mm-1119 === Subject: Re: Distribution Fitting - Generalised Gaussian based Distribution? >I wish to fit some distribution to my data ... I cant. >My data fits well to a modified form of the Generalised Gaussian >distribution (GG). The GG is; >f(x) = A e ^ (-|bx| ^ c) >where, >b = (1/sigma) * (Gamma(3/c) / Gamma(1/c)) ^ (1/2) >A = b c / (2Gamma(1/c)) >GG does not fit well. >However my data can be fitted nicely to; >f1(x) = P(m) e ^ (-|bx| ^ c) >where, >P(m) = probability of the modal value. Shouldnt be a probability: its a density. Maybe thats the problem. >Example params are; >b = 9.39 >c = 0.424 >P(m) = 0.2667 >Unfortunatley f1 is not a pdf! What are you using to fit your data to f1, and why do you say its a good fit? Its not a pdf because int_{-infty}^infty f1(x) dx is not 1 (with your numbers its about 0.162). I question how you could say your data fit this well when the integral is so far off: e.g. if you calculate the empirical cdf F(x) = fraction of data points that are <= x, F(x) = 1 for x >= the maximum of your data, but int_{-infty}^x f1(t) dt will be less than 0.162. At the median of your data, F will be approximately 1/2 and the integral will be even less. It certainly doesnt seem like a good fit to me. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Distribution Fitting - Generalised Gaussian based Distribution? Hi Robert, >>P(m) = probability of the modal value. > Shouldnt be a probability: its a density. Maybe thats the problem. Right. Its the relative frequency of the mode of the sample data. Its there to get the curve scaled nicely. > What are you using to fit your data to f1, and why do you say its > a good fit? I was curve fitting!; FIT with NonlinearLeastSquares in Matlab. I just was looking for a function that fits the sample relative frequency histogram. The reason I say its a nice fit is largely that it looks good visually (I could send an image), but theres also the RMS which is 0.0440 ... good enough when youre desperate enough. I just need a pdf that looks similar to f1. === Subject: Re: Classification Theorems for Banach Spaces? >Does anyone know if there are (even weak) >classification/representation theorems for various sorts of Banach >spaces? For example, can we say X Ôis an L^p space iff X and Y and >Z, etc? You might look at Lindenstrauss and Tzafriri, Classical Banach Spaces, Springer Lecture Notes in Mathematics #338. For example, Prop. II.3.1: A separable Banach space X is isometric to an L^p(mu) space iff X is the closure of the union of an increasing sequence of subspaces B_n where B_n is isometric to R^n with the p-norm. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: 4^n > n^4 > Show for eact natural number n >= 5, 4^n > n^4. Heres an interesting way to show it for all real t > 4 Start with exp(x) >= 1 + x for all x then exp( t/4 - 1 ) >= t/4 so exp(t/4) ^ 4log4 >= (t*e/4) ^ 4log4 where e = exp(1) which is 4^t >= t^4 * (t/4)^a where a = 4log4 - 4 > 0 and the result follows. === Subject: Re: how to show that set of solutions of sigma(from 0 to infinity) 1/2^n, belongs to (0,1] >other: >function f(A)=sum(1/2)^n. i want to show,that f:N >--on-->(0,1],N-natural numbers, and i m traying to show taht f is a >srriection,(means For all x belongs to (0,1],there is exist A belongs >to Natural taht: f(A)=x.so question is how to show taht f :N >--on-->(0,1] This still makes no sense. You say f(A) = sum(1/2)^n, and apparently A is supposed to represent a natural number, but the right side doesnt mention A. In any case there is no surjection of the natural numbers N to (0,1], because (0,1] has greater cardinality. But maybe I can guess what is behind your question. Perhaps f is supposed to be a mapping from 2^N, i.e. the set of all subsets of the natural numbers, defined by f(A) = sum_{n in A} (1/2)^n. That would be a surjection of 2^N on [0,1]. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: how to show that set of solutions of sigma(from 0 to infinity) 1/2^n, belongs to (0,1] >other: >function f(A)=sum(1/2)^n. i want to show,that f:N >--on-->(0,1],N-natural numbers, and i m traying to show taht f is a >srriection,(means For all x belongs to (0,1],there is exist A belongs >to Natural taht: f(A)=x.so question is how to show taht f :N >--on-->(0,1] > This still makes no sense. You say f(A) = sum(1/2)^n, and apparently > A is supposed to represent a natural number, but the right side doesnt > mention A. In any case there is no surjection of the natural numbers > N to (0,1], because (0,1] has greater cardinality. > But maybe I can guess what is behind your question. > Perhaps f is supposed to be a mapping from 2^N, i.e. the set of all > subsets of the natural numbers, defined by f(A) = sum_{n in A} (1/2)^n. > That would be a surjection of 2^N on [0,1]. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada of course, it will be a surjection,but i wonder how to show it. exactly i dunno if i show good: let x belongs to (0,1] (x=0 other case),now i need to show that there exist n, which belongs to N,that 1/2^n=x (or sum_(n=1 to oo) 1/2^n=x????), so n=log_(2) 1/x . is it enough to show taht it is a suriection, or i need to use 1/2^n is a descend (A_n+1 < A_n),so maximum is 1/2^0 = 1,and inf = lim_(n-->oo) 1/2^n=0. or soemthing more ?. tahnks for any help! === Subject: Pronunciation of Dirichlet? Since he was of French descent, but lived in Germany, I am unsure of how to pronounce Dirichlet. I would appreciate your input. ~~~~~~~~~~~ Professor Gauss ~~~~~~~~~~~ To hear is to forget, To see is to remember, To do is to understand. -- Ancient Chinese proverb Remove caps when replying. -- Modern American saying === Subject: Re: Pronunciation of Dirichlet? >Since he was of French descent, but lived in Germany, I am unsure of how to >pronounce Dirichlet. I would appreciate your input. See the current thread Pronouce Galois. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: MATLAB 7.0.1 Release 14 SP1, and eBooks MATLAB 7.0.1 Release 14 SP1, and eBooks MATLAB 7.0.1 Release 14 SP1 (c) MathWorks [3 CDs] MATLAB 7 Release 14 - MathWorks [2 CDs] CD NR 15 824, COMSOL_FEMLAB_V3.0A John.Wiley.Sons.Computational.Colour.Science.using.MATLAB.eBoo k Orchard.Publications.Signals.and.Systems.with.MATLAB.Applicati ons.eBook Orchard.Publications.Circuit.Analysis.II.with.MATLAB.Applicati ons.eBook 10/01/2001 WaveWarp Audio Toolbox v2.0.1 for MatLab please send e-mail, code_fu@pathfinder.gr, astra35@freemail.gr, === Subject: Re: Victor Guggenheim > Chicago . I see he isnt listed there among present > or emeritus faculty, but there must be someone who remembers him. information. Youd think hed get the name spelled right. :-p Chakolate -- I am extraordinarily patient, provided I get my own way in the end. --Margaret Thatcher === Subject: Re: Victor Guggenheim >> Victor Guggenheim > A Google search for mathematics and Victor and Guggenheim comes up > with over 1500 hits. spelling it wrong. Chakolate -- r === Subject: Re: Victor Guggenheim >> Im a research assistant for a topologist who wants to include some >> data about an obscure mathematician named Victor Guggenheim in a >> paper. I say Ôobscure because we cant find anything about him, >> including his birthdate. >> Does anyone know or know of him, or where I could start to look? >> Ive run smack out of ideas of my own. > First of all, its Gugenheim (no double g). The Library of Congress > lists his birthdate as 1923. The Mathematics Genealogy Project lists > him as obtaining his Ph.D. at Oxford in 1952. He was a professor at > the University of Illinois at Chicago during 1970-1990 or so. I have > a vague recollection that someone had posted to the algebraic topology > mailing list that he had died in the not too distant past, but I cant > find it in the archives of the mailing list. I would suggest writing > to Don Davis dmd1 @AT@ lehigh.edu, who maintains the algebraic > topology list, asking him to post an inquiry to the list. > Zig Fiedorowicz name spelled incorrectly. Chakolate -- I am extraordinarily patient, provided I get my own way in the end. --Margaret Thatcher === Subject: Re: Victor Guggenheim >> Victor Guggenheim >A Google search for mathematics and Victor and Guggenheim comes up >with over 1500 hits. Nearly all of which are irrelevant, of course: typically theyre pages that mention somebody named Victor and somebody in Mathematics and a Guggenheim fellow (and even if these three are the same person, its irrelevant). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: roots of polynomials that contain phi >Can someone please point me to a webpage that contains all polynomials that >have their roots containing the number phi in some way? >I have discovered a polynomial that has roots that contain phi in some way >and would like to know if it has already been discovered before. For starters: (x-phi)*(x+1/phi) = x^2 - x - 1 (x+phi)*(x-1/phi) = x^2 + x - 1 (x-i*phi)*(x+i/phi) = x^2 - i*x + 1 (x+i*phi)*(x-i/phi) = x^2 + i*x + 1 --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: roots of polynomials that contain phi days. My association with the Department is that of an alumnus. >Now: any monic polynomial of degree 2 with integer coefficients whose >discriminant is 5*(a square) will be expressible in terms of phi. In >face, it will be expressible as a + b*phi, where a and b are integers. To be more inclusive: any quadratic polynomial with rational coefficients whose discriminant is 5*(a rational square) is expressible in terms of phi, as (a+b*phi)/d, where a, b, and d are integers. (For those who have been following certain threads involving algebraic integers: if the discriminant is 5*(a rational square), then the splitting field is Q(sqrt(5)), whose ring of integers is Z[(1+sqrt(5))/2] = Z[(1+phi)] = Z[phi]. So every element of Q(sqrt(5)) can be written as an element of Z[phi] divided by a rational integer). -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Embedding Boolean Algebras > eventually Ill get around to reading and commenting, Dont worry William, I was just pointing out that since P(B) is isomorphic to P(A)xP(BA), the fact that P(A) can be embedded in P(B), can also be viewed as a particular case of the fact that F can be embedded in FxG... (for all F, G Boolean algebras) Noel. === Subject: Re: Embedding Boolean Algebras === Subject: Re: Embedding Boolean Algebras > I was just pointing out that since P(B) is isomorphic to > P(A)xP(BA), the fact that P(A) can be embedded in P(B), can > also be viewed as a particular case of the fact that F can be > embedded in FxG... (for all F, G Boolean algebras) I dont think that proof is correct for it assumed a maximal ideal. The Boolean algebra of clopen subsets of Q, is an example of an Boolean algebra without a maximal ideal. The existence of an atom, gives a coatom which generates a maximal ideal. Conversely in a Boolean algebra, a maximal ideal gives a coatom, hence an atom. Thus there is no difference between your proof and his proof, except he is explict about assuming an atom. Whats that crammed together a<=~b mean? -- How is P(B) isomorphic to P(A) x P(BA) ? P(A) x P(BA) = { (X,Y) | X subset A, Y subset BA } Ok, thats acceptable. disjoint A,B ==> P(A/B) isomorphic P(A) x P(B) As for your embedding of P(A) into P(B), its limited for as was pointed out, any surjective map f:B -> A gives an embedding of P(A) into P(B). I think you picked a particular easy surjection, did you not? ---- === Subject: Re: Embedding Boolean Algebras > I dont think that proof is correct for it assumed a maximal ideal. > The Boolean algebra of clopen subsets of Q, is an example > of an Boolean algebra without a maximal ideal. This is Krulls theorem: Every ideal in a commutative ring (with identity) is included in at least one maximal ideal. This applies in particular to Boolean algebras (which are just commutative rings in which every element is idempotent, i.e. a^2=a) I cannot see how Krulls theorem should fail to be true for the Boolean algebra of clopen sets in Q... What am I missing? > The existence of an atom, gives a coatom which generates a maximal ideal. Yes. > Conversely in a Boolean algebra, a maximal ideal gives a coatom, Why? There are numerous Boolean algebras with no atoms. Yet there always exists a maximal ideal (Krulls theorem). Not every (maximal) ideal is principal... > Whats that crammed together a<=~b mean? <= is the partial order on a Boolean algebra, defined by a<=b iff a=ab ~b is the negation of b, otherwise known as b, or b^c when dealing with sub-algebras of a power set... > As for your embedding of P(A) into P(B), its limited for as was pointed > out, any surjective map f:B -> A gives an embedding of P(A) into P(B). Yes, this is very nice and simple. I agree. My preferred method in fact :-) > I think you picked a particular easy surjection, did you not? Yes I did. === Subject: Re: Embedding Boolean Algebras === Subject: Re: Embedding Boolean Algebras > I dont think that proof is correct for it assumed a maximal > ideal. The Boolean algebra of clopen subsets of Q, is an example > of an Boolean algebra without a maximal ideal. > This is Krulls theorem: > Every *proper* ideal in a commutative ring (with identity) > is included in at least one maximal (proper) ideal. Quick result of Zorns lemma. > This applies in particular to Boolean algebras (which are just > commutative rings in which every element is idempotent, i.e. a^2=a) Itdoesnot,Booleanalgebrashave a + a = a. > I cannot see how Krulls theorem should fail to > be true for the Boolean algebra of clopen sets in Q... Because Boolean algebras arent rings. > What am I missing? The difference between Boolean algebra and Boolean ring. > The existence of an atom, gives a coatom which generates a maximal > ideal. > Yes. > Conversely in a Boolean algebra, a maximal ideal gives a coatom, > Why? There are numerous Boolean algebras with no atoms. Yet there > always exists a maximal ideal (Krulls theorem). Not every (maximal) > ideal is principal... Doesnt apply. > Whats that crammed together a<=~b mean? > <= is the partial order on a Boolean algebra, defined by a<=b iff > a=ab ~b is the negation of b, otherwise known as b, or b^c when > dealing with sub-algebras of a power set... > As for your embedding of P(A) into P(B), its limited for as was > pointed out, any surjective map f:B -> A gives an embedding of P(A) > into P(B). > Yes, this is very nice and simple. I agree. > My preferred method in fact :-) > I think you picked a particular easy surjection, did you not? > Yes I did. Rub is, its not useful for embedding A into direct product of two Boolean algebras AxB. BTW, its easier to prove A embedds AxB and then immediately observe if D isomorphic AxB, then A embedds D. ---- === Subject: Re: Embedding Boolean Algebras > This applies in particular to Boolean algebras (which are just > commutative rings in which every element is idempotent, i.e. a^2=a) > Itdoesnot,Booleanalgebrashave a + a = a. Sure, we have a / a = a , but if you define: a+b = ab / ab Then (B,+, /) is a commutative ring (with unit) in which every element is idempotent. Hence Krulls theorem applies. Every Boolean algebra has a maximal (proper) ideal. === Subject: Re: Embedding Boolean Algebras > This is Krulls theorem: Every *proper* ideal in a commutative ring (with identity) is included in at least one maximal (proper) ideal. Noel. === Subject: Re: New countable infiniity logic >not as equally infintesimally close Are you under the impression that that string of words has any meaning? >may also generate the infintesimals that you describe Nowhere in the text that you quote does he use the word infinitesimal. >I dont see why you do not believe this function >outputs all such actual rational repeating and irrational decimalic >numbers as well as those that approach. Because you defined it in such a manner that all outputs terminate in repeating zeros. A decimal expansion terminating in repeating zeroes cannot equal on terminating in repeating non-zero digits or digit sequences, other than repeating nines. >It seems to me you do not accept that the infinite set of naturals >(or the set of naturals of infinte extent) must produce a list >infinitely long in extent for this function. What gives you that idea? >I believe and assert that the only way to produce a list infinitely >long in extent from this function is if the rational repeating and >irrational decimalic numbers are included. Then your definition is wrong. >0.1, 0.01, 0.001, ... whose series includes forming the number I >call the lowest infintesimal; There is no such number. You are, of course, free to call a number, e.g., joe, but if you want to communicate then you have to define what number joe refers to. >0.1, 0.12, 0.123, ... whose series includes forming the >Champernownes irrational number ; No. It doesnt include any irrational numbers. You have defined it in such a way as to only include rational numbers, and not even all of them. You are confusing rearranging the sequence with rearranging a subsequence, and you are confusing the contents of a sequence with its limit. >In set notation, these individual finite decimalic numbers are >elements that each appear only once, nevertheless all such numbers >are used infinitely many times to form all of the series re No. >and further include what I call the infintesimals No. You havent defined infinitesimal. >(where the number of zeroes before the finite decimal becomes >infinite in extent). Meaningless. The only such sequence of digits is all zeros. >I guess where I am still confused Youre confused about what a list is, youre confused about what set theory is and youre confusing the number of entries in a list with what is in one entry. You also seem to be using infinite in extent to mean several unrelated things. >is why a function which should >generate a list that is infinite in extent - since it is based on >the naturals (which are themselves infinite in extent) - is >nevertheless somehow not allowed to generate an infinite number of >digits. SPECIFIC FUNCTION *YOU* DEFINED DOES NOT GENERATE DECIMAL FRACTIONS WITH AN INFINITE NUMBER OF NONZERO DIGITS. >Note: because the only way the list for this specified function >(where a new digit is added for each succesive power of ten >numbers inserted into the function) can be infinite in extent is >if the number of digits themselves are infinite in extent. As I said, youre confusing the number of entries in a list with what is in one entry. Each entry on the list that *YOU* defined has only a finite number of nonzero digits. The fact that there are an infinite number of entries doesnt change that. >After all the function keeps adding one more digit every time an >additional power of ten from the naturals are utilized. The number >of digits therefore have the same property as the number of >naturals; namely there is no largest number associated with them. Which has nothing to do with the number on nonzero digits in any individual output of your function. >In the same way No. >(where the ... means the decimalic expansion adds the next digit >for the intended number forever) What do intended numbers have to do with the output of the specific function that you defined? >Is it not obvious that the only way such infinite decimalic expanded >numbers are able not to be included in this function is if the >naturals themselves are not infinite in extent? No. It isnt even true. >While I can no more specify the value of x in X=f(x) that will >produce any particular decimal of infinite extent from this >specified function Its not that *you* cant produce it; its that IT DOESNT EXIST. >this fact has no bearing The fact that it doesnt exist, however, does have bearing. >or if you prefer - will not become a part Meaningless. > After all, the only way that this function would not produce an >infinite number of digits is if there was a largest finite number N >in the naturals No. The only way would be if you defined it that way, which you did. >One can no more expect a person to be able to name It has nothing to do with being able to name it. It doesnt exist. >if one accepts that the number of digits for this function What do you mean by the number of digits for this function? There are infinitely many occurences of nonzero digits in the list of numbers your function produces, but only finitely many IN EACH INDIVIDUAL OUTPUT. approach inifintesimally close What do you mean by that. >but the actual rational and irrational decimalic numbers are >included as expressed by the use of the ... in 0.333... No. There is nothing like that in your definition. >That number (as one example) must be present on the list No. Your definition excludes it. >because no other number from this particular series (0.3, 0.33, >0.333 etc.) expresses anything other than the finiteness of the >number of digits d No, thats irrelevant. >and they are known to be of infinite extent for this function. Again, you are confusing the entire list with an individual number in the list. >Did I miss anything? Everything. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: functional equation >We now that is possible to define the base $a>1$ logarithmic function as >the only function f:(0,+inf) --> R such that: >1. x f(x)2. f(a) = 1 >3. f(xy) = f(x)+f(y) >Now using only this theorem (thinks as nobody knows the proof), is it >possible to prove the existence and uniqueness of the restriction of the >log_a to [1,+inf). i.e. that exist and is unique the function f:[1,+inf) >--> R such that 1. 2. and 3. are satisfied? I dont know what thinks as nobody knows the proof means. You seem to be asking, if there is a unique f on (0,infty) satisfying 1,2 and 3, is there a unique f on [1,infty) satisfying the same conditions (but for x,y in [1,infty))? Hint: Given any such function f on [1,infty), define g on (0,infty) by g(x) = f(x) for x >= 1, g(x) = -f(1/x) for 0 < x < 1. Prove that g satisfies 1, 2 and 3... Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: functional equation >>We now that is possible to define the base $a>1$ logarithmic function as >>the only function f:(0,+inf) --> R such that: >>1. x f(x)>2. f(a) = 1 >>3. f(xy) = f(x)+f(y) >>Now using only this theorem (thinks as nobody knows the proof), is it >>possible to prove the existence and uniqueness of the restriction of the >>log_a to [1,+inf). i.e. that exist and is unique the function f:[1,+inf) >>--> R such that 1. 2. and 3. are satisfied? > I dont know what thinks as nobody knows the proof means. > You seem to be asking, if there is a unique f on (0,infty) satisfying > 1,2 and 3, is there a unique f on [1,infty) satisfying the same > conditions (but for x,y in [1,infty))? > Hint: Given any such function f on [1,infty), define g on (0,infty) by > g(x) = f(x) for x >= 1, g(x) = -f(1/x) for 0 < x < 1. Prove that > g satisfies 1, 2 and 3... Ok, let me try to use your hint. Let f1,f2 be to functions satisfying 1,2 and 3. Then I can consider the two extensions g1 and g2 putting: g1(x) = f1(x) for x >= 1, g1(x) = -f1(1/x) for 0 < x < 1 and g2(x) = f2(x) for x >= 1, g2(x) = -f2(1/x) for 0 < x < 1 and these exstension satisfies 1,2 and 3; so the guaranteed unicity: g1 = g2 and by this equality also the equality of their restriction to [1,inf); i.e. f1=f2. qed. Is this proof right? === Subject: Re: Elliptic Curves for Crypto - Order of a Point >when working with elliptic curves (crypto), we have to know the order of the >group (that is, the number of points on the elliptic curve E). >For example, E: y^2 = x^3 + ax + b (mod p) over F{q}. >Also, when we choose a random point, we test to see that it is a point on >the curve. >If so, we need to determine the points order. >We typically do this (the naive way) by taking the point P = (x, y) and >finding: >P, 2P, 3P, ..., kP until kP is the point at infinity. >Is there an efficient way of determining the order of a point? For starters, note that with a few dozen multiplications you can compute 80313433200 P . That multiplier is the lcm of all numbers less than 29, and so that multiple of P will be the point at infinity if P has order divisible only by low primes (to not-very-high orders). Obviously theres nothing special about 29 here; you can easily compute very high multiples of P, which will be the identity element of E unless the order of P is divisible by a largish prime. This approach works with any other group too, of course, like (Z/n)^* . Thats why cryptographic applications which rely on N = pq being hard to factor have to arrange it so that (p-1) and (q-1) are not products of small factors. dave === Subject: Fields, Q, Z/pZ Can a field contain a copy of both Q and Z/pZ? Clearly if the field is finite, it cant contain Q. But why cant an infinite field contain Z/pZ? A field has infinite cardinality if it has infinite amount of elements. Couldnt some of these elements have finite order? If F is a field, we have a unique homomorphism from Z to F sending 1 to 1. If the kernel is 0 then F contains Z and thus Q. But this doesnt show that F doesnt have a copy of Z/pZ in it (at least I dont think it does). Any help is appreciated, Isaac === Subject: Re: Fields, Q, Z/pZ days. My association with the Department is that of an alumnus. >Can a field contain a copy of both Q and Z/pZ? No. > Clearly if the field is >finite, it cant contain Q. But why cant an infinite field contain Z/pZ? >A field has infinite cardinality if it has infinite amount of elements. >Couldnt some of these elements have finite order? No, they cannot have finite additive order. >If F is a field, we have a unique homomorphism from Z to F sending 1 to 1. >If the kernel is 0 then F contains Z and thus Q. But this doesnt show that >F doesnt have a copy of Z/pZ in it (at least I dont think it does). Let F be any field, and assume that A = { n in Z: n>0 and there exists x in F, x<>0, with nx=0} (where we interpret nx as x + x + x + ... + x, n summands) is nonempty. Let d>0 be the smallest element of A. Then there exists an x in F with dx = 0. That is, x + x + ... + x = 0 (d summands). Since x is not zero, and F is a field, it has a multiplicative inverse y. Multiplying by y we have 1 + 1 + ... + 1 = 0 (d summands) or d = 0. Therefore, for every element z of F z + z + ... + z (d summands) = z(1+1+...