a;1,1,1,1...1) with j ones. For that primitive transformation I use a shorter form: a' = PC(a;N) with N steps so we have first example b = PC(a*2^i; J ) second example b' = 2^i* PC(a; J ) With that formalism it is simple to restate: 2^J b = a*2^i * ----- - PC(0; J) 3^J 2^J b' = 2^i*a * ---- - 2^i*(PC(0; J) 3^J the difference b - b' = a * ( 2^i * 2^J / 3^J - 2^i * 2^J/3^J) + PC(0; J)*( 2^i - 1) = a * 0 + PC(0;J) * (2^i - 1) = PC(0; J) * (2^i - 1) is independent of a , anc can be computed by evaluation of PC(0;J) which is 3^J - 2^J b - b' = ----------- * (2^i-1) 3^J (the exponent in the numerator may differ by one, don't have the exact formula at hand). This formalism is a very useful tool; I suggest , to use it for all computations. > Basically, this provides a static lattice all collatz trees will stick > to, independent of magnitude of the values of the tree in question. An > in-depth analysis of this structure leads to many quantifiable > properties of the tree, provides a solid model to analyze the problem > in 3d space, and provides indexing into the problem on the number of > times each of the two rules are applied. Those rules being the > standard collatz definition. Yes, it is a nice observation. > It also provides views of cycles of > repetition within the branches, allows them to be quantified and > mapped, explains the occurrence of consecutive integers, a higher > level of granularity for defining sets of collatz integers, and a > wealth of other concepts. It also provides a seemingly fractal view > into the core of the problem where every direction you look, you find > the identical pattern at every scale. I have a nice graphic of such trees on my site; also a fractal graphic. ( See http://www.uni-kassel.de/~helms/pmc/collatzgraphs.htm ) (hope that page is functioning, didn't look into it for long time...) > I know I rambled a bit, but I would very much like to have someone to > chat with on this result and someone to provide guidance in > documenting the idea. Gottfried Helms === Subject: Re: [Collatz] was : Re: Status of Waring-problem - Collatz - Sorry to intrude. For your interest, I want to add some more notes, relating the previous posting to the 1-cycle-problem, as an example: ------------ A 1-cycle is a PC()-transformation with one single descending step at its end: a' = PC(a,N:A) = C(a;1,1,1,1...1,A) with N-1 ones (in my previous post I omitted the trailing 1 in the PC()-notation, should have been PC(a;N:1), sorry) Now, a loop is, if the outout equals the input, thus a = PC(a;N:A) with N the number of steps, and S the sum of exponents = N+A-1 Now we have 2^S a = a * --- - PC(0;N:A) 3^N which is 2^S a( --- - 1 ) = PC(0; N:A) 3^N That is PC(0;N:A) a = ------------- * 3^N 2^S - 3^N Because 3^N - 2^N PC(0;N:A) = ----------- 3^N the previous becomes 3^N - 2^N a = -------------- (condition for a 1-cycle or primitve loop) 2^S - 3^N The solution is only valid if a is odd and is integer. For N=1 and A=2 we get the trivial loop of one step: 1 a = C(a;2) = --- = 1 1 and for two steps a = C(a;2,2) = C( C(a;2);2) = C(1;2) = 1 and for arbitrarily many of such steps a = C(a;2,2,2,2,2....) = 1 . No other combination of exponents is known to produce an integral a. ------------------- Back to the primitive loop. It can be shown, that for any a being an input value for a primitve loop of a number N of steps, a must have a simple structure: a = 2^N * i - 1 (where i is a free integer parameter >=0 ) so 2^S - 2^N 2^N * i - 1 = ----------- - 1 2^S - 3^N and 1 2^S - 2^N 2^(A-1) - 1 i = --- * ------------ = ---------------- 2^N 2^S - 3^N 2^(A-1+N) - 3^N No integral i is known except in that combination of i,N and A, which constitutes the trivial loop. I don't want to proceed here, but I think, the example shows, what the proposed notation for the collatz-transformation can be good for. Gottfried Helms === Subject: Re: [Collatz] was : Re: Status of Waring-problem - Collatz - Sorry === >Subject: Re: [Collatz] was : Re: Status of Waring-problem - Collatz - Sorry >Message-id: >For your interest, I want to add some more notes, relating >the previous posting to the 1-cycle-problem, as an example: >------------ >A 1-cycle is a PC()-transformation with one single descending >step at its end: > a' = PC(a,N:A) = C(a;1,1,1,1...1,A) with N-1 ones > (in my previous post I omitted the trailing 1 in the > PC()-notation, should have been PC(a;N:1), sorry) >Now, a loop is, if the outout equals the input, thus > a = PC(a;N:A) with N the number of steps, > and S the sum of exponents = N+A-1 >Now we have > 2^S > a = a * --- - PC(0;N:A) > 3^N >which is > 2^S > a( --- - 1 ) = PC(0; N:A) > 3^N >That is > PC(0;N:A) > a = ------------- * 3^N > 2^S - 3^N >Because > 3^N - 2^N > PC(0;N:A) = ----------- > 3^N >the previous becomes > 3^N - 2^N > a = -------------- (condition for a 1-cycle or primitve loop) > 2^S - 3^N >The solution is only valid if a is odd and is integer. For N=1 and A=2 we get >the trivial loop of one step: > 1 > a = C(a;2) = --- = 1 > 1 Does this only apply to positive integers? The reason I ask is my formulations span both positive and negative domains and I get two 1-cycle loops: 1 -> 4 -> 2 -> 1 and -1 -> -2 -> -1 >and for two steps > a = C(a;2,2) = C( C(a;2);2) = C(1;2) = 1 >and for arbitrarily many of such steps > a = C(a;2,2,2,2,2....) = 1 . >No other combination of exponents is known to produce an integral a. >------------------- >Back to the primitive loop. >It can be shown, that for any a being an input value for a primitve loop >of a number N of steps, a must have a simple structure: > a = 2^N * i - 1 (where i is a free integer parameter >=0 ) > 2^S - 2^N > 2^N * i - 1 = ----------- - 1 > 2^S - 3^N >and > 1 2^S - 2^N 2^(A-1) - 1 > i = --- * ------------ = ---------------- > 2^N 2^S - 3^N 2^(A-1+N) - 3^N >No integral i is known except in that combination of i,N and A, which >constitutes >the trivial loop. I don't want to proceed here, but I think, the example >shows, what the proposed notation for the collatz-transformation can be >good for. >Gottfried Helms -- Mensanator Ace of Clubs === Subject: Re: [Collatz] was : Re: Status of Waring-problem - Collatz - Sorry Am 28.08.04 06:48 schrieb Mensanator: > 3^N - 2^N > a = -------------- (condition for a 1-cycle or primitve loop) > 2^S - 3^N >The solution is only valid if a is odd and is integer. For N=1 and A=2 we get >the trivial loop of one step: > 1 > a = C(a;2) = --- = 1 > 1 > Does this only apply to positive integers? The reason I ask is my formulations > span both positive and negative domains and I get two 1-cycle loops: > 1 -> 4 -> 2 -> 1 > and > -1 -> -2 -> -1 Well, in my notation this would be 2^1 x = C(x;1) = x* --- + C(0;1) 3^1 3^1 3^1 - 2^1 3^1 x = C(0; 1) * ---------- = ------------* -------- = -1 2^1 - 3^1 3^1 2^1 - 3^1 You even can use rationals, if you do some more finetuned analysis. All these are just compacted notations, no new inventions. ---------------------------- For instance, another thing that you immediately see -if you use this notation- is, that infinetly many solutions in x' and x exist for a certain transformation- structure, say x' = C(x; A,B,C) This is 2^(A+B+C) x' = x * ---------- + C(0;A,B,C) 3^3 The C(0;...)- part is in general a fraction with the denominator 3^N, in this case of a three-step-transformation it is 3^3. So you see, that for all x with the same modular-class based on 3^3 you find an integral solution in x and x' for the transformation x->x' , or can try to find approriate exponents for a given pair (x,x'), for instance x' = 2x+1 or x'-x = 2^B or anything the like. Extended to the question of cycles it is simply demonstrable, that for a certain sequence of exponents only one solution for x is possible; that means, a search for possible loops over reduces to a search for configurations of exponents, and strong restrictions can be stated for such a sequence: the sum S of all exponents must between about 1.5*N and 2*N and the like. ----------------------------- Using a certain sequence of exponents A,B,C, x' = C(x; 2,2,1,2,1,...) you can construct numbers x->x' where x has an arbitray long trajectory , where the intermediate values never fall below x (I think, this is called glide ?). There are arbitrarily many transformations x' = C(x; A,B,C..), but to be applied to a pair of integral (x,x') it seems to me, that this sequence allows the smallest numbers (empirically and only few tests, maybe there are easy counterexamples except the trivial one). -------------------------------- and so on. As I said, I suggest it as a practical notation for the whole collatz-calculus. Gottfried Helms === Subject: Re: [Collatz] was : Re: Status of Waring-problem - Collatz - Sorry === >Subject: Re: [Collatz] was : Re: Status of Waring-problem - Collatz - Sorry >Message-id: >Am 28.08.04 06:48 schrieb Mensanator: > 3^N - 2^N > a = -------------- (condition for a 1-cycle or primitve loop) > 2^S - 3^N >The solution is only valid if a is odd and is integer. For N=1 and A=2 we >get >the trivial loop of one step: > 1 > a = C(a;2) = --- = 1 > 1 > Does this only apply to positive integers? The reason I ask is my >formulations > span both positive and negative domains and I get two 1-cycle loops: > 1 -> 4 -> 2 -> 1 > and > -1 -> -2 -> -1 >Well, in my notation this would be > 2^1 > x = C(x;1) = x* --- + C(0;1) > 3^1 > 3^1 3^1 - 2^1 3^1 > x = C(0; 1) * ---------- = ------------* -------- = -1 > 2^1 - 3^1 3^1 2^1 - 3^1 >You even can use rationals, if you do some more finetuned analysis. >All these are just compacted notations, no new inventions. Ok, I just wanted to be clear on that. Obviously, all correct algorithms must reach the same conclusion. In my system, the transform structure creates a function that is a straight line. This line cannot have a slope of 1, so it must intersect the line y=x. If the intersection is an integer, then you have a loop. The intersection can be positive or negative, it all depends on the transform structure. all 3x+C systems: +C, -C, -5C, -17C. All lot of people seem to consider the positive and negative domains to be seperate, but they are tied together when you consider transform structures. >---------------------------- >For instance, another thing that you immediately see -if you use this >notation- >is, that infinetly many solutions in x' and x exist for a certain >transformation- >structure, say In my system, the function is A' = (X*A - Z)/Y where A' is the result of the transform from A (this is going up the tree using x*2 and (x-1)/3 rules). Once you find the first A that gives an integer A', you can generate an infinite number of solutions by adding multiples of Y to A. To find A for any given transform, the function can be converted to a problem in linear congruence: X*A == Z (mod Y) which is solvable if GCD(X,Y) divides Z. SInce X is a power of 2 and Y is a power of 3, the GCD will always be 1, so the problem always has a solution. So not only does each solvable transform exist infinitely many times, _every_ possible legal transform exists somewhere on the Collatz tree. I don't know if that means anything, but it's interesting. > x' = C(x; A,B,C) >This is > 2^(A+B+C) > x' = x * ---------- + C(0;A,B,C) > 3^3 >The C(0;...)- part is in general a fraction with the denominator 3^N, in this >case of a three-step-transformation it is 3^3. >So you see, that for all x with the same modular-class based on 3^3 you >find an integral solution in x and x' for the transformation x->x' , or >can try to find approriate exponents for a given pair (x,x'), for instance >x' = 2x+1 or x'-x = 2^B or anything the like. >Extended to the question of cycles it is simply demonstrable, that >for a certain sequence of exponents only one solution for x is possible; >that means, a search for possible loops over reduces to a search for >configurations of exponents, and strong restrictions can be stated for >such a sequence: the sum S of all exponents must between about 1.5*N and >2*N and the like. >----------------------------- >Using a certain sequence of exponents A,B,C, > x' = C(x; 2,2,1,2,1,...) >you can construct numbers x->x' where x has an arbitray long trajectory , >where the intermediate values never fall below x (I think, this is called >glide ?). That's interesting. One of the things I was looking at is why Mersenne numbers, which have the highest excursion (largest value in sequence) don't ever seem to be sequence length record holders. Part of this seems to be related to the glide. The slope of the line up to the excursion is steep, but so is the fall from the excursion. The record holders have a gentler slope up to a lesser excursion, but also a gentler slope down from the excursion producing a longer glide. I haven't had much luck constucting sequence length record holders, so maybe the secret is focusing on the glide. >There are arbitrarily many transformations x' = C(x; A,B,C..), but to >be applied to a pair of integral (x,x') it seems to me, that this sequence >allows the smallest numbers (empirically and only few tests, maybe there >are easy counterexamples except the trivial one). >-------------------------------- >and so on. >As I said, I suggest it as a practical notation for the whole >collatz-calculus. Now I just have to try to understand it. >Gottfried Helms -- Mensanator Ace of Clubs === Subject: Casimir Force, Charg Clusters and Metric Engineering Quick comment I seem to recall that the Casimir force can also go repulsive and indeed that is the case for a charge cluster shell with the topology of a sphere? Casimir considered two models for the sphere, I and II. They are both discussed in my paper: H. E. Puthoff and M. A. Piestrup, Charge confinement by Casimir forces, http://arXiv.org/abs/physics/0408114 Model I is repulsive, Model II attractive. Hal In my theory Model I is attractive because of the extra term missing from Hal's incomplete model. OK Hal mentions Casimir's shell model I that did not work for stability of electron because it was repulsive. My exotic vacuum ZPF induced gravity term missing completely from Hal's model makes Casimir's model I work! This is a key idea. Hal then goes to adhoc model II, which is completely implausible in the light of the new dark energy/matter data from precision cosmology completely missing from Hal's thoughtscape. The point is that the Casimir force still has a very weak direct effect on the warping of space-time needed for metric engineering. To the extent that you can use Hal's favorite SED model for EM ZPF it only couples to space-time geometry Guv term via the Tuv source term and that defeats what Hal really wants to do ultimately. Also back to charge clusters, Ken Shoulders in The Good, The Bad and The Ugly is talking of LARGE EXOTIC ENERGY RELEASE related to cold fusion and even WMD potential! There is no way that the very tiny Casimir force can do that. Ian Peterson has shown that! The only source of exotic energy release larger than thermonuclear fusion is via the repulsive anti-gravity dark energy core inside the charge cluster holding it together! Hal never even mentions the term dark energy. Note that a repulsive dark energy core with a positive exponent r^2 dependence stabilizes the repulsive Coulomb energy barrier with 1/r dependence because they have opposing slopes. This works even in the Casimir shell model I where the Casimir force is repulsive exactly like the Coulomb barrier. Quick comment I seem to recall that the Casimir force can also go repulsive and indeed that is the case for a charge cluster shell with the topology of a sphere? Here is an elementary quick and dirty back of the envelope calculation on why Hal Puthoff's latest paper H. E. Puthoff and M. A. Piestrup, Charge confinement by Casimir forces, http://arXiv.org/abs/physics/0408114 is probably wrong. The repulsive Coulomb barrier potential self-energy per unit electron mass on a spherical shell of N electrons at radius r is of the form V(Coulomb) ~ N^2e^2/mr Notice that this is an inverse power law and it must be positive. Therefore if you plot V(Coulomb) vs r* you have a monotonic decreasing function. What basically kills Hal's argument is that the Casimir force is also an inverse power law! For example look at Hal's first equation for the Casimir pressure F/A ~ hc/r^4 OK, consider a model with an attractive Casimir force (which may not always be the case since the actual sign of the QED Casimir force seems to be very sensitive to the topology and perhaps actual shape of the boundary. V(total) = V(Coulomb) - V(Casimir) = |A(N)|/r - |B(N)|/r^n n ~ 3 but, in fact, the precise value of n does not matter as long as n > 1. First we need a critical point for the dynamical equilibrium of the charge cluster. dV(total)/dr* = 0 i.e. -|A|/r^2 + n|B|/r^(n+1) = 0 The critical point must be a stable minimum, therefore d^2V(total)/dr^2 > 0 i.e. +2|A|/r^3 -(n+1)n|B|/r^(n+2) > 0 Therefore, +2|A|/r^3 > (n+1)n|B|/r^(n+2) This cannot be automatically assumed in Hal's model. It needs to be computed with QED. So we need to check whether these conditions can even be obeyed in Hal's model for a realistic number for r* at the equilibrium that can be checked against the actual data by Ken Shoulders. In contrast my model is of the form V(total) = V(Coulomb) + V(Casimir) + V(Exotic Vacuum Core) = A(N)/r + B(N)/r^n + /*r^2 The third term from Einstein's general relativity for the direct warping of spacetime from zero point energy is a power law with a positive exponent, i.e. 2 where /* is a dynamical field that adjusts to make the dynamical equilibrium stable. Note if the charge cluster is rotating with orbital angular momentum L and if it is vibrating there will be additional terms. There will also be velocity dependent forces if there is an external magnetic field and the problem gets quite complicated. The stability condition is +2A/r^3 +(n+1)nB/r^(n+2) + /* > 0 Jack This is a revised version of what I sent to you yesterday. Ken Short-Range Electron Attractive Force by Ken Shoulders Statement of Action: There is an attractive force found between closely spaced, free electrons instead of the universally touted repulsion force. This attractive force is effective only at dimensions in the order of atomic spacing, being in the range of 10-10 meters, leaving older repulsion laws intact for large spacing. When this force binds two or more electrons, their expressed field at a distance is reduced. This is a The Effect: As shown by many writings of the author, as well as G. A. Mesyats in Russia (1), electrons easily cluster into complex structures having unique properties not available to single electrons. These electron clusters have been named EVs or EVOs by Shoulders in various papers available for download from: http://www.svn.net/krscfs/, and Ectons by Mesyats. There are several theories for their existence mentioned in associated literature references but a complete description is still lacking. Whatever the effect is that fosters this electron clustering action, it behaves like an unseen substance that enshrouds electron groups, partly masking their charge. It is a short-range force resembling a positive charge negating the effect of repulsive electronic charge and can further be defined as a near-field effect that seems to be an innate property of the electron occurring at the time of its creation. This local action is reminiscent of the induction field in electromagnetic theory. This attractive force is a property of the individual electron and not a large group effect, as it extends down in size to electron pairs as is seen in the electron accretion method of forming EVOs, described early on by Shoulders in an issued patent and the references cited above. Non Discovery Sequence: All references cited above show accumulated evidence for the existence of EVOs in various sizes and forms. Seeing the large total of accumulated evidence over such a long period of time brings up the question of how finding it was passed over for so long. A citation on how this likely happened is given below. amber through the more technically advanced era of silk and a glass rod, it was determined that like charges always repel. What should have been a temporary guideline using this data was erroneously cast in cement as a sacred truth and immutable law by fakirs crying from the scientific tower of Babel. This belief persisted throughout the very technical age of arc and spark investigation in spite of outstanding but unheeded evidence of charge accretion appearing everywhere in the so-called cathode spot phenomenon. The old law of like charge repulsion is good but not all-encompassing, because at any one time, there are likely more free electrons adhering to each other in this world than there are being repelled by each other. Electron clusters are ubiquitous. When the electron clustering effect was first found by the author, its mention to all others was treated as scientific sacrilege as the message from the fakir was still echoing through the halls after these many years. The message here is: Believe what your senses tell you and not what others say. What I see is that the like charge between electrons more often attracts than repels -- whenever the spacing between them is small. [1] Explosive Electron Emission by G. A. Mesyats, ISBN-7691-0881-5, 1998, URO-PRESS, Yekateringburg. === Subject: Algebraic simplification Point C (x,y) moves on segment of a circle through A(-a,0) and B(a,0) containing a constant angle gama. It is required to find an equation of the locus, as all circles through A and B calculated from cross product cos(gama)=AC.CB/(|AC||CB|)=(x^2+y^2-a^2)/sqrt[((x+a)^2+y^2)(( x-a)^2+y^2)] which does not seem to readily simplify to x^2+y^2-2 y A =a^2 form of circles centered on y-axis. Here A is an Arbitrary constant related to cos(gama). Got the result earlier, but somehow not getting it once again now :), except the obvious gama=Pi/2. Normally not inclined to post such things, but not readily vanishing fourth degree terms bother me. Please help, TIA. === Subject: Re: Algebraic simplification >Point C (x,y) moves on segment of a circle through A(-a,0) and B(a,0) >containing a constant angle gama. It is required to find an equation >of the locus, as all circles through A and B calculated from cross >product cos(gama)=AC.CB/(|AC||CB|)=(x^2+y^2-a^2)/sqrt[((x+a)^2+y^2)(( x-a)^2+y^2)] >which does not seem to readily simplify to x^2+y^2-2 y A =a^2 form of >circles centered on y-axis. Here A is an Arbitrary constant related to >cos(gama). Got the result earlier, but somehow not getting it once >again now :), except the obvious gama=Pi/2. Normally not inclined to >post such things, but not readily vanishing fourth degree terms bother >me. Please help, TIA. Let b = cos(gamma). Suppose that ((x+a)(x-a) + y^2)^2 = b^2((x+a)^2 + y^2) ((x-a)^2 + y^2) (x^2 - a^2 + y^2)^2 = b^2((x^2 + a^2 + y^2)^2 - (2ax)^2) (x^2 - a^2 + y^2)^2 = b^2((x^2 - a^2 + y^2)^2 + (2ay)^2) (1-b^2)(x^2 - a^2 + y^2)^2 = (2aby)^2 x^2 - a^2 + y^2 = +/-2cy [c = ab/sqrt(1-b^2) = a cot(gamma)] x^2 + (y-/+c)^2 = a^2 + c^2 = (a csc(gamma))^2 Thus, there are two circles of radius a |csc(gamma)| with centers at (0, +/-a cot(gamma)). Rob Johnson take out the trash before replying === Subject: Re: A question about multiplicity of generalized eigenvalues >Let A,B be two non singular real matrices of order N. >Can anyone confirm that the multiplicity of a >generalized eigenvalue lambda of the pair (A,B) is the dimension of the >null space of A-lambda *B, ie the nullity of A-lambda *B? What are you using as the definition of multiplicity here? There can be two distinct meanings: algebraic (multiplicity of lambda as a root of the polynomial det(A-x*B)) and geometric, which seems to be the one you want. >My proof is: If it's a definition, it needs no proof. >Let lambda be a generalized eigenvalue of (A,B). >Nullity( A-lambda*B ) = N - rank( A-lambda*B) > = N - rank( B*( inv(B)*A-lambda*I ) ) > = N - rank( inv(B)*A-lambda*I ) >So Nullity( A-lambda *B ) is the multiplicity of the eigenvalue of >inv(B)*A equal to lambda. The fact that the eigenvalues of inv(B)*A are >the generalized eigenvalues of A-lambda * B ends the proof. >Is there something wrong with my proof? The generalized eigensystem A x = lambda B x is indeed equivalent to the ordinary system B^(-1) A x = lambda x if B is invertible. Somewhat more interesting is when A and B are not invertible. ' === Subject: Lucas/Fibonacci bisections by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7RKGgZ21716; I realized that the sequence of ratios a(n+1)/a(n) from the sequence http://www.research.att.com/projects/OEIS?Anum=A097512 appears to approach the golden ratio phi + 1 = (3+sqrt(5))/2 This happens for Lucas(2n) and Fib(2n) as well. Might this suggest that another (possibly unknown) sequence X exists such that X(2n) = a(n)? C. Dement http://www.crowdog.de === Subject: Re: How Valid is This Proof? X-no-archive: yes > While researching the various math topics, I stumbled upon this: never mind that, what's the answer to this ? http://www.blessa.com/64-equals-65.gif sammi === Subject: Re: How Valid is This Proof? <2p9kb6Fig67sU2@uni-berlin.de> >never mind that, what's the answer to this ? >http://www.blessa.com/64-equals-65.gif The slope of the red and green shapes is 3/8 = .375 The slope of the orange and blue shapes (after rotation) is 2/5 = .4 So there is missing area in the form of a parallelogram in the center of the 5x13 rectangle. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Fibonacci number ending in 0000 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7RKWjK23091; Question: There is such a Fibonacci number among the first 100 000 001 that it ends with 0000 (four consecutive zeros) So far I have come sadly short in this problem. Is there any reasonable way of proving this statement true or false? joccis === Subject: Re: Fibonacci number ending in 0000 Fibonacci(n) where n = {7500, 15000, 22500, 30000, 37500, 45000, 52500, 60000,75000, 82500, 90000} ends is 0000 === Subject: Re: Fibonacci number ending in 0000 > Question: > There is such a Fibonacci number among the first 100 000 001 that it ends with 0000 (four consecutive zeros) > So far I have come sadly short in this problem. Is there any reasonable way of proving this statement true or false? > joccis While splitting the problem into divisibility by 2^4 and by 5^4 gives a faster answer (F(7500) is the first), and while computer-aided search gives the same answer in a few seconds, preceded by a few minutes of programming arithmetics modulo 10000, I suspect that the intention of the author of the problem was to use the pigeonhole principle. Why? It looks like a special case of the problem given K>1 integer, show that for some n between 1 and K^2+1, F(n) is divisible by K. (Consider the ordered pairs modulo K). === Subject: Re: Fibonacci number ending in 0000 > Question: > There is such a Fibonacci number among the first 100 000 001 that it > ends with 0000 (four consecutive zeros) > So far I have come sadly short in this problem. Is there any reasonable > way of proving this statement true or false? >While splitting the problem into divisibility by 2^4 and by 5^4 gives a >faster answer (F(7500) is the first), and while computer-aided search >gives the same answer in a few seconds, preceded by a few minutes of >programming arithmetics modulo 10000, I suspect that the intention of the >author of the problem was to use the pigeonhole principle. >Why? It looks like a special case of the problem >given K>1 integer, show that for some n between 1 and K^2+1, F(n) is >divisible by K. >(Consider the ordered pairs modulo K). Yes, I'm sure you're right. Another problem where the general case is easier because the specific cases have distractions. Of course the K^2+1 can be reduced somewhat, say to (K-1)^2+2. ' === Subject: Re: Fibonacci number ending in 0000 > Question: > There is such a Fibonacci number among the first 100 000 001 that it > ends with 0000 (four consecutive zeros) The answer is yes. I will give you two lemmas: (1) If 5^k | F_n, then 5^k+1 | F_{5n} (2) If 2^k | F_n then 2^k+1 | F_2n 8 | F_6 and 5 | F_5. So 5^4 | F_625. Now, F_a | F_{ab} , so we must have 8*5^4 | F_3750 and 16*5^4 | F_7500. You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: Fibonacci number ending in 0000 >While splitting the problem into divisibility by 2^4 and by 5^4 gives a >faster answer (F(7500) is the first), and while computer-aided search >gives the same answer in a few seconds, preceded by a few minutes of >programming arithmetics modulo 10000, I suspect that the intention of the >author of the problem was to use the pigeonhole principle. >Why? It looks like a special case of the problem >given K>1 integer, show that for some n between 1 and K^2+1, F(n) is >divisible by K. >(Consider the ordered pairs modulo K). I think the use of the pigeonhole principle here would be to say that if it doesn't happen in the first K^2 terms, it won't happen at all. Of course that's not the case for 0 mod 10000 as others have shown. If k=8, the sequence is 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0... the number 6 is completely left out, and only 12 of the 64 possible ordered pairs are ever visited. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Fibonacci number ending in 0000 > dated Fri, 27 >While splitting the problem into divisibility by 2^4 and by 5^4 gives a >faster answer (F(7500) is the first), and while computer-aided search >gives the same answer in a few seconds, preceded by a few minutes of >programming arithmetics modulo 10000, I suspect that the intention of the >author of the problem was to use the pigeonhole principle. >Why? It looks like a special case of the problem >given K>1 integer, show that for some n between 1 and K^2+1, F(n) is >divisible by K. >(Consider the ordered pairs modulo K). >I think the use of the pigeonhole principle here would be to say that if it >doesn't happen in the first K^2 terms, it won't happen at all. Of course >that's not the case for 0 mod 10000 as others have shown. >If k=8, the sequence is 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0... the number 6 >is completely left out, and only 12 of the 64 possible ordered pairs are >ever visited. The point is that if = mod K, then by induction = mod K, and in particular = = <0,1> mod K so F(n-m) is divisible by K. In fact we can say more: if = c mod K for some c, then F(n-m) is divisible by K. Hmm, that gives us a better bound: the ordered pairs in {0,...,K-1}^2 such that gcd(a,b,K) = 1 can be partitioned into equivalence classes where ~ if there is k such that gcd(k,K)=1 and = k mod K. Each equivalence class has phi(K) members. So we can replace K^2 + 1 by K^2/phi(K) + 1. ' === Subject: Re: Fibonacci number ending in 0000 joccis escribi.97: > Question: > There is such a Fibonacci number among the first 100 000 001 that it > ends with 0000 (four consecutive zeros) > So far I have come sadly short in this problem. Is there any > reasonable way of proving this statement true or false? > joccis http://math.smsu.edu/~les/Sol03_03.html? -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Fibonacci number ending in 0000 >There is such a Fibonacci number among the first 100 000 001 that it >ends with 0000 (four consecutive zeros) >So far I have come sadly short in this problem. Is there any reasonable >way of proving this statement true or false? Hint: look at the Fibonacci numbers mod 2^4 and mod 5^4. ' === Subject: Re: from every corner of the globe >is often heard in news and ads :).. the Earth still has some sharp corners. Sounds like a nice example of a mixed metaphor. But you could say it's using corner in the sense of definition #7 in the OED: Any part whatsoever, even the smallest, most distant or secluded. ' === Subject: Re: JSH: Curious, Wiles's work, null test > Now I'm being fair. If *anyone* actually understands Wiles's work > then what I'm asking should be a trivial and fun exercise for them!!! > Go through his work with an assumption opposite to the one he uses, > which is that there is in fact a non-modular elliptic curve, and see > if you find a point in his argument where it will squawk, NO!!! > That's how real math proofs work. I you understand Wiles' work then you go through it and point out where any mistake occurs. Actual real maths, not a new phrase you've learnt and want to repeat as often as you can, with all the capital letters possible. > My charges are specific and to the point. I say Wiles's argument > fails by Cum Hoc, Ergo Propter Hoc. That isn't specific. Specific would be pointing out the exact part of the proof which fails. === Subject: Re: JSH: Curious, Wiles's work, null test > Now I'm being fair. If *anyone* actually understands Wiles's work > then what I'm asking should be a trivial and fun exercise for them!!! > Go through his work with an assumption opposite to the one he uses, > which is that there is in fact a non-modular elliptic curve, and see > if you find a point in his argument where it will squawk, NO!!! > That's how real math proofs work. No, it really isn't. I'm not sure how genuinely interested in mathematics you are (and how much you'd rather just stir up Usenet), but reading real math proofs for a while ought to convince you this is wrong. Nontrivial proofs of nontrivial theorems often don't have a particular place where you can point to say, Aha! This argument excludes the possibility of such-and-such a counterexample. > If you challenge a mathematical proof with something contrary to its > conclusion then at some point in the argument you'll face a > contradiction. Unless the proof is divided into several cases, this point can really only come at the end. Why? Because at that point, you've actually proved the theorem! As an amateur completely unqualified to read Wiles' proof, I claim that the point in his argument that squawks, NO!! Your supposed non-modular semisimple elliptic curve cannot exist! should necessarily be the last line of the proof. === Subject: Re: JSH: Curious, Wiles's work, null test > Now I'm being fair. If *anyone* actually understands Wiles's work > then what I'm asking should be a trivial and fun exercise for them!!! > > Go through his work with an assumption opposite to the one he uses, > which is that there is in fact a non-modular elliptic curve, and see > if you find a point in his argument where it will squawk, NO!!! > > That's how real math proofs work. > No, it really isn't. I'm not sure how genuinely interested in mathematics > you are (and how much you'd rather just stir up Usenet), but reading > real math proofs for a while ought to convince you this is wrong. > Nontrivial proofs of nontrivial theorems often don't have a particular > place where you can point to say, Aha! This argument excludes the > possibility of such-and-such a counterexample. That's illogical. I see that you don't understand the *logic* of math proofs. Necessarily if something conßicts with the conclusion of a proof it will conßict with a step *before* that conclusion! Logically, it must. > If you challenge a mathematical proof with something contrary to its > conclusion then at some point in the argument you'll face a > contradiction. > Unless the proof is divided into several cases, this point can really > only come at the end. Why? Because at that point, you've actually > proved the theorem! As an amateur completely unqualified to read Wiles' > proof, I claim that the point in his argument that squawks, NO!! Your > supposed non-modular semisimple elliptic curve cannot exist! should > necessarily be the last line of the proof. I was expecting that one of you might try that angle, but it's bogus. Proofs are *proven* and aren't true just because someone says they are true. If you're proving a mathematical statement, how can that proof be just the conclusion? Necessarily you need the steps that lead TO that conclusion!!! Given a counterclaim to a proof--remember proof means validity--the proof will show at a step before its conclusion that the counterclaim is false. James Harris === Subject: Re: JSH: Curious, Wiles's work, null test Discussion, linux) > That's illogical. > I see that you don't understand the *logic* of math proofs. > Necessarily if something conßicts with the conclusion of a proof it > will conßict with a step *before* that conclusion! None of this makes a lick of sense. Suppose I have a valid proof of X. What sense does it make to assume NOT X and see what breaks? Each line in the proof of X follows from what comes before. The assumption of NOT X doesn't change that, because logical validity is a local, not global, condition. Assume the negation of a necessary truth is not a particularly useful exercise. You won't understand this because you don't grok the distinction between deductive arguments and their inductive counterparts (as found in empirical sciences). You will instead continue to shout that you're the only one in sci.math that understands mathematical proof. > Given a counterclaim to a proof--remember proof means validity--the > proof will show at a step before its conclusion that the counterclaim > is false. Wrong. A valid proof starts with axioms[1]. Assuming the negation of the conclusion doesn't change the list of axioms. A valid proof continues by applying rules of inference to previously derived formulas. Assuming the negation of the conclusion doesn't change the allowable rules of inference. Therefore (by induction on the length of the proof) assuming the negation of the conclusion of a proof does not invalidate the proof. Adding an extra assumption only adds the number of valid proofs, drastically adds in this case. It doesn't invalidate proofs. These are very simple and basic observations of mathematical logic. Can't imagine how you've gotten it all wrong. Footnotes: [1] Simplifying the discussion of proof syntax for obvious reasons. -- There was an accident in the air. There was a sign saying, ÔPlanes don't go here. The clouds have to be fixed.' -- Quincy P. Hughes applies the lessons of Dutch train travel to pretending about airplanes. === Subject: Re: JSH: Curious, Wiles's work, null test days. My association with the Department is that of an alumnus. > Just out of curiousity I went ahead and took another quick look at > Wiles's work which is now conveniently on-line to see if a null test > would end quickly, and I got to page 5 without seeing it end, as up to > that point Wiles is STILL saying to assume that rho (I think it's rho) > is modular. > So instead assume it's not. > That would definitely affect Theorem 0.2 but not necessarily affect > his *conclusion* in a way that you might suppose. >It won't be necessary for anyone to review Wiles' proof, because your >charge that it contains a logical error (Com Hoc, Ergo Propter Hoc) can be >immediately discarded by appeal to the logical truism -- Ex Harris, Ergo >Erratum. The copy I found, at www.eleves.ens.fr/home/rrichard/wiles.pdf provides evidence for the EH, EE argument. Theorem 0.2 does not refer to elliptic curves directly, but to a representation of the absolute Galois group of the rationals into GL_2(F_p). The conclusion is not about rho_0, but about a lifting of rho_0, called rho, into GL_2(O), where O is the ring of integers of a local field containing the p-adic rationals Q_p. Theorem 0.2 gives sufficient conditions for the representation rho to be induced by a modular form, the sufficient conditions being in terms of rho_0. The reason one discusses representations of the absolute Galois group of Q is that again, there is a connection between both modular forms and these representations, and between elliptic curves and the representation. If E is an elliptic curve over Q, and n>0, then we can consider the points of E in the algebraic closure of Q which are of irder n. This group is isomorphic to C_n x C_n, just as in the complex case. And the absolute Galois group G acts on this group: if P is a point of order n, with coordinates (x,y), then g in G maps P to the point (g(x),g(y)), which will also be a point in E (since E is defined over Q) of order n (since the operation on the points can be defined by rational functions). That means that, given an element g of the Galois group, we obtain an automorphism of C_n x C_n; this group is of course none other than GL_2(Z/nZ), the group of invertible 2 x 2 matrices with entries in Z/nZ. This holds for each n; so let us denote by g[n] the image of g under the map corresponding to n. A difficult theorem of Serre says that if E is a semistable elliptic curve over Q, and l is a prime such that g[l] is an irreducible representation, then g[l] is in fact surjective. Likewise there is a way to construct a representation of the Galois group from a given modular form. Representations that come from modular forms are, of course, modular. I don't really remember how one does this, however, and a quick perusal of Ribet's lectures in Arithmetic Algebraic Geometry did not seem to remind me of the method. One of the ways in which one can establish that an elliptic curve is modular (and the way in which this is done) has to do with an established connection through the absolute Galois group of rationals; in essence, one can use the representations as a Ôstepping stone' to show that the given curve is modular. This is what Wiles is talking about in his introduction and in theorems 0.2 and 0.3; in particular, 0.3 describes sufficient conditions for an elliptic curve to be modular. Then one shows that if the elliptic curve is semistable, then the sufficient conditions are met. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Curious, Wiles's work, null test > One of the ways in which one can establish that an elliptic curve is > modular (and the way in which this is done) has to do with > an established connection through the absolute Galois group of > rationals; in essence, one can use the representations as a Ôstepping > stone' to show that the given curve is modular. This is what Wiles is > talking about in his introduction and in theorems 0.2 and 0.3; in > particular, 0.3 describes sufficient conditions for an elliptic curve > to be modular. Then one shows that if the elliptic curve is > semistable, then the sufficient conditions are met. Not only implicit in Wiles's approach but directly stated *by* him in his paper is the assumption of modularity. Now I'm not going to pretend that I follow all the technicalities or even a significant number of them, but that's irrelevant to my request. The null test *requires* that you assume a non-modular elliptic curve and trace out his paper with that assumption looking for a contradiction. James Harris === Subject: Re: JSH: Curious, Wiles's work, null test Discussion, linux) > The null test *requires* that you assume a non-modular elliptic curve > and trace out his paper with that assumption looking for a > contradiction. I think we need some practice with this grand new proof-theoretic scheme of null tests before we try it out on Wiles. Can you help me out here? Here's a simple proof, one of my favorites. ,---- | Theorem: The square root of 2 (hereafter sqrt(2)) is irrational. | | Proof: Assume that sqrt(2) is rational. Let x and y be given such | that | | (1) sqrt(2) = x / y | (2) gcd(x,y) = 1 | | Then 2 = x^2 / y^2 so x^2 = 2 y^2. Now, if 2 divides x^2, then 2 | divides x, so let z be given such that x = 2 z. | | Then (2 z)^2 = 2 y^2, so 4 z^2 = 2 y^2 and hence 2 z^2 = y^2. | Therefore y^2 is divisible by 2 and hence y is also divisible by 2. | | But we have thus proven that both x and y are divisible by 2, | contradicting (2). `---- Is this proof valid? If so, then let's proceed and apply the null test to it. So, now we assume that sqrt(2) is in fact rational, right? Can you help me find where the contradiction occurs in that proof? Which formerly valid step is now invalid? -- Jesse Hughes Well, you know as soon as you have a new number I will be happy to add it to the list. Don't try those childish tit-for-tat games with me. -- Ross Finlayson on Cantor's theorem. === Subject: Re: JSH: Curious, Wiles's work, null test <87n00fvd2w.fsf@phiwumbda.org> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > The null test *requires* that you assume a non-modular elliptic curve > and trace out his paper with that assumption looking for a > contradiction. > I think we need some practice with this grand new proof-theoretic > scheme of null tests before we try it out on Wiles. Can you help me > out here? > Here's a simple proof, one of my favorites. > ,---- > | Theorem: The square root of 2 (hereafter sqrt(2)) is irrational. > | > | Proof: Assume that sqrt(2) is rational. Let x and y be given such > | that > | > | (1) sqrt(2) = x / y > | (2) gcd(x,y) = 1 > | > | Then 2 = x^2 / y^2 so x^2 = 2 y^2. Now, if 2 divides x^2, then 2 > | divides x, so let z be given such that x = 2 z. > | > | Then (2 z)^2 = 2 y^2, so 4 z^2 = 2 y^2 and hence 2 z^2 = y^2. > | Therefore y^2 is divisible by 2 and hence y is also divisible by 2. > | > | But we have thus proven that both x and y are divisible by 2, > | contradicting (2). > `---- > Is this proof valid? If so, then let's proceed and apply the null > test to it. > So, now we assume that sqrt(2) is in fact rational, right? Can you > help me find where the contradiction occurs in that proof? Which > formerly valid step is now invalid? This is too easy. You are just not capable of the mental powers needed to understand the mind of great thinkers. Here is your answer: the step marked as (2) is obviously the one that is contradicted if sqrt(2) is rational, as you yourself state in the last line of your proof (I am using quotation marks here in order to capture the Harrisian spirit appropriately). You start with let x and y be given such and of course, you have shown that they can't be given. So your tacit assumption that they can be given is invalid. Good grief, you are leaving gaping holes that complete morons can jump through (not many other people, though). -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Curious, Wiles's work, null test > >One of the ways in which one can establish that an elliptic curve is >modular (and the way in which this is done) has to do with >an established connection through the absolute Galois group of >rationals; in essence, one can use the representations as a Ôstepping >stone' to show that the given curve is modular. This is what Wiles is >talking about in his introduction and in theorems 0.2 and 0.3; in >particular, 0.3 describes sufficient conditions for an elliptic curve >to be modular. Then one shows that if the elliptic curve is >semistable, then the sufficient conditions are met. > Not only implicit in Wiles's approach but directly stated *by* him in > his paper is the assumption of modularity. Do you even know *what* is being assumed to be modular at the point in question? Of course you realize that the place where he assumed modularity is in the hypotheses of Theorem 0.2. Note that this is *not* a theorem about elliptic curves, but instead it's a theorem about liftings of Galois representations. The clue? Well, note the statement of the theorem itself: Theorem 0.2. Supposet that rho_o is irreducible and satisfies either condition (I) or (II) above. Suppose also that rho_o is modular and ...(the rest of the statement of Theorem 0.2)... Now, one wants to know what the theorem is saying, so one scans the preceding text for a clue. Here's what I found (note the author has already alerted the reader that he'll need to see what preceded the statement of the theorem): ................Excerpted from above............. Assume rho_o : Gal(Qbar/Q) --> GL_2(Fbar_p) is a continuous representation with values in the algebraic closure of a finite field of characteristic p and that det(rho_o) is odd. and of course, I was able to find what he meant by (I) and (II): (I) rho_o is ordinary (at p) : technical conditions on rho_o (II) rho_o is ßat (at p) I won't bore you with the details, except to point out that he has outlined the plan of attack in the paragraph immediately preceding these two technical conditions: Our point of view will be to assume that rho_o is modular and then to attempt to give conditions under which a representation rho lifting rho_o comes from a modular form ... This spells out the purpose for Theorem 0.2: it's there to give us ways to determine whether a lifting of a modular representation is itself Galois representation, and quite another to deduce that your elliptic curve is itself modular. Oh, and I just gathered from the language of the theorem itself, as well as from the context, that we weren't really talking at about elliptic curves per se. At least, not yet. As far as the assumption of modularity, your remark suggests that you believe he's attempting to assume just what it is he wants to prove. Again, Wiles is not discussing any elliptic curve just yet. Note further, that when discussing elliptic curves directly, he does *not* assume that they are modular, but proves that certain representation *are* modular, and invokes known theorems to show the curve itself is modular. If you were to page forward to Chapter 5, you would find where Theorem 0.2 is invoked (in the proof of the main theorem, Theorem 5.2). In the scheme of things, this is a piece of the whole: the case in which the representation rho-bar_(E,3) on E[3] (the set of points of period 3 in the group E) is irreducible. By a result of Langlands, mentioned by the author in the top paragraph of the second page of the paper, this *guarantees* that that particular representation is modular. Theorem 0.2 then shows that certain liftings are also modular, and Serre's theorem on isogenies then shows that the elliptic curve E is itself modular. The next paragraph addresses the case that rho-bar_(E,3) is reducible. It takes more proof, and again (this time, looking at rho-bar_(E,5) on the points of period 5 in the curve) proves that the hypotheses of Theorem 0.2 are met. He applies Theorem 0.2 *only* after verifying that it is legitimate to do so, and in each case, it includes a proof that the representation that is lifted is in fact modular. > Now I'm not going to pretend that I follow all the technicalities or > even a significant number of them, but that's irrelevant to my > request. That's silly. You pretend to be reading a paper, but your ignorance of the language places you in the situation John Cleese's character finds himself in Monty Python's Hungarian Phrasebook sketch. *Of course* it is relevant whether one knows what's being discussed. > The null test *requires* that you assume a non-modular elliptic curve > and trace out his paper with that assumption looking for a > contradiction. As I understand your intent, it appears that you believe that Wiles's proof would apply equally well to a non-modular, semistable elliptic curve. Perhaps you should go to the point where there *is* in fact an elliptic curve present in the discussion! *THEN* assume the curve is not modular. > James Harris Dale. === Subject: Re: JSH: Curious, Wiles's work, null test > > One of the ways in which one can establish that an elliptic curve is > modular (and the way in which this is done) has to do with > an established connection through the absolute Galois group of > rationals; in essence, one can use the representations as a Ôstepping > stone' to show that the given curve is modular. This is what Wiles is > talking about in his introduction and in theorems 0.2 and 0.3; in > particular, 0.3 describes sufficient conditions for an elliptic curve > to be modular. Then one shows that if the elliptic curve is > semistable, then the sufficient conditions are met. > Not only implicit in Wiles's approach but directly stated *by* him in > his paper is the assumption of modularity. > Now I'm not going to pretend that I follow all the technicalities or > even a significant number of them, but that's irrelevant to my > request. > The null test *requires* that you assume a non-modular elliptic curve > and trace out his paper with that assumption looking for a > contradiction. Ok, since you chose to ignore what Arturo has shown you, I'll give you some more stuff to ignore. Since you say you don't understand the technicalities, I'll pick one of the simpler parts of Wiles paper. In the introduction (p. 444), Wiles refers to the fact that it is known that E is modular if and only if the associated 3-adic representation is modular. Clearly due to the if and only if