+1) (d summands) = zd = dz = 0z = 0. In particular, F cannot contain a copy of Q. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Fields, Q, Z/pZ >>Can a field contain a copy of both Q and Z/pZ? > No. >> Clearly if the field is >>finite, it cant contain Q. But why cant an infinite field contain Z/pZ? >>A field has infinite cardinality if it has infinite amount of elements. >>Couldnt some of these elements have finite order? > No, they cannot have finite additive order. >>If F is a field, we have a unique homomorphism from Z to F sending 1 to 1. >>If the kernel is 0 then F contains Z and thus Q. But this doesnt show >>that >>F doesnt have a copy of Z/pZ in it (at least I dont think it does). > Let F be any field, and assume that > A = { n in Z: n>0 and there exists x in F, x<>0, with nx=0} > (where we interpret nx as x + x + x + ... + x, n summands) > is nonempty. > Let d>0 be the smallest element of A. Then there exists an x in F with > dx = 0. > That is, x + x + ... + x = 0 (d summands). > Since x is not zero, and F is a field, it has a multiplicative inverse > y. Multiplying by y we have > 1 + 1 + ... + 1 = 0 (d summands) > or d = 0. > Therefore, for every element z of F > z + z + ... + z (d summands) = z(1+1+...+1) (d summands) > = zd > = dz = 0z = 0. > In particular, F cannot contain a copy of Q. I understand this, but then how do you on the other hand prove that if F contains a copy of Q, it doesnt contain a copy of Z/pZ? > -- > Its not denial. Im just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu === Subject: Re: Fields, Q, Z/pZ days. My association with the Department is that of an alumnus. >Can a field contain a copy of both Q and Z/pZ? >> No. > Clearly if the field is >finite, it cant contain Q. But why cant an infinite field contain Z/pZ? >A field has infinite cardinality if it has infinite amount of elements. >Couldnt some of these elements have finite order? >> No, they cannot have finite additive order. >If F is a field, we have a unique homomorphism from Z to F sending 1 to 1. >If the kernel is 0 then F contains Z and thus Q. But this doesnt show >that >F doesnt have a copy of Z/pZ in it (at least I dont think it does). >> Let F be any field, and assume that >> A = { n in Z: n>0 and there exists x in F, x<>0, with nx=0} >> (where we interpret nx as x + x + x + ... + x, n summands) >> is nonempty. >> Let d>0 be the smallest element of A. Then there exists an x in F with >> dx = 0. >> That is, x + x + ... + x = 0 (d summands). >> Since x is not zero, and F is a field, it has a multiplicative inverse >> y. Multiplying by y we have >> 1 + 1 + ... + 1 = 0 (d summands) >> or d = 0. >> Therefore, for every element z of F >> z + z + ... + z (d summands) = z(1+1+...+1) (d summands) >> = zd >> = dz = 0z = 0. >> In particular, F cannot contain a copy of Q. >I understand this, but then how do you on the other hand prove that if F >contains a copy of Q, it doesnt contain a copy of Z/pZ? Because if it contains a copy of Z/pZ, then it cannot contain a copy of Q. You do know that A implies B is equivalent to not(B) implies not(A), right? So: If F contains Z/pZ THEN F does not contain Q So: not(F does not contain Q) THEN not(F contains Z/pZ) So: F contains Q THEN F does not contain Z/pZ. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Fields, Q, Z/pZ === Subject: Re: Fields, Q, Z/pZ > Can a field contain a copy of both Q and Z/pZ? Clearly if the field is > finite, it cant contain Q. But why cant an infinite field contain Z/pZ? > A field has infinite cardinality if it has infinite amount of elements. > Couldnt some of these elements have finite order? There is a unique minimal subfield of K, containing 0 and 1. It is contained in any subfield of K (why?). Any subfield of K including K itself is a vector space over this minimal subfield. === Subject: Re: Fields, Q, Z/pZ >> Can a field contain a copy of both Q and Z/pZ? Clearly if the field is >> finite, it cant contain Q. But why cant an infinite field contain >> Z/pZ? >> A field has infinite cardinality if it has infinite amount of elements. >> Couldnt some of these elements have finite order? > There is a unique minimal subfield of K, containing 0 and 1. It is > contained in any subfield of K (why?). Any subfield of K including K > itself is a vector space over this minimal subfield. I see a little problem. Why is there a unique minimal subfield of K containing 0 and 1? Is it just by definition the intersection of all subfields of K? But then what if the minimal subfield (called it E) contained a copy of Q and Z/pZ? Then youd say that its not minimal, but how would you find the minimal subfield of E? Take the intersection of its subfields and I dont see how you arrive at a contradiction. === Subject: Re: Fields, Q, Z/pZ Can a field contain a copy of both Q and Z/pZ? Clearly if the field is >> finite, it cant contain Q. But why cant an infinite field contain >> Z/pZ? >> A field has infinite cardinality if it has infinite amount of elements. >> Couldnt some of these elements have finite order? > There is a unique minimal subfield of K, containing 0 and 1. It is > contained in any subfield of K (why?). Any subfield of K including K > itself is a vector space over this minimal subfield. > I see a little problem. Why is there a unique minimal subfield of K > containing 0 and 1? Every subfield of a field F contains 0 and 1. Now if Z_p and Q are both subfields F, then order of 1 is p, since 1 in Z_p and order of 1 is oo, since 1 in Q, thus a contradiction. === Subject: Re: Fields, Q, Z/pZ days. My association with the Department is that of an alumnus. > Can a field contain a copy of both Q and Z/pZ? Clearly if the field is > finite, it cant contain Q. But why cant an infinite field contain > Z/pZ? > A field has infinite cardinality if it has infinite amount of elements. > Couldnt some of these elements have finite order? >> There is a unique minimal subfield of K, containing 0 and 1. It is >> contained in any subfield of K (why?). Any subfield of K including K >> itself is a vector space over this minimal subfield. >I see a little problem. Why is there a unique minimal subfield of K >containing 0 and 1? Is it just by definition the intersection of all >subfields of K? Not by definition, but simply because the intersection of two subfields is a subfield, so the intersection of all subfields must itself be a subfield, and it is trivial to see then that it is the smallest one (hence minimal). >But then what if the minimal subfield (called it E) >contained a copy of Q and Z/pZ? Then youd say that its not >minimal, Yes: because then it would contain a SMALLER subfield (namely, both Q and Z/pZ), so it cannot possibly be the smallest, now, can it? > but >how would you find the minimal subfield of E? You intersect all the subfields. The smallest one cannot contain both Q and Z/pZ, because then it would contain an even smaller one, and you cannot have something which is the smallest and yet have even smaller things. >Take the intersection of its >subfields and I dont see how you arrive at a contradiction. You need to think a bit more; you seem to be giving up way too easily. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: anlysis probem I have the following Problem during analysis of the runtime of an algorithm. Consider you have a bowl with N white balls. You draw one and if it isnt marked you mark it an throw it back in the bowl. You redo this s times. The number of marked balls is m. 1.) My problem is I need the expected value of the function f(s,N)=E[m/N] 2.) second (this is maybe easy, if I have the function in a closed form) that the function converges to 1/s. This should be easy, if Im not wrong. What I have so far: the probability in one experiment, that I draw a marked ball should be (from here I use a little bit of the latex) p_N(s)=m/N the probability that I got a marked ball in the s draw then f(s,N)=(N-sum_{i=1}^{s-1}p_N(i))/N =1-frac{sum_{i=1}^{s-1} p_N(i)}{N} If we redo the last step we get =(sum_{i=1}^{s} (-1)^{i+1} frac{N}{N^i} )+(-1)^{s+2} g(N,s) where g(N,s)=1/N^i so we get finally = N * sum_{i=1}^{s+1} ( (-1)^{i+1} /(N^i)) and thats the point where I have no idea who to go on to get the desired result (I tried some like to calculate the inverse fourier where a_i= (-1)^{i+1}/ (N^i) and evaluate it at 0 or to check wether 1/N^s=1/N^{s+1}= 1/s - 1/(s+2) but also with that I didnt made any advances) === Subject: Re: anlysis probem >I have the following Problem during analysis of the runtime of an algorithm. >Consider you have a bowl with N white balls. >You draw one and if it isnt marked you mark it an throw it back in the >bowl. What do you do if it is marked? Also throw it back? >You redo this s times. >The number of marked balls is m. >1.) My problem is I need the expected value of the function f(s,N)=E[m/N] Hint: Lets write M_j for the number of marked balls after j drawings, to indicate the dependence on j. What is E[M_{j+1} | M_j] ? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: anlysis probem > What do you do if it is marked? Also throw it back? yes > Hint: Lets write M_j for the number of marked balls after j drawings, > to indicate the dependence on j. What is E[M_{j+1} | M_j] ? qutientenkriterium (dont know how it is called in englisch, but if a_{n+1}/a_nin (0,1) than a_i convergece absolute) because M_{j+1}=M_j+M_j/N, so M_{j+1}/M_j=1+1/N maybe I loose the sight but Iam not very fit in those things anymore === Subject: Re: anlysis probem > What do you do if it is marked? Also throw it back? >yes >> Hint: Lets write M_j for the number of marked balls after j drawings, >> to indicate the dependence on j. What is E[M_{j+1} | M_j] ? >qutientenkriterium (dont know how it is called in englisch, but if >a_{n+1}/a_nin (0,1) >than a_i convergece absolute) >because M_{j+1}=M_j+M_j/N, so M_{j+1}/M_j=1+1/N No, E[M_{j+1} | M_j] = M_j + (1-M_j/N) = (1-1/N) M_j + 1 So E[M_{j+1}] = (1-1/N) E[M_j] + 1 Now look for solutions of the form E[M_j] = a r^j + b. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Two worlds complementing each other Are you familiar with Abraham Ribinsons non-standard analysis and his notion of hyperreal numbers and does it have any relevance for you? === Subject: Re: Roots of x^3 = 1; at 09:53 PM, saju.pillai@gmail.com (Saju) said: >I am trying to find roots of x^3 = 1 Google for primitive roots of unity. In general, cos(n2Pi/N) + I sin(n2Pi/N) is an Nth root of unity; you only need to consider 0I am able to generate 1 real root and infinite complex roots No, because e^{i2Pi}=1. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Duel to Settle 0.999... =? 1 > There are people that have won Nobel Prizes In Physics with > schizophrenia. I > think Sir Isaac Newton had it too. Einstein had problems too. Its > like the formation of a pearl. It starts out as a problem and can turn > into something beautiful in time as they work around the problem. Well > look at Prof. S. Hawkins, a great thinker in higher level physics. He > tries to work around his problems. You should watch that move, A > Beautiful Mind. He won a Nobel Prize too. > So I wouldnt judge people too harshly, because just because you > dont > understand something, doesnt mean they are wrong or a stupid troll. > I never judged you, I was hoping you read the post and perhaps seek help. As for Hawkins, his problem is not with thinking clearly, it is a physical disability. Schizophrenia is a mental condition that sometimes does not allow you to think clearly, or to follow logic. If it is left untreated, it becomes worse. Trust me on this one, Ive dealt with myself. I watched the movie, A Beautiful Mind. It was trash. You should *read* *the* *book*, A Beautiful Mind. It is a much more accurate portrayal of the life of John Nash. It has not been done up for Hollywood and is based on his autobiography. Oh, and by the way, Nash did not win the Nobel prize. - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Smart the Schizo (was 0.9999 =/= 1) >>I watched the movie, A Beautiful Mind. It was trash. You should >>*read* *the* *book*, A Beautiful Mind. It is a much more accurate >>portrayal of the life of John Nash. It has not been done up for >>Hollywood and is based on his autobiography. >>Oh, and by the way, Nash did not win the Nobel prize. >> - Tim > Oh, yes he did. > http://www.popular-science.net/nobel/nash.html Silly, silly little man. John Nash won a prize in Economics (for his Nash Equilibria Points, I believe). There is no such thing as a Nobel Prize in Economics. Alfred Nobel made no such request in his will. The Nobel Prize is an international award given yearly since 1901 for achievements in physics, chemistry, medicine, literature and for peace. There is nothing in there about Economics. Im sorry, but you cannot win something that does not exist. Learn the facts, read a book. It is all perfectly well explained in A Beautiful Mind, the book. Hmm, Hollywood seemed to skip over that part in the movie. Please, learn the facts before you try to post things of which you know nothing. > You are a dang lie. John Nash was ( or still is) a mathematical > genius and I never said he wasnt a mathematical genius. I just said he never won the Nobel Prize. There simply is no Nobel Prize in mathematics nor in economics. > did have schizophrenia and so did alot of other mental geniuses of > physics, like Newton and Einstein. I also suffer from mild schizophrenia, thus I wasnt judging you. But I have seen what happens if it is not treated. That is why I was trying to help you. But then, I guess this means I am a mental genius but you have not claimed to actually have schizophrenia. Where does that leave you? > The problem is you atheists, which I believe are so full of crap > you cant > see straight enough to see you are mentally sick. I am not an atheist. I do believe in a God. I just do not believe *you* are a god or a creator. > Oh BTW, > I have shown you, > 1) DIM oo that .9999... == 1 > 2) Accelerated the derivative state to infinitesimal areas of > smallness to show, .999999.... == 1 > 3) Have proven Partial Sums only uses a limited amount of significant > figure to make a finite convergence of .9999... --> 1 > 4) And others are starting to agree with me. > 5) Maybe you should seek help. You cant see beyond about 16 > significant figures and you have 20 years of experience in math????? Actually, I can see to an infinite amount of significant figures. In fact, I was the one who posted computations up 999 significant figures. Maybe you should seek help. You cant see beyond 999 significant figures and you obviously have zero years of experience in math. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 The above link takes you to a site full of mindless ramblings. You would be wise not to visit. > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ Maybe I should start my own journal since no one else will publish by rubbish. Or maybe if I just produce quality material, respectable journals will publish it. - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Smart the Schizo (was 0.9999 =/= 1) >>I watched the movie, A Beautiful Mind. It was trash. You should >>*read* *the* *book*, A Beautiful Mind. It is a much more accurate >>portrayal of the life of John Nash. It has not been done up for >>Hollywood and is based on his autobiography. >>Oh, and by the way, Nash did not win the Nobel prize. >> - Tim > > Oh, yes he did. > > http://www.popular-science.net/nobel/nash.html > Silly, silly little man. John Nash won a prize in Economics (for his > Nash Equilibria Points, I believe). There is no such thing as a Nobel > Prize in Economics. Alfred Nobel made no such request in his will. > The Nobel Prize is an international award given yearly since 1901 for > achievements in physics, chemistry, medicine, literature and for peace. > There is nothing in there about Economics. Im sorry, but you cannot > win something that does not exist. Except this In 1968, the Bank of Sweden instituted the Prize in Economic Sciences in Memory of Alfred Nobel, founder of the Nobel Prize. This prize is run by the same organisation, and is in the name of Nobel, though, as you say, he didnt found it himself. Regeards Tim === Subject: Re: Smart the Schizo (was 0.9999 =/= 1) >I watched the movie, A Beautiful Mind. It was trash. You should >*read* *the* *book*, A Beautiful Mind. It is a much more accurate >portrayal of the life of John Nash. It has not been done up for >Hollywood and is based on his autobiography. >>Oh, and by the way, Nash did not win the Nobel prize. >> - Tim > >> Oh, yes he did. >> >> http://www.popular-science.net/nobel/nash.html >> Silly, silly little man. John Nash won a prize in Economics (for his >> Nash Equilibria Points, I believe). There is no such thing as a Nobel >> Prize in Economics. Alfred Nobel made no such request in his will. >> The Nobel Prize is an international award given yearly since 1901 for >> achievements in physics, chemistry, medicine, literature and for peace. >> There is nothing in there about Economics. Im sorry, but you cannot >> win something that does not exist. >Except this In 1968, the Bank of Sweden instituted the Prize in >Economic Sciences in Memory of Alfred Nobel, founder of the Nobel >Prize. >This prize is run by the same organisation, and is in the name of >Nobel, though, as you say, he didnt found it himself. The point remains John Nash did in fact win the Nobel Prize in economics. >Regeards >Tim Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ > ...and it goes downhill from there. > No it doesnt, no matter how many times you ßush, > occasionally some turds will remain. > To quote myself: > Focus well on negative Ôresponses. Are they vicious ranting? Are the > replies actually responsive? All of this from Elasticus, one of the most negative, ranting idiots on this NG. > Do they rant about gravity, or how Relativity > is proved correct a million times each day, or some other Ôwe are proved > right rave that doesnt deal in details about the debunking done here? Or more importantly You are wrong > is typically General Relativity or items about the energy and mass of moving > objects that are being waved at you, and such items are completely > irrelevant to coordinate transformations and invariance.. They are completely relevant to why Relativity is such an important theory BECAUSE IT PREDICTS PHENOMENA THAT CAN BE TESTED BY EXPERIMENT. > Just ask them for a list of all the observations that have been made of the > shortening (contraction) of moving objects that Special Relativity says > always occurs. The detection of muons The behaviour of electrons in synchotrons The detection of neutrons from distant supernovae All of the above involve the contraction of space and time dilation and prove that you are a doofus. > The detection of neutrons from distant supernovae COOL. Cite? I have never heard of that! >> The detection of neutrons from distant supernovae > COOL. > Cite? I have never heard of that! TitanPoint may mean this... URL:http://arxiv.org/PS_cache/astro-ph/pdf/0405/0405006.pdf (plenty more with Google Advanced, requiring the word supernova and the exact phrase detection of neutrons) The neutron is what is detected, in the decay/interaction of *neutrinos* from supernovae. With a halßife of 10 minutes, not many neutrons would survive from distant supernovae. Or it might have been a boo-boo. David A. Smith >> The detection of neutrons from distant supernovae > COOL. > Cite? I have never heard of that! > TitanPoint may mean this... > URL:http://arxiv.org/PS_cache/astro-ph/pdf/0405/0405006.pdf > (plenty more with Google Advanced, requiring the word supernova and the > exact phrase detection of neutrons) > The neutron is what is detected, in the decay/interaction of *neutrinos* > from supernovae. With a halßife of 10 minutes, not many neutrons would > survive from distant supernovae. > Or it might have been a boo-boo. Or the OP was an even more loathsome spewing idiot that you are. Hey stupid, what is the nominal half-life of a neutron? http://pdg.lbl.gov/ Hey stupid, what is the nominal half-life of a TeV neturon? A 10 TeV neutron? How would you (or somebody with a usable brain) make 10 TeV neutrons? -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf > The detection of neutrons from distant supernovae >> COOL. >> Cite? I have never heard of that! >> TitanPoint may mean this... >> URL:http://arxiv.org/PS_cache/astro-ph/pdf/0405/0405006.pdf >> (plenty more with Google Advanced, requiring the word supernova and >> the >> exact phrase detection of neutrons) >> The neutron is what is detected, in the decay/interaction of *neutrinos* >> from supernovae. With a halßife of 10 minutes, not many neutrons would >> survive from distant supernovae. >> Or it might have been a boo-boo. > Or the OP was an even more loathsome spewing idiot that you are. > Hey > stupid, what is the nominal half-life of a neutron? I think I said 10 minutes above, didnt I? Are you due for a dipstick check? > http://pdg.lbl.gov/ > Hey stupid, what is the nominal half-life of a TeV neturon? A 10 TeV > neutron? How would you (or somebody with a usable brain) make 10 TeV > neutrons? I think Red Sox will be involved. Or some lucky 12 TeV gamma radiation... Quit yer grousing. David A. Smith > The detection of neutrons from distant supernovae >> COOL. >> Cite? I have never heard of that! >> TitanPoint may mean this... >> URL:http://arxiv.org/PS_cache/astro-ph/pdf/0405/0405006.pdf >> (plenty more with Google Advanced, requiring the word supernova and >> the >> exact phrase detection of neutrons) >> The neutron is what is detected, in the decay/interaction of *neutrinos* >> from supernovae. With a halßife of 10 minutes, not many neutrons would >> survive from distant supernovae. >> Or it might have been a boo-boo. > Or the OP was an even more loathsome spewing idiot that you are. > Hey > stupid, what is the nominal half-life of a neutron? > I think I said 10 minutes above, didnt I? Are you due for a dipstick > check? Neutron half-life is about 885.6 seconds, idiot. Pull your thumb out when you spew information. > http://pdg.lbl.gov/ > Hey stupid, what is the nominal half-life of a TeV neturon? A 10 TeV > neutron? How would you (or somebody with a usable brain) make 10 TeV > neutrons? > I think Red Sox will be involved. Or some lucky 12 TeV gamma radiation... > Quit yer grousing. Whats the matter, idiot, cant you do arithmetic with Beta and mass-equivalent? If you cannot do it, why are you spewing crap disguised as knowledge? -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf > Neutron half-life is about 885.6 seconds, Relatively to whom?, You of course. Put a neutron in a powerful g-field or something similiar, inside a nucleus (think super-gravity) and the 1/2 life gets extended. Likely the neutron is stabilized from decay by the Strong Force aside from time dilation effects. Ken > Neutron half-life is about 885.6 seconds, > Relatively to whom?, You of course. > Put a neutron in a powerful g-field or > something similiar, inside a nucleus > (think super-gravity) and the 1/2 life > gets extended. > Likely the neutron is stabilized from > decay by the Strong Force aside from > time dilation effects. > Ken Idiot. Interstellar space is Minkowski. Tell us how many gees you would need to extend a 14.76 *minute* half-life to 100 million years. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf > > > Neutron half-life is about 885.6 seconds, > > Relatively to whom?, You of course. > > Put a neutron in a powerful g-field or > something similiar, inside a nucleus > (think super-gravity) and the 1/2 life > gets extended. > > Likely the neutron is stabilized from > decay by the Strong Force aside from > time dilation effects. > Ken > Idiot. Interstellar space is Minkowski. Tell us how many gees you > would need to extend a 14.76 *minute* half-life to 100 million years. Aunt Alice, Ref, G_uv =k*T_uv where T_uv has a density of billions of tons/cubic inch. Care to use your abacus to determine time dilation in that nuclear density? I do not know if the dilation is sufficient to impede the decay of the neutron in that density field, do you? Ken S. Tucker VOTE CREATIVE >> The detection of neutrons from distant supernovae >> COOL. >> Cite? I have never heard of that! >> TitanPoint may mean this... > URL:http://arxiv.org/PS_cache/astro-ph/pdf/0405/0405006.pdf > (plenty more with Google Advanced, requiring the word supernova and > the > exact phrase detection of neutrons) >> The neutron is what is detected, in the decay/interaction of > *neutrinos* > from supernovae. With a halßife of 10 minutes, not many neutrons > would > survive from distant supernovae. >> Or it might have been a boo-boo. >> Or the OP was an even more loathsome spewing idiot that you are. >> Hey >> stupid, what is the nominal half-life of a neutron? >> I think I said 10 minutes above, didnt I? Are you due for a dipstick >> check? > Neutron half-life is about 885.6 seconds, idiot. Pull your thumb out > when you spew information. A free neutron will decay with a half-life of about 10.3 minutes I dont know where your information comes from, but it is apparently not correct. > http://pdg.lbl.gov/ >> Hey stupid, what is the nominal half-life of a TeV neturon? A 10 TeV >> neutron? How would you (or somebody with a usable brain) make 10 TeV >> neutrons? >> I think Red Sox will be involved. Or some lucky 12 TeV gamma >> radiation... >> Quit yer grousing. > Whats the matter, idiot, cant you do arithmetic with Beta and > mass-equivalent? If you cannot do it, why are you spewing crap > disguised as knowledge? Did you mean gamma Uncle Al? Beta is an electron released in nuclear decay... 12 TeV for a neutron would be for a velocity that is very close to c, such that on a trip from a star 30,000 ly away, the light arrive about 48 minutes before the neutron pulse containing this neutron. I take it you did not bet on the Red Sox... David A. Smith >> The detection of neutrons from distant supernovae >> COOL. >> Cite? I have never heard of that! >> TitanPoint may mean this... > URL:http://arxiv.org/PS_cache/astro-ph/pdf/0405/0405006.pdf > (plenty more with Google Advanced, requiring the word supernova and > the > exact phrase detection of neutrons) >> The neutron is what is detected, in the decay/interaction of > *neutrinos* > from supernovae. With a halßife of 10 minutes, not many neutrons > would > survive from distant supernovae. >> Or it might have been a boo-boo. >> Or the OP was an even more loathsome spewing idiot that you are. >> Hey >> stupid, what is the nominal half-life of a neutron? >> I think I said 10 minutes above, didnt I? Are you due for a dipstick >> check? > Neutron half-life is about 885.6 seconds, idiot. Pull your thumb out > when you spew information. > correct. > http://pdg.lbl.gov/ Idiot. 14.76 minutes. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf > The detection of neutrons from distant supernovae >> COOL. >> Cite? I have never heard of that! >> TitanPoint may mean this... >> URL:http://arxiv.org/PS_cache/astro-ph/pdf/0405/0405006.pdf >> (plenty more with Google Advanced, requiring the word supernova >> and >> the >> exact phrase detection of neutrons) >> The neutron is what is detected, in the decay/interaction of >> *neutrinos* >> from supernovae. With a halßife of 10 minutes, not many neutrons >> would >> survive from distant supernovae. >> Or it might have been a boo-boo. >> Or the OP was an even more loathsome spewing idiot that you are. > Hey > stupid, what is the nominal half-life of a neutron? >> I think I said 10 minutes above, didnt I? Are you due for a > dipstick > check? >> Neutron half-life is about 885.6 seconds, idiot. Pull your thumb out >> when you spew information. >> A free neutron will decay with a half-life of about 10.3 minutes >> I dont know where your information comes from, but it is apparently not >> correct. >> http://pdg.lbl.gov/ > Idiot. 14.76 minutes. You should *not* get your information from idiots, Uncle Al. Your citation points to a different number for the lifespan of the neutron, the mean life, and a host of values are presented: .. choosing the entry for neutrons: PostScript PDF (17 pages) n Mean life [...snip...] 885.7 +/- 0.8 886.8 +/- 1.2 885.4 +/- 0.9 889.2 +/- 3.0 882.6 +/- 2.7 888.4 +/- 3.1 887.6 +/- 3.0 891 +/- 9 888.4 +/- 2.9 893.6 +/- 3.8 878 +/- 27 877 +/- 10 876 +/- 10 903 +/- 13 937 +/- 18 875 +/- 95 881 +/- 8 Note that the mean life and the half life can be mathematically correlated. So Uncle Al, who is the spewing idiot? *You* stated half life, but quoted figures for mean life. I think you need another beer... I hope this doesnt mean your eotvos project has been sidetracked? David A. Smith >> The detection of neutrons from distant supernovae >> COOL. >> Cite? I have never heard of that! >> TitanPoint may mean this... > URL:http://arxiv.org/PS_cache/astro-ph/pdf/0405/0405006.pdf > (plenty more with Google Advanced, requiring the word supernova and > the > exact phrase detection of neutrons) >> The neutron is what is detected, in the decay/interaction of > *neutrinos* > from supernovae. With a halßife of 10 minutes, not many neutrons > would > survive from distant supernovae. >> Or it might have been a boo-boo. >> Or the OP was an even more loathsome spewing idiot that you are. >> Hey >> stupid, what is the nominal half-life of a neutron? >> I think I said 10 minutes above, didnt I? Are you due for a dipstick >> check? > Neutron half-life is about 885.6 seconds, idiot. Pull your thumb out > when you spew information. > correct. > http://pdg.lbl.gov/ >> Hey stupid, what is the nominal half-life of a TeV neturon? A 10 TeV >> neutron? How would you (or somebody with a usable brain) make 10 TeV >> neutrons? >> I think Red Sox will be involved. Or some lucky 12 TeV gamma >> radiation... >> Quit yer grousing. > Whats the matter, idiot, cant you do arithmetic with Beta and > mass-equivalent? If you cannot do it, why are you spewing crap > disguised as knowledge? > Did you mean gamma Uncle Al? Beta is an electron released in nuclear > decay... > 12 TeV for a neutron would be for a velocity that is very close to c, such > that on a trip from a star 30,000 ly away, the light arrive about 48 > minutes before the neutron pulse containing this neutron. > I take it you did not bet on the Red Sox... > David A. Smith Is it just me, or has Uncle Al hit the sauce again? > The detection of neutrons from distant supernovae COOL. Cite? I have never heard of that! > TitanPoint may mean this... > URL:http://arxiv.org/PS_cache/astro-ph/pdf/0405/0405006.pdf > (plenty more with Google Advanced, requiring the word supernova and > the > exact phrase detection of neutrons) > The neutron is what is detected, in the decay/interaction of > *neutrinos* > from supernovae. With a halßife of 10 minutes, not many neutrons > would > survive from distant supernovae. > Or it might have been a boo-boo. >> Or the OP was an even more loathsome spewing idiot [than] you are. Logic check: How do you tell the difference between a neutron that travelled the entire way, from a neutron that was a local product of a neutrino that travelled the entire way (see the link I provided)? Arrival times perhaps? David A. Smith >> The detection of neutrons from distant supernovae > > COOL. > > Cite? I have never heard of that! TitanPoint may mean this... > URL:http://arxiv.org/PS_cache/astro-ph/pdf/0405/0405006.pdf > (plenty more with Google Advanced, requiring the word supernova and > the > exact phrase detection of neutrons) The neutron is what is detected, in the decay/interaction of > *neutrinos* > from supernovae. With a halßife of 10 minutes, not many neutrons > would > survive from distant supernovae. Or it might have been a boo-boo. >> Or the OP was an even more loathsome spewing idiot [than] you are. > Logic check: > How do you tell the difference between a neutron that travelled the entire > way, from a neutron that was a local product of a neutrino that travelled > the entire way (see the link I provided)? Arrival times perhaps? The neutron never made it and there is no way to sufficiently accelerate it on its way so it would. -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf >> The detection of neutrons from distant supernovae > COOL. > Cite? I have never heard of that! > TitanPoint may mean this... > URL:http://arxiv.org/PS_cache/astro-ph/pdf/0405/0405006.pdf > (plenty more with Google Advanced, requiring the word supernova and the > exact phrase detection of neutrons) > The neutron is what is detected, in the decay/interaction of *neutrinos* > from supernovae. With a halßife of 10 minutes, not many neutrons would > survive from distant supernovae. > Or it might have been a boo-boo. > David A. Smith Bingo. My source was Paul Davis excellent book About Time Time dilation must have been huge since they had to travel some 30,000 light years in less than 10 minutes of the neutrons proper time. I can predict the Henri Wilson reply: Mere Einsteiniana. And if that doesnt work then But this is what is predicted by the Ballistic Theory bull. Every piece of experimental evidence gets swatted away by the Wilson Ignorance Shield. === Subject: Re: Energy of Gravity is Nonlocal >So the loss here is in peoples conceptual models of the world, not >anything related to Nature. >> All the contacts we have with `nature are conceptual models. They are >> regularly proven wrong and exchanged for new `conceptual models. >Well, OK. There are nuanced differences in what we each meant by >conceptual model. No matter. All that has happened is that one >conceptual model (Newtonian mechanics) that contains global conservation >laws has been replaced by GR, which contains only local conservation >laws. But these local laws are indistinguishable from the old global >laws for all cases they have been examined, AFAIK, except for the binary >pulsar data. That data agrees with GR, not NM (Hulse and Taylor received >a Nobel Prize for that work). > [Yes, Im oversimplifying. But Im not omitting anything > essential to _this_ discussion.] >What I was trying to say is that the global conservation laws have an >intellectual appeal to humans, but such an appeal is irrelevant in >formulating models of Nature. >Tom Roberts tjroberts@lucent.com So we formulate our `conceptual models with the naive assumptions that the underlying basis for our `model has existed since time = 0 until the present and into the eternity of the future. Only the brave speculate about 0 - 10^ -43 sec. Do you have a conceptual model for that period of time? Always conserved (as far as known) : momentum and energy, angular momentum, electric charge/color/weak isospin, baryon number, lepton number, and a few other quantum statistics. And we confidently award a prize for the negative vacuum energy that we know little about, other than it is allegedly accelerating our slow (in our frame of reference) fade to black. We even conceptualize chaos as orderly disorder. A wrinkled space and accordion unidirectional time cones, confident that the speed of electromagnetic radiations is a constant limit. Sounds closer to hubris than intellectual appeal. I tend to be a bit cynical. IMHO, even our invented language of mathematics has more `intellectual appeal than the above melange of assumptions that appear to have worked for the anecdotal period of a few hundred years in vicinity of a small star. But carry on, it gives you something to do while DNA plays out its game. === Subject: Re: Energy of Gravity is Nonlocal >So the loss here is in peoples conceptual models of the world, not >anything related to Nature. >> All the contacts we have with `nature are conceptual models. They are >> regularly proven wrong and exchanged for new `conceptual models. >Well, OK. There are nuanced differences in what we each meant by >conceptual model. No matter. All that has happened is that one >conceptual model (Newtonian mechanics) that contains global conservation >laws has been replaced by GR, which contains only local conservation >laws. But these local laws are indistinguishable from the old global >laws for all cases they have been examined, AFAIK, except for the binary >pulsar data. That data agrees with GR, not NM (Hulse and Taylor received >a Nobel Prize for that work). > [Yes, Im oversimplifying. But Im not omitting anything > essential to _this_ discussion.] >What I was trying to say is that the global conservation laws have an >intellectual appeal to humans, but such an appeal is irrelevant in >formulating models of Nature. >Tom Roberts tjroberts@lucent.com > So we formulate our `conceptual models with the naive assumptions > that the underlying basis for our `model has existed since time = 0 > until the present and into the eternity of the future. Only the brave > speculate about 0 - 10^ -43 sec. So we have brave souls in physics - why is that bad? > Do you have a conceptual model for > that period of time? Maybe - see http://nedwww.ipac.caltech.edu/level5/Guth/Guth_contents.html. > Always conserved (as far as known) : momentum and > energy, angular momentum, electric charge/color/weak isospin, baryon > number, lepton number, and a few other quantum statistics. And we > confidently award a prize for the negative vacuum energy that we know > little about, other than it is allegedly accelerating our slow (in our > frame of reference) fade to black. Yea because it solves a number of puzzells eg from the above link ÔWhile it may be too early to say that inßation is proved, I claim that the case for inßation is compelling. It is hard to even conceive of an alternative theory that could explain the basic features of the observed Universe. Not only does inßation produce just the kind of special bang that matches the observed Universe, but quantum ßuctuations during inßation could have produced nonuniformities which served as the seeds of cosmic structure. These nonuniformities can be observed directly in the cosmic background radiation, with an amplitude of about one part in 100,000. So far the measurements of the spectrum have been beautifully consistent with the predictions of inßation, although it must be admitted that nonuniformities created by cosmic strings are also consistent with the observations. Cosmic strings, however, cannot explain the large-scale homogeneity or the ßatness of the Universe. Of course that is not he work Hulse and Taylor received Nobel Prize for. > Hulse and Taylor received >a Nobel Prize for that work). >We even conceptualize chaos as > orderly disorder. Have not heard that one before. > A wrinkled space and accordion unidirectional time > cones, confident that the speed of electromagnetic radiations is a > constant limit. We are not 100% confident of that at all - but what really going on is probably beyond your ken. > Sounds closer to hubris than intellectual appeal. I > tend to be a bit cynical. Your cynicism is fine - simply do not confuse it with science. > IMHO, even our invented language of mathematics has more `intellectual > appeal than the above melange of assumptions that appear to have > worked for the anecdotal period of a few hundred years in vicinity of > a small star. > But carry on, it gives you something to do while DNA plays out its > game. It is obvious the above poster understands nothing or at the most little about science. His philosophical evaluation of a theories foundations are irrelevant - what is relevant is correspondence with experiment. Now I wonder if he has any concerns on those grounds? Bill === Subject: Re: The real numbers, and general comments >What makes you say that? It is this minimalist assumption that creates >interesting questions such as whether there are inaccessible cardinals. >> This question is only interesting if their existence is not dependent >> on an arbitrary choice of model, so we must go outside ZFC to answer >> ir. >> The are interesting questions precisely because it is not known if they >> are dependent on the choice of the model. If a model exists where >> inaccessible cardinals do exist, and another exists where they dont >> exist, then you can create a new axiom about the existence of >> inaccessible cardinals. On the other hand, if all known models do or do >> not contain inaccessible cardinals, then perhaps there is another >> theorem available to be proven. Given any model, one can construct one with no inaccessible cardinals. One cannot construct a model with inaccessible cardinals unless one assumes that this can be done; it cannot be proved consistent with ZF, as this would enable a proof of consistency of ZF within itself. >Well, to me, this shows either that ZF is too limited or that these >questions really are not interesting. For, if they are, why no >develope a better foundation that can answer them? There are theories with inaccessible cardinals. There is a large amount of literature on this. > Larger models can have additional properties that are not necessary >consequences of the axioms. One can add the axioms about existence of inaccessible cardinals, and even of different types of them. >> They are, however, necessary consequences of the model. >> Usually, at least in my experience, the model is of less interest than >> the axioms and theorems. Models are useful primarily to prove that >> axioms are consistent or to help suggest new theorems. >Sorry, but I am still not clear on exactly what a Ômodel is. I cant >imagine its anything other than new axioms (which may be propsed >theorems, of course). A model is a realization of the axioms. Godels proof of consistency of V = L, and hence of AC and GCH, consisted of constructing a model within a given model which satisfied this. Skolems proof of the existence of countable models, and its generalizations, consist of construction of such models given other models. Other consistency proofs involve such constructions of models from assumed models, including proofs about consistency of negations. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: The real numbers, and general comments >> The are interesting questions precisely because it is not known if they >> are dependent on the choice of the model. If a model exists where >> inaccessible cardinals do exist, and another exists where they dont >> exist, then you can create a new axiom about the existence of >> inaccessible cardinals. On the other hand, if all known models do or do >> not contain inaccessible cardinals, then perhaps there is another >> theorem available to be proven. > Given any model, one can construct one with no inaccessible > cardinals. One cannot construct a model with inaccessible > cardinals unless one assumes that this can be done; it > cannot be proved consistent with ZF, as this would enable a > proof of consistency of ZF within itself. Then, it is more parsimonious to assume no inaccessible, unless we want to use a different theory than ZFC, right? >> They are, however, necessary consequences of the model. >> Usually, at least in my experience, the model is of less interest than >> the axioms and theorems. Models are useful primarily to prove that >> axioms are consistent or to help suggest new theorems. >Sorry, but I am still not clear on exactly what a Ômodel is. I cant >imagine its anything other than new axioms (which may be propsed >theorems, of course). > A model is a realization of the axioms. Godels proof of > consistency of V = L, and hence of AC and GCH, consisted of > constructing a model within a given model which satisfied > this. Skolems proof of the existence of countable models, > and its generalizations, consist of construction of such > models given other models. Other consistency proofs involve > such constructions of models from assumed models, including > proofs about consistency of negations. Yes. But I dont see why a model is anything different from a set of axioms defining it. Andrew Usher === Subject: Re: The real numbers, and general comments > Sorry, but I am still not clear on exactly what a Ômodel is. I cant > imagine its anything other than new axioms (which may be propsed > theorems, of course). A model of a system of axioms is a class of objects that satisfy those axioms. A model exists if and only if the axioms are consistent. Back in junior high school geometry, the teacher said we could start from a few undefined terms (point, line, etc.) and some axioms about them, to prove interesting geometrical theorems. She also drew some pictures and gave some explanations about the meaning of these terms such that I felt I could visualize them, so I was confused by the meaning of undefined. Later on I learned about non-euclidean geometry. I read that there is more than one geometry that satisfies the standard geometrical axioms minus the parallel postulate. These are actually different models of those axioms. For example, you can construct a non-euclidean plane geometry on the surface of a sphere, where lines are modeled by great circles, and there exist no parallel lines. Again leaving out the parallel postulate, this model satisfies the same geometrical axioms as does the model of plane geometry I visualized in school. If we prove a theorem from the axioms, it applies to any model of the axioms, whether a line is a great circle or one of those infinitely long & straight things in my imagination. point of view, a model is what you get when you define all the undefined terms and get something that satisfies the axioms. However, these two models differ in a number of properties (such as the existence of parallel lines) that arent determined by the axioms. A statement about these properties would therefore be *undecidable* from the axioms; there are models that answer it both ways. This example exhibits a property of all (consistent & sufficently rich) axiom systems--they have multiple models. Adding more axioms can decide more undecidable propositions, but as long as your system stays consistent it will have multiple models and undecidable propositions. This phenomenon shows up all the time in set theory, general topology, and other foundational branches of mathematics. There are many many statements that have been shown to be consistent with and independent of ZFC. What this means is that someone has, starting from the assumption that ZFC is consistent, given proofs of the existence of a model of ZFC in which the statement is true and another in which it is false. And theres nothing special about ZFC in this regard. Bertrand Russell wasnt really kidding when he said, Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true. === Subject: Re: The real numbers, and general comments <41726a73$1_5@newsfeed.slurp.net> <417652a7_5@newsfeed.slurp.net> <417cea9c$1_3@newsfeed.slurp.net> >Later on I learned about non-euclidean geometry. I read that there is >more than one geometry that satisfies the standard geometrical >axioms minus the parallel postulate. These are actually different >models of those axioms. For example, you can construct a >non-euclidean plane geometry on the surface of a sphere, where lines >are modeled by great circles, and there exist no parallel lines. >Again leaving out the parallel postulate, this model satisfies the >same geometrical axioms as does the model of plane geometry I >visualized in school. Actually not: a pair of distinct great circles intersects in two points rather than in one. To get Riemannian Geometry you must identify antipodes so that two lines intersect in a single point. But your basic point is valid; there are models for Elliptic Geometry and for Hyperbolic Geometry in Euclidean Geometry. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Basis for Q(pi) over Q? What is a basis for Q(pi) over Q? Well clearly the extension is infinite. Also, Q(pi) = { f(pi)/g(pi) where f, g are in Q[x]}, so the extension is countable. Furthermore, since pi is transcendental over Q, Q(pi) = Q(x), just rational functions in x with coefficients in Q. Well I thought that x^n where n is a natural number was a basis for Q(x) over Q, but then I realized that 1/(1+x) cant be written as a linear combination of these guys. My suspicion is that a basis at least contains the following set : { (1/h(x))^n | h(x) is irreducible in Q[x] and n is a natural number } We need powers here because if we didnt, then we cant get the function 1/x^2 I dont think. Any ideas? Isaac === Subject: Re: Basis for Q(pi) over Q? > Any ideas? partial fractions === Subject: Re: Basis for Q(pi) over Q? days. My association with the Department is that of an alumnus. >What is a basis for Q(pi) over Q? As a vector space over Q? > Well clearly the extension is infinite. >Also, Q(pi) = { f(pi)/g(pi) where f, g are in Q[x]}, so the extension is >countable. >Furthermore, since pi is transcendental over Q, Q(pi) = Q(x), just rational >functions in x with coefficients in Q. Well I thought that x^n where n is a >natural number was a basis for Q(x) over Q, but then I realized that 1/(1+x) >cant be written as a linear combination of these guys. {x^n : n>0} is a basis for Q[x], not for Q(x). >My suspicion is that a basis at least contains the following set : { >(1/h(x))^n | h(x) is irreducible in Q[x] and n is a natural number } We >need powers here because if we didnt, then we cant get the function 1/x^2 >I dont think. Yes. But they arent enough: how do you write x/(1+x^2) ? >Any ideas? If you think back to partial fractions, youll see that any expression of the form f(x)/g(x) can be written as a polynomial, plus fractions of the form u(x)/v(x)^n, with deg(u)0. So {1} U {x^n : n>0 } U {u(x)/[v(x)]^n: u,v in Q[x], v irreducible, deg(u)0} generates as a Q-vector space. You can make it a bit better by writing every polynomial over Q as c*f(x) with f(x) primitive and with integer coefficients, and c a rational. So then we can replace the above with {1} U {x^n : n > 0} U {u(x)/[v(x)]^n : u,v in Z[x] primitive, v irreducible over Z[x]} : deg(u)0 } Can you show whether or not they are linearly independent? -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: How does one pronoune Dirichlet? How does one pronounce Dirichlet? Btw, does anyone here have inßuence at MacTutor (http://www-gap.dcs.st-and.ac.uk/~history/)? If so could they drop a hint that guides to pronunciation there would be handy? === Subject: Re: How does one pronoune Dirichlet? > How does one pronounce Dirichlet? Most repsonses have given the French-type pronunciation. There are those who maintain that it should have a German-type pronunciation... For example, the final T should be pronuounced. See... http://mathforum.org/discuss/sci.math/m/456731/647838 -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: How does one pronoune Dirichlet? > How does one pronounce Dirichlet? > Most repsonses have given the French-type pronunciation. > There are those who maintain > that it should have a German-type pronunciation... For example, > the final T should be pronuounced. See... > http://mathforum.org/discuss/sci.math/m/456731/647838 > -- D> G. A. Edgar http://www.math.ohio-state.edu/~edgar/ Actually only mine was the real French pronunciation, where the others all had long ea-sounds, which would be how English speaking people incorrectly pronounce it. The i-sounds should be ultra-short, like in dish, rick, and cleptomania... certainly not like dih-reesh-lay or DeaReacley. Horror! Knowing where Dirichlets family originated (Belgium, the French speaking part), the name should of course be pronounced the French way. See http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/ Dirichlet.html Dirk Vdm === Subject: Re: How does one pronoune Dirichlet? ... > Knowing where Dirichlets family originated (Belgium, the French > speaking part), the name should of course be pronounced the French > way. I do not see why. Frequently, when a family moves to another country, the pronunciation of the name (also by the family itself) will change. A prime example is Dvorak, inventor of the Dvorak keyboard, compared to his family in Czechoslovakia (with the composer as a member). But the US branch of the family also dropped the hacek from the Ôr. Most Americans would not have any idea how to deal with that. See also van Beethoven who himself frequently Germanised his name to von Beethoven. Or Roosevelt, where the pronunciation has been anglified (the name is of Dutch origin, the meaning is field of roses). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: How does one pronoune Dirichlet? > ... > > Knowing where Dirichlets family originated (Belgium, the French > > speaking part), the name should of course be pronounced the French > > way. > I do not see why. Frequently, when a family moves to another country, the > pronunciation of the name (also by the family itself) will change. A prime > example is Dvorak, inventor of the Dvorak keyboard, compared to his family > in Czechoslovakia (with the composer as a member). But the US branch of > the family also dropped the hacek from the Ôr. Most Americans would not > have any idea how to deal with that. Just like Americans and English (and jullie Nederlanders) have a problem pronouncing a really real French Ôr ;-) > See also van Beethoven who himself > frequently Germanised his name to von Beethoven. Or Roosevelt, where > the pronunciation has been anglified (the name is of Dutch origin, the > meaning is field of roses). > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Well, being raised with virtually two languages (Dutch and French), hearing a clearly French name pronounced the English way, makes us smile with a feeling of... compassion at least. Knowing that the family moved to Germany though, I could live with the German way, but the proposed DeaReaCley really *does* sound very silly. Germans would probably pronounce the ch as such in stead of as k, and they would indeed pronounce the final t. Op je gezondheid! Dirk Vdm === Subject: Re: How does one pronoune Dirichlet? ... > Just like Americans and English (and jullie Nederlanders) have > a problem pronouncing a really real French Ôr ;-) No, only some of them. In Dutch about three or four pronunciations of the Ôr are common. I cannot pronounce a French Ôr, but my daughter does (she does not know how to pronounce the real, truly, rolling Dutch labial Ôr, which is also so obvious in the speech of Bavarian actors). > Well, being raised with virtually two languages (Dutch and French), > hearing a clearly French name pronounced the English way, makes > us smile with a feeling of... compassion at least. > Knowing that the family moved to Germany though, I could live with > the German way, but the proposed DeaReaCley really *does* > sound very silly. > Germans would probably pronounce the ch as such in stead of > as k, and they would indeed pronounce the final t. Strange enough, I learned it with the ichlaut for ch but with omitted t. But mathematicians are at least trying. Listen to some sports commentators. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: How does one pronoune Dirichlet? > ... > > Just like Americans and English (and jullie Nederlanders) have > > a problem pronouncing a really real French Ôr ;-) > No, only some of them. In Dutch about three or four pronunciations of the > Ôr are common. I cannot pronounce a French Ôr, but my daughter does (she > does not know how to pronounce the real, truly, rolling Dutch labial Ôr, > which is also so obvious in the speech of Bavarian actors). Labial Ôr? Alveolar, yes. Uvular, yes. Velar, yes. Labial, woh! In English there is a bilabial trill that is only found before Ôrs, but its use if often indicative of a speech impediment or immaturity (e.g. Ôbroom broom /br[um br[um/ for car.) > > Well, being raised with virtually two languages (Dutch and French), > > hearing a clearly French name pronounced the English way, makes > > us smile with a feeling of... compassion at least. > > Knowing that the family moved to Germany though, I could live with > > the German way, but the proposed DeaReaCley really *does* > > sound very silly. > > Germans would probably pronounce the ch as such in stead of > > as k, and they would indeed pronounce the final t. > Strange enough, I learned it with the ichlaut for ch but with omitted t. > But mathematicians are at least trying. Listen to some sports commentators. However, feeding Ôcanonicalised (a -> Ôae, etc.) names to sports commentators in a country with a phonetic language is just cruel. Hearing a Finnish commentator naively trying to pronounce all four vowels in Ôaeae is painful. Give them the original long vowel aa and they do just fine. Just dropping the diacriticals would be accurate than multiplying the number of letters. Phil -- They no longer do my traditional winks tournament lunch - liver and bacon. Its just what you need during a winks tournament lunchtime to replace lost === Subject: Re: How does one pronoune Dirichlet? > ... > > Just like Americans and English (and jullie Nederlanders) have > > a problem pronouncing a really real French Ôr ;-) > > No, only some of them. In Dutch about three or four pronunciations of the > Ôr are common. I cannot pronounce a French Ôr, but my daughter does (she > does not know how to pronounce the real, truly, rolling Dutch labial Ôr, > which is also so obvious in the speech of Bavarian actors). > Labial Ôr? > Alveolar, yes. Uvular, yes. Velar, yes. Labial, woh! > In English there is a bilabial trill that is only found before Ôrs, > but its use if often indicative of a speech impediment or immaturity > (e.g. Ôbroom broom /br[um br[um/ for car.) Sorry, I was asleep indeed, I meant alveolar. > Strange enough, I learned it with the ichlaut for ch but with omitted > t. > But mathematicians are at least trying. Listen to some sports > commentators. > However, feeding Ôcanonicalised (a -> Ôae, etc.) names to sports > commentators in a country with a phonetic language is just cruel. Indeed, that is a bit silly. > Give them the original long vowel > aa and they do just fine. Just dropping the diacriticals would be > accurate than multiplying the number of letters. I think you meant *more* accurate... But dropping diacriticals is just what creates wrong pronunciations in many cases. One example I actually have heard is quite some time ago about a Romanian football player who also played some time in the Netherlands. When playing elswhere a letter Ôa was changed to a letter Ôi. The commentator correctly noted that his name had changed, and switched from using a Dutch Ôa at that place to a Dutch Ôi. Indeed his name was changed, due to a change of orthography in Romanian, but the pronunciation was not changed. The change was that a-breve was subsequently written as i-circonßex, the pronunciation remained schwa-like. And how about that commentator that thought two Chinese participants where sisters because they shared the first name? As a guide for pronunciation t English speakers. Use for the vowels in foreign names the pronunciation as is used in German, French, Italian, Spanish or whatever other language except English. You will already be much closer. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: How does one pronoune Dirichlet? Sorry, but I too am interested in the answer to this question; however, I do not speak French or Dutch. I do speak German; however, even if the correct pronunciation of Dirichlet is German, I would not know which syllable to accent. So, could one of you please enlighten me as to how *you* think it should be pronounced? Perhaps you can describe it using some kind of English-based phonetic scheme (I know, I know, thats an oxymoron). I have been pronouncing it dear-ISH-lett, with an anglicized sh instead of German -- ~~~~~~~~~~~ Professor Gauss ~~~~~~~~~~~ To hear is to forget, To see is to remember, To do is to understand. -- Ancient Chinese proverb Remove caps when replying. -- Modern American saying > ... > Just like Americans and English (and jullie Nederlanders) have > a problem pronouncing a really real French Ôr ;-) > No, only some of them. In Dutch about three or four pronunciations of the > Ôr are common. I cannot pronounce a French Ôr, but my daughter does > (she > does not know how to pronounce the real, truly, rolling Dutch labial Ôr, > which is also so obvious in the speech of Bavarian actors). > Well, being raised with virtually two languages (Dutch and French), > hearing a clearly French name pronounced the English way, makes > us smile with a feeling of... compassion at least. > Knowing that the family moved to Germany though, I could live with > the German way, but the proposed DeaReaCley really *does* > sound very silly. > Germans would probably pronounce the ch as such in stead of > as k, and they would indeed pronounce the final t. > Strange enough, I learned it with the ichlaut for ch but with omitted > t. > But mathematicians are at least trying. Listen to some sports > commentators. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, > +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; > http://www.cwi.nl/~dik/ === Subject: Re: How does one pronoune Dirichlet? > Sorry, but I too am interested in the answer to this question; however, I do > not speak French or Dutch. I do speak German; however, even if the correct > pronunciation of Dirichlet is German, I would not know which syllable to > accent. So, could one of you please enlighten me as to how *you* think it > should be pronounced? Perhaps you can describe it using some kind of > English-based phonetic scheme (I know, I know, thats an oxymoron). I have > been pronouncing it dear-ISH-lett, with an anglicized sh instead of German I think that stress on the middle syllable (as you use) is extremely unlikely. In German it would be the first or last syllable. French barely uses stress, but I think there is in French a very slight stress on the last syllable (as in all French words). Considering this I think you should put the stress on the last syllable. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: How does one pronoune Dirichlet? >Sorry, but I too am interested in the answer to this question; however, I do >not speak French or Dutch. I do speak German; however, even if the correct >pronunciation of Dirichlet is German, I would not know which syllable to >accent. So, could one of you please enlighten me as to how *you* think it >should be pronounced? Im German and I always pronounced it French. French used to be the language of the upper class in Germany. So undoubtedly his name was pronounced French (with German accent of course) then. A truly German pronunciation would be considered a little ignorant. >Perhaps you can describe it using some kind of >English-based phonetic scheme (I know, I know, thats an oxymoron). I have >been pronouncing it dear-ISH-lett, with an anglicized sh instead of German quite correct though, as was pointed out already. Try to remove the i-sound at the end of it, keep the sound even. Thomas === Subject: Re: How does one pronoune Dirichlet? >Sorry, but I too am interested in the answer to this question; however, I do >not speak French or Dutch. I do speak German; however, even if the correct >pronunciation of Dirichlet is German, I would not know which syllable to >accent. In French, the accent should go to the last syllable. My feeling of German (my 3rd language) tells me the accent should be on the first or last, but certainly not on the second. > So, could one of you please enlighten me as to how *you* think it >should be pronounced? > Im German and I always pronounced it French. French used to be the > language of the upper class in Germany. So undoubtedly his name was > pronounced French (with German accent of course) then. A truly German > pronunciation would be considered a little ignorant. >Perhaps you can describe it using some kind of >English-based phonetic scheme (I know, I know, thats an oxymoron). I have >been pronouncing it dear-ISH-lett, with an anglicized sh instead of German Like I said, keep all the syllables *short* di-ri-kle di like in DIsh ri like in RIch kle like in CLEptomania Short syllables only. Dirk Vdm > quite correct though, as was pointed out already. Try to remove the > i-sound at the end of it, keep the sound even. > Thomas === Subject: Re: How does one pronoune Dirichlet? > How does one pronounce Dirichlet? > Btw, does anyone here have inßuence at MacTutor > (http://www-gap.dcs.st-and.ac.uk/~history/)? If so could they drop a > hint that guides to pronunciation there would be handy? Gee... the 3 replies you already got... Aw! Pronounce it like di-ri-kle, where all parts are pronounced as follows: di like in dish ri like in rich but with a Frrrrench rrrrrrrr kle like in cleptomania Dirk Vdm === Subject: Re: How does one pronoune Dirichlet? Poincare got me... Try Pisot or Yoccoz or Benoit. or Fatou or Douady... The French have a big hand in Modern Mathematics! I hate it when a snob turns his nose up at you for misprouncing something youve only read about in books, ha, ha... And they so love to make you feel uneducated in the Eastern Ivy sense. Id only read his name in number theory books , never heard it in a lecture. DeaReacley... awful. >How does one pronounce Dirichlet? >Btw, does anyone here have inßuence at MacTutor >(http://www-gap.dcs.st-and.ac.uk/~history/)? If so could they drop a >hint that guides to pronunciation there would be handy? -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : alternative email: rlbtftn@netscape.net URL : http://home.earthlink.net/~tftn === Subject: Re: How does one pronoune Dirichlet? Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >How does one pronounce Dirichlet? Dih-reesh-lay -- Richard === Subject: Re: How does one pronoune Dirichlet? > How does one pronounce Dirichlet? Like DeaReacley; ea like in tea, french (german) R, cl as in close and ey as in ... hmm ... hard to say exactly, almost as in Rooney I guess. === Subject: Re: How does one pronoune Dirichlet? > How does one pronounce Dirichlet? > Like DeaReacley; ea like in tea, french (german) R, cl as in close and > ey as in ... hmm ... hard to say exactly, almost as in Rooney I guess. knows that === Subject: Re: Dilworths Lemma > By induction? Yicks, this is looking messy. Whats youre version of > Diworths lemma and what obscure text whence it comes? >> Surely there must be nicer way to work the proof. Ill give what I >> have upto where I first apply PHP. > Whats PHP? I dont see any reason for invading this quagmire unless > you can point out to me where in my reasonings of the previous posts > Im wrong about use of Dilworths lemma (as given above from > referenced paper) My version of Dilworths Lemma is as follows: In any partial order on a set X of m*n + 1 elements, there exists a chain of length m+1 ir an antichain of size n+1. (Introduction to Combinatorics, by Martin J. Erickson, 1996) My thought was that before you can apply Dilworths Lemma, you must make sure that the order partitions X into sets, some with property you are looking for and some without. As an extremely obvious example of what I mean: Let my partial ordering be a ~ b if a My version of Dilworths Lemma is as follows: > In any partial order on a set X of m*n + 1 elements, there exists a > chain of length m+1 ir an antichain of size n+1. Thats a weak version of the Dilworths Lemma I sent with proof hint (if you look closely). > My thought was that before you can apply Dilworths Lemma, you must > make sure that the order partitions X into sets, some with property > you are looking for and some without. As an extremely obvious > example of what I mean: How does an order partition its set of reßexive elements? > Let my partial ordering be a ~ b if a Using that ordering, I cannot prove the following. > Given any set of n^2 + 1 integers, either n+1 of them are > divisible by 3 or n+1 of them are not divisible by 3. Where did 3 come from? > It is a true statement, and can be proven. However, the partial > order I stated cannot be used to prove it because that order does not > preserve the property of divisibilty by 3. Of course not, it talked about even numbers. What do you mean by an order preserving a property? > And so to prove this, my partial ordering would somehow have > to create chains or antichains that show divisibility by 3. Why not do same thing, the suborder of N ordering only 3N ? Ie a <<= b when a <= b and a,b in 3N or a = b. Namely <<= = (<= / 3Nx3N) assuming you can deciper the C++ opaque terseness of the notation. Another interesting order is a <<= b when a <= b, a = b (mod n). Thus the residual groups of Z_n, ie the cofactor of nZ are chains and any representive set of the cofactors is an anti chain. In otherwords, its n parallel lines. > Maybe I am wrong about this and you can explain it to me. Maybe I > am overthinking on this problem. But to me, the above example shows > how you must make sure the partial order in some sense preserves > the property you are interested in, such as divisibility by 3 or > being disjoint sets. Yah, it just seems like common sense, that the order has to consider the property of concern for it to conclude anything about the propery. However, why belabor the point? Did not your order give you the discernment you needed to prove the problem? I still dont understand why the induction. The important points that need to be established was that the sets of a chain were pairwise disjoint and that the intersection of all the sets of an anti-chain was nonnul. The first was straight forward after your correction. The other I didnt check. Is that what your induction was about? Again I suggest you do a wright up, defining and proving your order. Show chains are sets of pairwise disjoint intervals. Show the intersection of all the intervals of an anti-chain is nonnul. This last is the harder part. So lets see what you can do. ---- === Subject: Re: intersection of two cones >here is a simple question: given two circular cones such that their >axes crossing, what is the volume of their intersection. > Probably an awful mess in general, if its finite and nonzero. > May be doable in special cases (e.g. maybe if the axes intersect > at right angles, at the same distance from the vertices, and the opening > angles are the same). > In case anybody has trouble visualising this intersection, here is a > picture: > http://www.pisquaredoversix.force9.co.uk/Cones.png > I have rendered the intersection a little over-size so that it bulges out, > very slightly, from the cones. Nice 3D visualization. BTW, need the cone vertex angles and, distance between cone axes intersection point to vertex be same ?? === Subject: Re: intersection of two cones > >here is a simple question: given two circular cones such that their >axes crossing, what is the volume of their intersection. > > Probably an awful mess in general, if its finite and nonzero. > May be doable in special cases (e.g. maybe if the axes intersect > at right angles, at the same distance from the vertices, and the opening > angles are the same). > In case anybody has trouble visualising this intersection, here is a > picture: > http://www.pisquaredoversix.force9.co.uk/Cones.png > I have rendered the intersection a little over-size so that it bulges out, > very slightly, from the cones. > Nice 3D visualization. > BTW, need the cone vertex angles and, distance > between cone axes intersection point to vertex be same ?? Not really. To get a nice surface (with two intersecting ellipses in the surface) I think that we only need the two cones to circumscribe some common sphere. Their vertex angles can be different but their axes (which must intersect) need not be perpendicular. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: probability..... >A_1 : the event that interviewer A put white ball into urn. >A_2 : the event that interviewer A put blact ball into urn. >E_1 : the event that one ball which take out from urn is white ball. >E_2 : the evetn that the remaining ball from urn is white ball. >P(E_1 ^ E_2) <= i must find this probability. but i cant....... E_1 ^ E_2 = A_1 therefore P(E_1 ^ E_2) = P(A_1) --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: probability..... >suppose that >interviewer A put unknown ball into urn and >interviewer B put white ball into urn. >and >one ball which take out from urn is white ball. >find that the probability of remaining ball from urn is white ball Does this help? Suppose the probability that the candidate passes interviewer As test is p. Then the probability that he fails interviewer As test is (1-p). W = white ball, B = black ball. The probabilities for what the urn contains: P(urn = WW) = p P(urn = BW) = 1-p The probabilities for choosing a white ball given the contents of the urn (assuming that both balls are equally likely to be chosen): P(W chosen | urn = WW) = 1 P(W chosen | urn = BW) = 1/2 The probability of chosing a white ball: P(W chosen) = P(W chosen | urn = WW)* P(urn = WW) + P(W chosen | urn = BW)* P(urn = BW) = 1*p + (1/2)*(1-p) = (1+p)/2 It is now possible to find the probabilities you seek with Bayes Theorem: P(urn = WW | W chosen) = P(W chosen | urn = WW)*P(urn = WW)/P(W chosen) = 1* p / ((1+p)/2) = (2*p)/(1+p) If you know p, you can determine the value of this probability. For instance, if p=1/2 then P(urn = WW | W chosen)=2/3. >------------------------------------------------------- >um.....i think..... >A_1 : the event that interviewer A put white ball into urn. >A_2 : the event that interviewer A put blact ball into urn. >E_1 : the event that one ball which take out from urn is white ball. >E_2 : the evetn that the remaining ball from urn is white ball. >P(A_1) = 1/2 >P(A_2) = 1/2 >P(E_1)=P(A_1)P(E_1 I A_1)+P(A_2)P(E_1 I A_2)=(1/2)*1+(1/2)*(1/2)=3/4 >P(E_2 I E_1) = P(E_1 ^ E_2) / P(E_1) (^ is intersection) >P(E_1 ^ E_2) <= i must find this probability. but i cant....... >i need your advice. >thank you very much. === Subject: Re: Roger Penrose Nonlocality of Gravity Energy & Other Matters >> Absolute time and absolute velocity measurements relative to >> this Hubble ßow are now routine and provide the basis for the reliable >> navigation of our coming warp drive Space Navy as we fulfill our >> Manifest Destiny Matrix Out There in the Never-Ending High Frontier of >> the Infinity of Worlds in Super Cosmos. >As ever, Jack Sarfatti mixes perfectly good science with complete bollocks. Titty old bean, what leads you to conclude that perfectly good science cant be complete bollocks? Some see wonderful apparitions in shadows on the walls of their cave - about all I ever see look like fuzzy chickens. === Subject: Re: Roger Penrose Nonlocality of Gravity Energy & Other Matters >Absolute time and absolute velocity measurements relative to >this Hubble ßow are now routine and provide the basis for the reliable >navigation of our coming warp drive Space Navy as we fulfill our >Manifest Destiny Matrix Out There in the Never-Ending High Frontier of >the Infinity of Worlds in Super Cosmos. Yip Yip Yahoo, Cadet Jack lets get it on, Pilgrim !! Only thing holding me back is that my wetsuit was developed in a warm sea on earth protected by quite a few miles of atmosphere. How many feet of lead thickness will I need in my space suit when I set out to surf space ripples? It occurs to me that Hiroshima may be a picnic compared to the cosmic and gamma breezes and space dust I may run into. Perhaps my manifest destiny is to serve with those who stand and wait. Did I forget to mention, Im an Oxygen/nitrogen/protein/carbohydrate/ lipid addict?. Er...what leads you to term it the `high frontier? Got some frame of reference? === Subject: expected values given randomly perturbed dynamical system how do I obtain them? sorry for newbie question. but i find it difficult to do this on my own as im not a mathematician. consider a system such as, { x[t] == y[t] - v x[t], y[t] == x[t] - d y[t] } and let v and d be Random numbers from 0 to 1. and i can simulate the system over random parameter space many times and i can average the variables trajectories. things i want to know are... 1. is this an example of monte carlo simulation? if so, what reference(s) suggest that? 2. there is an expected values for v and d? what does it mean intuitively? 3. is there expected values for x and y as well? 4. do x,y as well as the parameters v and d have expected values? > stated the questions very clearly. Hope that the following is more > clear. > ///////////////////////////////////////////////////////////// ///////// > Input: k,A,b > //k is integer; A is integer; b is integer > Output:X > X = 0.0 > for k = 1..A Youre contradicting yourself here. Either k is an input or k is a loop index; it cannot be both. Ill assume you meant it to be a loop index. > for all possible set (m_1,m_2,...,m_k) satisfying m_1+m_2+...+m_k = > X = X + f_1(m_1) * f_2(m_2) * ... * f_k(m_k) > end > end > //in this algorithm, f_i(m_i) is a function with m_i as input. > The difficulty lies in the fact that there are too many possible > combination of set (m_1,m_2,...,m_k) satisfying m_1+m_2+...+m_k = b. > So, the number of cycle is ver high. > Can you plz give some suggestions in reduce the computation time and > -- > Y.ZHANG I assume that you want the m_j to be either positive or nonnegative integers as well (?). The first thing to do is to tabulate f_1(x) for x=0,...,b and save that, then tabulate f_2(x) for x=0,...,b, etc. That way you wont be wasting time recomputing them each time they are used. (This assumes you have enough memory to retain the tables.) To do the summation, use recursion. Define a function g(bb,kk) (where bb is the right hand side of the constraint, and kk is the index of the last f function being used), as follows: g(bb,kk) = sum {x=0...bb} f_kk(x)*g(bb-x,kk-1) g(bb,1) = f_1(bb) Your k-th sum then should be g(b,k). To get additional speed reductions (at the cost of increased memory use), you can put each computed value of g in a table, so that you can look it up if/when it is used again later. How much total work is this? We have A*(b+1) function evaluations f_j(x) up front. After that, we tabulate g(bb,kk) for bb=0...b and kk=2...A, so we compute (A-1)(b+1) values of g(). Computing g(bb,kk) involves bb+1 multiplications, bb+1 additions (assuming you initialized the running total to 0), and 2*(bb+1) lookups. So the total work for g(bb,kk) is proportional to bb+1, which means the work to tabulate g(bb,kk) for all bb=0...b is O(b^2), and so the work to tabulate all values of g is O(A*b^2). -- Paul ************************************************************* *************** ************** Paul A. Rubin Phone: (517) 432-3509 Department of Management Fax: (517) 432-1111 The Eli Broad Graduate School of Management E-mail: rubin@msu.edu Michigan State University http://www.msu.edu/~rubin/ East Lansing, MI 48824-1122 (USA) ************************************************************* *************** ************** Mathematicians are like Frenchmen: whenever you say something to them, they translate it into their own language, and at once it is something entirely different. J. W. v. GOETHE === Subject: Re: (Another) Nim game problem > > Ive tried to solve this simple Nim problem: > > . Two players; > . Just one pile of 22 fruits; > . I can only take one, two or three fruits at a time; > . Whoever takes the last fruit, looses; > > Ive seen a lot of binary theories applied to the original Nim but I think > they cannot be applied in this case. > > Can anyone give me a help? > > > > > RV > The first player takes two fruit. Game over! > Suppose that sum of successive draws could not be four. If A takes 2, > B can take 1 or 3, but not 2. If B takes 3, A can take 2 or 3, but > not 1. Could there be a winning strategy for either player? Just a > thought. Whoops, the first player takes only one fruit initially, not 2. Sorry. === Subject: fermat (Message posted from unknown at 4.21.185.2) - explanation In Reply to: Youve never explained anything about physics to me... ..and youre full of since you have no idea who is a decent physicist since youre a major pea-brain. waite sure isnt a decent physicist. Hes a hack at best But I see that youre unable to answer a simple question - figures. See? This is why youre unabled to learn. Youre mind is slo closed that not even air can get in. ------------------------------------------------------------- --------------- ---- 67551 - Schutz ------------------------------------------------------------- --------------- ---- Reload Page | Topic Thread | Post a Followup | The Hawking Forum | Search | FAQ Back in Thread | Next in Thread | Back by Number | Next by Number | Back to Entry Point ------------------------------------------------------------- --------------- ---- (Message posted from unknown at 4.21.185.2) - explanation In Reply to: another dumb comment from the resident asshole = pew By the way moron - I recommend that you stay away from Schutzs new text. Your head will most likely explode if you read it. Besides, from what youre always ßaming about and from the fact that youre unable to understand how mass and proper mass are defined its quite apparent that you wouldnt be able to understand it if you actuall did attempt to read it. And you might as well stay away from Rindler, Wald, MTW, DInverno, French, Mould, Peacock, Schutz (his GR text) etc. No wonder youre so ignorant in SR/GR. You either cant read, you refuse to read or that youre unable to comprehend what youre reading. Youd better stick with that toilet paper that waite calls his book - ROTFL!!! ------------------------------------------------------------- --------------- ---- *-----------------------* www.GroupSrv.com *-----------------------* === Subject: fermat ..and youre full of since you have no idea who is a decent physicist since youre a major pea-brain. waite sure isnt a decent physicist. Hes a hack at best But I see that youre unable to answer a simple question - figures. See? This is why youre unabled to learn. Youre mind is slo closed that not even air can get in. (Message posted from unknown at 68.8.194.133) - explanation In Reply to: If so then its because youre unable to learn.. posted I can tell the difference between a physiscist and a Ôshut in wannabe such as you brown. I dont chose to answer questions for you brown. Historically it leads to long protracted arguments where you refuse to acknowledge your errors. ----------------------------- In Reply to: Fermat22222222222222222222222222222222222222222222222222222222 22 Humboldt State University Presidents Office hsupres@humboldt.edu The members of the physics Stephen Hawking Forum, http://www.psyclops.com/hawking/forum would like to inform you that your computer network is being used to disrupt this forum. The tech email listed below has been notified, along with a photo of the instigator: Ben Ito (a resident of Arcata). He is believed not to be a student at HSU, but is using your Library/Computer Facilities to disrupt many physics venues (with numerous lengthy non-physics postings). He has disrupted this forum many times in the past from an HSU location, causing much frustration and loss of valued membership. Legal recourse against your institution is being discussed. Please assist us in this matter. OrgName: Humboldt State University Address: 1 Harpst St City: Arcata StateProv: CA PostalCode: 95521 Country: US TechPhone: +1-707-826-5000 TechEmail: netops@humboldt.edu *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Fermat (Message posted from unknown at 4.21.185.2) - explanation In Reply to: Itos fascination with bogus solutions for Fermats And you claim that Im the one always trolling. So youre not just a royal scumbag and an incredible idiot but youre also a hypocrit as well. And I see that youre still the pathetic little weasle that yoluve always proven yourself to be huh scumbag? pew - you let that moron waite brainwash that very small thing that you try to pass off as a brain for far too long. Its quite obvious from the fact that youve been unable, for all these years, to comprehend the most basic concept that can be found in all of relativity - i.e. mass. And whats worse is that even when the difference between proper mass and (relativsitic mass) was explained to you in detail, that little rotten cabbage that you try to pass off as a brain is still incapable of grasping it. This is probably why youve been too scared to read the physics literature outside what waite tells/allows you to read. One might be able to understand your stupidity if what as been explained to you a million times couldntbe found everywhere in the relativity community and in the most respectable of sources. Tell you what moron - See if that pea brain of yours can answer figure this problem out (HA! As if he could!! LOL!!) Let there be a rod, of rest mass m[sub]0[/sub] and length L, at rest lying on the x-axis in the inertial frame S[sub]0[/sub]. As observed in S[sub]0[/sub] there is a force on each end of equal magnitude and opposite direction so as to impose stress in the rod but not to accelerate it. Let S be in standard configuration with S[sub]0[/sub] and moving in the -x direction with speed v. What is the momentum, p, of the rod as measured in S? What is the energy, E, of the rod as measured in S? Define the quantity m as m = p/v (I assume that youre not so stupid as to confuse m with proper mass .. or do I need to explain THAT to you as well???). Does E = mc[sup]2[/sup]? If not then do you think that m can be replaced with E whenever its found? If so then prove it. Lets see your pea brain even *attempt* to respond to that. My guess is that youll weasle out of it by claiming that its not a valid question. Thats how your loser buddy waite would weasle of it. ------------------------------------------------------------- --------------- ---- Back in Thread | Next in Thread | Back by Number | Next by Number | Back to Entry Point ------------------------------------------------------------- --------------- ---- Topic Thread: Stephen Hawkings forum *-----------------------* www.GroupSrv.com *-----------------------* === Subject: fermat ..and youre full of since you have no idea who is a decent physicist since youre a major pea-brain. waite sure isnt a decent physicist. Hes a hack at best But I see that youre unable to answer a simple question - figures. See? This is why youre unabled to learn. Youre mind is slo closed that not even air can get in. (Message posted from unknown at 68.8.194.133) - explanation In Reply to: If so then its because youre unable to learn.. posted I can tell the difference between a physiscist and a Ôshut in wannabe such as you brown. I dont chose to answer questions for you brown. Historically it leads to long protracted arguments where you refuse to acknowledge your errors. ----------------------------- In Reply to: Fermat22222222222222222222222222222222222222222222222222222222 22 Humboldt State University Presidents Office hsupres@humboldt.edu The members of the physics Stephen Hawking Forum, http://www.psyclops.com/hawking/forum would like to inform you that your computer network is being used to disrupt this forum. The tech email listed below has been notified, along with a photo of the instigator: Ben Ito (a resident of Arcata). He is believed not to be a student at HSU, but is using your Library/Computer Facilities to disrupt many physics venues (with numerous lengthy non-physics postings). He has disrupted this forum many times in the past from an HSU location, causing much frustration and loss of valued membership. Legal recourse against your institution is being discussed. Please assist us in this matter. OrgName: Humboldt State University Address: 1 Harpst St City: Arcata StateProv: CA PostalCode: 95521 Country: US TechPhone: +1-707-826-5000 TechEmail: netops@humboldt.edu *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Is the earth still ßat??? Yes or no? What did they think of him. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: Is the earth still ßat??? _____________________ /| /| | | ||__|| | Do not feed the | / | --Mgt. | / |_____________________| / _ || / |____ || / | | | |____/ || / |_|_|/ | _|| / / |____| || / | | | --| | | | |____ --| * _ | |_|_|_| | -/ *-- _-- _ | || / _ | / ` * / _ /- | | | * ___ c_c_c_C/ C_c_c_c____________ === Subject: Re: Is the earth still ßat??? A threat from Bin Laden is all that is required to provide a landslide > _____________________ > /| /| | | > ||__|| | Do not feed the | > / | --Mgt. | > / |_____________________| > / _ || > / |____ || > / | | | |____/ || > / |_|_|/ | _|| > / / |____| || > / | | | --| > | | | |____ --| > * _ | |_|_|_| | -/ > *-- _-- _ | || > / _ | / ` > * / _ /- | | | > * ___ c_c_c_C/ C_c_c_c____________ May we borrow this image artwork as a generic reply to all troll postings in this NG? Shedar === Subject: Ben ito I read the posts about once a month. I find it very funny dont you. Why would they write so much about me if they dont like me. I think Im getting them hysterical great for hallowen mask. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: ben ito They tried to ßatten him. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: ben ito _____________________ /| /| | | ||__|| | Do not feed the | / | --Mgt. | / |_____________________| / _ || / |____ || / | | | |____/ || / |_|_|/ | _|| / / |____| || / | | | --| | | | |____ --| * _ | |_|_|_| | -/ *-- _-- _ | || / _ | / ` * / _ /- | | | * ___ c_c_c_C/ C_c_c_c____________ === Subject: Re: Easy real analysis proof (help needed) > But I need to show that if a set A has a non-measurable subset E, then > A has positive outer measure, i.e., m*(A) > 0. Or you could show the contrapositive, i.e. if A has measure 0 then A does not have a nonmeasurable subset. Which is what I did. === Subject: Re: Easy real analysis proof (help needed) > But I need to show that if a set A has a non-measurable subset E, then > A has positive outer measure, i.e., m*(A) > 0. > Or you could show the contrapositive, i.e. if A has measure 0 then A > does not have a nonmeasurable subset. Which is what I did. === Subject: DOH! Replied to wrong post again... >> But I need to show that if a set A has a non-measurable subset E, then >> A has positive outer measure, i.e., m*(A) > 0. > Or you could show the contrapositive, i.e. if A has measure 0 then A > does not have a nonmeasurable subset. Which is what I did. Should have been a reply to mjmorrisons second post. === Subject: Re: Easy real analysis proof (help needed) > But I need to show that if a set A has a non-measurable subset E, then > A has positive outer measure, i.e., m*(A) > 0. So it is enough to show that every non-measurable set has positive outer measure. === Subject: fermat111111111111111111111111111111111111111111111111111111 Fermats Last Theorem Ben Ito 10-28-04 I will solve Fermats last theorem. l. Introduction I will show that Fermats (n=4) and Wiles proofs are invalid then prove that Fermats equation x^n + y^n = z^n (equ 1) only forms integer solutions when n>2 using a transformation. 2. Fermats Proof (n=4) The following equations are used to describe the integer solutions of Fermats equation (n=2), A = 2uv, B = u^2 - v^2, and C = u^2 + v^2 (equ 2) (Shanks, p.141). The varibales A, B and C represent integer solutions of Fermats equation (equ l). Using u = 2 and v = 1, in equation 2, A = 4, B = 3 and C = 5 which are integer solutions when n=2. Fermats proof for n=4 is described. Fermat implies that by proving that, A^4 + B^4 = C^2 (equ 3). does not form integer solutions also proves that A^4 + B^4 = C^4 (equ 4) does not form integer solutions. Fermat uses the following equations to prove that equation 3 does not form interger solutions, A^2 = 2uv, B^2 = u^2 - v^2, and C = u^2 + v^2 (equ 5), However, A and B of equations 2 and 5 are not equal, A^2 =/ A and B^2 =/ B. (equ 6) Consequently, Fermat is improperly using equation 2; therefore, Fermats proof for n=4 is invalid. 3. Wiles Proof Wiles proof of Fermats Last Theorem is based on the elliptic curve equation, y^2 = ax^3 + bx + c (equ 7) where a^p + b^p = c^p (equ 8) Wiles assumed that since equation 8 is similar to equation l that Fermats last theorem can be described using elliptic curves; however, Fermats equation is not dependent on an ellilptic curve; therefore, equation 8 is not Fermats equation (equ l) as implied by Wiles. In addition, there are an infinite number of higher order equations of x and y, when n>3, that are not represented with Wiles ellipitic curves. Wiles uses a deception by implying that equation 8 represent the higher orders of x and y using elliptics curves; however, Wiles ellipitic curve only represent lower orders of x and y; therefore, Wiles proof of Fermats Last Theorem is incomplete and therefore invalid. Wiles ignores that Fermats equation is not dependent on an ellipitc curve and that elliptic curves do not represent higher orders of x and y. 4. Itos Proof. I will form the Proof of Fermats Last Theorem by showing why right triangles form integer solutions. I will use x, y and z to represent the sides of a right triangle when n=2. Using a transformation let z = c (integer), Fermat equation becomes an equation of a circle (n=2), x^2 + y^2 = c^2. (equ 9) In the circle transformation, the hypotenus of the right triangle becomes the radius of the circle. Consequently, a circle of radius r and the right triangle with a hypotenus z can be represented together on the x and y plane which forms the primary alignment. Only when n=2 forms the primary that is required in forming the integer solution of Fermats equation. 5. Conclusion I have shown that Fermats derivation of n=4 is base on a mathematical error. I then show that Wiles proof of Fermats Last theorem only describe lower orders of x and y with ellipitic curves; therefore, Wiles proof is incomplete. I then show that only n=2 forms the primary alignment that forms the of integer solutions of Fermats equation. Consequently, only n=2 forms the condition where the integer solutions can be formed. ************************* ... 6. References Robert Osserman. Fermats Last Theorem (a supplement to the video). MSRI Berkeley. 1994 Marilyn vos savant. The Worlds Most Famoous Math Problem. St Martins Press. 1993 Daniel Shanks. Solved and Unsolved Problems in Number Theory. Chelsea Pub. 1985. 7. Acknownlegment Science forum, and About Physics forum, HSU, CSUS, CR, SCC, USC, Hiram Johnson HS Sacramento (Mrs Larson), UCD, Stanford, MIT, Harvard and UCLA mathematics Dept. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: fermat (revised 10-31-04) Nate, There are an infinite number of probems with Fermats n=4 proof. Heres another one. Ben Fermats Last Theorem Ben Ito 10-31-04 I will solve Fermats last theorem. l. Introduction I will show that Fermats (n=4) and Wiles proofs are invalid then prove that Fermats equation x^n + y^n = z^n (equ 1) only forms integer solutions when n>2 using a transformation. 2. Fermats Proof (n=4) The following equations are used to describe the integer solutions of Fermats equation (n=2), A = 2uv, B = u^2 - v^2, and C = u^2 + v^2 (equ 2) (Shanks, p.141). The varibales A, B and C represent integer solutions of Fermats equation (equ l). Using u = 2 and v = 1, in equation 2, A = 4, B = 3 and C = 5 which are integer solutions when n=2. Fermats proof for n=4 is described. Fermat implies that by proving that, A^4 + B^4 = C^2 (equ 3). does not form integer solutions also proves that A^4 + B^4 = C^4 (equ 4) does not form integer solutions. Fermat uses the following equations to prove that equation 3 does not form interger solutions, A^2 = 2uv, B^2 = u^2 - v^2, and C = u^2 + v^2 (equ 5), Fermat is using the equations that describe the integer solutions of n=2 (equ 2) to prove that equation 3 does not form integer solutions; however, the equation derived from n=2 x^2 + y^2 = z^2 (equ 6) is completely different form equation 3; therefore, Fermat is violating logic by implying that the equations that describe the integer solutions of n=2 (equ 2) can be used to prove that equation 3 does not form integer solutions. Therefore, Fermats proof for n=4 is incomplete and therefore invalid. 