part of that statement, if we assume that there exists that there is a non-modular elliptic curve E, then there will be a non-modular 3-adic representation. He also points out that if ?3 is irreducible, then it is also modular. This forms the starting point of his work which is done with the Galois representations (e.g. lifting ?3) to show that the 3-adic representations of an elliptic curve is modular. Thus we have a contradiction if we assume that there exist a non-modular elliptic curve. -Saint Cad === Subject: Re: JSH: Curious, Wiles's work, null test > > One of the ways in which one can establish that an elliptic curve is > modular (and the way in which this is done) has to do with > an established connection through the absolute Galois group of > rationals; in essence, one can use the representations as a Ôstepping > stone' to show that the given curve is modular. This is what Wiles is > talking about in his introduction and in theorems 0.2 and 0.3; in > particular, 0.3 describes sufficient conditions for an elliptic curve > to be modular. Then one shows that if the elliptic curve is > semistable, then the sufficient conditions are met. > > Not only implicit in Wiles's approach but directly stated *by* him in > his paper is the assumption of modularity. > Now I'm not going to pretend that I follow all the technicalities or > even a significant number of them, but that's irrelevant to my > request. > The null test *requires* that you assume a non-modular elliptic curve > and trace out his paper with that assumption looking for a > contradiction. > Ok, since you chose to ignore what Arturo has shown you, I'll give you some > more stuff to ignore. Since you say you don't understand the > technicalities, I'll pick one of the simpler parts of Wiles paper. In the > introduction (p. 444), Wiles refers to the fact that it is known that E is > modular if and only if the associated 3-adic representation is modular. > Clearly due to the if and only if part of that statement, if we assume > that there exists that there is a non-modular elliptic curve E, then there > will be a non-modular 3-adic representation. He also points out that if ?3 > is irreducible, then it is also modular. This forms the starting point of > his work which is done with the Galois representations (e.g. lifting ?3) > to show that the 3-adic representations of an elliptic curve is modular. > Thus we have a contradiction if we assume that there exist a non-modular > elliptic curve. > -Saint Cad Well at least you're trying. Really now, was that so hard? Isn't it even kind of fun? Why wouldn't going through Wiles's work in this way not just give another perspective that might help understanding? Yet some of you act like I'm trying to pull teeth! Now then, it seems you need to work a bit more at it. Like, why can't you lift with the assumption of a non-modular elliptic curve? What if the associated 3-adic representation is NOT modular? So what? Your last sentence doesn't follow from what came before, and it looks like you tossed out a few basic things, and then thought you could just tack on what you *believe* and no one would be the wiser. But you're cheating yourself. Have fun with this! Assume a non-modular elliptic curve all the way through, and see if something jumps out and yelps in Wiles's work. If it exists you will be able to give the paragraph and page number and amazingly enough, it will be at a single logical step. That's how math proofs work. James Harris === Subject: Re: JSH: Curious, Wiles's work, null test > >One of the ways in which one can establish that an elliptic curve is >modular (and the way in which this is done) has to do with >an established connection through the absolute Galois group of >rationals; in essence, one can use the representations as a Ôstepping >stone' to show that the given curve is modular. This is what Wiles is >talking about in his introduction and in theorems 0.2 and 0.3; in >particular, 0.3 describes sufficient conditions for an elliptic curve >to be modular. Then one shows that if the elliptic curve is >semistable, then the sufficient conditions are met. >Not only implicit in Wiles's approach but directly stated *by* him in >his paper is the assumption of modularity. >Now I'm not going to pretend that I follow all the technicalities or >even a significant number of them, but that's irrelevant to my >request. >The null test *requires* that you assume a non-modular elliptic curve >and trace out his paper with that assumption looking for a >contradiction. > Ok, since you chose to ignore what Arturo has shown you, I'll give you some > more stuff to ignore. Since you say you don't understand the > technicalities, I'll pick one of the simpler parts of Wiles paper. In the > introduction (p. 444), Wiles refers to the fact that it is known that E is yield anything promising looking... alex === Subject: Re: JSH: Curious, Wiles's work, null test > > One of the ways in which one can establish that an elliptic curve is > modular (and the way in which this is done) has to do with > an established connection through the absolute Galois group of > rationals; in essence, one can use the representations as a Ôstepping > stone' to show that the given curve is modular. This is what Wiles is > talking about in his introduction and in theorems 0.2 and 0.3; in > particular, 0.3 describes sufficient conditions for an elliptic curve > to be modular. Then one shows that if the elliptic curve is > semistable, then the sufficient conditions are met. >Not only implicit in Wiles's approach but directly stated *by* him in >his paper is the assumption of modularity. >Now I'm not going to pretend that I follow all the technicalities or >even a significant number of them, that's encouraging. next question: are you going to pretend you understand -any- of them? for example, can you explain in your own words rxactly what it means for an elliptic curve to be modular? >but that's irrelevant to my >request. >The null test *requires* that you assume a non-modular elliptic curve >and trace out his paper with that assumption looking for a >contradiction. this is not as ridiculous as some people are saying. but your claim that your lack of understanding of Ôthe technicalities', ie the entire proof, is irrelevant is very funny. since you don't have -any- idea what any of the paper actually -means- how could you possibly know that the contradiction you mention isn't there? >James Harris ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: JSH: Curious, Wiles's work, null test if you have a hypothesis, monsieur Harris, then you should make it. I don't see how it could, since the *requirement* of modularity excludes more of the points of the plane then meet it. also, inductive proofs, including those by contradiction, are one-to-one with deductive proofs, as shown in 2-1/2 page proof in Mathematics Monthly, which I finished to my satisfaction; I mean, it was very simple! > Just out of curiousity I went ahead and took another quick look at > Wiles's work which is now conveniently on-line to see if a null test > would end quickly, and I got to page 5 without seeing it end, as up to > that point Wiles is STILL saying to assume that rho (I think it's rho) > is modular. > So instead assume it's not. > That would definitely affect Theorem 0.2 but not necessarily affect > his *conclusion* in a way that you might suppose. --Chairman George! http://tarpley.net/bush12.htm === Subject: Re: JSH: Curious, Wiles's work, null test > Just out of curiousity I went ahead and took another quick look at > Wiles's work which is now conveniently on-line to see if a null test > would end quickly, and I got to page 5 without seeing it end, as up to > that point Wiles is STILL saying to assume that rho (I think it's rho) > is modular. This snippet is just *so* James - to paraphrase I haven't read the complete work, and I'm not sure I can remember what I have read, but hey, I'm right, and if you evil, lying mathematicians would just do all the work, I'll be rich and famous (and get babes). If you don't, I'll unleash the Army and The Hammer and all you miserable lying mathematicians will be on the dole and Mathematics as we know it will be finished - because Maths only cares about The Truth. -- Min I blame the jelly === Subject: Re: JSH: Assocation does not prove > > Comparisons are ripe for the logical error Cum Hoc, Ergo Propter > Hoc. > > No, James, comparisons aren't how the fallacy occurs. Causal > explanations are the setting for this fallacy. Mathematical proofs > are not causal explanations. > Oh you mean that math proofs don't follow from logical reasons? > No, I mean what I said. Mathematical proofs are not causal > explanations. > Remember though cause and effect also have to do with time in your > sorry little life, ultimately the idea behind cause and effect is that > there is a *reason* for why things happen. > So move beyond time and you can say that cause and effect has to do > with the reasons why starting from point A you end up at point B then > point C. > Like in a mathematical proof you begin with a truth and proceed by > logical steps to a conclusion that then MUST be true. > The effect, a true conclusion, follows from the cause, a perfect > logical path from truth to truth to final truth. > You're babbling, son. Mathematical justifications are not causal > explanations in the sense relevant to Cum Hoc. You're being obstinate. Read what I said again. Carefully this time. Better yet, go try your argument on someone else, not online or in a post, just talk it out with someone who knows logic. You need to understand what the logical fallacy is about, and I've given you all the information you need, like this subject line says a lot. But if that's not enough, you need to go out, do some research and talk it out with others versus being hard-headed and relying on your own understanding. James Harris === Subject: Re: JSH: Assocation does not prove <87zn4hyvf6.fsf@phiwumbda.org> <87vff4stop.fsf@phiwumbda.org> Discussion, linux) > You're babbling, son. Mathematical justifications are not causal > explanations in the sense relevant to Cum Hoc. > You're being obstinate. Read what I said again. > Carefully this time. > Better yet, go try your argument on someone else, not online or in a > post, just talk it out with someone who knows logic. James, you are a presumptuous twit and a silly man. Just this once, I will mention my background. I don't care to play compare-the-credentials, but your insult is remarkably misplaced. I have an oversized diploma saying that I've received a PhD in Logic, Computability and Methodology (from DeVry Business School and Logic Academy). I know that my background doesn't compare to yours. You took an undergraduate course in logic, huh? Wow. -- Jesse Hughes Certainly he who can digest a second or third ßuxion need not, methinks, be squeamish about any point in divinity. George Berkeley, 1734 === Subject: Re: JSH: Assocation does not prove > > Comparisons are ripe for the logical error Cum Hoc, Ergo Propter > Hoc. > > No, James, comparisons aren't how the fallacy occurs. Causal > explanations are the setting for this fallacy. Mathematical proofs > are not causal explanations. > > Oh you mean that math proofs don't follow from logical reasons? > No, I mean what I said. Mathematical proofs are not causal > explanations. > Remember though cause and effect also have to do with time in your > sorry little life, ultimately the idea behind cause and effect is that > there is a *reason* for why things happen. > > So move beyond time and you can say that cause and effect has to do > with the reasons why starting from point A you end up at point B then > point C. > > Like in a mathematical proof you begin with a truth and proceed by > logical steps to a conclusion that then MUST be true. > > The effect, a true conclusion, follows from the cause, a perfect > logical path from truth to truth to final truth. > You're babbling, son. Mathematical justifications are not causal > explanations in the sense relevant to Cum Hoc. >You're being obstinate. Read what I said again. >Carefully this time. >Better yet, go try your argument on someone else, not online or in a >post, just talk it out with someone who knows logic. you really must have no idea, even after all these years, what sort of impression it makes when somone points out how you're simply wrong about something and you reply this way. >You need to understand what the logical fallacy is about, and I've >given you all the information you need, like this subject line says a >lot. >But if that's not enough, you need to go out, do some research and >talk it out with others versus being hard-headed and relying on your >own understanding. >James Harris ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: JSH: Assocation does not prove <87zn4hyvf6.fsf@phiwumbda.org> <87vff4stop.fsf@phiwumbda.org> Discussion, linux) > Knowing about logic is not the same as being in touch with reality. My ears are burning. -- Jesse F. Hughes I can't tell you how many times she left me. I lost count the very first time that she did. -- The Flatlanders, I Thought the Wreck Was Over === Subject: Re: JSH: Assocation does not prove <87zn4hyvf6.fsf@phiwumbda.org> <87vff4stop.fsf@phiwumbda.org> <874qmozapc.fsf@phiwumbda.org> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Knowing about logic is not the same as being in touch with reality. > My ears are burning. Vulcans. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Assocation does not prove > It turns out that I wondered a bit about Wiles's approach when I first > heard about it, as yes, I actually did take a course in logic while in > college! > One thing that is telling is that mathematicians don't have a ready > answer for the charge that Wiles approach fails by Cum Hoc, Ergo > Propter Hoc, and my guess is that most mathematicians don't bother > taking logic courses in college. [More raving snipped] Uh-oh, methinks here comes James' next obsession... I gather the progression goes something like this: 1) James makes an outrageous and fallacious claim. 2) Sci.Math posters scoff and demand proof. 3) James provides incoherent and/or false mathematical arguments. 4) Sci.Math posters repeatedly and clearly refute these arguments. 5) Iterate between 3) and 4) for a while as James insults and whines. 6) James raves like a lunatic and talks about conspiracies. 7) James eventually gives up, finds a new obsession, and goes back to 1). And as always, vindication for James is on the horizon (i.e. The Hammer), but it never seems to occur. his simple proof of FLT, followed by the core error in mathematics, and the significance of his prime counting algorithm. Am I missing anything? Now we have a claim that Wiles' proof of FLT fails, but unlike his previous delusions, it's not backed-up by any math at all -- just a lot more lunatic raving. Come on James, let's see some irrelevant, incoherent, and incorrect algebra! Spooner === Subject: Re: JSH: Assocation does not prove ... > Come on James, let's see some irrelevant, incoherent, and incorrect > algebra! Yeah, bring it on! === Subject: The mathematics of Triangulation Ok, I have a question (actually several questions) about triangulation. First, when I say triangulation I mean determining the location of a source with three receivers at known points. I once saw on a TV show a system being used to determine the location of a gun shot in cities with a lot of crime. It used three microphones that were positioned around the city that would pick up a gunshot and triangulate the location of the gun shot and report it to the police. Now at first impression I thought that this is simply the same idea used in GPS, but I later realized it's different. In a GPS system satellites emit a signal at accurate times. These signals are picked up by a receiver and the location of the receiver is determined. But in the gunshot scenario its opposite, the source is the single point and the receiver is the three known points. Now, for my real question: In the gunshot scenario, the receivers do not know the time the gunshot occurred (this of course is impossible). The three receivers ONLY now which receiver heard the shot first, second, and third and the times it took in between. How can the location be determined by this info alone? For a simplified example, let's say that the gunshot is equidistant from two receivers. The shot would be heard by the two receivers at the same time like in the diagram below (receiver 1 and 2 hear at the same time). (Gunshot) (reciever1) (reciever2) (reciever3) The only other information that is known is the additional time it took for receiver 3 to hear the shot. BUT wouldn't this be the same whether the gunshot occurred at its current position or any distance behind (or in front of) its current position since the space between the first two receivers and the third stays the same, that is to say, the delay will stay the same? so and all elaborate. === Subject: Re: The mathematics of Triangulation > Ok, I have a question (actually several questions) about > triangulation. First, when I say triangulation I mean determining the > location of a source with three receivers at known points. > I once saw on a TV show a system being used to determine the > location of a gun shot in cities with a lot of crime. It used three > microphones that were positioned around the city that would pick up a > gunshot and triangulate the location of the gun shot and report it to > the police. > Now at first impression I thought that this is simply the same > idea used in GPS, but I later realized it's different. In a GPS system > satellites emit a signal at accurate times. These signals are picked > up by a receiver and the location of the receiver is determined. But > in the gunshot scenario its opposite, the source is the single point > and the receiver is the three known points. > Now, for my real question: In the gunshot scenario, the receivers > do not know the time the gunshot occurred (this of course is > impossible). The three receivers ONLY now which receiver heard the > shot first, second, and third and the times it took in between. How > can the location be determined by this info alone? > For a simplified example, let's say that the gunshot is > equidistant from two receivers. The shot would be heard by the two > receivers at the same time like in the diagram below (receiver 1 and 2 > hear at the same time). > (Gunshot) > (reciever1) (reciever2) > > > (reciever3) > The only other information that is known is the additional time > it took for receiver 3 to hear the shot. BUT wouldn't this be the same > whether the gunshot occurred at its current position or any distance > behind (or in front of) its current position since the space between > the first two receivers and the third stays the same, that is to say, > the delay will stay the same? > so and all elaborate. Okay, I apologize before hand for bringing in co-ordinates when I'm sure with a little more thought and a little more sleep I could have found a way to solve the problem without them. In your city, let the radios that heard the gunshot be A, B, C, in that order. Apply a co-ordinate plane (rectangular/Cartesian/the regular one) to your city. You know the co-ordinates of radio A are (x_A, y_A), radio B's are (x_B, y_B), and radio C's are (x_C, y_C). You know all of these co-ordinates because you know in your city where the radios are. Then, radio A hears a gunshot. You don't know how far the gunshot is, so you let the variable D_A represent how far the sound is (in seconds). Then, radio B hears the gunshot. You know that it heard the gunshot b seconds after radio A heard it, so D_B = D_A + b Then, radio C hears the gunshot. You know that it heard the gunshot c seconds after radio A heard it, so D_C = D_A + c Thus, you know that the shot's location (x, y) is one of the points D_A seconds from (x_A, y_A). (x, y) is one of the points (D_A + b) seconds from (x_B, y_B). And you know (x, y) is one of the points (D_A + c) seconds from (x_C, y_C). So, algebraically, you have: (x - x_A)^2 + (y - y_A)^2 = (D_A)^2 (x - x_B)^2 + (y - y_B)^2 = (D_A + b)^2 (x - x_C)^2 + (y - y_C)^2 = (D_A + c)^2 There are three equations, and three variables. The solution (x, y) will give you the location of the gunshot. === Subject: Re: The mathematics of Triangulation > For a simplified example, let's say that the gunshot is >equidistant from two receivers. The shot would be heard by the two >receivers at the same time like in the diagram below (receiver 1 and 2 >hear at the same time). > (Gunshot) >(reciever1) (reciever2) > > > (reciever3) > The only other information that is known is the additional time >it took for receiver 3 to hear the shot. BUT wouldn't this be the same >whether the gunshot occurred at its current position or any distance >behind (or in front of) its current position since the space between >the first two receivers and the third stays the same, that is to say, >the delay will stay the same? No, not unless the distance between receiver1 and receiver2 is very small. plane equidistant from receiver1 and receiver2. Let's say that the distance between the receivers is 2*a (they are arranged in an equilateral trangle). You know the time-lag for receiver3, and using weather conditions to determine the speed of sound, you can convert that into a distance difference, d. Call the distance from the gunshot to receiver3 b. If you assume all points are in a ßat plane, you can use the Pythagorean theorem to give the equation: (b-d)^2 = a^2 + (b - sqrt(3)/2 * a)^2 and then you solve for b. Of course, if the points aren't lined up nicely it's a more complicated system of equations, and you can only do it with 3 mics if you make the assumption that the shot is coming from the surface. For 3-D you need 4 mics, I think. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: The mathematics of Triangulation > Ok, I have a question (actually several questions) about > triangulation. First, when I say triangulation I mean determining the > location of a source with three receivers at known points. > I once saw on a TV show a system being used to determine the > location of a gun shot in cities with a lot of crime. It used three > microphones that were positioned around the city that would pick up a > gunshot and triangulate the location of the gun shot and report it to > the police. > Now at first impression I thought that this is simply the same > idea used in GPS, but I later realized it's different. In a GPS system > satellites emit a signal at accurate times. These signals are picked > up by a receiver and the location of the receiver is determined. But > in the gunshot scenario its opposite, the source is the single point > and the receiver is the three known points. > Now, for my real question: In the gunshot scenario, the receivers > do not know the time the gunshot occurred (this of course is > impossible). The three receivers ONLY now which receiver heard the > shot first, second, and third and the times it took in between. How > can the location be determined by this info alone? > For a simplified example, let's say that the gunshot is > equidistant from two receivers. The shot would be heard by the two > receivers at the same time like in the diagram below (receiver 1 and 2 > hear at the same time). > (Gunshot) > (reciever1) (reciever2) > (reciever3) > The only other information that is known is the additional time > it took for receiver 3 to hear the shot. BUT wouldn't this be the same > whether the gunshot occurred at its current position or any distance > behind (or in front of) its current position since the space between > the first two receivers and the third stays the same, that is to say, > the delay will stay the same? > so and all elaborate. One ends up with delta time measurements 2 sensors, one delta t, 3 sensors, 3 delta ts 4 sensors, 6 delta ts and so on the solution sets are hyperbolic sets of lines for each two points where they intersect are possable solutions. the case you mention has one delta t zeroed out, so one needs another sensor. Checkout LORAN C too. Similar timing === Subject: Does commutativity imply associativity? I know that there are mathematical objects which are associative but not commutative with respect to some binary operation. Matrix multiplication is an example, associative but not commutative. My question is: Is there some set of objects which is commutative but not associative with respect to some binary operation, or does commutativity imply associativity? === Subject: Re: Does commutativity imply associativity? >I know that there are mathematical objects which are associative but >not commutative with respect to some binary operation. Matrix >multiplication is an example, associative but not commutative. >My question is: Is there some set of objects which is commutative but >not associative with respect to some binary operation, or does >commutativity imply associativity? A classical example arises in Jordan algebras: you might for example define a new produce & on the set of square matrices of a given size by A & B = (AB + BA) , which is clearly commutative but not associative (try e.g. A & ( A & B ) and ( A & A ) & B ). dave === Subject: Re: Does commutativity imply associativity? > I know that there are mathematical objects which are associative but > not commutative with respect to some binary operation. Matrix > multiplication is an example, associative but not commutative. > My question is: Is there some set of objects which is commutative but > not associative with respect to some binary operation, or does > commutativity imply associativity? Take any non-commutative associative ring and redefine multiplication by x*y = xy+yx. This gives a commutative, non-associative (with trivial exceptions) ring, called a Jordan ring and there is a large theory of such rings. === Subject: Re: Does commutativity imply associativity? > I know that there are mathematical objects which are associative but > not commutative with respect to some binary operation. Matrix > multiplication is an example, associative but not commutative. > My question is: Is there some set of objects which is commutative but > not associative with respect to some binary operation, or does > commutativity imply associativity? Practically anything you can think of. Lets define x % y = 2x + 2y . It's commutative, right? But is it associative? x % (y % z) = 2x + 4y + 4z , (x % y) % z = 4x + 4y + 2z , not equal in general. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Does commutativity imply associativity? days. My association with the Department is that of an alumnus. >I know that there are mathematical objects which are associative but >not commutative with respect to some binary operation. Matrix >multiplication is an example, associative but not commutative. >My question is: Is there some set of objects which is commutative but >not associative with respect to some binary operation, or does >commutativity imply associativity? You can easily construct one. Take the following operation on the set of positive integers: a*b = max{a,b} + 10. The operation is clearly commutative. But (25*17)*10 = (35)*10 = 45 25*(17*10) = 25*(27) = 37. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Null test, Wiles's work jstevh@msn.com > Here's an attempt at getting some interest by suggesting a null test > of Wiles's work. > > Why would you think about a null test? That is more related to > statistics, > but the mathematics going on here are *not* statistics, the show (or > attempt to show) an absolute truth. > > Basically someone needs to post how Wiles's approach could have shown > a result *opposite* to what many think it did. > > I think what James is looking for is that Wiles proof is not only consistent > with FLT being true but also FLT being false. Since everyone here accepts > Wiles proof, it is up to James to show that the assumption of a solution to > FLT along with Wiles' logic leads to a nonmodular elliptic curve. > Actually what I'm looking for is not complicated. > What I'm saying is, assume that a non-modular elliptic curve DOES > exist, could Wiles's approach have shown that result? A null test is meaningless. Mensantor's Theorem: all Mersenne numbers with an even number of bits are divisible by 3 (i.e., == 0 (mod 3)) Yes, it is proved by induction, so it is a theorem. But let's apply the null test: there is an even bit Mersenne number == 2 (mod 3). The successor rule for a Mersenne number m is: succ(m) -> 2*m + 1 2 * [2 (mod 3)] = 1 (mod 3) 1 + [1 (mod 3)] = 2 (mod 3) therefore, succ(2 (mod 3)) -> 2 (mod 3) The null test shows that _every_ Mersenne number is == 2 (mod 3). This is trivially false as none of 1, 3, 7, 15, 31, ... are == 2 (mod 3). And the reason it's false is I left out that all important induction step of demonstrating that it is true for the initial value upon which all successors depend. In my original proof, I demonstrated that, for 2*m + 1 succ(0 (mod 3)) -> 1 (mod 3) succ(1 (mod 3)) -> 0 (mod 3) and the intial value is 1 (mod 3). So all Meresenne numbers alternate between 0 (mod 3) and 1 (mod 3) and since the first number (1) has an odd bit count, all the even bit count numbers are 0 (mod 3), i.e., divisible by 3. QED. That is sufficient proof. Although it may be interesting as a curiosity, the null hypothesis is completely irrelevant to the proof. Why don't you try proving the Collatz Conjecture? That way _I_ could argue actual math with you. This crap you're into is of no interest whatsoever. > That is, the point of the null test is to see if inherent in his > approach is the conclusion he desired such that the truth is > irrelevant to it. > That is, if it were in fact true that a non-modular elliptic curve > does exist, would the approach that Wiles took show it? > What I'm trying to get you to do is drop your assumption that Wiles is > correct and take a *critical* eye to his approach! > The difficulty with even communicating that idea shows just how deeply > the belief is embedded. > Remember, math proofs don't need your allegiance, don't need you to > protect them, and don't need you to worry over them if challenged. > They are perfect, inviolable, and absolute. > A math proof resists all challenges, and can handle a null test!!! > I want some of you to turn into critics versus true believers and ask > the simple question, if a non-modular elliptic curve actually did > exist would Wiles's approach have shown it? > James Harris === Subject: Re: Null test, Wiles's work days. My association with the Department is that of an alumnus. [.snip.] > Actually what I'm looking for is not complicated. > What I'm saying is, assume that a non-modular elliptic curve DOES > exist, could Wiles's approach have shown that result? >A null test is meaningless. Especially when we drop assumptions. Wiles's work shows that ->semistable<- elliptic curves are modular. Where is the problem if we assume a semistable elliptic curve is not modular? I'm using the copy of the paper at www.eleves.ens.fr/home/rrichard/wiles.pdf You will find that after the introduction is complete, the first appearance of the phrase elliptic curve is on pp. 87 of the manuscript., where it is mentioned only that a particular grossencharacter is associated to a unique elliptic curve. After that, the next occurrence is on pp. 94, again mentioning the existence of an elliptic curve once you know what the grossencharacter is. After that, it only appears again on pp. 100, once we are in the very short section 5. Before this, there is nothing about elliptic curves, modular or otherwise, but in fact about representations of the absolute Galois group. Thus the first theorem which mentions elliptic curves in its statement is Theorem 5.2, on pp. 100 of the file (pp. 542 of the journal). No assumption there is made of modularity. The next appearance on a theorem is Theorem 5.3 two pages later. Again, no assumption of modularity is made. And that's the end of the paper prior to the appendices, which begin on pp. 545. Pick ANY semistable elliptic curve over Q, and call it E; if you insist, assume for the sake of argument it is not modular. The fact that you have an elliptic curve over Q (and only that fact) gives you automatically a representation rho of the absolute Galois group of Q on the group of 3-torsion of the points over the algebraic closure of Q. We have two possibilities: either the representation is irreducible, or else it is not irreducible. If it is irreducible, then it must in fact be absolutely irreducible, or else there would exist a nontrivial abelian extension of Q(sqrt(-3)) unramified outside 3 and of order prime to 3; but no such extension exists. So if the representation rho is irreducible then it is absolutely irreducible. Then Theorem 0.2 implies that the representation comes from a modular form. Serre's Isogeny Theorem guarantees that if the representation we obtain in this manner from E comes from a modular form, then E itself will be a modular curve; so if we assume that E is not modular, we may conclude that the representation we obtained cannot be irreducible. So the representation is reducible. In that case, the claim is that in fact if instead of looking at the representation induced by the action on the 3-torsion we look at the representation induced by the 5-torsion, then the resulting representation will also be reducible. This follows from a known result. Now, this representation either comes from a modular form or not. If it does, then again Serre's Isogeny Theorem would guarantee that E is modular, so we must conclude that this representation does NOT come from a modular form, given our hypothesis. Because of Theorem 0.2, that means that if we restrict it to the absolute Galois group of Q(sqrt(5)), this result cannot be absolutely irreducible. This in turn means that the representation we obtain this way must be an induced representation over Q(sqrt(5)) (that's the only way it can fail to be absolutely irreducible). That means that there is a point in the corresponding curve of moduli which codifies this information; and that point allows the construction of an elliptic curve E' which induces the same representation in the 5 torsion as E does. Because of the way E' is obtained, the representation induced by the 5-torsion of E' ->does<- satisfy the hypothesis of Theorem 0.2, and so the induced representation comes from a modular form. (So E' is modular). However, the construction of E' was such that the induced representation by the 5-torsion of E' was the same as by the five torsion of E as representations of the absolute Galois group. And our hypothesis was the the representation induced by the 5-torsion of E does NOT come from a modular form, and we have now concluded that, in fact, it ->does<-. This contradiction arises from our assumption that E is not modular. If we were not assuming, at this point we would be able to invoke Serre's Isogeny Theorem again to deduce, from the fact that the representation induced by the 5-torsion of E comes form a modular form, that E itself is modular. The approach would have revealed the lack of modularity when we analyze the representation induced by the 5-torsion; it would then be impossible to construct the curve E' with the right properties. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Does commutativity imply associativity? Of course, there is such sets. Suppose the following algebraic structure: <{a, b, c}, +> with the operation + define as: a+a=a a+b=c a+c=b b+a=c b+b=b b+c=a c+a=b c+b=a c+c=c Clearly, the operation + is commutative but not associative: (a+b)+c = c+c = c a+(b+c) = a+a = a === Subject: Re: What is the expert opinion on deriving SR with a Shubertian clock? > : > : > : > : The two are totally incompatible as even a passing acquaintance > : > : with algebra would show > : > > : > Please explain why persons who claim to be able to read Landau > : > and Lifz can't identify a specific algebraic error in my > : > opening post or full-length paper? > : > > : > http://www.everythingimportant.org/relativity/ > : Your numerous errors have been explaned counless times eg you try > : and pull a shonky by not defining an inerial reference frmae and > : leaving such up in the air. > The paper http://www.everythingimportant.org/relativity has been > greatly simplified to correct those countless misperceptions. What misconceptions? Did a scan on the word inertial - it is used once and not defined. > What's a shonky? Not defining terms properly. > Why must I use the words, inertial reference frame? Do those > words have magical power? Because by definition that is what SR is about. Why use the words line and points in Euclidian geometry? Why define what they mean? > Why is it not sufficient to ask the reader > to imagine two pristine, frictionless rulers and to imagine one of > them sliding on another with constant velocity? Because the rulers need to be contained in something and by definition that is a frame. If you wish to discuss SR learn what SR is. Bill Why are you so upset > with my attempt to explain special relativity so that algebra students > can understand it? > : > : > I don't understand your gripe against mathematics. All > : > : > mathematicians make up definitions when it's required to > : > : > do so. > : > : > > : > : They do not make them up after standard definitions have been > : > : agreed eg they do not decide to call Euclidian geometry the > : > : geometry of a sphere after it is well established what > : > : Euclidian geometry means. > : > > : > Where are the conßicts and contradictions between standard > : > definitions and my definitions in the opening post and full-length > : > paper? > : As explained above you try an pull a shonky by not making it clear > : SR deals with inertial reference frames and defining what they are. > It sounds to me that you're angry that I'm not teaching others to > parrot conventional ideas and phraseology. Again, what's a shonky? > : Your claim ÔEducated physicists have testified of the genius > : in deriving the Lorentz transformation from the Galilean > : transformation' is the ratings of an obvious brain dead idiot > Is Tom Snyder a brain dead idiot? > Tom Snyder has affirmed my presumably impossible, yet apparently > magical derivation of the Lorentz transformation from the Galilean > transformation: > http://www.everythingimportant.org/relativity > You need to confront your hypocrisy and at least be consistent and > start condemning people who agree with me as brain dead idiots. > I've said very clearly that: > I have discovered a great way to teach special relativity: > http://www.everythingimportant.org/relativity. What makes > my approach intuitive, easy and interesting is that I derive the > Lorentz transformation from the Galilean transformation using > magic, elementary algebraic manipulation and a clear definition > of time. > Tom Snyder responded favorably and agreeably: > I enjoyed your derivation. It reminded me of some comments that Pauli > made in his book _Theory of Relativity_ (1921), page 11. > Tom Snyder then goes on to compare my axioms, point by point, with the > assumptions of Ignatowsky, Frank and Rothe. Those assumptions are > listed in Pauli's book. Tom Snyder's endorsement is that he makes > explicit reference to my use of the Galilean transformation > T=T'=(x-x')/u and compares it to Assumption (c): > Assumption (c) of Pauli is similar to your taking the expressions for > T and T' [given just above your statement of Theorem 1] as identical > to one another with the same denominator [mu]. > (c) the contraction of lengths at rest in K' and observed in K is > equal to that of lengths at rest in K and observed in K'. > Tom Snyder isn't suggesting that my derivation is asinine, ludicrous > or impossible. He is clearly supporting the reasonableness of my > derivation and my use of the Galilean transformation in deriving the > Lorentz transformation. > : a simple glance at the Lorentz transformations show the Galilean > : transformations result if c is infinity. A quantity can not be > : infinity and finite at the same time. > Obviously true but very irrelevant. > Everyone understands that magicians really don't saw women in half. > It only appears that they do to a great majority. Likewise, I'm only > claiming to do magic beyond your power of understanding. > Eugene Shubert > http://www.everythingimportant.org/relativity === Subject: Re: What is the expert opinion on deriving SR with a Shubertian clock? > : Your numerous errors have been explaned counless times > The paper http://www.everythingimportant.org/relativity has been > greatly simplified to correct those countless misperceptions. > : eg you try and pull a shonky by not defining an inerial reference > : frmae and leaving such up in the air. As I've pointed out, the meaning is clear from the context: http://www.everythingimportant.org/relativity > Did a scan on the word inertial - it is used once and not defined. I used the word inertial in the context of frictionless sliding rulers. Use a dictionary to learn what the word inertial means. http://dictionary.com says: inertial adj : of or relating to inertia inertia n. 1. Physics. The tendency of a body to resist acceleration; the tendency of a body at rest to remain at rest or of a body in straight line motion to stay in motion in a straight line unless acted on by an outside force. > What's a shonky? > Not defining terms properly. Then you're a shonky writer because the words you use (e.g. shonky) aren't even in the dictionary. > Why is it not sufficient to ask the reader to imagine two pristine, > frictionless rulers and to imagine one of them sliding on another > with constant velocity? > Because the rulers need to be contained in something and by definition > that is a frame. If you wish to discuss SR learn what SR is. Eugene Shubert http://www.everythingimportant.org/relativity === Subject: Re: What is the expert opinion on deriving SR with a Shubertian clock? > THis is something of an aside. but I am not convinced that an inertial > frame is a terribly well-defined concept. > I think you mean ill defined or not well defined. And yes would seem it is > not possible to completely define operationally an inertial frame (I am open > to be proven wrong here) - the best probably is an infinitesimal freely > falling frame. But the logical structure of SR does not depend on such. Well, apparently you know what I meant :). The problem with freely falling strikes me as the same as the one whether a coordinate system is freely falling if I wanted to use this to define something else on top of it. But how would I test that? I undertsand that I can always compare my motion to some very distant fixed-star frame but that would put me squarely into the realm of a global (rather than a local) concept. >I mean I understand what it > *is*, but spelling it out appears rather difficult. As an example let > me use this one: > It usually > the frame. In the coordinate system in which the moon is at rest, the moon moves as if it were the only thing in the universe (namely not at all). It course the reference fram in which the moon is at rest is not an inertial system. I'm posting all this, since I have found that the definition of an inertial reference frame is about the hardest part of special relativity. Everything else is just a bit of straightforward math. > Which of the many dozen books that carry Mr Landau's name are you > referring to here? > Page 5 Landau - Mechanics. === Subject: Re: What is the expert opinion on deriving SR with a Shubertian clock? > THis is something of an aside. but I am not convinced that an inertial > frame is a terribly well-defined concept. > > I think you mean ill defined or not well defined. And yes would seem it is > not possible to completely define operationally an inertial frame (I am open > to be proven wrong here) - the best probably is an infinitesimal freely > falling frame. But the logical structure of SR does not depend on such. > Well, apparently you know what I meant :). Of course - just being pedantic :). When I reread it I laughed and thought it was not really required. > The problem with freely falling strikes me as the same as the one > whether a coordinate system is freely falling if I wanted to use this > to define something else on top of it. But how would I test that? I > undertsand that I can always compare my motion to some very distant > fixed-star frame but that would put me squarely into the realm of a > global (rather than a local) concept. I said it was the best I could think of - not it was perfect. As I stated I think it is operationally very difficult, perhaps even impossible, to operationally define an inertial frame. But one can conceptually define it similar to conceptually defining lines and points in Euclidian geometry. >I mean I understand what it > *is*, but spelling it out appears rather difficult. As an example let > me use this one: > > It usually > in > the frame. > In the coordinate system in which the moon is at rest, the moon moves > as if it were the only thing in the universe (namely not at all). But such a frame is not inertial. > course the reference fram in which the moon is at rest is not an > inertial system. As I mentioned above before reading on. > I'm posting all this, since I have found that the definition of an > inertial reference frame is about the hardest part of special > relativity. Everything else is just a bit of straightforward math. Ok here is my take. A frame of reference is a conceptual standard of rest against which experiments can be conducted. An inertial frame is one in which space and time are homogeneous and space is isotropic and stationary lines and points obey Euclid's axioms. Thus we can construct a Cartesian coordinate system. Sync clocks by say having two guns fire bullets from midway between the clocks and set them to zero when the clock is hit by the bullet. By homogeneity and isotropy if the guns are exactly the same the bullets have exactly the same transit time and hence the clocks are synced. By homogeneity identical clocks must remain in sync. Imagine such a conceptual clock at each point. A space-time event is something that occurs at a conceptual clock thus is assigned a position (the position of the clock) and a time (the time read by the clock). By homogeneity it can be shown that the transformation between the same space-time events recorded in different inertial frames must be linear and hence the frames must be moving at constant velocity wrt each other. The POR now comes into play - the laws of physics are the same in all inertial frames and ones moving at constant such a frame - by definition it is free. Consider an infinitesimal time such that it has constant velocity during that time. Go to a frame where it is instantaneously at rest. By isotropy it must remain at rest. Thus free one can derive the Lorentz transformations eg http://arxiv.org/abs/physics/0110076. Bill > Which of the many dozen books that carry Mr Landau's name are you > referring to here? > Page 5 Landau - Mechanics. === Subject: Re: What is the expert opinion on deriving SR with a Shubertian clock? > THis is something of an aside. but I am not convinced that an inertial > frame is a terribly well-defined concept. > > > I think you mean ill defined or not well defined. And yes would seem it > is > not possible to completely define operationally an inertial frame (I am > open > to be proven wrong here) - the best probably is an infinitesimal freely > falling frame. But the logical structure of SR does not depend on such. > Well, apparently you know what I meant :). > Of course - just being pedantic :). When I reread it I laughed and thought > it was not really required. > The problem with freely falling strikes me as the same as the one > whether a coordinate system is freely falling if I wanted to use this > to define something else on top of it. But how would I test that? I > undertsand that I can always compare my motion to some very distant > fixed-star frame but that would put me squarely into the realm of a > global (rather than a local) concept. > I said it was the best I could think of - not it was perfect. As I stated I > think it is operationally very difficult, perhaps even impossible, to > operationally define an inertial frame. But one can conceptually define it > similar to conceptually defining lines and points in Euclidian geometry. >I mean I understand what it > *is*, but spelling it out appears rather difficult. As an example let > me use this one: > > It usually velocity > > > in > the frame. > In the coordinate system in which the moon is at rest, the moon moves > as if it were the only thing in the universe (namely not at all). > But such a frame is not inertial. > course the reference fram in which the moon is at rest is not an > inertial system. > As I mentioned above before reading on. > I'm posting all this, since I have found that the definition of an > inertial reference frame is about the hardest part of special > relativity. Everything else is just a bit of straightforward math. > Ok here is my take. > A frame of reference is a conceptual standard of rest against which > experiments can be conducted. An inertial frame is one in which space and > time are homogeneous and space is isotropic and stationary lines and points > obey Euclid's axioms. Thus we can construct a Cartesian coordinate system. > Sync clocks by say having two guns fire bullets from midway between the > clocks and set them to zero when the clock is hit by the bullet. By > homogeneity and isotropy if the guns are exactly the same the bullets have > exactly the same transit time and hence the clocks are synced. By > homogeneity identical clocks must remain in sync. I want to elaborate this a bit more. First, my mistake, one requires both homogeneity and isotropy. Suppose we have two clocks with one at the origin. Suppose the other clock (clock 2 say) gets out of sync by delta t ie clock 1 - clock 2 = delta t. Suppose we orientate the coordinate system so that the other clock is as the origin and we have exactly the same experiment with the roles of the clocks reversed (this will require rotating the axis hence the need for isotropy) we have exactly the same experiment so they must get out of sync by delta t ie clock 2 - clock 1 = delta t so delta t = 0. Bill > Imagine such a conceptual > clock at each point. A space-time event is something that occurs at a > conceptual clock thus is assigned a position (the position of the clock) and > a time (the time read by the clock). By homogeneity it can be shown that > the transformation between the same space-time events recorded in different > inertial frames must be linear and hence the frames must be moving at > constant velocity wrt each other. The POR now comes into play - the laws of > physics are the same in all inertial frames and ones moving at constant is > such a frame - by definition it is free. Consider an infinitesimal time > such that it has constant velocity during that time. Go to a frame where it > is instantaneously at rest. By isotropy it must remain at rest. Thus free > one can derive the Lorentz transformations eg > http://arxiv.org/abs/physics/0110076. > Bill > Which of the many dozen books that carry Mr Landau's name are you > referring to here? > > Page 5 Landau - Mechanics. > === Subject: Explaining the foundations of math Sometimes I think that profs and text writers fail students. I recall that I was taught about the epsilon-delta formulation of limits and proofs with the attitude of we do this because this is how one does it, with no explanation of why it is important, the logical problems it solves, the importance of putting things on a firm logical foundation. For example, Weierstrass's main lasting contribution to math was to define the limit of f(x) at x_o as follows: For any e > 0 there exists d_o such that if 0 < d < d_o, then | f(x_o +/- d) - L | < e, then L = lim(f(x)) as x --> x_o. So what? The point is that there is no mention of any infinitely small quantities, differentials (or Newton's ßuxions, etc.), only real numbers R with + and < (order). Van === Subject: Re: Explaining the foundations of math > Sometimes I think that profs and text writers fail students. > I recall that I was taught about the epsilon-delta formulation > of limits and proofs with the attitude of we do this because > this is how one does it, with no explanation of why it > is important, the logical problems it solves, the importance > of putting things on a firm logical foundation. > For example, Weierstrass's main lasting contribution to math > was to define the limit of f(x) at x_o as follows: > For any e > 0 there exists d_o such that if 0 < d < d_o, then > | f(x_o +/- d) - L | < e, then L = lim(f(x)) as x --> x_o. Huh? Oh, an interesting variant of L = lim(x->a) f(x) when for all e > 0, some d with for all x, 0 < |x - a| < d ==> |f(x) - L| < e > So what? The point is that there is no mention of any > infinitely small quantities, differentials (or Newton's ßuxions, > etc.), only real numbers R with + and < (order). Oh, that's where he's coming from, which explains the unusual accent or antique expression of his definition. Hurray Weierstrass! === Subject: Re: Explaining the foundations of math > Sometimes I think that profs and text writers fail students. > I recall that I was taught about the epsilon-delta formulation > of limits and proofs with the attitude of we do this because > this is how one does it, with no explanation of why it > is important, the logical problems it solves, the importance > of putting things on a firm logical foundation. > For example, Weierstrass's main lasting contribution to math > was to define the limit of f(x) at x_o as follows: > For any e > 0 there exists d_o such that if 0 < d < d_o, then > | f(x_o +/- d) - L | < e, then L = lim(f(x)) as x --> x_o. > So what? The point is that there is no mention of any > infinitely small quantities, differentials (or Newton's ßuxions, > etc.), only real numbers R with + and < (order). I don't disagree with your statement of the point. However, it is also important that the definition uses | |, a notion of a metric. Limits can be defined in terms of topology; they don't need a metrical notion. I don't know if it is a failure of teaching not to start at the more general topological definition. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: Explaining the foundations of math >Sometimes I think that profs and text writers fail students. >I recall that I was taught about the epsilon-delta formulation >of limits and proofs with the attitude of we do this because >this is how one does it, with no explanation of why it >is important, the logical problems it solves, the importance >of putting things on a firm logical foundation. >For example, Weierstrass's main lasting contribution to math >was to define the limit of f(x) at x_o as follows: >For any e > 0 there exists d_o such that if 0 < d < d_o, then >| f(x_o +/- d) - L | < e, then L = lim(f(x)) as x --> x_o. >So what? The point is that there is no mention of any >infinitely small quantities, differentials (or Newton's ßuxions, >etc.), only real numbers R with + and < (order). > I don't disagree with your statement of the point. However, it > is also important that the definition uses | |, a notion of a > metric. Limits can be defined in terms of topology; they don't > need a metrical notion. I don't know if it is a failure of > teaching not to start at the more general topological > definition. While the more general one is nice in many ways, even the standard epsilon-delta definition is often a bit abstract for Calc I students. -Ron === Subject: Re: Explaining the foundations of math >Sometimes I think that profs and text writers fail students. >I recall that I was taught about the epsilon-delta formulation >of limits and proofs with the attitude of we do this because >this is how one does it, with no explanation of why it >is important, the logical problems it solves, the importance >of putting things on a firm logical foundation. > >For example, Weierstrass's main lasting contribution to math >was to define the limit of f(x) at x_o as follows: > >For any e > 0 there exists d_o such that if 0 < d < d_o, then >| f(x_o +/- d) - L | < e, then L = lim(f(x)) as x --> x_o. > >So what? The point is that there is no mention of any >infinitely small quantities, differentials (or Newton's ßuxions, >etc.), only real numbers R with + and < (order). > I don't disagree with your statement of the point. However, it > is also important that the definition uses | |, a notion of a > metric. Limits can be defined in terms of topology; they don't > need a metrical notion. I don't know if it is a failure of > teaching not to start at the more general topological > definition. > While the more general one is nice in many ways, even the standard > epsilon-delta definition is often a bit abstract for Calc I students. > -Ron Whilst I find myself spectacularly unqualified to speak on the competency of Calc I students, the concept that you are measuring a length (metric) on a shape that is deformed continuously (ie a function) is really the only way to understand epsilon - delta, which to me seems abstact in comparison. === Subject: notation q. hello group, I was wondering, What does the following notation mean in study of networks? P(k) ~ k^(-gamma). which is from a paper, Topology of Evolving Networks: Local Events and Universality by R.8eka Albert and Albert-L.87szl.97 Barab.87si to giuve a context, following is an exerpt from that paper. ...recent results on the Internet topology [10] found that, independently of the nature of the system and the identity of its constituents, P(k) decays as a power law, following P(k) ~ k^(-gamma). These results offered the first evidence that some large networks can self-organize into a scale-free state, a feature unexpected by previous network models [6,7]... === Subject: Re: notation q. > hello group, > I was wondering, > What does the following notation mean in study of networks? > P(k) ~ k^(-gamma). > which is from a paper, Topology of Evolving Networks: Local Events and > Universality by R.8eka Albert and Albert-L.87szl.97 Barab.87si > to giuve a context, following is an exerpt from that paper. > ...recent results on the Internet topology [10] found that, > independently of the nature of the system and the identity of its > constituents, P(k) decays as a power law, following P(k) ~ k^(-gamma). > These results offered the first evidence that some large networks can > self-organize into a scale-free state, a feature unexpected by > previous network models [6,7]... I believe it means that P(k) is asymptotically proportional to k^(-gamma), which is to say that after a while, P(k) looks like c*k^(-gamma) for some constant c. === Subject: additive characters over finite fields I am trying to get to grips with characters over finite groups (Goodman & Wallach) Let F be a finite field of characteristic p, F has p^n elements. Let S^1 be the multiplicative group of complex nos. of absolute value 1. Chi : F -> S^1 is such that Chi(x+y) = Chi(x)Chi(y) (additive character) and Chi(0)=1. The smallest subfield of F, K is isomorphic with Zp. Define: nu(z)=exp(2.pi.i.z/p). Then nu defines an additive character of Zp and hence of K. If a (- F (here (- is for belongs to), then define a linear transformation : la : F -> F by la(x) = ax. Set Chi_1(a) = nu(Trace(la)) Show that if eta is an additive character of F then there exists a unique u such that eta(x) = Chi_1(x) for all x (- F. Is this merely a statement about the periodicity of eta ? Or something more. Also, does not a character in general depend on the representation chosen? === Subject: Differential equation question What is the differential equation whose solution is h(t) = (A/t) exp(B/t), where A and B are constants. I am not sure about the initial condition, probably h(0) = 0. Bora === Subject: Re: Differential equation question >What is the differential equation whose solution is >h(t) = (A/t) exp(B/t), where A and B are constants. I am not sure about the >initial condition, probably h(0) = 0. >Bora No initial condition is needed to answer the question and h(0) = 0 is obviously senseless === Subject: Re: Differential equation question >What is the differential equation whose solution is If you're asking the question you seem to be asking, the answer is to differentiate both sides of the solution and you'll have (one form of) the differential equation. Is that really what you mean? >h(t) = (A/t) exp(B/t), where A and B are constants. h'(t) = (A/t)' exp(B/t) + (A/t) [exp(B/t)]' h'(t) = -A * exp(B/t) / t^2 + A exp(B/t) * (-B/t^2) / t h'(t) = (-A/t^3) * exp(B/t) * (t - B) is one form of the differential equation whose solution is the h(t) function you gave. >I am not sure about the initial condition, probably h(0) = 0. Doesn't matter. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: Differential equation question >What is the differential equation whose solution is >h(t) = (A/t) exp(B/t), where A and B are constants. I am not sure about the >initial condition, probably h(0) = 0. I assume you mean A and B are arbitrary constants, so this is the general solution. It's probably best to start with g(x) = A x exp(B x). Then g'(x) = A (1 + B x) exp(B x), so g'(x)/g(x) = (1 + B x)/x, and B = (x g'/g - 1)/x = g'/g - 1/x. Thus 0 = B' = (g'/g - 1/x)' = (g g x^2 - (g')^2 x^2 + g^2)/(g^2 x^2) so a differential equation whose general solution is g(x) is g g x^2 - (g')^2 x^2 + g^2 = 0 Now apply the change of variables x = 1/t, and I get h h t^2 + 2 h' h t - t^2 (h')^2 + h^2 = 0 Since it's a second order DE, you need two initial conditions to specify A and B. However, you don't want an initial condition where either t = 0 or h(t) = 0, as the DE is singular there. ' === Subject: Re: Berechnung modulo groer Zahl > sie n=pq das Produkt zweier ungerader, groer Primzahlen. Sei a E Z_n^19. > Des weiteren sei folgender Wert gegeben: > b=a*n^18 mod n^19. > Ist es nun irgendwie m.9aglich aus dem gegebenen b den Wert a zu berechnen > ohne, dass ich alle Elemente von Z_n^19 durchprobieren muss? Ich wei, dass > ich es nicht direkt berechnen kann, da n^18 und n^19 nicht teilerfremd sind > und dementsprechend es kein Inverses zu n^18 gibt. Jedoch existiert > vielleicht irgendein Algorithmus, mit dem man soetwas effizient berechnen > kann wie z.B. bin.8are Suche oder etwas .8ahnliches? > a * n^18 is a multiple of n^18, and b = (a * n^18) mod n^19 is still a > multiple of n^18. Divide b/n^18, and you get b/n^18 = a (mod n). But a is an element of Z_n^19 and therefore it can originally be for example: a=23 + 1111*n^7, but if I calculate a mod n then I will get as result 23 and not the value 23 +1111*n^7. Or am I thinking wrong? Julia === Subject: Re: Berechnung modulo groer Zahl > sie n=pq das Produkt zweier ungerader, groer Primzahlen. Sei a E > Z_n^19. > Des weiteren sei folgender Wert gegeben: > b=a*n^18 mod n^19. > Ist es nun irgendwie m.9aglich aus dem gegebenen b den Wert a zu berechnen > ohne, dass ich alle Elemente von Z_n^19 durchprobieren muss? Ich wei, > dass > ich es nicht direkt berechnen kann, da n^18 und n^19 nicht teilerfremd > sind > und dementsprechend es kein Inverses zu n^18 gibt. Jedoch existiert > vielleicht irgendein Algorithmus, mit dem man soetwas effizient > berechnen > kann wie z.B. bin.8are Suche oder etwas .8ahnliches? > a * n^18 is a multiple of n^18, and b = (a * n^18) mod n^19 is still a > multiple of n^18. Divide b/n^18, and you get b/n^18 = a (mod n). > But a is an element of Z_n^19 and therefore it can originally be for > example: > a=23 + 1111*n^7, but if I calculate a mod n then I will get as result 23 and > not the value 23 +1111*n^7. Or am I thinking wrong? By multiplying by n^18, you just lost a lot of the available information. Take a = 23 + 1111*n^7. a*n^18 = 23*n^18 + 1111*n^25. b = (23*n^18 + 1111*n^25) mod n^19 = 23n^18 mod n^19. Given b, you can only determine the original value of a up to a multiple of n, the remaining information is lost. === Subject: =?iso-8859-1?q?Re:_Berechnung_modulo_gro=DFer_Zahl?= !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > sie n=pq das Produkt zweier ungerader, groer Primzahlen. Sei a E > Z_n^19. > Des weiteren sei folgender Wert gegeben: > b=a*n^18 mod n^19. > Ist es nun irgendwie m.9aglich aus dem gegebenen b den Wert a zu berechnen > ohne, dass ich alle Elemente von Z_n^19 durchprobieren muss? Ich wei, > dass > ich es nicht direkt berechnen kann, da n^18 und n^19 nicht teilerfremd > sind > und dementsprechend es kein Inverses zu n^18 gibt. Jedoch existiert > vielleicht irgendein Algorithmus, mit dem man soetwas effizient > berechnen > kann wie z.B. bin.8are Suche oder etwas .8ahnliches? > a * n^18 is a multiple of n^18, and b = (a * n^18) mod n^19 is still a > multiple of n^18. Divide b/n^18, and you get b/n^18 = a (mod n). > But a is an element of Z_n^19 and therefore it can originally be for > example: > a=23 + 1111*n^7, but if I calculate a mod n then I will get as result 23 and > not the value 23 +1111*n^7. Or am I thinking wrong? a will not be uniquely determined. Adding _any_ multiple of n to a will not change b. 1111*n^7 is just one example. 3*n would be another. You can't determine more than a (mod n). -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: lottery bet (combinatorics) It is easy to see that in my example there must be at least 5 lines. There are 84 different selections of 3 from 9, but each line of six numbers can only account for 20 selections, since that is the number of ways of selecting three from 6. So 84 divided by 20 is 4.2 which means that you must have 5 lines at least. === Subject: Re: lottery bet (combinatorics) I have discussed this kind of problem before. If I select lines of six, and I want to be guaranteed three from 9 numbers then the solution looks something like this. 123456 123457 123458 123459 456789 356789 256789 156789 Choose any three numbers between 1 and 9 and they will appear somewhere in these 8 lines. E.g. 379 in line 6. Proving that 8 is the minimal nbr of lines required is hard and was done for me by a professor, as I remember. === Subject: Re: lottery bet (combinatorics) >MOTIVATION >There is a lottery that chose 10 numbers (draw) from 70 numbers. >People can bet on their 10 numbers ticket (arbitrary numbers). How much >particular person wins depends on how many numbers are the same in the ticket >and in the lottery draw. >Now one wants to make him sure that if any 6 numbers from his favourite 20 >numbers will be in the draw he will have at least one ticket in which at least 3 >drawn numbers are. >Q1: He wants to make the cheapest bet how many 10 numbers ticket he should >bet? >Q2: Which numbers have to be written on the tickets? In this example, he buy 2 tickets with 10 numbers on each. If 6 numbers are winning, at least 3 must be on one ticket or the other. >For future purpose denote the numbers in above story as follows >70 N >10 T (ticket) >20 L (set of loved numbers) >06 C (loved numbers in draw) >03 M (guaranteed minimal win) >I want to know answer to Q1 and Q2 in general (for any numbers). N doesn't matter. If M/C <= T/L, you only need to cover each number once, so buy ceil(L/T) tickets. There are choose(L,C) ways to have C loved numbers, and each ticket full of those numbers covers choose(T,M) of those ways. So the answer to Q1 must always be at least (choose(L,C)/choose(T,M)). [choose(x,y) = x!/y!/(x-y)!] I'll leave Q2 alone for now. >in the next trivial examples the loved numbers are 1, 2, 3, , L >N = 10; L = 4; T = 3; C = 3; M =2 >The minimal number of tickets is 1 f( N, L, T, C, M) = 1 >The tickets (e.g.): >1 2 3 >N = 10; L = 5; T = 3; C = 2; M =2 >f(10, 5, 3, 3, 2) = 4 >The tickets (e.g.): >1 2 3 >1 4 5 >2 4 5 >3 4 5 >N = 10; L = 5; T = 4; C = 3; M =2 >f( 10, 5, 4, 3, 2) = 1 >The tickets (e.g.): >1 2 3 4 >N = 10; L = 6; T = 4; C = 3; M =2 >f( 10, 6, 4, 3, 2) = 2 >The tickets (e.g.): >1 2 3 4 >1 4 5 6 >N = 10; L = 8; T = 4; C = 3; M =2 >f( N = 10, L = 8, T = 4, C = 3, M =2) = 2 >The tickets (e.g.): >1 2 3 4 >5 6 7 8 --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Theorem concerning the internal structure of uncountable subsets of P(N). For any collection C of subsets of N (positive whole numbers), if for no x in C do there exist y1,y2,y3, in C such that x is a subset of the union of y1,y2,y3,, then C is countable. Proof. We build an enumeration of C as follows. If the union of sets in C is not equal to N then we extend C to a set C* by putting W = N C and C* = C union {W}. Since W is disjoint from all the other sets in C, this extension does not affect our hypothesis. Now let F map C* onto a subset of N as follows. Select any y1 from C* and let F(y1) = the least number in y1. Now select any y2 from C* and let F(y) = the least number in y2 y1. Now select any y3 from C* and let F(y3) = the least number in y3 (y2 union y1). Generally, at the nth stage we select any yn in C* and let F(yn) = the least number in yn (union of y1,y2,yn-1). By the hypothesis, at no stage will yn (union of y1,y2,yn-1) be undefined, since yn is never the subset of the union of y1,y2,yn-1. We need to show that F is one to one. Suppose that for some z1 and z2 in C* we have F(z1) = F(z2) = w, say, with z1 appearing earlier in the construction than z2. Now, w obviously belongs to z1, or else it could not have been chosen to represent z1, hence w does not belong to z2 (union of y1,y2,y3,z1,), so w cannot be selected from z2 to represent F(z2). Hence F(z2) cannot be equal to F(z1). Thus F is always defined and is one to one. So C* is mapped onto a subset of N and hence is countable. It follows that if a collection C of subsets of N is uncountable, then for some x in C there must exist y1,y2,y3, in C such that x is a subset of the union of y1,y2,y3, === Subject: Re: Theorem concerning the internal structure of uncountable subsets of P(N). I'll take a look at this but first explain what the special characters are as they don't show up as you expect. As it is, Cantor showed a countable number of countable sets is countable, which is a generalization of your problem. He listed them as a1, a2, a3, ... b1, b2, b3, ... c1, c2, c3, ... ... He counted them by the upper left corner first a1 then the diagonal b1, a2, ie the upper left corner with a1 shaved off then the diagonal c1, b2, a3, etc There is a simple polynomial relating the position (n,m) in the infinite matrix and the position in the sequence. Let's see, something like (n + m - 1)(n + m - 2)/2 + m Which, when it's the right formula, is straight forward to show is bijection. > For any collection C of subsets of N (positive whole numbers), if for > no x in C do there exist y1,y2,y3,.85 in C such that x is a subset of What's the character after y3, ? in C Is it ... ? > the union of y1,y2,y3,.85, then C is countable. > Proof. > We build an enumeration of C as follows. If the union of sets in C is > not equal to N then we extend C to a set C* by putting W = N .96 C and What's the character after = N ? Is it union, ie / or perhaps U ? > C* = C union {W}. Since W is disjoint from all the other sets in C, > this extension does not affect our hypothesis. > Now let F map C* onto a subset of N as follows. > Select any y1 from C* and let F(y1) = the least number in y1. Now > select any y2 from C* and let F(y) = the least number in y2 .96y1. Now > select any y3 from C* and let F(y3) = the least number in y3 .96 (y2 > union y1). Generally, at the nth stage we select any yn in C* and let > F(yn) = the least number in yn .96 (union of y1,y2,.85yn-1). By the What do you mean by yn-1 ? y_n - 1 or y_(n-1) ? x_n for x sub n > hypothesis, at no stage will yn .96 (union of y1,y2,.85yn-1) be undefined, > since yn is never the subset of the union of y1,y2,.85yn-1. > We need to show that F is one to one. Suppose that for some z1 and z2 > in C* we have F(z1) = F(z2) = w, say, with z1 appearing earlier in > the construction than z2. Now, w obviously belongs to z1, or else > it could not have been chosen to represent z1, hence w does not belong > to z2 .96 (union of y1,y2,y3,.85z1,.85), so w cannot be selected from z2 to > represent F(z2). Hence F(z2) cannot be equal to F(z1). > Thus F is always defined and is one to one. So C* is mapped onto a > subset of N and hence is countable. > It follows that if a collection C of subsets of N is uncountable, then > for some x in C there must exist y1,y2,y3,.85 in C such that x is a > subset of the union of y1,y2,y3,.85 === Subject: Re: Theorem concerning the internal structure of uncountable subsets of P(N). > There is a simple polynomial relating the position (n,m) in the infinite > matrix and the position in the sequence. Let's see, something like > (n + m - 1)(n + m - 2)/2 + m > Which, when it's the right formula, is straight forward to show is > bijection. > For any collection C of subsets of N (positive whole numbers), if for > no x in C do there exist y1,y2,y3,? in C such that x is a subset of > What's the character after y3, ? in C Is it ... ? ******** Yes ******** > the union of y1,y2,y3,?, then C is countable. > Proof. > We build an enumeration of C as follows. If the union of sets in C is > not equal to N then we extend C to a set C* by putting W = N ? C and > What's the character after = N ? Is it union, ie / or perhaps U ? ********************************** It is a minus sign, meaning the intersection of N and C complement. I just felt it would be less messy if the union of all the sets to be counted was equal to N. ************************************ > C* = C union {W}. Since W is disjoint from all the other sets in C, > this extension does not affect our hypothesis. > Now let F map C* onto a subset of N as follows. > Select any y1 from C* and let F(y1) = the least number in y1. Now > select any y2 from C* and let F(y) = the least number in y2 ?y1. Now > select any y3 from C* and let F(y3) = the least number in y3 ? (y2 > union y1). Generally, at the nth stage we select any yn in C* and let > F(yn) = the least number in yn ? (union of y1,y2,?yn-1). By the ***************************************************** yn-1 means y subscript n-1 I note the ? for ... again. The reason for this is that I prepared the message in Word. ***************************************************** > What do you mean by yn-1 ? y_n - 1 or y_(n-1) ? x_n for x sub n > hypothesis, at no stage will yn ? (union of y1,y2,?yn-1) be undefined, > since yn is never the subset of the union of y1,y2,?yn-1. > We need to show that F is one to one. Suppose that for some z1 and z2 > in C* we have F(z1) = F(z2) = w, say, with z1 appearing earlier in > the construction than z2. Now, w obviously belongs to z1, or else > it could not have been chosen to represent z1, hence w does not belong > to z2 ? (union of y1,y2,y3,?z1,?), so w cannot be selected from z2 to > represent F(z2). Hence F(z2) cannot be equal to F(z1). > Thus F is always defined and is one to one. So C* is mapped onto a > subset of N and hence is countable. > It follows that if a collection C of subsets of N is uncountable, then > for some x in C there must exist y1,y2,y3,? in C such that x is a > subset of the union of y1,y2,y3,? === Subject: Current favorite undergrad textbooks Which textbooks are currently trendy for the main undergrad algebra and analysis courses? TIA - AA === Subject: Re: Current favorite undergrad textbooks > Which textbooks are currently trendy for the main undergrad algebra > and analysis courses? For Algebra, the big two were Hungersford and Herstein. When I went back for my masters, the professor said they were still the most popular ones. === Subject: Re: Why do different ways of calculating the same thing in fact yield the same answer? > When I calculated one thing several different ways (a very simple > example being the order of addition), I found it amazing and wonderful > that every time (barring mistakes), I found the same answer. Of > course, having studied mathematics, I know that's the way it's > supposed to be, but still it often inspired me with wonder. If you start with the statement x + 3 = 10 and one of your methods yields a conclusion of the form x = [something other than 7], there would be something wrong with that method. Likewise, if you try to answer the question Who was President of the United States in 1983? using several different references, a disagreement between references implies that at least one is, again, wrong. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Why do different ways of calculating the same thing in fact yield the same answer? X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 11:24 PM, nickprillewits@yahoo.com (Tijmen Tieleman) said: >When I calculated one thing several different ways (a very simple >example being the order of addition), I found it amazing and >wonderful that every time (barring mistakes), I found the same >answer. Depending on how I parse that, there are two obvious answers. By definition if you calculate the same thing two different ways you get the same answer; otherwise you would be calculating two different things. It gets more interesting if you are calculating approximations, because their you can get different answers. If, instead, you are referring to such properties as the commutative law, a*b=b*a, the associative law, a*(b*c)=(a*b)*c, or the distributive law, a*(b+c) = (a*b)+(a*c), then the answer is that they are properties of the real numbers but not of all algebraic systems. In particular, many important algebraic systems do not[1] obey the commutative law. [1] More precisely, have two binary operations only one of which is commutative. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Why do different ways of calculating the same thing in fact yield the same answer? >When I calculated one thing several different ways (a very simple >example being the order of addition), I found it amazing and >wonderful that every time (barring mistakes), I found the same >answer. > Depending on how I parse that, there are two obvious answers. > By definition if you calculate the same thing two different ways you > get the same answer; otherwise you would be calculating two different > things. It gets more interesting if you are calculating > approximations, because their you can get different answers. It's perfectly interesting to consider the case of exact identity - although its interest-value isn't specifically mainstream mathematical in nature. That's not quite true, now that I think about it - the sub-field of algorithm analysis is perfectly well mathematical, and also goes to town on the notion of calculating one thing in several different ways. Lots more interesting examples of exploiting this stuff abound in logic - lambda calculus, and AIT, for example. But that's not what I had originally intended to remark on. For me at least, the more obvious interest of the OP's observation was conceptual/philosophical/logical - which essentially means Fregean (note that Church was hugely inßuenced by Frege). I'm aware there are quite a number of people who equate that triumvirate with uninteresting - indeed such a person has already popped up in this thread - but I find such an inference to be unmotivated and uselessly elitist (I'm so glad I'm a Delta...). The philosophical issue that one might reasonably find interesting is the nature of conceptual (specifically mathematical, in this case) content. As Frege noticed, a=a is not an informative statement, while a=b is often the stuff major discoveries are made of. I find it perfectly interesting to take on the project of explaining (among other things) exactly what the difference between the two statements consists in. However - all this does raise the question: Did the OP intend to be interpreted mathematically (commutativity, algebraic rules, etc.), or conceptually (as I've mentioned above). lol - I have no idea. cdj === Subject: Post-sci-math cheap oem s0ftware shipping worldwide boundary=--31796107029318906214 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) with ESMTP id i7S2eWa18838; by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, $Revision: 1.6 secondary) with SMTP id i7S2dXX2010827; ------------------------------------------------------------- -------- TOP quality software:

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? 2N47A3zVjC9Xk2yarmenia|post-sc i-math@mathforum.org>or un*su*bs*cr*ibe ----31796107029318906214-- === Subject: Re: math, physics, and languages, art > at 04:14 PM, DaLoveRhino@hotmail.com (Love Rhino) said: >I know a few mathematicians and physicists, and one thing I found >common amoung them is their ability to pick up languages. > I haven't noticed that, but I have noticed an assoication with music. >But, I'm wondering how many mathematicians/physicists are naturally >gifted painters, sketchers or naturally gifted sculptors? >(Naturally gifted, not trained) > How do you tell the difference? The nature versus nurture debate has > been going on for a long time. Was, e.g., Leonardo Da Vinci talented > because of his environment or because of his genetics? He was talented because when he lived *talent* was *by definition* composing oil paintings of fictious women. But unfortunely for Leonardo, and his wild-ass scientoid spectulars, the most talented person to have ever lived in Europe is still ranked throughout the world as Johann Gutenburg, not an artist. And the most talented people to have ever lived in the Americas are still ranked as Thomas Jefferson and James Madison, not a artist, or scientist. === Subject: How to calculate average longitude? My brain must have gone soft. I am struggling to figure out the proper way to get the average longitude of a set of longitudes. The problem is of course the discontinuity at 180 degrees. Please help me while I go eat some fish. Yours in embarrassment, Tom === Subject: Re: How to calculate average longitude? Convert to x,y,z coordinates, average, convert back and read off the latitude and longitude. -- Ron Hardin rhhardin@mindspring.com On the internet, nobody knows you're a jerk. === Subject: Re: How to calculate average longitude? > My brain must have gone soft. I am struggling to figure out the proper > way to get the average longitude of a set of longitudes. The problem is > of course the discontinuity at 180 degrees. How about the following? Think of unit vectors pointing from the center of the earth toward the various longitudes. Add the vectors together. If their sum is nonzero, the resulting vector points to the desired average longitude; if their sum is zero, it simply does not make sense to think of an average longitude. David Cantrell === Subject: Re: How to calculate average longitude? > My brain must have gone soft. I am struggling to figure out the proper > way to get the average longitude of a set of longitudes. The problem is > of course the discontinuity at 180 degrees. > Please help me while I go eat some fish. > Yours in embarrassment, > Tom The problem is that the average depends upon where you define the origon (longitude = 0) to be. As this is essentially arbitrary, there isn't really an average longitude. For convenience, lets just consider points on the equator. Lets have the longitude running from 0 to 360. Pick two points close together on the equator, at say 10 and 20 degrees longitude. The average longitude would be 15. Now consider two points the same distance apart, at say 355 and 5 degrees longitude. In this case, the average would be 180 degrees. But these are still two points 10 degrees apart. How come in one case the average lies between them (at 15) and in the other case on the opposite side of the world (at 180?). Its because there isn't really an average position, in that many positions can be an average depending upon where you pick the origon (it depends upon the co-ordinate system you choose). It gets worse with more than 2 points. Here's a similar example - what temperature is twice as hot as 50 degrees Fahrenheit? (100 degrees F ?) What temperature is twice as hot as 10 degrees Centigrade (20 degrees Centigrade?) But 50 degress F is about the same as 10 degrees C. But 100 degrees F is about 40 degress C. So twice as hot also depends upon where you pick the origon, and the zero in Fahrenheit is at a different place to the zero in Centigrade. (And yes, I know, there is an absolute zero from which we can measure temperature and get one canonical meaning for twice as hot - but its only an analogy, dammit). === Subject: Re: How to calculate average longitude? >My brain must have gone soft. I am struggling to figure out the proper >way to get the average longitude of a set of longitudes. The problem is >of course the discontinuity at 180 degrees. >Please help me while I go eat some fish. >Yours in embarrassment, >Tom It's late and this is just off the top of my head but say you convert your longitudes to 0 <= L < 2Pi. Say your two measurements are L1 and L2 with L2 >= L1. Try this: If L2 - L1 <= Pi then ave = (L2 + L1)/2 otherwise ave = {(L2 + L1)/2 + Pi} mod 2Pi I'm sure someone will correct it if it doesn't work :-0 --Lynn === Subject: Re: McNally illustrates corrupt time xform. was: sci.math vs True Believers > Explain the experiments that support SR rather than Newton, then. > >Lordy! You True Believer cretins might as well have been born brain dead. > >In response to Ôin newtonian/glilean' theory' you say ÔExplain the >experiments that support SR rather than Newton, then.' > >LOL! LOL! LOL! LOL! LOL! LOL! LOL! > >So you admit that you already knew that you were wrtong since you >cannot explain the experiments that falsify Newtonian mechanics. > >Explain how experiment can show a simple math operation to be wrong. >So you admit that you were never talking about Newtonian Mechanics at >all, just algebra that is in no way connected to the physical world. > I can't tell if he is just plain ol' ing stupid or maliciously > stupid. > First, he will babble on about how well Galilean transforms work. Then > you explain to him, and mabey point to evidence saying, that the > universe is Lorentzian and not Galilean. > His response to that is HOW CAN EXPERIMENT DISPROVE SIMPLE MATH? > Then he starts applying it to reality, as if what was told to him > never existed. All standard tactics of the crank and crackpot. Bill === Subject: Re: Set Theory homework - please help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7S51GV29462; > Hello! > I just got some homework from the professor dealing with set theory, > only our books are not due in the book store until monday. I need help > with a problem if possible. I just started the class this week and > this is due tomorrow. I have done a little of it, but i don't have > enough info in order to finish it. I don't know where to go next. My > teacher hasn't gone through it thuroughly yet. >If you have done Venn Diagramms, try drawing a General 3-circle Venn >Diagram, labelling each region, and then working through using A,B,C as >sets of regions. Whether that would be accepted by your professor as a >proof is another matter, but it's an idea. > the question is this > (A-B) - (B-C) = A-B > so far i have this for the proof > A, B, AND C are all sets > & = INTERSECT > + = UNION > Ô = COMPLEMENT > LHS=(A-B) - (B-C) > =(A&B') - (B&C') //difference of sets in each set of parenthesis > =(A&B') & (B'+C) //difference of sets in the whole set problem > =((A&B') + B') & (A&B')+C) //distributive properties of the sets >This line should be: =((A&B')&B') + ((A&B')&C) >=(A&(B'&B')) + ((A&B')&C) // associative property of & >=(A&B') + ((A&B')&C) // S&S=S for all sets S >=((A&B')&U) + ((A&B')&C) // S&U=S for all sets S, universe U >=(A&B') & (U+C) // distributive properties >=(A&B') & U // U+S=U for all sets S, universe U >=(A&B') // S&U=S for all sets S, universe U >=A-B > --------------------------------- > ANOTHER PROBLEM > (A-B) + (B-A) = (A+B) - A&B > my proof so far > LHS = (A-B+ + (B-A) >Should be (A-B) + (B-A) > = (A&B') + (B&A') > =((A&B') & B) + ((A&B') &A') >This should be: =((A&B')+B) & ((A&B')+A') >=((A+B)&(B'+B)) & ((A+A')&(B'+A')) >=((A+B)&U) & (U&(B'+A')) >=(A+B) & (B'+A') > that it for that one too! > My teacher suggested that i use the distributive set in the last step > of both problems, which i did, and then i simplify, but i can't figure > out how in the world, that equals out to A-B >RHS = (A+B) - (A&B) >= (A+B) & (A&B)' >= (A+B) & (A'+B') >= ((A+B)&A') + ((A+B)&B') >= ((A&A') + (B&A')) + ((A&B')+(B&B')) >= (0+(B&A')) + ((A&B')+0) >= (B&A') + (A&B') >= ((B&A')+A) & ((B&A')+B') >= ((B+A)&(A'+A)) & ((B+B')&(A'+B')) >= ((B+A)&U) & (U&(A'+B')) >= (B+A) & (A'+B') >Done. > Can anyone suggest a Venn Diagram to explan it to me, or just let me > know in words. I have never taken this course before and we just > started it up this week. >Venn diagrams usually are much clearer and simpler, once you know how to >draw them. Working from both sides to a common middle for showing >equality is another standard trick. >-- >Will Twentyman >email: wtwentyman at copper dot net thank you so much. Hopefully the book store will have the books in on monday, and i will be able to understand it better by reading. Erica === Subject: Re: JSH: Curious, Wiles's work, null test by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7S51Gr29443; >... > > If I'm right, you can trace through Wiles's entire work--dropping the > assumption he puts forward so much himself about a certain something > being modular--and see that you don't invalidate the conclusion, which > should be a bit weird given some of the the other works he cites. > > It seems to me that it'd be rather odd in fact for his work to fail > this null test, but hey, I didn't see anything that jumped out at me > that saves him. > > Maybe one of you will read through and find it. > > Remember to give the actual page, and some words of support for your > claim! > > James Harris > It won't be necessary for anyone to review Wiles' proof, because your > charge that it contains a logical error (Com Hoc, Ergo Propter Hoc) can be > immediately discarded by appeal to the logical truism -- Ex Harris, Ergo > Erratum. > -- > There are two things you must never attempt to prove: the unprovable -- > and the obvious. > -- > Democracy: The triumph of popularity over principle. >Now I'm being fair. If *anyone* actually understands Wiles's work >then what I'm asking should be a trivial and fun exercise for them!!! >Go through his work with an assumption opposite to the one he uses, >which is that there is in fact a non-modular elliptic curve, and see >if you find a point in his argument where it will squawk, NO!!! >That's how real math proofs work. >If you challenge a mathematical proof with something contrary to its >conclusion then at some point in the argument you'll face a >contradiction. >If sci.math'ers refuse to be even minimally reasonable, why should >anyone doubt my charges against them? >It looks to me like I make specific charges and face political attack >ads masquerading as posts in return! >Maybe some of you should go work for the Bush campaign, like they need >any help! >Come on! A null test of Wiles's work should just have been done, may >have been done, and if you can give a reference, fine! >But making replies here with more propaganda posts just makes it look >like, yes, mathematicians are trying to hide what may be one of the >biggest intellectual boon-doggles in human history: a bogus proof of >Fer,at's Last Theorem. >My charges are specific and to the point. I say Wiles's argument >fails by Cum Hoc, Ergo Propter Hoc. >It just so happens that a good way to handle b.s. from people who try >to attack that specific charge is to ask for a null test. >Now then, if you say my charge is false, then you can end the >discussion forthwith by answering the null test challenge!!! >Don't you think sci.math can do better than leave the question open? >Or is the newsgroup too lacking in *real* versus imagined mathematical >ability to handle it? >Notice I'm challenging your real math and logic skills versus allowing >you to play like you're a mathematician on a newsgroup! >If you wish to just keep playing at being a mathematician, post in >another thread, or show yourself as a poser like C. Bond with a weak >post in this one. >Here b.s. like from C. Bond is just an indication that you think >talk is all math is about. >Here, deliver or fail. >James Harris May I suggest you read the extremely accessible book, _How to Read and Do Proofs_ by Daniel Solow, so that you can speak from knowledge rather than repeating this gibberish ad nauseum? Only you can prevent yourself from looking like an utter moron. Tom === Subject: Re: JSH: Curious, Wiles's work, null test >... > > If I'm right, you can trace through Wiles's entire work--dropping the > assumption he puts forward so much himself about a certain something > being modular--and see that you don't invalidate the conclusion, which > should be a bit weird given some of the the other works he cites. > > It seems to me that it'd be rather odd in fact for his work to fail > this null test, but hey, I didn't see anything that jumped out at me > that saves him. > > Maybe one of you will read through and find it. > > Remember to give the actual page, and some words of support for your > claim! > > James Harris > > It won't be necessary for anyone to review Wiles' proof, because your > charge that it contains a logical error (Com Hoc, Ergo Propter Hoc) can be > immediately discarded by appeal to the logical truism -- Ex Harris, Ergo > Erratum. > > -- > There are two things you must never attempt to prove: the unprovable -- > and the obvious. > -- > Democracy: The triumph of popularity over principle. >Now I'm being fair. If *anyone* actually understands Wiles's work >then what I'm asking should be a trivial and fun exercise for them!!! >Go through his work with an assumption opposite to the one he uses, >which is that there is in fact a non-modular elliptic curve, and see >if you find a point in his argument where it will squawk, NO!!! >That's how real math proofs work. >If you challenge a mathematical proof with something contrary to its >conclusion then at some point in the argument you'll face a >contradiction. ... >But making replies here with more propaganda posts just makes it look >like, yes, mathematicians are trying to hide what may be one of the >biggest intellectual boon-doggles in human history: a bogus proof of >Fer,at's Last Theorem. >My charges are specific and to the point. I say Wiles's argument >fails by Cum Hoc, Ergo Propter Hoc. >It just so happens that a good way to handle b.s. from people who try >to attack that specific charge is to ask for a null test. ... >Here, deliver or fail. >James Harris > May I suggest you read the extremely accessible book, _How to > Read and Do Proofs_ by Daniel Solow, so that you can speak > from knowledge rather than repeating this gibberish ad nauseum? > Only you can prevent yourself from looking like an utter moron. > Tom You failed. Maybe you should read that book more carefully yourself. Math proofs necessarily contradict claims contrary to their conclusions or what would be the point of a proof? If you challenge a math proof, the proof always wins. And note, a proof of something necessarily is correct, though some people tend to call *claims* of proof, proofs, which can lead to confusion. Your suggestion is a classic dodge where you think you can make the issue about me, when it's about whether or not Wiles's work fails. My charge is simple: Wiles's approach fails by Cum Hoc, Ergo Propter Hoc. Some have tried to say that doesn't apply so I simply requested the next thing that follows from my charge: a null test. That is, if you say that someone is using Cum Hoc, Ergo Propter Hoc, and they deny it, but refuse to be logical, you can request a null test. A null test is just to go through Wiles's work with the assumption that a non-modular elliptic curve exists and see if that contradicts with anything in his argument! So why do *I* need to read a book on proofs? It seems to me that you wish to make this about me when I'm making it about Wiles's work. If any of you have information at this point that Wiles's work does fail but you're trying to defend it in any way then you are engaging in fraud. Now then, either Wiles's work does or does not fail by Cum Hoc, Ergo Propter Hoc, and what I know about proofs is irrelevant to that truth. Too many of you seem to think that math is a social game. It's not. Now then, if Wiles's work is correct then assuming the existence of a non-modular elliptic curve and going through it step-by-step to see what happens, is not going to hurt it! What are you people afraid of? James Harris === Subject: Re: JSH: Curious, Wiles's work, null test > If you challenge a math proof, the proof always wins. And note, a > proof of something necessarily is correct, though some people tend to > call *claims* of proof, proofs, which can lead to confusion. I seem to remember something like that. Don't you always come up with ingenious proofs, and every single time, maths wins and your proof loses? Of course, that is where the fine distinction comes in: You have never been able to show a proof, only _claims_ of proofs... === Subject: Re: JSH: Curious, Wiles's work, null test > What are you people afraid of? Suggested study for James Harris: www.tinyurl.com/5exhx === Subject: Re: JSH: Curious, Wiles's work, null test > What are you people afraid of? > Suggested study for James Harris: > www.tinyurl.com/5exhx Projection preserves cross-ratios. Yes, indeed: everyone gets cross, and no one remains rational. -- Chris Henrich The total lack of evidence is the surest sign that the conspiracy is working. === Subject: Re: JSH: Curious, Wiles's work, null test >... > > > If I'm right, you can trace through Wiles's entire work--dropping the > assumption he puts forward so much himself about a certain something > being modular--and see that you don't invalidate the conclusion, which > should be a bit weird given some of the the other works he cites. > > It seems to me that it'd be rather odd in fact for his work to fail > this null test, but hey, I didn't see anything that jumped out at me > that saves him. > > Maybe one of you will read through and find it. > > Remember to give the actual page, and some words of support for your > claim! > > James Harris > > It won't be necessary for anyone to review Wiles' proof, because your > charge that it contains a logical error (Com Hoc, Ergo Propter Hoc) can be > immediately discarded by appeal to the logical truism -- Ex Harris, Ergo > Erratum. > > -- > There are two things you must never attempt to prove: the unprovable -- > and the obvious. > -- > Democracy: The triumph of popularity over principle. > >Now I'm being fair. If *anyone* actually understands Wiles's work >then what I'm asking should be a trivial and fun exercise for them!!! > >Go through his work with an assumption opposite to the one he uses, >which is that there is in fact a non-modular elliptic curve, and see >if you find a point in his argument where it will squawk, NO!!! > >That's how real math proofs work. > >If you challenge a mathematical proof with something contrary to its >conclusion then at some point in the argument you'll face a >contradiction. > > ... > >But making replies here with more propaganda posts just makes it look >like, yes, mathematicians are trying to hide what may be one of the >biggest intellectual boon-doggles in human history: a bogus proof of >Fer,at's Last Theorem. > >My charges are specific and to the point. I say Wiles's argument >fails by Cum Hoc, Ergo Propter Hoc. > >It just so happens that a good way to handle b.s. from people who try >to attack that specific charge is to ask for a null test. > > ... > >Here, deliver or fail. > > >James Harris > May I suggest you read the extremely accessible book, _How to > Read and Do Proofs_ by Daniel Solow, so that you can speak > from knowledge rather than repeating this gibberish ad nauseum? > Only you can prevent yourself from looking like an utter moron. > Tom > You failed. Maybe you should read that book more carefully yourself. > Math proofs necessarily contradict claims contrary to their > conclusions or what would be the point of a proof? > If you challenge a math proof, the proof always wins. And note, a > proof of something necessarily is correct, though some people tend to > call *claims* of proof, proofs, which can lead to confusion. > Your suggestion is a classic dodge where you think you can make the > issue about me, when it's about whether or not Wiles's work fails. > My charge is simple: Wiles's approach fails by Cum Hoc, Ergo Propter > Hoc. > Some have tried to say that doesn't apply so I simply requested the > next thing that follows from my charge: a null test. > That is, if you say that someone is using Cum Hoc, Ergo Propter Hoc, > and they deny it, but refuse to be logical, you can request a null > test. > A null test is just to go through Wiles's work with the assumption > that a non-modular elliptic curve exists and see if that contradicts > with anything in his argument! > So why do *I* need to read a book on proofs? > It seems to me that you wish to make this about me when I'm making it > about Wiles's work. > If any of you have information at this point that Wiles's work does > fail but you're trying to defend it in any way then you are engaging > in fraud. > Now then, either Wiles's work does or does not fail by Cum Hoc, Ergo > Propter Hoc, and what I know about proofs is irrelevant to that truth. > Too many of you seem to think that math is a social game. > It's not. > Now then, if Wiles's work is correct then assuming the existence of a > non-modular elliptic curve and going through it step-by-step to see > what happens, is not going to hurt it! > What are you people afraid of? > James Harris James, It's OBVIOUS you don't know the first thing about proofs. You are a childish arrogant jerk and aren't much of a man if you act like this. Get a life. Dave === Subject: Re: JSH: Curious, Wiles's work, null test > James, It's OBVIOUS you don't know the first thing about proofs. You are a > childish arrogant jerk and aren't much of a man if you act like this. Get a > life. He has a life. A part of it is the provocation of statements like this. The rest of us posters to sci.math do not have the power to change that. (Come to think of it, we don't have the right, either.) -- Chris Henrich God just doesn't fit inside a single religion. === Subject: Re: topics/problems in mathematics and the law interesting thread about mathematics and the law: >In the UK there appeared to be clusters of childhood cancers near nuclear >power stations. How this cluster analysis was done was a subject of much >debate as I believe it was not certain if they were real or not. >Ditto power lines. > Something like that happens with GSM base stations. Although thourough statistical analysis was not > performed yet, research shows clusters of cancer around those GSM base stations. Because it is > already shown that the radiation of the new technology UMTS actually leads to feelings of discomfort > in human beings, things will probably get much worse in the near future. Well, radiation of the new technology UMTS actually leads to feelings of discomfort in human beings, things will probably get much worse in the near future sounds more like hand-waving than proving to me. I'll tell what I have felt from my GSM mobile phone, which I have used in two countries while living an unknown distance from the nearest base station. I feel 1) the convenience of being to adjust appointments and meetings by communication while away from any land-line telephone, 2) the reduced pollution of less driving because of 1), 3) the security of knowing I can reach the police or my loved ones in possibly dangerous situations, and 4) the comfort of knowing I can medical help rapidly for my young children while taking walks in the neighborhood. > In this example, it is easily seen how the government prevales money above health. My point number 4) points out one way (other people can point out other ways, I'm sure) that mobile phones improve health, by getting people out of the house for physical exercise without risk of losing communication with emergency care. Mobile phone networks are only profitable business ventures if lots of consumers buy and use mobile phones, and at least in my country nobody in the government is forcing anyone to use a mobile phone. I got my first mobile phone in Taiwan (that was required by my [private sector] employer there) and learned to love using it. Once my wife had