3. Wiles Proof Wiles proof of Fermats Last Theorem is based on the elliptic curve equation, y^2 = ax^3 + bx + c (equ 7) where a^p + b^p = c^p (equ 8) Wiles assumed that since equation 8 is similar to equation l that Fermats Last Theorem can be described using elliptic curves; however, Fermats equation is not dependent on an ellilptic curve; therefore, equation 8 is not Fermats equation (equ l) as implied by Wiles. In addition, there are an infinite number of higher order equations of x and y, when n>3, that are not represented with Wiles ellipitic curves. Wiles is implying that equation 8 represent the higher orders of x and y using elliptics curves; however, Wiles ellipitic curve only represent lower orders of x and y; therefore, Wiles proof of Fermats Last Theorem is incomplete and therefore invalid. Wiles ignores that Fermats equation is not dependent on an ellipitc curve and that elliptic curves do not represent higher orders of x and y. 4. Itos Proof. I will form the Proof of Fermats Last Theorem by showing that only right triangles form integer solutions. I will use x, y and z to represent the sides of a right triangle when n=2. Using a transformation, let z = c (integer), Fermat equation becomes the equation of a circle (n=2), x^2 + y^2 = c^2. (equ 9) In the circle transformation, the hypotenus of the right triangle becomes the radius of the circle. Consequently, a circle of radius r and the right triangle with a hypotenus z can be represented together on the x-y plane which forms the primary alignment. Only n=2 forms the x, y and z lengths on the x-y plane which allows for the possible formation of integer solution of Fermats equation when n=2. The equation describe with Fermats equation when n>2 never forms a transformed structure (z=c) that forms the x, y and z lengths on the integer solutions. 5. Conclusion I have shown that Fermats derivation of n=4 is base on questionable logic. I then show that Wiles proof of Fermats Last theorem only describe lower orders of x and y with ellipitic curves; therefore, Wiles proof is incomplete and therefore invalid. I then show that the transformation (z=c) forms the x, y and z lengths on the plane only when n=2;consequently, only n=2 forms the condition where the integer solutions can be formed. ************************* ... 6. References Robert Osserman. Fermats Last Theorem (a supplement to the video). MSRI Berkeley. 1994 Marilyn vos savant. The Worlds Most Famoous Math Problem. St Martins Press. 1993 Daniel Shanks. Solved and Unsolved Problems in Number Theory. Chelsea Pub. 1985. 7. Acknownlegment Science forum, and About Physics forum, HSU, CSUS, CR, SCC, USC, Hiram Johnson HS Sacramento (Mrs Larson), UCD, Stanford, MIT, Harvard and UCLA mathematics Dept. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: fermat 1564.33 in reply to 1564.32 No, I just happened to read the messages from Nate to you that you made a point of deleting (because they undermine your argument, I suppose). Nate is right. You obviously dont understand Fermats proof for n = 4. + - - - peter Options Reply *-----------------------* www.GroupSrv.com *-----------------------* === Subject: fermat === Subject: Re: Fermat Hi Ben, I am sorry I didnt finish reading your paragraph so here is the rest of my response. > In addition, the solution (equation 2) of n=2 are used to prove that > n=4 does not form integer solution when n=4 which violates logic. There > are an infinite many problems with Fermats n=4 proof. First of what do you mean by defies logic? Just because you dont understand something doesnt mean it is incorrect or invalid. The only way which something can be proved FALSE in mathematics is if you can show that if it were true it would lead to a contradiction (i.e. a statement of the form A and not A (i.e. 1 + 1 =2 and 1+1 not =2)). equation A^4 + B^4 = C^2 is perfectly valid as A and A are DIFFERENT. Nate *-----------------------* www.GroupSrv.com *-----------------------* === Subject: fermat Ben, The idea is that A, and A are place holders for DIFFERENT NUMBERS! For exampl A could be 4 while A is 2. Or A could be 9 while A is 3, ect. The idea is that while they are place holders our assumptions place some restictions on what values they can be (specifically we requre (A)^2 = A) Nate > If I can prove that for all integers A, > B, > C, (A)^4 + (B)^4 not = (C)^2. Then I have proved for all integers > A, > B, C A^4+ B^4 not = C^2. This is becasue A, B, C, A, B, C are just > PLACEHOLDERS for integers (that is what it means to be a variable). > If A, B, C, A, B, C are just placeholders then A=A ect. Therefore, > Fermats proof is invalid. In addition, the solution (equation 2) of n=2 > are used to prove that n=4 does not form integer solution when n=4 which > violates logic. There are an infinite many problems with Fermats n=4 > proof. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: fermat111111111111111111111111111111111111111111111111111111 http://mygate.mailgate.org/mynews/sci/sci.math/ 3326c2efb70140bd04fb3f686619e1 d4.48257%40mygate.mailgate.org === > Subject: fermat111111111111111111111111111111111111111111111111111111 As of that point, anything else you had to say was already instantly dismissable as simply additional jumping up and down seeking attention by you. FYI xanthian. -- === Subject: Re: fermat111111111111111111111111111111111111111111111111111111 === > Subject: fermat111111111111111111111111111111111111111111111111111111 > As of that point, anything else you had to say was already > instantly dismissable as simply additional jumping up and > down seeking attention by you. > FYI> xanthian. Sorry , your number:(10^54 - 1)/9 is not a prime number. Prove with (10^1093 -1 )/9 . Ludovicus === Subject: fermat Hi *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Fermat Nate I disagree since Fermat is using u and v in both derivations; therefore, one cannot assume that A^2 = A. Assume (A)^4 + (B)^4 = C^2, A, B, C all integers Therefore, A=A since both A and A are integers; therefore, your assumption, Nate, invalid. *-----------------------* www.GroupSrv.com *-----------------------* === Subject: fermat === Subject: Re: Fermat Hi Ben, So close! You were so close to putting it in the form I asked for Your sections 1 and 2, while not completely in the form were close enough (at least for now). But then you went and reverted back to your sloppy, unclear, inaccurate writing :( But to be fair I will tell you what your mistakes were in Secions I and II. So first of Section I. You didnt actually prove anything in section I as it was more an introduction (which is fine). So I dont have any comments on section I. As for section II... > 2. Fermats Proof (n=4) > The following equations are used to describe the integer solutions of > Fermats equation (n=2), > A = 2uv, B = u^2 - v^2, and C = u^2 + v^2 (equ 2) > (Shanks, p.141). The varibales A, B and C represent integer solutions > of Fermats equation (equ l). Using u = 2 and v = 1, in equation 2, A = > 4, B = 3 and C = 5 which are integer solutions when n=2. > Fermats proof for n=4 is described. Fermat implies that by proving > that, > A^4 + B^4 = C^2 (equ 3). > does not form integer solutions also proves that > A^4 + B^4 = C^4 (equ 4) > does not form integer solutions. Fermat uses the following equations to > prove that equation 3 does not form interger solutions, *Okay, so far you are fine*. > However, A and B of equations 2 and 5 are not equal, > A^2 =/ A and B^2 =/ B. (equ 6) This is where your mistake is. If you want to be a real stickler, it is possible that the proof you were given wasnt TECHNICALLY perfect, but most proofs assume some form of reasoning on the part of the reader (otherwise the simplest proof would be hundreds of pages long!). This particular assumption was that you can change the meaning of a letter depeneding on the context (for example, if I have a friend named Mark and who lives in London, and you have a friend named Mark who lives in New York, I can just use the term Mark if the one I am refering to is clear from the context.) If you want to be a real stickly about different variable names in different context what chould have been written was something like the following: Assume (A)^4 + (B)^4 = C^2, A, B, C all integers Let (A)^2 = A (and so (A)^4 = A^2) Let (B)^2 = B (and so (B)^4 = B^2) So we have A^2 + B^2 = C^2 We therefore have (A)^2 = A = 2uv, (B)^2 = B = u^2 - v^2, and C = u^2 + v^2. Then I bet if you continue on with the proof you were given you wont run into problems. As for the rest, I am sorry to say that it isnt in the form I asked for and so I wont comment on it. Nate *-----------------------* www.GroupSrv.com *-----------------------* === Subject: Re: fermat111111111111111111111111111111111111111111111111111111 _____________________ /| /| | | ||__|| | Do not feed the | / | --Mgt. | / |_____________________| / _ || / |____ || / | | | |____/ || / |_|_|/ | _|| / / |____| || / | | | --| | | | |____ --| * _ | |_|_|_| | -/ *-- _-- _ | || / _ | / ` * / _ /- | | | * ___ c_c_c_C/ C_c_c_c____________ === Subject: Re: commuting?/non-group cipher? > I dont know the right word, semigroup and groupoid are wrong, so Ill > use sabo for a set-and-binary-operation, with no implications of closure > or associativity, where the binary operation can be, and can only be, > applied to any two members of the set, producing an output. > We are investigating a generalised cipher sabo which also possesses a > particular property, and want to know whether it must be a group. > The cipher part means that the sabo must function as a cipher. We will > also want to investigate ciphers which are not sabos, eg where the > encryption and decryption operations differ, to see if a cipher with the > particular property which is not a sabo can exist. > The particular property is : Given a set S and a binary operation *; the property that for all a, b in S there exists c in S such that, for all d in S, c*d = a*(b*d). In crypto terms, the property that a double encryption under two keys is always equivalent to a single encryption under some different key. I can only think of three ciphers which have the property - Caesar, Pohlig-Hellman and Vernam/otp. All are groups, where the group set is the set of texts and keys, and the group operation is the encryption/decryption operation. All are also permutation groups, where the group set is the set of permutations (regarding an encryption under a specific key as a permutation) and the group operation is composing^ the permutations. Kristian Gjsteen has shown that any cipher with the property must be a permutation group. (^doubly permuting) > The elements of the set, S of the sabo consists of all the possible > messages, ciphertexts and keys. The binary operation * takes any two members > of the set and produces an output, and is the encryption/decryption > operation. > We want to know whether _that sabo_ must be a group. We are not interested > in whether there is an associated permutation group unless it tells us > something about the sabo. > On first glance I thought obviously it doesnt have to be a group; but now > Im not so sure. So far I have: the sabo must have closure. >> Denoting the operation on K by #, we get an induced operation # on X >> given by f(k1, ) # f(k2, ) = f(k1#k2, ). > Nope. lost me there. -- Peter Fairbrother === Subject: Re: differentiating integrals sci.math: comment: ] >Incorrect. Well, arent _you_ a helpful little thing! -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: What is so hard about standard deviations? > Sci.math seems as good a place as any to vent about this. > It probably never makes sense to ask a purely rhetorical question, > least of all when people are blinded by impatience or vitriol or > something, but: Why do we get this kind of thing from comparatively > reputable firms when they speak of sensitive matters? The polling firms dont want to admit their more serious problem: nonresponse. (When is the last time YOU took time out of your busy day to answer poll questions from some stranger who phones you up randomly?) Polls have been very bad at handicapping elections in my high-turnout jurisdiction in recent years, and I figure I have NO IDEA how even the popular vote will turn out in the current United States presidential election. Have you seen the reviews of statistics textbooks at http://mathpc04.plymouth.edu/MAAFIXED.PDF (which was recently recommended to me)? What do you think of the reviews? -- Karl M. Bunday P.O. Box 1456, Minnetonka MN 55345 Learn in Freedom (TM) http://learninfreedom.org/ remove .de to email === Subject: 1st derivative test. I just think something is missing here and if its ok then I am a bit confused. Can someone please clear things up for me? Suppose f(x) is continuous on [a,b] and differentiable on (a,b) except possibly at c in (a,b) s/t f Ô (x) >0 for all x in (a,c) and f Ô(x) <0 for all x in (c,b). I need to show that has a max on (a,b) at x =c. 1st I agree with this statement. Pf: for all x in (a,c) we have [f(c) - f(x)]/(c-x) >0 => f(c)>f(x) (since c-x>0) {I dont see why this is exactly true but it must be true for the statement to be true. I know that I need for f(x) to be continuous at x=c-and this is given- but how do I use this fact???}. I get a similar statement for the interval (c,b) and I have the same question. Can someone please fill in this missing step so I can move on to other problems! === Subject: Re: 1st derivative test. > I just think something is missing here and if its ok then I am a bit > confused. Can someone please clear things up for me? > Suppose f(x) is continuous on [a,b] and differentiable on (a,b) except > possibly at c in (a,b) s/t f Ô (x) >0 for all x in (a,c) and f Ô(x) <0 for > all x in (c,b). I need to show that has a max on (a,b) at x =c. > 1st I agree with this statement. > Pf: for all x in (a,c) we have [f(c) - f(x)]/(c-x) >0 => f(c)>f(x) (since > c-x>0) {I dont see why this is exactly true but it must be true for the > statement to be true. I know that I need for f(x) to be continuous at This is a consequence of the Mean Value Theorem - since f is continuous on [x,c] and differentiable on (x,c), there is some x* in (x,c) such that (f(c) - f(x))/(c - x) = f(x*) and f(x*) > 0 (by assumption). > x=c-and this is given- but how do I use this fact???}. > I get a similar statement for the interval (c,b) and I have the same > question. ... which has a similar answer :-) > Can someone please fill in this missing step so I can move on to other > problems! Note that theres really a bit more to say here, if you intend a *complete* proof. Since f is continuous on the closed interval [a,b], you know that f actually _attains_ a maximum on [a,b] -- the argument hinted at above shows that that maximum cant occur at a point in either of the open intervals (a,c), (c,b). In order to conclude that it *must* occur at c, you have to rule out a and b as candidates ... doing so requires another couple of appeals to the Mean Value Theorem ... === Subject: Re: 1st derivative test. > Suppose f(x) is continuous on [a,b] and differentiable on (a,b) except > possibly at c in (a,b) s/t f Ô (x) >0 for all x in (a,c) and f Ô(x) <0 for > all x in (c,b). I need to show that has a max on (a,b) at x =c. > 1st I agree with this statement. If f(x)> 0 (f(x)<0) for all x in (a,x) (in (c,b)), what follows for the behavior of the values of f(x) in these intervals? Combine the results and you have the solution? Ciao Karl === Subject: Re: 1st derivative test. > Suppose f(x) is continuous on [a,b] and differentiable on (a,b) except > possibly at c in (a,b) s/t f Ô (x) >0 for all x in (a,c) and f Ô(x) <0 for > all x in (c,b). I need to show that has a max on (a,b) at x =c. > 1st I agree with this statement. > If f(x)> 0 (f(x)<0) for all x in (a,x) (in (c,b)), what follows for > the behavior of the values of f(x) in these intervals? Combine the > results and you have the solution? > Ciao > Karl Karl, But what if f is not continuous at x=c? Say f (c) < f (a) and f (c) < f(b)? Then clearly f does not have a max at x = c. Steven === Subject: Re: 1st derivative test. > Karl, > But what if f is not continuous at x=c? Say f (c) < f (a) and f (c) < f(b)? > Then clearly f does not have a max at x = c. > Steven Hi In your first post you write: >Suppose f(x) is continuous on [a,b] and differentiable on (a,b) except >possibly at c in (a,b) s/t f Ô (x) >0 for all x in (a,c) and f Ô(x) <0 This means that f(x) is also continuous at c. The except means only that maybe it is not differentiable at c. Ciao karl === Subject: Re: 1st derivative test. Correct! > Karl, > But what if f is not continuous at x=c? Say f (c) < f (a) and f (c) < f(b)? > Then clearly f does not have a max at x = c. > Steven > Hi In your first post you write: > >Suppose f(x) is continuous on [a,b] and differentiable on (a,b) except > >possibly at c in (a,b) s/t f Ô (x) >0 for all x in (a,c) and f Ô(x) <0 > This means that f(x) is also continuous at c. The except means only > that maybe it is not differentiable at c. > Ciao > karl === Subject: Re: 6 === >Subject: Re: 6 >6 = 1+2+3 >6 = 1x2x3 >6 = sqr-root of (1cube + 2cube + 3cube) >Any other way? 6 = sum((1+2+3)^2) / (3*((1*2*3)^2 + 1)) -- Mensanator Ace of Clubs === Subject: Hilbert-adjoint operator Theorem. Let V be an inner product space whose dimension is finite and let T:V->V be a linear operator on V. Consider the Hilbert-adjoint operator T*:V->V and an orthonormal basis B of V. Using this basis, the matrix that represents the operator T is the complex conjugate transpose of the matrix that represents T*. In this theorem, it is very important that a given basis B is orthonormal. But there exists a similar result if we do not consider this hypothesis that B is orthonormal? === Subject: I need curve fitting explained Im taking physics in college and the lab assignments require me to fit my data to a curve and come up with a formula using the polynomial method - can someone reccomend a book which explains the basics of curve fitting for We are suppose to come up with a power formula === Subject: Re: I need curve fitting explained don don@panix.com wants help finding polynomial function to fit data: >Im taking physics in college and the lab assignments require me to fit my >data to a curve and come up with a formula using the polynomial method - can >someone reccomend a book which explains the basics of curve fitting for >We are suppose to come up with a power formula You can pick what degree of polynomial you want; this refers to the exponent of your variable in your terms. The simplest understanding of linear algebra and matrices can give you what you want. Basically, your variables and function values are known, and they come from your data directly. What you do not know but need to find are the coefficients. You CAN do it! I have done it and I never had a course of linear algebra. Look at your data, plot some of it, and decide what degree polynomial function you want (requires having studied intermediate or college algebra). As example, you may have desire to make a fit for something like: a.x^3 + b.x^2 + c.x^1 = y (where y= functionValue - constant) You can then pick the data you want to use, plug them in for x and y, and .... use software or a graphing calculator with matrix capabilities, and let it determine your a, b, c values. Be sure to study about matrices, at least from a college algebra book so that you know what you need and what you want to find. You could create your own small software program to find all the x^(whatever) and y values in order to create your matrices; and you can, as I remind, pick what ever degree polynomial you want for your data. You need as many data points as equations for your matrix. You might actually want to use a software program such as CurveExpert which lets you put in your raw data and will give you a polynomial fit to whatever degree you want. You would not need to set up a matrix. The program lets you examine how well your curve fits your data. G C === Subject: Re: I need curve fitting explained > don don@panix.com wants help finding polynomial function to fit data: >Im taking physics in college and the lab assignments require me to fit my >data to a curve and come up with a formula using the polynomial method - can >someone reccomend a book which explains the basics of curve fitting for >We are suppose to come up with a power formula > You can pick what degree of polynomial you want; this refers to the exponent of > your variable in your terms. > The simplest understanding of linear algebra and matrices can give you what you > want. Basically, your variables and function values are known, and they come > from your data directly. What you do not know but need to find are the > coefficients. > You CAN do it! I have done it and I never had a course of linear algebra. > Look at your data, plot some of it, and decide what degree polynomial function > you want (requires having studied intermediate or college algebra). > As example, you may have desire to make a fit for something like: > a.x^3 + b.x^2 + c.x^1 = y (where y= functionValue - constant) > You can then pick the data you want to use, plug them in for x and y, and .... > use software or a graphing calculator with matrix capabilities, and let it > determine your a, b, c values. Be sure to study about matrices, at least from > a college algebra book so that you know what you need and what you want to > find. > You could create your own small software program to find all the x^(whatever) > and y values in order to create your matrices; and you can, as I remind, pick > what ever degree polynomial you want for your data. You need as many data > points as equations for your matrix. > You might actually want to use a software program such as CurveExpert which > lets you put in your raw data and will give you a polynomial fit to whatever > degree you want. You would not need to set up a matrix. The program lets you > examine how well your curve fits your data. > G C MSExcel does curve fitting too. Darren === Subject: Re: I need curve fitting explained > Im taking physics in college and the lab assignments require me to fit my > data to a curve and come up with a formula using the polynomial method - can > someone reccomend a book which explains the basics of curve fitting for Let (x_i, y_i), i=1..n represent your data points. Let f(x,a) represent the function youre trying to fit, where a is the set of unknown parameters. For instance, for a straight line fit f(x,a) = m*x + b and a = {m,b} are the unknown parameters. Then the least squares problem is that you want to find the set a of parameters (represented as a vector) which minimizes sum(i=1 to n) (y_i - f(x_i, a))^2 For straight lines and polynomials, you can solve explicitly for the best fitting parameters. The minimization problem translates into a set of linear equations where the coefficients and right hand side are sums such as sum(x_i), sum (x_i*y_i), sum(x_i^2).. For a straight line its a 2 x 2 system and the algebraic solution is not too complicated. For polynomials with m coefficients its a m x m system, and usually references will just give you the expression for the system, not the solution. To solve it youll want software that can solve systems of linear equations. Here is one website which derives the systems of equations for the different cases: http://www.efunda.com/math/leastsquares/leastsquares.cfm > We are suppose to come up with a power formula I dont know what that means. - Randy === Subject: Re: I need curve fitting explained posted until at least 13 hours later. delayed, etc? - Randy === Subject: Re: I need curve fitting explained === >Subject: Re: I need curve fitting explained >posted until at least 13 hours later. >delayed, etc? Ive been using Google Groups 2. The posting time is minutes instead of hours. Sometimes my post shows up by the time I get back to the table of contents. But its still beta, so they havent got everything worked out yet. When I first started using it, you couldnt get a tree view of the thread. Now you can but the sort by date option isnt functional yet. I occaisionally get an OOPS, server error message when posting. Often this error is inconsequential as the message does indeed get posted. But it gets lost often enough to make me save a copy of what I typed just in case. At least with Groups 2 you dont have to wait 13 hours to find out if its really lost. A week or so ago the Groups 2 server crashed and it took a week to get caught up once it was back online. Using regular Groups for a week made me realize how much I prefer Groups 2 even as beta. At least with Groups 2 I havent gotten any of those stupid youve made too many posts, try again in a few hours messages that Groups is so fond of. > - Randy -- Mensanator Ace of Clubs === Subject: VonNeumann Gametheory applied to StockMarket; election factor Portfolio of PAF as of 29OCT04 BCE 7,000 BLS 10 BMY 100 MRK 10 Q 50,000 SBC 480 SGP 100 cash of approx $173,000. ready to invest realestate land 3APR03 of 3 lots $19,000. science-art of pictures,porcelain etc starting JAN03 for $16,133. realestate land 30JUL03 another lot $11,500. Back then I : Today I decided to: Several weeks ago I unloaded about 98% of the drug stocks in the portfolio and with the proceeds bought Qwest and had 100,000 shares up to today. Because I figured the drug sector is in the doldrums of an election and a close election at that and because of the Merck woes on Vioxx. But yesterday I realized that this election is so close that it may not be decided Nov 3 but be contentious and litiguous as was 2000 and stocks do very poorly in a period of uncertainty. And another factor is that Qwest reports earnings soon and since Verizon recently reported more landline loss and the last time Qwest reported a shortfall of earnings its stock sank far below the $3. mark. So deeming the probabilities and likelihoods of bad news in the upcoming week for stocks I decided why not take a profit of approx $4,500. and sit back and relax for the upcoming election week. It is rare that the portfolio is in cash for any length of time because cash is money that is *not at work*. But the negatives of the upcoming news is more likely to weigh down the market. Of course when the election results become clear there is likely to be a rally and I shant miss the rally. Also I want to note that today BCE crossedover that of the price of BMY brießy. The first time I had seen that in perhaps going back more than 3 or 4 or 5 years in the fact that today BCE was 23.25 at a time when BMY was below 23.25 but it was not much of a crossover for it was too small and too brief. It is rare for the portfolio to be in much cash at all, for generally it is fully invested. But the circumstances of election and perhaps a election mayhem of a tie bodes ill for the stockmarket for the next several days. We see the evidence in the troops of lawyers on watch in the swing states. So this uncertainty is just begging me to reap whatever profit is there and put a sizable chunk into cash for the next several days. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin) > Recently I had a private correspondence with Alexander Zenkin; among many > things I brought to his attention *this* thread discussing his publications > on Cantor. > > I am forwarding the following message on behalf of Dr. Zenkin. > > (You will find most of his comments below marked with AZ == Alexander > Zenkin) > > had made an elementary mistake in the paper, such that we must be > highly skeptical of it. Dr. Zenkins explanations make it clear that > the criticisms were not quite realistic. > > > Good heavens! Assuming that you are serious in trying to understand > the mathematical basis for Cantors argument, you could hardly do > worse than to accept his responses as addressing the criticisms that > have been leveled here. > > You should indeed be highly skeptical of this paper; and if you wish > to proceed with your studies of mathematical philosophy, you should > assure yourself of why this skepticism is justified. > I wish to proceed with my studies of mathematical philosophy. Excellent. It is an interesting field. In my opinion, however, mathematical philosophy should not be the study of, and exposition on, whatever one _as a philosopher_ thinks mathematics is, but what one as a philosopher thinks about what _mathematicians_ think mathematics is. > The paper is a philosophical one. I cannot myself say I endorse each > and every claim, but Zenkin makes criticism, and some of it I find of > substance. > Some people, like Rob, claimed elementary mistakes in the paper (like > that Zenkin says there are only n n-bit bitstrings, rather than 2^n) > which was not the case apparently. This would apparently make him a > crank. In that instance, I think it just makes Zenkin a poor explainer - but his logical errors are still quite apparent. > Anyway, its a shame if people claimed elementary mistakes > where there was none, only because the guy said Cantor, he must be a > crank. If one wishes to study, and discourse upon, the operations of a dairy, then it would practically fruitless to attempt this without first understanding the needs of a dairy farm; and how dairy farmers go about their activities. Otherwise, the philosophy of dairys would resemble a guy at a bar, who has never been on a farm, and knows little about a dairy either first- or second-hand, saying Yknow, they ought to feed those cows chocolate, then theyd get chocolate milk!. You should notice that no credible mathematician here respects the _arguments_ made by Zenkin; not because hes a crank, but because they fail the test of being about whatever it is that _mathematicians_ consider mathematics. The reason why hes considered a crank is that, by professing to be a philosopher of mathematics while displaying an ignorance of mathematical logic, he displays the willful ignorance characteristic of cranks. He sits in a bar and tells the dairy farmer Feed those cows chocolate, I want chocolate milk! > There is one obvious place that Dr. Zenkin does not sound correct. The > diagonal method is not the only means to establish a proof of Cantors > theorem that there are more reals than there are natural numbers. > Cantors first proof is an example for this, ... Okay. So you know this, which is good. It is the work of a moment (okay, maybe a bit more) for a _motivated_ person to discover this fact, as you did. For a philosopher of mathematics, one would think that this would be a known fact. But, as the saying goes, ignorance is no excuse before the law! If you sent Zenkin an e-mail pointing out this gaping error in his justification of his arguments, how should he respond? Shouldnt he discard this line of argument as invalid? Do you think that you would be the first person who had noticed this, and brought it to his attention? Thats one reason to suspect him as a crank - when corrected, cranks do not retract their arguments. And it beggars belief to imagine someone trained in the history and philosophy of mathematics _not_ knowing the afore-mentioned fact. Conclusion: Zenkin is a likely crank. I will examine his arguments below. > ...and we can find other > proofs which do not necessarily use infinitary reasoning. More on infinitary reasoning below - but I cannot imagine a reasonable definition of infinitary reasoning that applies to Cantors proof that the power set of a set cannot be put into bijection with that set. > He could elaborate his point by arguing that any such proof which does > not use infinitary reasoning (which he argues is invalid), must > necessarily use abstraction of actual infinity, Okay, heres where you should really, _really_ try to understand what is different about how mathematicians think about mathematics. First, he introduces a term, infinitary reasoning, _without defining it_. He gives what he calls an _example_ of it (his failure to find a counterexample); and he then attempts to makes an argument that reasoning invalidates Cantors arument. This is like saying consider the number 5, which is prime. Any number which is like 5 I shall call Ôfish-like. Therefore, 2^10001 - 1 is Ôfish-like, and therefore prime. I cant even evaluate this argument, because I dont know if 2^10001 - 1 really is Ôfish-like, let alone verify that being Ôfish-like implies being prime. A mathematician would never accept Zenkins argument as is, primarily because there is no definition given for infinitary reasoning that allows us to evaluate his (incorrect for other reasons) argument. We can only _guess_ whether, for example, he considers the standard argument proving that there is no largest integer also to be an example of infinitary reasoning. Maybe he thinks that it is, and thefore the argument is false; and maybe he thinks it isnt - but theres no way we can tell _from his argument_ because he hasnt been specific enough in defining what he means by infinitary reasoning. reasoning to prove proposition P means that this proof of proposition P involves an unbounded number of steps. That would meet the standard definition - a proof must have a _finite_ number of steps. But the next logical problem is, the proof he demonstrates as requiring Ôinfinitary reasoning is his own attempt to _refute_ Cantors argument by showing the existence of a counterexample - so it is _he_ who is using an infinite number of steps in a (failed, for still other reasons) attempt to prove the existence of a counterexample. If he means to say that any proof of P, which can be disproven by a proof T which is invalid because it contains an infinite number of steps, is then not a valid proof, then hes not going to have a lot of valid proofs sitting around. If a proof P is a _valid_ proof, then _every_ disproof T _must_ be invalid for some reason - thats the whole _point_ of consistency. But this all just shows Zenkin doesnt understand Cantors argument in the first place; nor mathematical concerns in general. Even if we allowed his infinitary reasoning constructions to be valid, it _still_ doesnt yield a counterexample. Thats the whole beauty of Cantors proof - the existence of a counterexample will _always_ be a contradiction, no matter how we construct it. So Zenkins logic is invalid for another reason - one unrelated to whether or not one accepts the existence of an infinite set. Since he gives no definition of infinitary reasoning, and his argument is certainly no mathematical argument, and really no logical argument at all - it all boils down to just his thoughts about what bothers him about his own lack of understanding of Cantors results. As a meta-mathematician, he is a man in a bar giving advice to a dairy farmer. If you want to understand mathematics as _mathematicians_ understand it, you dont need to understand why Wiles proof of Fermats last theorem is true (only a few probably could); but you _must_ get to the point where you understand why _no_ mathematician worthy of the name would accept Zenkins arguments. > which is a point that > might be impossible to settle for two reasons: > 1. It might turn out that we are unable to define what abstraction > of actual infinity is precisely enough, because we do not really know > what abstraction is, It may turn out that you are unable to _agreee_ with how abstraction of actual infinity is defined, but for a mathematician, there is a perfectly satisfactory precise definition: It is equivalent to the existence of the set defined by the axiom of infinity. You may be thinking that actual infinity is that which cannot be completed, made complete, something bigger than big, the unfathomable. Thats fine - but then youre not talking about what _mathematicians_ talk about when they talk about actual infinity, and youre therefore youre not doing _mathematical_ philosophy. What a mathematician would find relevant is : given the above definition, what would we do in the _absence_ of the axiom of infinity? Which proofs still hold, and does the result correspond to the set of problems that mathematicians are interested in? Does it correspond to their intuitions? And, more philosophically, should we then as rational beings accept the axiom of infinity or reject it? Mathematicians who found the implications of Cantors approach distasteful to their intuitions did not cease doing math. They eliminated those axioms and forms of logic which they found did not correspond to their intuitions, but they still did what mathematicians do - they investigated the logical consequences of their favorite assumptions. That is not what Zenkin is doing. > or because actual infinity is illogical (and > thus necessarily leads to paradoxes when trying to define its > abstraction) Without getting into Godel-ish complexities, if it could be easily proven that the assumption of the axiom of infinity in the standard set of axioms caused a contradiction (i.e. was inconsistent), believe me, wed have heard about it. Just as dairy farmers work with cows every day, mathematicians work with the logical implications of infinity all the time, and have done unlikely that Id uncover one. Is it possible? Sure, humans are fallible; but an awful lot of pitching of hay, filling of feed troughs, and just plain milking was done by set theoretic mathematicians in order to assure themselves that these foundations lacked any contradicitions. Read a good book on axiomatic set theory. In any case, to accept that the axiom of infinity created contradictions, Id _still_ have to see it proven - and certainly, nothing in Zenkins paper proves this. If you want to be a meta-mathematician, you should learn to see why that is - why a _mathematician_ would reject his paper as hopelessly ßawed, regardless of whether it has resonance to you as a philosopher. > 2. One might argue that Cantors first proof does not contain > abstraction of actual infinity. verify the correctness of Cantors proof that the power set of a set cannot be put into bijection with that set. So that is _not_ why a mathematician rejects Zenkins arguments. In the absence of the axiom of infinity, there is the problem of being very specific about what you mean by the naturals and the reals, when your system lacks an axiom saying that the naturals actually _do_ exist as a set, and without assuming that the naturals exist as a set as part of your argument. But that problem is not at all addressed by Zenkin. > (I cannot say what I believe about 1, but 2 I disagree with) On what basis? If you disagree with 2, but you dont know how you would define abstraction of actual infinity, how can you be possibly be doing mathematical philosophy? If you want to know how can mathematicians describe the naturals, as a set, without implicitly invoking the axiom of infinity? then that is a reasonable (and interesting) question; but you must expect that, to understand the answer, you will need to follow a mathematical argument. And to learn how to follow a mathematical argument, you should assure yourself that Zenkins argument is logically, irreparably invalid. Dont take my word for it; follow his argument and _prove_ it is wrong. > However, if he can settle the following metamathematical theorem: > Any proof of Cantors theorem either requires infinitary reasoning or > abstraction of actual infinity. (Whatever it is!) > Then, he would have put forward a much stronger argument. That would be trying to settle the following theorem: Any proof of Cantors theorem is invalid in the absence of the axiom of infinity. Then, he would have put forward a stronger, but equally false argument. As Barbie once said, Math is hard. To study mathematical philosophy, one must also study math. Study why Zenkin is wrong. === Subject: Re: Burgers equation >my problem concerns the inviscid burgers equation : >u_t + u * u_x =0 > u_l if x < 0 >with initial values u(x,0)= where u_l, u_r are >constant. > u_r if x > 0 >I would to show that the following function is a weak solution of this >equation. > u_l if x < st >u(x,t) = > u_r if x > st >where s=1/2 (u_l + u_r) (shock speed) Write down the definition of weak solution. Recall how the definition was arrived at -- especially how integration by parts put all derivatives on a test function. Break the integrals in the defining equation into left and right parts. Substitute your putative solution. Reverse the above integrations by parts. Be careful about what the normals to the boundaries are. Voila! You may like to carry this out for the more general equation u_t+f(u)_x=0; Burgers equation is the special case f(u) = (1/2)u^2. I say this because the simple expression for s is part of the specialness. May the force be with you. /dan === Subject: Reals without infinity While responding to Eray, I found myself wondering, if we abandon the axiom of infinity... I think I can see how to construct something isomorphic to the ring Z in ZF, and from that almost imagine how to construct Q as the (provable) lower bound of any finite collection of quotients of Z (elements of ZxZ) meeting some conditions. But in the absence of the axiom of infinity, how _does_ one construct an ordered field F such that F is isomorphic to R in ZF? I suppose the same way... but here my imagination fails me. I presume it involves some re-definitions of the idea of convergence... TIA - Chas === Subject: Re: Reals without infinity (Resending my previous reply, which seems to be have been lost.) > While responding to Eray, I found myself wondering, if we abandon the > axiom of infinity... > I think I can see how to construct something isomorphic to the ring Z > in ZF, and from that almost imagine how to construct Q as the > (provable) lower bound of any finite collection of quotients of Z > (elements of ZxZ) meeting some conditions. > But in the absence of the axiom of infinity, how _does_ one construct > an ordered field F such that F is isomorphic to R in ZF? I suppose the > same way... but here my imagination fails me. > I presume it involves some re-definitions of the idea of > convergence... I have implemented a simplified version of ZF in the form of a PC-based proof checker (DC Proof 1.0). It has only 4 set axioms (schemas actually) for: subsets, power sets, set equality and Cartesian products -- with no axiom of infinity. There are, of course, a number of definitions and justifiable shortcuts built in to make life easier. Though very much a work in progress, I have set out to construct the real numbers in my system starting with a first order version of Peanos Axioms for N as an initial premise. Here is an outline of how I plan to proceed: 1. Using the Cartesian product and subset axioms, construct the add function (+) for N, prove it is unique, and prove that it has the required algebraic properties (associativity, etc.). 2. Using the Cartesian product and subset axioms, and + for N, construct the multiplication function (*) for N, prove it is unique, and prove that it has the required algebraic properties. 3. Using the Cartesian product, power set and subset axioms, and + for N, construct the set of integers Z -- roughly speaking, an equivalence relation on NxN. Informally, (a,b)=(c,d) <=> a+d=c+b 4. Using the Cartesian product and subset axioms, and + for N, construct the add function (+) for Z, prove it is unique, and prove that it has the required algebraic properties. 5. Using the Cartesian product and subset axioms, and + for Z, construct the multiplication function (*) for Z, prove it is unique, and prove it has the required algebraic properties. (This about where I am now.) 6. Using the Cartesian product, power set and subset axioms, and * for Z, construct the set of rational numbers Q -- roughly speaking, an equivalence relation on Zx(non-zero Z). Informally, (a,b)=(c,d) <=> a*d=c*b. 7. Using the Cartesian product and subset axioms, and + and * for Z, construct the addition function (+) for Q, prove it is unique, and prove it has the required algebraic properties. 8. Using + on Q, define a strict ordering (<) on Q. 9. Using the power set and the subset axioms, and < on Q, construct the set of Dedekind cuts on Q -- the set of real numbers R. Comments welcome. Dan dc@dcproof.com Download DC Proof 1.0 at http://www.dcproof.com === Subject: Re: Reals without infinity > While responding to Eray, I found myself wondering, if we abandon the > axiom of infinity... > I think I can see how to construct something isomorphic to the ring Z > in ZF, and from that almost imagine how to construct Q as the > (provable) lower bound of any finite collection of quotients of Z > (elements of ZxZ) meeting some conditions. > But in the absence of the axiom of infinity, how _does_ one construct > an ordered field F such that F is isomorphic to R in ZF? I suppose the > same way... but here my imagination fails me. > I presume it involves some re-definitions of the idea of > convergence... An approach I have implemented in the form of a proof checker (downloadable at my homepage) is based very loosely on ZF and does not include an axiom of infinity. My set axioms consist include only schemas for: subsets, power sets, set equality and Cartesian products. There are, of course, several definitions and justifiable shortcuts built in to make life easier. Though still very much a work in progress, I have done some work toward constructing the real numbers starting from a first order version of Peanos Axioms as an intitial premise. Informally, here is an outline of how I plan to proceed: 1. Construct the addition and multiplication functions for N using the Cartesian product, power set and subsets axioms. Prove that these functions are unique and have the required algebraic properties. 2. Construct the set of integers Z as a kind of equivalence class on the power set of NxN using the Cartesian product, power set and subset axioms, as well as addition for N. 3. Construct the addition and multiplication functions for Z using the Cartesian product, power set and subset axioms, as well as the addition and multplication functions for N. Prove that these functions are unique and have the required algebraic properties. (I am at about this point.) 4. Construct the set of rational numbers Q as a kind of equivalence class on the power set of Zx(non-zero integers) using the Cartesian product, power set and subset axioms, as well as the multiplication function for Z. 5. Construct the addition function for Q using the Cartesian product, power set and subsets axioms, as well the addition and multiplication function for Z. Prove that these functions are unique and have the required algebraic properties. 