one (she won a drawing for which a mobile phone was the prize) I experienced the joy of his-and-hers mobile phones. I think my mobile phone has easily added a year to my life, or at least a year of PLEASURE to my life, by the convenient communication it has made possible. > It is clear that Ôsomething bad is happening' in the GSM/UMTS cases, Is it? Where is the evidence? > but expansion of the networks carries on because > of the high prices the companies have to pay for the necessary licenses. In all developed countries I know of, the companies might simply go bankrupt after paying for the licenses, if they don't find willing customers. Nobody is forcing the companies to build base stations even after they pay for electromagnetic spectrum licenses, although surely the investor-owners of a publicly traded company want it to do something (e.g., resell the license) that makes money from the license. > Compare this with > medication: many years of testing are required before a new type of medicin is allowed on the > market. Only if it's absolutely Ôproven' that no permamanent damage to a human's health can result > from using the medicine, it passes the test. No, that is not the usual legal standard in the countries I know about, not anything as strict as absolutely Ôproven' that no permamanent damage to a human's health can result. In the United States, the legal standard is proof of safety and effectiveness, that the drug doesn't cause MAJOR harm (ALL drugs have side effects) and does do what it is purported to do to treat disease. And many observers of the United States drug approval process (including physicians I knew in Taiwan) criticize it for being too strict, too slow to allow useful new drugs to market. I think the United States strikes the balance quite well between doubting and approving new drugs, but always there are trade-offs involved in these issues in every country. > The government can require this because there is no easy > money in it for them by allowing new medicines on the market faster. Many public policy economists would point out that this vastly underestimates the inßuence that regulated industries have on regulatory authorities. > We can wait for lawsuits regarding health damage due to GSM/UMTS base stations where statsitical > analysis will be part of the evidence. This post illustrates why mathematical training ALONE is not enough to reach sound decisions on public policy (as I can easily recognize as a lawyer). On the whole, I agree with others who have posted in this thread that most lawyers are abysmally innumerate, and that results in wrong decisions in the legal system. But I have found to my dismay that many (I didn't say most) mathematicians are so used to thinking about numbers as pure mathematicians do that they rarely reasonably consider costs, uncertainty of information, incentives to human behavior built into different decision-making systems, and other important issues that every lawyer and legislator has to consider every day. By the way, there are trained professionals who consider problems of mathematics and the law every day. They are called economists. Their writings on legal issues are often very interesting and helpful, and hardly ever innumerate. === Subject: Re: topics/problems in mathematics and the law > Compare this with > medication: many years of testing are required before a new type of medicin is allowed on the > market. Only if it's absolutely Ôproven' that no permamanent damage to a human's health can result > from using the medicine, it passes the test. > No, that is not the usual legal standard in the countries I know > about, not anything as strict as absolutely Ôproven' that no > permamanent damage to a human's health can result. In the United > States, the legal standard is proof of safety and effectiveness, > that the drug doesn't cause MAJOR harm (ALL drugs have side > effects) and does do what it is purported to do to treat disease. > And many observers of the United States drug approval process > (including physicians I knew in Taiwan) criticize it for being too > strict, too slow to allow useful new drugs to market. I think the > United States strikes the balance quite well between doubting and > approving new drugs, but always there are trade-offs involved in > these issues in every country. For those who think that the USA's drug approval is too long, I remind them that thalidomide was NEVER approved in the US. American women went to foreign countries that had fast drug approval systems to get the drug. > We can wait for lawsuits regarding health damage due to GSM/UMTS base stations where statsitical > analysis will be part of the evidence. > This post illustrates why mathematical training ALONE is not > enough to reach sound decisions on public policy (as I can easily > recognize as a lawyer). On the whole, I agree with others who have > posted in this thread that most lawyers are abysmally innumerate, > and that results in wrong decisions in the legal system. But I > have found to my dismay that many (I didn't say most) > mathematicians are so used to thinking about numbers as pure > mathematicians do that they rarely reasonably consider costs, > uncertainty of information, incentives to human behavior built > into different decision-making systems, and other important issues > that every lawyer and legislator has to consider every day. I think that this is not so much due to the field of study as it is that many of the mathematicians have purely academic lives. Let me have a conversation with two other school-teachers - one who has lived in the real world and is teaching as a second career and the other got a BA in child psych and immediately got their MA in education with a credential and started teaching. It will take me about 30 seconds to know which is which. === Subject: Re: How to calculate average longitude? In my case I am trying to find a suitable geographical average of the cities in a country. My algorithm has to work for countries straddling zero degrees and for those straddling 180 degrees. Perhaps I can take the mathematical average and compare it to something, then add 180 degrees if necessary. I had wondered if there was a known, elegant, simple algorithm... === Subject: Re: How to calculate average longitude? > In my case I am trying to find a suitable geographical average of the > cities in a country. My algorithm has to work for countries straddling > zero degrees and for those straddling 180 degrees. > Perhaps I can take the mathematical average and compare it to > something, then add 180 degrees if necessary. > I had wondered if there was a known, elegant, simple algorithm... Firstly, no country straddles your line at longitutude 180, so just average the lattitudes and the longitudes of the cities. If you want this to be meaningful, you should weight by multiplying each lat and long with the population of that city, so Ney York counts more than Portland, Oregon. Then divide by the population of all the cities added together to get the proper average. If you want to do it with an imaginary country that crossed long=180, then just subtract or add 180 and measure from there. Add the 180 back at the end. If it crosses lat=180 and lat=0, then you can't meaningfully do it (as I posted before). === Subject: Re: Current favorite undergrad textbooks === Subject: Re: Current favorite undergrad textbooks X-RFC2646: Original > Which textbooks are currently trendy for the main undergrad algebra > and analysis courses? >For Algebra, the big two were Hungersford and Herstein. When I went back >for my masters, the professor said they were still the most popular ones. When I was at Princeton, they used Herstein's Topics in Algebra for first semester undergraduate abstract algebra. I used it my junior year. which is an easier/shorter version, if that's what you're interested in. A friend of mine who went to Michigan State for undergrad said they used Hungerford (no s) there for their graduate course, but I had heard it was for undergrads, so I'm not sure about that one. I glanced at it once, and it looked nowhere near as well-written as Herstein's book. Michael === Subject: Re: tomorrow? > Hi. > > Elizabeth is not going to dance tomorrow, so I can come up > then. I could be there about 1:45 or so. Will you be out > of your meeting by then? > > Peggy > > Sure will be. > > What planet will we meet on? > I like soft boiled eggs, but not too soft! That's odd - you didn't used to. === Subject: Re: Left/Right Rotation Permutations > puzzler/solution: >Question... >Is there an easy way to directly calculate the final position >of k in the permutation of [1,2,3,...,m]? Final position of k in the permutation of [1,2,3,...,m]: what do you mean? -- Alex Vinokur http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn === Subject: Re: Math of metaballs > Should that be meatballs? Mmh, someone call them meatball, yes.. but is a mistake. But i'm talking about isosurfaces. Bye! -- Elwood (Peter S.) === Subject: Re: The groups Z*_n with n = 2^k -- Easy number theory problem > What I am looking for now is to show that for a field with p^k >elements, q = p^k -1 of them non-zero, the solns. x^q - 1 are the >elements of the field >How does this help you find b such that o(b) = p-1? Z_p is the spliting field of x^p - x = 0, hence by field theorem Z_p^* is cyclic. Let b be any generator of Z_p^*. >phi(n) = p^(k-1)(p-1) = |Z*_n|, and G = Z*_n is a cyclic p^(k-1)(p-1) ambiguous. Use at least p^(k-1) (p-1) >group with |G| = phi(n). But a finite field over Z_p, GF(p^k) has >q = p^k - 1 non-zero elements, which gives a cyclic field F* >with |F*| = p^k - 1. I see no relation between Z*_n and F*. >x in F* has x^q = 1, x in Z*_n has x^phi(n) = 1. There isn't. Instead of jumping to a new topic, I included a section upon previous unresolved topics. The most outstanding being For odd prime p, k in N, n = p^k, if g generates Z_n^*, then g generates Z_p^* This I can proof. The same method for the converse doesn't work. Is the converse true? Have you found a proof? Can your program find a counterexample? Curious how the generators were 2 or 3. Any explaination? Any that were found other than 2 or 3 ? -- Z_n^*, n = 2^k > 7.5.9. Lemma. Let k be an integer with k >= 2, and let n=2^k. > (a) In Z*_n, the element [5] has order 2^k >Should be order 2^(k-2). . > (b) In Z*_n, the elements [ +- 5^n ] are distinct, if n is a > nonnegative integer less than 2^k. >Should be n = 1,2,...,2^(k-2), i.e n < 2^(k-2), if n = 0 is allowed. So what? > 7.5.10. Theorem. If k >= 3, and n=2^k, then Z*_n is isomorphic to > the direct product of a cyclic group of order 2 and a cyclic group > of order 2^k. >Should be order 2^(k-2). Indeed. Neither of these however are needed to show 7.5.11. Z_n^* cyclic iff n in { 1, 2, 4, p^k, 2p^k | p odd prime, k in N } Z_2^* = { 1 }; Z_4^* = { 1,3 }, 3^2 = 1 (Z_2p^k)^* = Z_2^* x (Z_p^k)^* = Z_p^k^* 0 <= k, n = 2^(k+3) ==> Z_n^* not cyclic { 1, 2^(k+2) - 1, 2^(k+2) + 1, -1 }_n not cyclic subgroup 2 < m,n ==> Z_m^* x Z_n^* not cyclic { (1,1), (-1,1), (1,-1), (-1,-1) }_m,n not cyclic subgroup >Theorem: Z*_2^k =~ Z_2 x Z_2^(k-2). >Proof; k >= 2 >Let n = 2^k. Then there are n/2 = 2^(k-1) = order of Z*_n >odd elements x (with (2,x) = 1). We can deal with this group >by using 2 levels of mod. Write the elements of Z*_n as >x = +/- (2m - 1), with m = 1,2,...,2^(k-2). >Also, o(x) = o(-x) for all x != 1. The +/- in >x = +/- (2m - 1) gives a group of the form >Z*_n = Z_2 x G, where G =~ Z_2^(k-2) = Z_(n/4) is cyclic. Write This may be a result of Sylow theorems. As for G = H x G/H the closest I've found is 7.1.3. Theorem. Let G be a group with normal subgroups H, K such that HK = G and H / K = {e}. Then G is isomorphic to H x K. So have you proof for normal H or for Abelian G or whatever, that G = H x G/H ? >G = Z*_n/Z_2, where Z_2 = <-1> = (-1,1), and we do our >arithmetic 1) mod n = 2^k ; and 2) mod (+/-). The ordered pair (1,-1) isn't the set { -1,1 }. >proving that G is cyclic, using the mod(2^k) mod(+/-) >arithmetic, which I will denote MOD(2^k). That needs explaining or at least clarification or formula. This working thru the problem by thinking out loud is hard to follow. Seems you've intuited and visualized the situtation for Z_2^(k+1)^* = Z_2 x Z_2^K Is an isomorphism (adjust to cosets) something like f:Z_2 x Z_2^K -> Z_2^(k+1)^*, (a,n) -> (1 - 2a)(2n + 1) ? Anyway would you pull your insights together into a readable proof? Somewhere below are you considering a representative set for the cosets of Z_n^* / { -1,1 } ? That would be a nice picture to have in clear focus. >|G| = 2^(k-2), and the elements of G are >x = +/- (2m - 1), with m = 1,2,...,2^(k-2). >Let m_i = 2^(k-i-1), so that x_i = 2^(k-i) - 1, for i = 1,...,k-2. >For m = m_1 = 2^(k-2), (x_1)^2 = [2^(k-1) - 1]^2 MOD(2^k) > = {2^[2(k-1)] - 2^k + 1} MOD(2^k) = 1. >For m_2 = 2^(k-3), (x_2)^2 = [2^(k-2) - 1]^2 MOD(2^k) > = {2^[2(k-2)] - 2^(k-1) + 1} MOD(2^k) > = [2^(k-1) - 1] MOD(2^k) = x_1 MOD(2^k). >Continuing in this way gives for m_i, x_i = 2m_i - 1, and > (x_i)^2 MOD(2^k) = x_(i-1) MOD(2^k). >This is the critical eqn. of this argument. >Finally we get to m_(k-2) = 2, and x_(k-2) = 3, and > 3^2 MOD(2^k) = x_(k-3). >Thus we have taken k-2 steps, >dividing m by 2 each time, and increasing the order of >x by 2 each time, so clearly x_(k-2) = 3 has order 2^(k-2), >so G must be cyclic, G =~ Z_2^(k-2). >So G is always generated by 3, as well as by the x associated >with odd values of m. So m = 2,3,5,...,2^(k-2) - 1 gives 2^(k-3) >generators x = 3,5,9,..., 2^(k-1) - 3. QED. ---- === Subject: Re: Uncountable sets in CZF? > For myself, the easiest way to understand generic extensions is through > Boolean-valued models. >[rest snipped] >I do have one question, though. >With this notion of extension, is it possible that an existing set in V >gets new elements in V[G]? Can you give me an example? No, an existing set in V does not get new elements in V[G]. For a set x in V, since dom(hat x) = {hat y : y in V} and hat x(hat y) = 1 for all y in V, then for any generic ultrafilter G, i_G(hat x) = {i_G(hat y) : hat y in dom(hat x), hat x(hat y) in G} = {i_G(hat y) : y in x, 1 in G} = {i_G(hat y) : y in x}. By e-induction, i_G(hat x) = x for all x in V, and the elements of x are unchanged in going from V to V[G]. If the cardinality of A is greater than 1 and B is infinite, then the set of functions from A to B can actually change between V and V[G], but the set of those functions which are elements of V remains a set in V[G], and this is the set which is identified with the original set of functions. >That is, suppose V is a (set-)model of ZFC and V[G] an extension as you >describe, and X is a set in V. So hat X will be in V[G]. Actually, hat X is in V^B, and its image in V[G] is i_G(hat X), which is equal to X. The elements of V^B are not sets in a model of ZFC, but their images under i_G form a model of ZFC. Formulae and sentences are not taken to be true or false in V^B, but are instead assigned an element of B as the value. V[G] satisfies a formula iff the value of the formula in V^B is an element of G. >Can it occur >that for some x in V[G], x e hat X, but for all y in V, x =/= hat y? >I suppose it is intended that this is possible, but from the definition >you gave (of the embedding of V into V^B), i'm a little confused whether >this can occur or not. I think that it is probably best to think of V^B as a skeleton on which to hang a model of ZFC. It is not itself a model of ZFC (in the model- theoretic sense), but the introduction of a generic ultrafilter gives it the structure of a model. So V^B is an intermediary structure, built from the model V and a Boolean algebra B in V (so that every element of V^B is an element of V, all elements being functions). The choice of a generic ultrafilter in B clothes the skeleton that is V^B and converts it into a model V[G] of ZFC. This is done by mapping each element x of V^B to the set of images of elements of dom(x) which x maps into an element of G, i.e., for x in V^B, i_G(x) = {i_G(y) : y in dom(x), x(y) in G}. The model is given by the class V[G] = {i_G(x) : x in V^B}. The identification of hat x in V^B corresponding x in V is just a means of identifying an element of V^B which necessarily maps to x in V[G], independently of the choice of G. David > For x in V, we can define hat x in V^B inductively by > dom(hat x) = {hat y : y in x}, > hat x(hat y) = 1 for all y in x. > This gives an embedding of V in V^B as a subclass. > For x and y in V^B, ||x in y||, ||x subset y|| and ||x = y|| can be > defined inductively by > ||x in y|| = sup{ y(z).||x = z|| : z in dom(y) }, > ||x subset y|| = inf{ x(z)'+||z in y|| : z in dom(x) }, > ||x = y|| = ||x subset y||.||y subset x||, > so that the truth values of x in y, x subset y and x = y are given > as elements of B. > Specifically, it is demonstrable that > ||hat x in hat y|| = 1 if x, y in V and x in y, > ||hat x in hat y|| = 0 if x, y in V and x not in y, > For all x in V, i_G(hat x) = x, so that V is a submodel of V[G]. >-- >Herman Jurjus === Subject: Re: Uncountable sets in CZF? >How is it that anothing non-empty exists in ZF? There is not an axiom >saying exists empty, There is, actually. Or it is a theorem following from the Axiom of Separation and the Axiom of Infinity (the latter being only invoked to guarantee the existence of at least one set - the Axiom of Infinity can be rewritten in such a manner that does not require the prior existence of an empty set). The existence of an empty set also follows from the Axiom of Infinity (which guarantees the existence of at least one set) and an appropriate statement of the Axiom of Replacement. >exists set containing empty, .... Since ZF guarantees the existence of an empty set (which is unique in ZF by the Axiom of Extensionality), the Axiom of the Power Set guarantees the existence of a set which has the empty set as an element. So the Axioms of ZF guarantee the existence opf an empty set and the existence of sets which are not empty. >Now you >might demand an axiom to that effect, if so, then congratulations, you >require a new axiom. I don't. No new axiom required. ZF is already capable of guaranteeing the existence of these sets. >Now, here's a problem for you, in V[G], name an integer not in V. As Jesse pointed out, there are none. There are no new natural numbers, so there are no new integers. David ----- === Subject: Re: Uncountable sets in CZF? >Because you're a moron, preferring unjustified and implicit assumption >rather than explicitly stating your axioms. > Now, here's a problem for you, in V[G], name an integer not in V. >If I understood David's explanation, there ain't any. That's correct. There are no new natural numbers in the extension, so there are no new integers. David ----- === Subject: Re: Uncountable sets in CZF? > > There are no axioms, that way it can be complete in extension of > Goedel. It just uses zero/empty, tautology, and excluded middle, > paraconsistently. > > There are infinitely many and/or no axioms, in extension of Goedel and > myself. > You're making no sense. I've written many words to sci.math, some of them are not even meaningless. -- Jesse quoting me Jesse, if set theory is to be the foundation, then a priori in set theory there are only sets. If a set is a container, then we have two points at which it can be considered: the container of nothing and the container of everything. One or the other or both is the only possible thing to consider, with no other knowledge. You know I've expressed preference for plain language statements of logic, and that goes along with the concept that any explanation that requires more than a single sentence or two is overcomplex, keep it simple. the one containing that, and so on and so forth, ad infinitum. Then as well we can consider the set of empty set and the set of empty set, that's often called the ordinal one. It is easy to continue with saying that every counting number is represented as an ordinal by some construction of sets. How is it that anothing non-empty exists in ZF? There is not an axiom saying exists empty, exists set containing empty, .... Now you might demand an axiom to that effect, if so, then congratulations, you require a new axiom. I don't. Now in these ways of constructing ordinals from the empty set, we've seen two simple methods, with 4 = {{{}}} + { {}, { {}, {} } }. Jesse, 2 + 2 = 4. Another method to consider is a binary bitstring representing each ordinal, as everything is an ordinal. 00000000... Then, one set that is not necessarily the same as that is 11111111... While that is so, in some meaning they represent the same value. So anyways, I think that there are ways to have those equal, and as well 11000000... = 2 and 01000000... = 2 In that way, I can rationally declare that in a theory of ubiquitous ordinals, infinite sets are equivalent. That's good for me, because then there are more reasons for EF to be a bijective function between N and the unit interval of R, because it is, and not not. Now, here's a problem for you, in V[G], name an integer not in V. Ross === Subject: Re: Uncountable sets in CZF? <2otdsjFdqemsU1@uni-berlin.de> <87llg5uisd.fsf@phiwumbda.org> <873c2ctk7u.fsf@phiwumbda.org> <873c2915w8.fsf@phiwumbda.org> Discussion, linux) > You know I've expressed preference for plain language statements of > logic, and that goes along with the concept that any explanation that > requires more than a single sentence or two is overcomplex, keep it > simple. If you won't enumerate the axioms clearly and completely, we have nothing at all to discuss. You're diddling, pretending to think about deep issues when you don't actually make explicit any of your fundamental assumptions. > How is it that anothing non-empty exists in ZF? There is not an axiom > saying exists empty, exists set containing empty, .... Now you > might demand an axiom to that effect, if so, then congratulations, you > require a new axiom. I don't. Because you're a moron, preferring unjustified and implicit assumption rather than explicitly stating your axioms. > Now, here's a problem for you, in V[G], name an integer not in V. If I understood David's explanation, there ain't any. -- Jesse Hughes [I]f gravel cannot make itself into an animal in a year, how could it do it in a million years? The animal would be dead before it got alive. --The Creation Evolution Encyclopedia === Subject: Why to call it pseudorandom? If an algorithm produces a infinte sequence of digits that never falls into a loop and passes the most stringent statistical tests about its uniformity, why to call it pseudorandom? If genius discover a system of equations for the roulette, the old random experiments made with roulettes will appear at daybreak as pseudorandom? === Subject: Re: Why to call it pseudorandom? >If an algorithm produces a infinte sequence of digits that never >falls into a loop and passes the most stringent statistical tests >about its uniformity, why to call it pseudorandom? maybe because the digits are not random? if they're produced by an algorithm then they're not unpredictable. >If genius discover a system of equations for the roulette, the >old random experiments made with roulettes will appear at daybreak >as pseudorandom? ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Why to call it pseudorandom? >If an algorithm produces a infinte sequence of digits that never >falls into a loop and passes the most stringent statistical tests >about its uniformity, why to call it pseudorandom? > maybe because the digits are not random? if they're produced by > an algorithm then they're not unpredictable. >If genius discover a system of equations for the roulette, the >old random experiments made with roulettes will appear at daybreak >as pseudorandom? > ************************ > David C. Ullrich > sorry about the inelegant formatting - typing > one-handed for a few weeks... And at a practical level, all computer rand() generators loop. They take a seed n and calculate f(n), then f(f(n)) etc. So they loop over whatever range n can take (at most). If n is a 32 bit integer - and it often is - it will loop over 4 billion numbers, probably enough to generate enough mineseweeper boards and bridge hands to keep most people happy. Many rand() functions are however random according to all other statistical criteria. === Subject: Re: Why to call it pseudorandom? >If an algorithm produces a infinte sequence of digits that never >falls into a loop and passes the most stringent statistical tests >about its uniformity, why to call it pseudorandom? > maybe because the digits are not random? if they're produced by > an algorithm then they're not unpredictable. >If genius discover a system of equations for the roulette, the >old random experiments made with roulettes will appear at daybreak >as pseudorandom? >[...] >And at a practical level, all computer rand() generators loop. They take a >seed n and calculate f(n), then f(f(n)) etc. >So they loop over whatever range n can take (at most). If n is a 32 bit >integer - and it often is - it will loop over 4 billion numbers, probably >enough to generate enough mineseweeper boards and bridge hands to keep most >people happy. >Many rand() functions are however random according to all other statistical >criteria. of course any sequence generated by an actual physical computer loops. but this is not true of sequences generated by -algorithms- [for example an algorithm generating the digits of pi] ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Why to call it pseudorandom? You can make any pseudorandom sequence even more pseudorandom by adding another random number generator mod the identical range of them all. The whole thing is _ruined_ by an actual random number anywhere in the sum, though, and it becomes actually random. Back when I needed decent random numbers, I added a lot of them. Then you get no worse than the quality of the best one. I imagine that still works. As the unix manual said before grown-ups took it over RAND(3A) SYNOPSIS (seed in r0) jsr pc,srand /to initialize jsr pc,rand /to get a random number srand(seed) int seed; rand() WARNING The author of this routine has been writing random-number generators for years and have never been known to write one that worked. BUGS The low-order bits are not very random. CB-UNIX January 15, 1973 -- Ron Hardin rhhardin@mindspring.com On the internet, nobody knows you're a jerk. === Subject: Re: Why to call it pseudorandom? It is called pseudorandom because it is not completely random but almost. It has many qualities of a random sequence but is not truly random, as it is generated by an algorithm. Most all PRN sequences loop. === Subject: Product of functions in W^{1,oo} Just a question: if f,g are functions of the Sobolev space W^{1, oo}, can you say that f*g is in W^{1,oo} ? I suppose that it's true, using the fact that given a function f in W^{1, oo}, there's a lipschitz function h such that f=h almost everywhere. === Subject: Serge Lang's Political Activism Masterpiece RATS IN A BOX from Vietnam Era I wish Lang would allow his manuscript RATS IN A BOX to be scanned in the computer to a pdf file since it can be used to fight against this generation's wars. If you haven't read this, you should because it is so relevent today. I first read this 30 years ago as a young apathetic professor and it is impossible to find today but it helped me change my mind about Vietnam and politics in general! I can't track down Serge Lang because he does not use email. I hope somebody who knows him will ask him and tell him it made a big difference to me at least back then. === Subject: Re: Fibonacci number ending in 0000 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7SFaJL10098; Best Wishes joccis === Subject: Re: Explaining the foundations of math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7SFaJs10117; >Sometimes I think that profs and text writers fail students. >I recall that I was taught about the epsilon-delta formulation >of limits and proofs with the attitude of we do this because >this is how one does it, with no explanation of why it >is important, the logical problems it solves, the importance >of putting things on a firm logical foundation. >For example, Weierstrass's main lasting contribution to math >was to define the limit of f(x) at x_o as follows: >For any e > 0 there exists d_o such that if 0 < d < d_o, then >| f(x_o +/- d) - L | < e, then L = lim(f(x)) as x --> x_o. >So what? The point is that there is no mention of any >infinitely small quantities, differentials (or Newton's ßuxions, >etc.), only real numbers R with + and < (order). > I don't disagree with your statement of the point. However, it > is also important that the definition uses | |, a notion of a > metric. Limits can be defined in terms of topology; they don't > need a metrical notion. I don't know if it is a failure of > teaching not to start at the more general topological > definition. >While the more general one is nice in many ways, even the standard >epsilon-delta definition is often a bit abstract for Calc I students. > -Ron A necessary condition to understanding it is surely to have grasped why the quantors exists and for all can't simply be swapped. blackboard in Analysis I, he asked Well, is there anyone out there who has never seen this before and understands it now? Whoever that is will be a very good mathematician because, generally, you will have to think about for a while before it clicks, first. That comment didn't help me much, on the contrary- it was quite intimidating. Our *student* teacher helped much by saying There is a big difference between: ÔFor every man there is a woman' and ÔThere is a woman for every man!' However, strangely enough, due to my colloquial street English from the past, I was not able to discern the difference in the two statements immediately - although the back of my head did immediately know what he must be getting at. C. Dement http://www.crowdog.de === Subject: Re: How to calculate average longitude? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i7SFaLs10166; >In my case I am trying to find a suitable geographical average of the >cities in a country. My algorithm has to work for countries straddling >zero degrees and for those straddling 180 degrees. >Perhaps I can take the mathematical average and compare it to >something, then add 180 degrees if necessary. >I had wondered if there was a known, elegant, simple algorithm... Here is a trick that will work: Add 1000 to each longitude figure. Then take the arithmetical average. The subtract 1000 from the result. phil === Subject: Re: How to calculate average longitude? >In my case I am trying to find a suitable geographical average of the >cities in a country. My algorithm has to work for countries straddling >zero degrees and for those straddling 180 degrees. >Perhaps I can take the mathematical average and compare it to >something, then add 180 degrees if necessary. >I had wondered if there was a known, elegant, simple algorithm... > Here is a trick that will work: > Add 1000 to each longitude figure. Then take the arithmetical > average. The subtract 1000 from the result. A couple of test cases: Longitude is usually defined in the range -180 to +180 (west positive). Consider two locations, +170 and -170. Obviously by inspection, the average is 180. Your method gives (1180 + 820)/2 - 1000 = 0. Alternatively, define longitude over 0 to 360. Then consider two locations, 10 and 350. The obvious average is 0. Your method gives (1010 + 1350)2 - 1000 = 180. === Subject: S^3 as the union of solid tori Can anyone explain why S^3 can be expressed a a union of two solid tori (S^1 x 2-Disc) with an embedded torus (S^1 x S^1) as a common boundary? The book says it follows easily by expressing R^3- {solid torus} as a unions of circle and a straight line and then add a point at the infinity. How can I use the hint of the book? === Subject: S^3 as a union of two solid tori Can anyone explain why S3 can be expressed a a union of two solid tori (S1 x 2-Disc) with an embedded torus (S1 x S1) as a common boundary? The book says it follows easily by expressing R3- {solid torus} as a unions of circle and a straight line and then add a point at the infinity. How can I use the hint of the book? === Subject: Re: S^3 as a union of two solid tori > Can anyone explain why S3 can be expressed a a union of two solid >tori (S1 x 2-Disc) with an embedded torus (S1 x S1) as a common >boundary? The book says it follows easily by expressing R3- {solid >torus} as a unions of circle and a straight line and then add a point at >the infinity. How can I use the hint of the book? I think an easier way to see this is to consider the cartesian product D of two round 2-disks. On the one hand, D is homeomorphic to the round 4-disk D^4 (you can take D to be the unit bidisk in R^4, and then easily construct the homeomorphism explicitly), so its boundary is homeomorphic to the boundary of D^4, namely, S^3. On the other hand, the given product structure on D gives you an expression of the boundary of D as the union of two solid tori intersecting along their common boundary: one solid torus is the Cartesian product of a round circle with a round 2-disk, and the other is the Cartesian product of a round 2-disk with a round circle. Since you're at UBC, why don't you go talk to Dale Rolfsen? Lee Rudolph === Subject: Re: Null test, Wiles's work | That the existence of a solution to FLT leads to a non-modular elliptic curve | was an older result, and in fact the curve is semi-stable. The T-S | conjecture is that every elliptic curve is modular. What Wiles did show | was a subset of the T-S conjecture: every semi-stable elliptic curve is | modular. This has later been improved by Ribet (I think) that indeed | T-S is a theorem. As I recall, somebody observed that the same method was enough for curves semistable except at 3 and 5. According to Mathworld, Brian Conrad, Fred Diamond, and Richard Taylor were the ones who then improved Wiles' proof to show that it holds when the conductor is not 27; then the three of them together with Christophe Breuil further improved it to prove the general conjecture. Ribet announced that there was a proof at a conference, but those four should be given due credit. And of course Richard Taylor had collaborated with Andrew Wiles on the original proof for the semistable case. Keith Ramsay === Subject: Re: Null test, Wiles's work > Here's an attempt at getting some interest by suggesting a null test > of Wiles's work. > Why would you think about a null test? That is more related to > statistics, > but the mathematics going on here are *not* statistics, the show (or > attempt to show) an absolute truth. > Basically someone needs to post how Wiles's approach could have shown > a result *opposite* to what many think it did. > I think what James is looking for is that Wiles proof is not only consistent > with FLT being true but also FLT being false. Since everyone here accepts > Wiles proof, it is up to James to show that the assumption of a solution to > FLT along with Wiles' logic leads to a nonmodular elliptic curve. Actually what I'm looking for is not complicated. What I'm saying is, assume that a non-modular elliptic curve DOES exist, could Wiles's approach have shown that result? That is, the point of the null test is to see if inherent in his approach is the conclusion he desired such that the truth is irrelevant to it. That is, if it were in fact true that a non-modular elliptic curve does exist, would the approach that Wiles took show it? What I'm trying to get you to do is drop your assumption that Wiles is correct and take a *critical* eye to his approach! The difficulty with even communicating that idea shows just how deeply the belief is embedded. Remember, math proofs don't need your allegiance, don't need you to protect them, and don't need you to worry over them if challenged. They are perfect, inviolable, and absolute. A math proof resists all challenges, and can handle a null test!!! I want some of you to turn into critics versus true believers and ask the simple question, if a non-modular elliptic curve actually did exist would Wiles's approach have shown it? James Harris === Subject: Re: Null test, Wiles's work jstevh@msn.com > Here's an attempt at getting some interest by suggesting a null test > of Wiles's work. > > Why would you think about a null test? That is more related to > statistics, > but the mathematics going on here are *not* statistics, the show (or > attempt to show) an absolute truth. > > Basically someone needs to post how Wiles's approach could have shown > a result *opposite* to what many think it did. > > I think what James is looking for is that Wiles proof is not only consistent > with FLT being true but also FLT being false. Since everyone here accepts > Wiles proof, it is up to James to show that the assumption of a solution to > FLT along with Wiles' logic leads to a nonmodular elliptic curve. > Actually what I'm looking for is not complicated. > What I'm saying is, assume that a non-modular elliptic curve DOES > exist, could Wiles's approach have shown that result? Yes! As I discuss in one of your other threads, Wiles shows that there is an if and only if relationship between modular elliptic curves and modular Galois representation. In fact, there were a class of representations that Wiles could not easily show to be modular. He then examined those in more detail. Ultimately, they were shown to be modular (in the technically most difficult part of the paper), but had he shown them to be non-modular, then this would have disproven FLT. -Saint Cad === Subject: Re: Null test, Wiles's work > Actually what I'm looking for is not complicated. > What I'm saying is, assume that a non-modular elliptic curve DOES exist, > could Wiles's approach have shown that result? Uhm...but that doesn't really have any bearing on whether or not Wiles' approach works. You basically seem to be asking if Wiles' approach would have been able to disprove FLT if FLT had been false. While that might be an interesting question, whatever the answer to that question turns out to be doesn't say anything about whether the approach works in the case where FLT is true. Let's try a very simple example. Conjecture: 19981928192 is not prime. I claim this conjecture is true. My proof is to divide the number by 2 and see that there is no remainder. If the number HAD been prime, my method of proof could not have shown that. It would fail to show the number is not prime, but it would not establish that it is prime. -- --Tim Smith === Subject: Re: Null test, Wiles's work > Actually what I'm looking for is not complicated. > What I'm saying is, assume that a non-modular elliptic curve DOES exist, > could Wiles's approach have shown that result? > Uhm...but that doesn't really have any bearing on whether or not Wiles' > approach works. Sure it does. I can see that some of you lack real math experience. Sometimes someone can put something forward that they say is a proof, and people looking over it can't seem to find anything obviously wrong but there's this vague sense that something is not right. But, a null test--assuming the opposite of the argument's conclusion--does not change a single step in the supposed proof! Obviously, something must be wrong and the argument is not a proof. > You basically seem to be asking if Wiles' approach would have been able to > disprove FLT if FLT had been false. While that might be an interesting > question, whatever the answer to that question turns out to be doesn't say > anything about whether the approach works in the case where FLT is true. Supposedly math people are into specifics and objectivity, so why are you talking about FLT? The discussion is over whether or not assuming the existence of a non-modular elliptic curve will make any difference in Wiles's work. It seems to me that you have stumbled into here focusing on the pop issue with his work, which is FLT, but I'm not talking about pop culture here. > Let's try a very simple example. Conjecture: 19981928192 is not prime. > I claim this conjecture is true. My proof is to divide the number by 2 > and see that there is no remainder. And therefore it follows that it is not prime as a prime number does not have any other natural except 1 as a factor, and you can conclude that your conjecture is true. > If the number HAD been prime, my method of proof could not have shown that. > It would fail to show the number is not prime, but it would not establish > that it is prime. And you wouldn't have a proof then, now would you? To null test assume that 19981928192 IS prime. But it has a factor of 2, so it's not prime, null test passed. Get it? Aren't any of you trained in basic techniques like assuming the opposite of your conclusion to check an argument? It seems to me that possibly there's a clear separation going on between the mathematically trained and the people who simply are drawn to sci.math because they like the idea but have no real math ability or mathematical training. My B.Sc. is in physics so I got that kind of training from my physics classes. I assume that the mathematicians among you got that kind of training in your courses as well, which leaves those others who either lack the training or were sleeping through important parts of their classes! James Harris === Subject: Re: Null test, Wiles's work > Aren't any of you trained in basic techniques like assuming the > opposite of your conclusion to check an argument? A valid argument does not require a check. You're demanding a Ôsecond opinion' on an issue which does not consist of subjective evaluation. > It seems to me that possibly there's a clear separation going on > between the mathematically trained and the people who simply are drawn > to sci.math because they like the idea but have no real math ability > or mathematical training. > My B.Sc. is in physics so I got that kind of training from my physics > classes. That may explain your peculiar frame of reference. Mathematics is an exact science and physics is an empirical science. Wiles' proof has nothing to do with physics. > I assume that the mathematicians among you got that kind of training > in your courses as well, which leaves those others who either lack the > training or were sleeping through important parts of their classes! Sleeping through Ôimportant parts'of classes is one thing, but remaining stupid throughout all classes *your* specialty. NOTA BENE: The above conclusion was reached after carfully reviewing JSH's posting record. The Ônull' test has been applied to confirm it. Ex Harris, Ergo Erratum -- QED -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Null test, Wiles's work >For instance, *assume* a non-modular elliptic curve exists, and point >out WHERE in Wiles's work that would show up. > Someone else mentioned Pythagoras's theorem, and I think it's an > instructive example. Assume that a right-angled triangle with sides > 4, 5, and 6 exists, and point out where in the usual proof of > Pythagoras's theorem that would show up. I think you'll find there is > no such point, because it's not that kind of proof. > -- Richard You're wrong. Most proofs of it are geometric, and an early step is that the triangle is a right one. However, 4, 5 and 6 do not give a right triangle, so my point remains. If you wish to push the point, give an algebraic proof, and I'll point out to you where your claim will draw a contradiction *before* the conclusion. It seems odd to me that any of you would dare to question such a basic point of mathematics that math proofs will show a contradiction with a false claim *before* their conclusion, and a proof will ALWAYS reveal a claim contradictory to its conclusion to be false. It's basic. Up front, yes, I can show you the obvious, but upfront, simply by questioning basics of mathematical proofs you reveal you're not a mathematician. No matter what you may call yourself, if you don't understand math proofs, then how can you actually be a mathematician? James Harris === Subject: Re: Null test, Wiles's work Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >You're wrong. Most proofs of it are geometric, and an early step is >that the triangle is a right one. However, 4, 5 and 6 do not give a >right triangle, so my point remains. I already posted a reply to this, but I realise it doesn't really get to the point. Let me try again. When you prove the theorem geometrically, you don't have to use any particular right-angled triangle. You don't actually have to use a right-angled triangle at all, because the proof doesn't rely on measuring the triangle, but let's suppose you use a 3-4-5 triangle in your diagram. Now when you've done the proof, you haven't just proved it for 3-4-5 triangles, but for *any* right-angled triangle. So the existence of a 4-5-6 right-angled triangle is not refuted by any particular step of the proof, but by the proof as a whole. >If you wish to push the point, give an algebraic proof, and I'll point >out to you where your claim will draw a contradiction *before* the >conclusion. Ok, consider Euclid's proof as given about half way down this page: http://mathworld.wolfram.com/PythagoreanTheorem.html (the windmill proof). You'll see that it has a triangle with sides roughly 5-12-13. Which step would fail if a 4-5-6 right-angled triangle existed? -- Richard === Subject: Re: Null test, Wiles's work Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >You're wrong. Most proofs of it are geometric, and an early step is >that the triangle is a right one. However, 4, 5 and 6 do not give a >right triangle, so my point remains. Of course they don't give a right-angled triangle. But if you draw a triangle, label its sides as being 4, 5, and 6, and the angle between 4 and 5 as a right angle, where does the proof break down? The proof uses only straight-edge and compass, so you can't *measure* the lines or angles. -- Richard === Subject: Re: Null test, Wiles's work Discussion, linux) > If you wish to push the point, give an algebraic proof, and I'll point > out to you where your claim will draw a contradiction *before* the > conclusion. I find the following proof easier to present, so would you please apply the test to it? (I've presented this in another post, but I repeat it here since you offered.) ,---- | Theorem: The square root of 2 (hereafter sqrt(2)) is irrational. | | Proof: Assume that sqrt(2) is rational. Let x and y be given such | that | | (1) sqrt(2) = x / y | (2) gcd(x,y) = 1 | | Then 2 = x^2 / y^2 so x^2 = 2 y^2. Now, if 2 divides x^2, then 2 | divides x, so let z be given such that x = 2 z. | | Then (2 z)^2 = 2 y^2, so 4 z^2 = 2 y^2 and hence 2 z^2 = y^2. | Therefore y^2 is divisible by 2 and hence y is also divisible by 2. | | But we have thus proven that both x and y are divisible by 2, | contradicting (2). `---- Now let's see how the null test works. It's like this, right? We assume (contrary to the above theorem) that the square root of 2 is rational and then we see which step in the proof is invalid, right? Okay, so I'm assuming... Golly. Every step still seems to follow from previous steps or previously proven theorems. Uh oh. Help me out here. -- Jesse F. Hughes I have written many words to sci.math, some of them are not even meaningless. --Ross Finlayson === Subject: Re: Null test, Wiles's work yeah; the excerpted paragraph was a wonder of concision. > Okay, as I understand it, Taniyama-Shimura (andhence Wiles's work) > does not establish an isomorphism... > a great exposition for the general mathematical public (as opposed --Chairman George! http://tarpley.net/bush12.htm === Subject: Re: Null test, Wiles's work an arbitrary curve wouldn't likely to be modular; eh? are most eliptical curves also not modular? the requirement of modularity is a big restriction, which I thinbk has to do with points of curves, both of whose planar coordinates are rational. talk about tilting at windmills. if you're going to an attempt an elementary proof of Fermat's last theorem e.g., you'd better have a new insight, not just undemonstrated arithmetical abilities! > That is, assume there is an elliptic curve out there that is not > modular, what makes you think Wiles's approach would have indicated > that it exists? > It would not. He showed that for semi-stable stuff (and those were > important at that point) no counter example exists. --Chairman George! http://tarpley.net/bush12.htm === Subject: classifying degenerate critical points I'm looking for some help in classifying degenerate critical points as minima, maxima, or saddles. Assuming a function f : R^n --> R, the critical points are those where the gradient of f is zero. The standard technique for classifying critical points involves examination of the eigenvalues of the Hessian at the critical points. This is basically an extension of the Second Derivative Test to the multivariate case, and is a consequence of Taylor's theorem. However, when the Hessian is not invertible (i.e. when the critical point is degenerate or non-Morse), this test breaks down, much as the Second Derivative Test breaks down for, say, f(x)=x^4. For this scalar example, we have to examine higher order derivatives to classify the critical points. Is there a similar technique for the multivariate case? Here are three examples I've been struggling with, all of which have degenerate critical points at (0,0): f1(x,y) = x^4 + y^4 f2(x,y) = x^3-3xy^2 f3(x,y) = (x-y^2)(x-2y^2) The first is clearly a maximum, the second is the so-called monkey saddle, and the third is also a saddle. Historical note: f3 was the counter-example used by Peano to disprove a proposition by Lagrange -- see pg 33 of Hancock's 1917 text Theory of Maxima and Minima, available free online at this long url -- http://www.hti.umich.edu/cgi/t/text/text-idx?c=umhistmath; idno=acl8264.0001. 001 The books I've looked in all seem to just gloss over the issue of degenerate critical points, and they state something like, if the Hessian is singular, then the test is indeterminate. The End. So, any help in finding a technique to classify degenerate critical points would be great, and a reference on this topic would also be wonderful. I'm not even sure what topic this would fall under, or where to begin looking. monkey_saddle@yahoo.com === Subject: Re: classifying degenerate critical points >I'm looking for some help in classifying degenerate critical points as >minima, maxima, or saddles. I suspect this is much harder than you think it is. (I mean, I know it's very hard, and I suspect you don't think that--yet.) >Assuming a function f : R^n --> R, the >critical points are those where the gradient of f is zero. The >standard technique for classifying critical points involves >examination of the eigenvalues of the Hessian at the critical points. >This is basically an extension of the Second Derivative Test to the >multivariate case, and is a consequence of Taylor's theorem. However, >when the Hessian is not invertible (i.e. when the critical point is >degenerate or non-Morse), this test breaks down, much as the >Second Derivative Test breaks down for, say, f(x)=x^4. For this >scalar example, we have to examine higher order derivatives to >classify the critical points. Is there a similar technique for the >multivariate case? Even in one variable, if by higher order derivatives in the phrase examine higher order derivatives you mean only the values of the higher order derivatives at the critical point, then you *cannot* find *necessary and sufficient* conditions in terms of higher order derivatives for a critical point to be, or not to be, a local extremum: there are, as you know, smooth functions which are infinitely ßat at some critical point (i.e., all the derivatives vanish there), and you can readily convince yourself that such examples exist which are local minima, others which are local maxima, and yet others which are neither. Things only get worse (much worse) in higher dimensions. Now, if you restrict your attention to *real analytic*, or further to *real polynomial*, functions, then you may be able to get somewhere. A good keyword phrase to look for is finitely determined singularity or finitely determined germ. Maybe Wall's fairly recent _magnum opus_ on singularities of real mappings would have pointers in its bibliography. Lee Rudolph