6. Define a strict ordering (<) on Q using the addition function for Q. 7. Construct the set on Dedekind cuts on the set Q using the power set and subset axioms, as well as this strict ordering (<) on Q. Readers commments would be appreicated. Dan Download DC Proof 1.0 at http://www.dcproof.com === Subject: Re: Reals without infinity > An approach I have implemented in the form of a proof checker > (downloadable at my homepage) is based very loosely on ZF and does not > include an axiom of infinity. My set axioms consist include only > schemas for: subsets, power sets, set equality and Cartesian products. > There are, of course, several definitions and justifiable shortcuts > built in to make life easier. > Though still very much a work in progress, I have done some work > toward constructing the real numbers starting from a first order > version of Peanos Axioms as an intitial premise. Informally, here is > an outline of how I plan to proceed: So youre assuming the set of natural numbers exists? Thats basically the same thing as the axiom of infinity. === Subject: Re: Reals without infinity >While responding to Eray, I found myself wondering, if we abandon the >axiom of infinity... >I think I can see how to construct something isomorphic to the ring Z >in ZF, I doubt that. >and from that almost imagine how to construct Q as the >(provable) lower bound of any finite collection of quotients of Z >(elements of ZxZ) meeting some conditions. >But in the absence of the axiom of infinity, how _does_ one construct >an ordered field F such that F is isomorphic to R in ZF? I suppose the >same way... but here my imagination fails me. >I presume it involves some re-definitions of the idea of >convergence... >TIA - Chas ************************ David C. Ullrich === Subject: Re: Reals without infinity >While responding to Eray, I found myself wondering, if we abandon the >axiom of infinity... >I think I can see how to construct something isomorphic to the ring Z >in ZF, > I doubt that. Youre no doubt correct (and the curtness of your reply makes me think Im either missing something extremely basic or extremely abstract); but I still cant see whats wrong with this approach (feel free to jump in at the first glaring error): We define something like the naturals via the usual successor function; so a set has the property is a natural iff it is the result of a finite sequence of applications of the successor function to the empty set. This is not the same as asserting that the _set_ of all naturals exists; it just asserts that I can tell you whether any (finite) set S is, or is not, a natural. Attach a sign via cross product with the finite set (Z_2), and call any set which can be so constructed an integer. It seems clear to me that we can always determine, in a finite number of steps, whether or not a given finite set S satisfies the property integer. We can then construct, for any two sets satisfying the above definition of integer, a new, finite set which also provably has the property integer for each of multiplication and addition. Given any set S with the property integer, I can prove the existence of a set -S with property integer which is its additive inverse. To show an isomorphism: in ZF, we have the axiom of infinity, and its obvious that each element with the property integer is a member of Z. Conversely each member of Z in ZF is clearly the result of a finite number of applications of the successor function (and a sign); and so has the property integer. The operations multiplication and addition are preserved in both directions. Whats wrong with this approach (aside from the fact that I dont actually have any problems with the axiom of infinity, and this alternative seems like a tremendous pain in the neck)? It seems somewhat like the idea of proper classes; I specify the class of integers by specifying a property, without asserting that the resulting collection of all integers actually exists as a set. === Subject: Re: Reals without infinity >>While responding to Eray, I found myself wondering, if we abandon the >>axiom of infinity... >>I think I can see how to construct something isomorphic to the ring Z >>in ZF, >> I doubt that. >Youre no doubt correct (and the curtness of your reply makes me think >Im either missing something extremely basic or extremely abstract); >but I still cant see whats wrong with this approach (feel free to >jump in at the first glaring error): >We define something like the naturals via the usual successor >function; so a set has the property is a natural iff it is the >result of a finite sequence of applications of the successor function >to the empty set. >This is not the same as asserting that the _set_ of all naturals >exists; it just asserts that I can tell you whether any (finite) set S >is, or is not, a natural. You can certainly do _that_ - thats not what I thought you meant by construct something isomorphic to Z. (Probably because thats not what construct something isomorphic to Z _means_.) >Attach a sign via cross product with the finite set (Z_2), and call >any set which can be so constructed an integer. It seems clear to me >that we can always determine, in a finite number of steps, whether or >not a given finite set S satisfies the property integer. But I have no idea what construction youre talking about here, sorry. >We can then construct, for any two sets satisfying the above >definition of integer, a new, finite set which also provably has the >property integer for each of multiplication and addition. Not sure whether I believe that or not (hard to say since I dont know what your integers are.) But one can define positive integers to be finite ordinals and define the sum of two ordinals without the axiom of infinity (I think). >Given any >set S with the property integer, I can prove the existence of a set >-S with property integer which is its additive inverse. >To show an isomorphism: in ZF, we have the axiom of infinity, and its >obvious that each element with the property integer is a member of >Z. Conversely each member of Z in ZF is clearly the result of a finite >number of applications of the successor function (and a sign); and so >has the property integer. The operations multiplication and addition >are preserved in both directions. >Whats wrong with this approach (aside from the fact that I dont >actually have any problems with the axiom of infinity, and this >alternative seems like a tremendous pain in the neck)? >It seems somewhat like the idea of proper classes; I specify the >class of integers by specifying a property, without asserting that >the resulting collection of all integers actually exists as a set. ************************ David C. Ullrich === Subject: Re: Reals without infinity Originator: joshp@xoxy.net (joshp) >It seems somewhat like the idea of proper classes; I specify the >class of integers by specifying a property, without asserting that >the resulting collection of all integers actually exists as a set. Yes. For a fully-worked out formal proof that the class of natural numbers can be constructed in ZF without the axiom of infinity, see: through . -- Josh Purinton === Subject: Re: Reals without infinity >>It seems somewhat like the idea of proper classes; I specify the >>class of integers by specifying a property, without asserting that >>the resulting collection of all integers actually exists as a set. > Yes. > For a fully-worked out formal proof that the class of natural > numbers can be constructed in ZF without the axiom of infinity, see: > through > . That theorem doesnt prove the existence of the naturals, but says that if S is a set which includes 0 and is closed under the successor operation, then the set of naturals (omega), is a subset of S. How do you prove the existence of sets which satisfy that theorem (withut AoI) ? Also, how do you get omega without the axiom of infinity? -- Replace Roman numerals with digits to reply by email === Subject: Re: Reals without infinity Originator: joshp@xoxy.net (joshp) >It seems somewhat like the idea of proper classes; I specify the >class of integers by specifying a property, without asserting that >the resulting collection of all integers actually exists as a set. Yes. For a fully-worked out formal proof that the class of natural numbers can be constructed in ZF without the axiom of infinity, see: through . === Subject: Re: Reals without infinity > While responding to Eray, I found myself wondering, if we abandon the > axiom of infinity... > I think I can see how to construct something isomorphic to the ring Z > in ZF, and from that almost imagine how to construct Q as the > (provable) lower bound of any finite collection of quotients of Z > (elements of ZxZ) meeting some conditions. > But in the absence of the axiom of infinity, how _does_ one construct > an ordered field F such that F is isomorphic to R in ZF? I suppose the > same way... but here my imagination fails me. > I presume it involves some re-definitions of the idea of > convergence... > TIA - Chas All axioms of ZF other than infinity also hold in HF (the hereditarily finite sets), so youll have a hard time even constructing the set of natural numbers, never mind Z, Q or R. === Subject: [perfect numbers] In what I wrong? [paradox!] cant see in what step: By eq. num (31) of http://mathworld.wolfram.com/Euler-MascheroniConstant.html, we have that sigma(n)/n - [ln n +2gamma -1] tends to 0 as n tends to infinity. In particular, if we take some succession {a_n} of natural numbers, it happens to applied to that succession. If we take {a_n} = {P_n} where P_n is the n-th even perfect number, we have that: sigma(P_n)/P_n - [ln P_n + 2gamma -1] tends to 0 as n tends to infinity. But by being P_n perfect, we have that sigma(P_n)=2P_n. So 2 - [ln P_n + 2 gamma -1] --> 0 as n--> infinity But the second summand tends to infinity, so there is a contradiccion!. Please, help me, I will thank you many times. Xan. === Subject: Re: [perfect numbers] In what I wrong? [paradox!] > cant see in what step: > By eq. num (31) of http://mathworld.wolfram.com/Euler-MascheroniConstant.html, > we have that > sigma(n)/n - [ln n +2gamma -1] tends to 0 as n tends to infinity. I do not have access to MathWorld right now, but that formula that you sigma(n) = #{divisors of 1} + #{divisors of 2} + ... + #{divisors of 1} Jose Carlos Santos === Subject: Re: [perfect numbers] In what I wrong? [paradox!] >> By eq. num (31) of >> http://mathworld.wolfram.com/Euler-MascheroniConstant.html, >> we have that >> sigma(n)/n - [ln n +2gamma -1] tends to 0 as n tends to infinity. > I do not have access to MathWorld right now, but that formula that you > sigma(n) = #{divisors of 1} + #{divisors of 2} + ... + #{divisors of 1} Of course, it should have been divisors of n at the end of the previous line. Furthermore, now that I have again access to MathWorld I can that thats what written there. Jose Carlos Santos === Subject: Re: [perfect numbers] In what I wrong? [paradox!] >> By eq. num (31) of >> http://mathworld.wolfram.com/Euler-MascheroniConstant.html, >> we have that >> sigma(n)/n - [ln n +2gamma -1] tends to 0 as n tends to infinity. > > > I do not have access to MathWorld right now, but that formula that you > > sigma(n) = #{divisors of 1} + #{divisors of 2} + ... + #{divisors of 1} > Of course, it should have been divisors of n at the end of the > previous line. Furthermore, now that I have again access to MathWorld I > can that thats what written there. > Jose Carlos Santos Okay. Stupid error!. another time, I will read first the text more carefully. Xan. === Subject: Re: Simple group(s) of order 504? >[...] > As for existence, it would not be impossibly difficult to show > directly > that the group constructed above really does have order 504 > and is > simple. I am sure I can come up with a proof of that if you are > interested. >>Very! Especially if its elegant. :-) >> OK - here goes, but very brießy! >> We have G = < x,w,t > with >> x=(2,3)(4,5)(6,7)(8,9), w=(3,4,6,5,8,9,7), t=(1,2)(4,7)(6,9)(5,8). >> We constructed these generators to make w induce an automorphism of >> order 7 >> by conjugation on < x, y=x^w, z=x^(w^2) >, so H := < x,w > has order >> 56. >> Now t^-1 H t fixes point 2, so H ^ t^-1 H t = < w > has order 7, and >> the >> double coset HtH has order |H|^2 / |H ^ t^-1 H t| = 56 * 8. So, to >> prove >> that G has order 504 = 56 * 9, it suffices to show that G = H U HtH. >> To do this, we need to show that tHt <= H U HtH. >Im probably missing something obvious, but why does >tHt <= H U HtH imply that G = H U HtH? > H = < H, t >, so to show that G = H U HtH, it is enough to show that > H U HtH is a subgroup of G. It is clearly closed under inversion, so need > to check closure, and that follows easily from tHt <= H U HtH. -- Jim Heckman === Subject: Re: Simple group(s) of order 504? [...] >> As for existence, it would not be impossibly difficult to show directly >> that the group constructed above really does have order 504 and >> is >> simple. I am sure I can come up with a proof of that if you are >> interested. >Very! Especially if its elegant. :-) > OK - here goes, but very brießy! > We have G = < x,w,t > with > x=(2,3)(4,5)(6,7)(8,9), w=(3,4,6,5,8,9,7), t=(1,2)(4,7)(6,9)(5,8). [snip proof that G is a group of order 504] > Since G acts 3-transitively, a proper nontrivial normal subgroup could > only have order 9 or 72. Again Im probably missing something obvious, but how/why does 3-transitivity limit the normal subgroups? > A normal subgroup of order 9 would be a minimal > normal subgroup, and hence elementary abelian, but we know that the > Sylow 3-subgroups are cyclic. A normal subgroup of order 72 would contain > all 9 Sylow 2-subgroups of G, but then it would have only 9 elements left > of order a power of 3, so we would get a normal subgroup of order 9 again. > So G is simple. [...] -- Jim Heckman === Subject: Re: Simple group(s) of order 504? >[...] > As for existence, it would not be impossibly difficult to show directly > that the group constructed above really does have order 504 and > is > simple. I am sure I can come up with a proof of that if you are > interested. >>Very! Especially if its elegant. :-) >> OK - here goes, but very brießy! >> We have G = < x,w,t > with >> x=(2,3)(4,5)(6,7)(8,9), w=(3,4,6,5,8,9,7), t=(1,2)(4,7)(6,9)(5,8). >[snip proof that G is a group of order 504] >> Since G acts 3-transitively, a proper nontrivial normal subgroup could >> only have order 9 or 72. >Again Im probably missing something obvious, but how/why does >3-transitivity limit the normal subgroups? You can deduce my statement about normal subgroups having order 9 or 72 from the theorem that any nontrivial normal subgroup of a 2-transitive group is transitive (which itself follows from the more general result that a nontriival normal subgroup of a primitive group is transitive). Derek Holt. >> A normal subgroup of order 9 would be a minimal >> normal subgroup, and hence elementary abelian, but we know that the >> Sylow 3-subgroups are cyclic. A normal subgroup of order 72 would contain >> all 9 Sylow 2-subgroups of G, but then it would have only 9 elements left >> of order a power of 3, so we would get a normal subgroup of order 9 again. >> So G is simple. >[...] >-- >Jim Heckman === Subject: Re: Simple group(s) of order 504? >[...] > As for existence, it would not be impossibly difficult to show > directly > that the group constructed above really does have order 504 > and is > simple. I am sure I can come up with a proof of that if you are > interested. >>Very! Especially if its elegant. :-) >> OK - here goes, but very brießy! >> We have G = < x,w,t > with >> x=(2,3)(4,5)(6,7)(8,9), w=(3,4,6,5,8,9,7), t=(1,2)(4,7)(6,9)(5,8). >[snip proof that G is a group of order 504] >> Since G acts 3-transitively, a proper nontrivial normal subgroup could >> only have order 9 or 72. >Again Im probably missing something obvious, but how/why does >3-transitivity limit the normal subgroups? > You can deduce my statement about normal subgroups having order 9 or 72 > from the theorem that any nontrivial normal subgroup of a 2-transitive > group is transitive (which itself follows from the more general result > that a nontriival normal subgroup of a primitive group is transitive). Ah, right. So a normal subgroup N must have order a multiple of 9, and all but 9 and 72 are excluded by the fact that N must contain all 63 conjugate elements of order 2, or all 216 elements of order 7, or both. OK, got it. -- Jim Heckman === Subject: Re: Simple group(s) of order 504? ... > Since G acts 3-transitively, a proper nontrivial normal subgroup could > only have order 9 or 72. >>Again Im probably missing something obvious, but how/why does >>3-transitivity limit the normal subgroups? >> You can deduce my statement about normal subgroups having order 9 or 72 >> from the theorem that any nontrivial normal subgroup of a 2-transitive >> group is transitive (which itself follows from the more general result >> that a nontriival normal subgroup of a primitive group is transitive). >Ah, right. So a normal subgroup N must have order a multiple of >9, and all but 9 and 72 are excluded by the fact that N must >contain all 63 conjugate elements of order 2, or all 216 >elements of order 7, or both. OK, got it. OK! But what I was thinking of was, N must have order a multiuple of 9. Either it is exactly 9, or the intersection of N with the point stabilizer is a nontrivial normal subgroup of a 2-transitive group of degree 8 and order 56, so it must be divisible by 8. Hence 9 and 72 are the only possibile orders for N. Derek Holt. === Subject: Re: Simple group(s) of order 504? > I dont know, but personally I dont see much similarity between > the structures of L(2,7) =~ L(3,2), of order 168, and L(2,8), of > order 504. Obviously they are both special linear groups, and they are the two smallest Hurwitz groups, meaning they have a presentation which adds relations to a^2=b^3=(ab)^7=1. L2(7) is the automorphism group of the Klein quartic, and L2(8) of the Fricke-Macbeath curve/Riemann surface of degree seven. This is kind of neat, but whether you think it amounts to much by way of similarity is up to you. === Subject: Re: Enumerating automorphisms of a group > Im going to ask a very general question. Im hoping that if someone > can enlightened me on this, then I will be able to see how the various > theorems of group theory fall together. The question is this: > Given G a finite group. Find the group of automorphisms of G. > For some groups, there is a certain answer. For example, for Sn the > symmetric group of order n!, we know Aut Sn = Inn Sn = Sn (at least for > n != 6). But in general, how should one go about enumerating all the > automorphisms of any finite group? Is there even a way to just count > the *number* of automorphisms? Im sorry you didnt really get any informative replies. I was hoping to learn something. The way I myself do this is: If you have a presentation for G and know its conjugacy classes and subgroup structure, then you can count the number of automorphisms by counting how many ways you can choose the first generator, multiply by the number of ways you can choose the second generator given that youve already chosen the first, etc. For example, A_4 has a presentation . There are 3 ways to choose x, then 2 ways to choose y given x, then 4 ways to choose z given x and y, so |Aut(A_4)| = 3*2*4 = 24. -- Jim Heckman === Subject: Re: =/= ? >Why is =/= used in place of =? >http://mysite.verizon.net/vze8adrh/news.html (profile) --Tim923 My email is >valid. Because one says its not equal and the other says, what is it equal to. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Weak convergence In the following, C denotes the complex numbers. Let E be a complex Banach space, and A a subset of E. My book calls A weakly closed whenever for any sequence (a_n) of elements of A, if lim x*(a_n)=x*(a) for any bounded linear functional x*: E->C, then a is in A. A natural question -that is not mentioned in my book- would be to determine whether a weakly closed set A is closed for the weak topology, whose subbasis is given by {x*^(-1)(U); x*: E->C bounded, U open set of C}. I fail when trying to prove that whenever a is in Adh(A)-A, then there exists a sequence (a_n) of elements of A weakly converging to a. The fact is I can control the behaviour of a_n according to a finite number of x* only. Any suggestion ? - a one word answer would suffice, Id just want to know if I should try proving it or not -. -- Julien Santini === Subject: Re: Weak convergence >In the following, C denotes the complex numbers. >Let E be a complex Banach space, and A a subset of E. My book calls A >weakly closed whenever for any sequence (a_n) of elements of A, if lim >x*(a_n)=x*(a) for any bounded linear functional x*: E->C, then a is in A. >A natural question -that is not mentioned in my book- would be to determine >whether a weakly closed set A is closed for the weak topology, whose >subbasis is given by {x*^(-1)(U); x*: E->C bounded, U open set of C}. >I fail when trying to prove that whenever a is in Adh(A)-A, then there >exists a sequence (a_n) of elements of A weakly converging to a. The fact is >I can control the behaviour of a_n according to a finite number of x* only. >Any suggestion ? - a one word answer would suffice, Id just want to know if >I should try proving it or not -. Not. ************************ David C. Ullrich === Subject: Re: Weak convergence > In the following, C denotes the complex numbers. > Let E be a complex Banach space, and A a subset of E. My book calls A > weakly closed whenever for any sequence (a_n) of elements of A, if lim > x*(a_n)=x*(a) for any bounded linear functional x*: E->C, then a is in A. > A natural question -that is not mentioned in my book- would be to determine > whether a weakly closed set A is closed for the weak topology, whose > subbasis is given by {x*^(-1)(U); x*: E->C bounded, U open set of C}. > I fail when trying to prove that whenever a is in Adh(A)-A, then there > exists a sequence (a_n) of elements of A weakly converging to a. The fact is > I can control the behaviour of a_n according to a finite number of x* only. > Any suggestion ? - a one word answer would suffice, Id just want to know if > I should try proving it or not -. > -- > Julien Santini If you use weak convergence of nets (generalized sequences), then weakly closed described that way is equivalent to closed in the weak topology. But in general it is not. If your subset A is convex, then sequences are enough. But in general, sequences are not enough. In the Banach space l^1 of summable sequences, it is surprising that a squence converges weakly if an only if if converges in norm. (a gliding hump argument...) But the weak and norm topologies are different, so there are sets A that are norm closed but not weakly closed. Such as the set of all x with norm >=1 . -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Weak convergence >> In the following, C denotes the complex numbers. >> Let E be a complex Banach space, and A a subset of E. My book calls A >> weakly closed whenever for any sequence (a_n) of elements of A, if lim >> x*(a_n)=x*(a) for any bounded linear functional x*: E->C, then a is in A. >> A natural question -that is not mentioned in my book- would be to determine >> whether a weakly closed set A is closed for the weak topology, whose >> subbasis is given by {x*^(-1)(U); x*: E->C bounded, U open set of C}. >> I fail when trying to prove that whenever a is in Adh(A)-A, then there >> exists a sequence (a_n) of elements of A weakly converging to a. The fact is >> I can control the behaviour of a_n according to a finite number of x* only. >> Any suggestion ? - a one word answer would suffice, Id just want to know if >> I should try proving it or not -. >> -- >> Julien Santini >If you use weak convergence of nets (generalized sequences), then weakly >closed described that way is equivalent to closed in the weak topology. >But in general it is not. If your subset A is convex, then sequences are >enough. Because then A weakly closed actually implies A norm-closed. >But in general, sequences are not enough. >In the Banach space l^1 of summable sequences, it is surprising that >a squence converges weakly if an only if if converges in norm. Aargh. Seemed like l^1 should be an exampple since the dual is far from separable, but I couldnt recall the argument so I had to rely on the fact that he said a one-word answer would suffice. Hence my compulsion to explain why l^1 has this surprising property: Lets write x(1), x(2), ... for the coordinates of the sequence x in l^1, and suppose that x_n is a sequence in l^1 that tends to 0 weakly but not in norm. We can assume that ||x_n|| > 5 for all n. There is a finite set F_1 such that sum_{j not in F_1}|x_1(j)| < 1, and hence sum_{j in F_1} |x_1(j)| > 3. Now since x_n -> 0 weakly we must have lim_n x_n(j) = 0 for all j. In particular there exists x_n such that the sum of |x_n(j)| for j in F_1 is < 1; to simplify the typing we assume that x_2 has this property. There is a finite set F_2 containing F_1 such that sum_{j not in F_2}|x_2(j)| < 1, and hence sum_{j in F_2F_1} |x_1(j)| > 3. And so on - relabeling the subsequence of the x_n were constructing as x_n we get increasing finite sets F_n such that sum_{j in F_{n-1}|x_n(j)| < 1, sum_{j not in F_n}|x_n(j)| < 1, and hence sum_{j in F_nF_{n-1}} |x_n(j)| > 3. Now choose y in l^infinity with |y(j)| = 1 for all j and such that y(j)x_n(j) = |x_n(j)| for j in F_nF_{n-1} and it follows that || >= 3 - 1 - 1 > 0, so its not true that x_n -> 0 weakly. >(a gliding hump argument...) But the weak and norm topologies are >different, so there are sets A that are norm closed but not weakly >closed. Such as the set of all x with norm >=1 . ************************ David C. Ullrich === Subject: Re: Weak convergence > If you use weak convergence of nets (generalized sequences), then weakly > closed described that way is equivalent to closed in the weak topology. > But in general it is not. terminology weakly closed by the author was quite awkward ... I had to check ! -- Julien Santini === Subject: Relativistic field theory for ßuids posting-account=jcZk7AwAAADXpPEyHtVyWC264SxtppRB FIELD THEORY FOR A PERFECT FLUID IN 2D SPACETIME Consider a CMCS (T,z) = (z^0,z^1) in 2D spacetime. Let the ßuid have pressure p given by its internal or thermal energy q(n). The 1st law of thermo is T_e dS = dq + p d(1/n) ; Assume the ßow is isentropic; S = const. dS = 0 ==> p = n^2 dq/dn ; dp = n df or dr = f dn where r = n(1 + q) = total proper energy density T_e = temperature ; S = entropy ; f = 1 + q + p/n = relativistic enthalpy. Define c = fu. We also have L = - r. The Euler Lagrange can be dealt with by absorbing n_o into z, so that j = dz, which follows from dj = 0. The Euler-Lagrange (EL) eq. is div(dL/d(dz)) = 0 ; - dL/d(dz) = dr/d(dz) = f dn/d(dz) = f d sqrt(dz|dz)/d(dz) = f dz/n = *(fu) = *c ; where c = fu; The EL eq. is div(*c) = *d**(fu) = *d(fu) ==> d(fu) = dc = 0 ==> c = dT = fu ; This suggests that we should choose z^0 = T, e^0 = dz^0 = dT , u = dT/f ;and z^1 = z, e^1 = dz = j = n*u , *u = dz/n . So in the CMCS (T,z) the e^i are not unit vectors, though they are orthogonal (in 2D)., since g^01 = g(e^0,e^1) = (dT|dz) = nf *u(u) = 0. g^00 = (dT|dT) = - f^2 ; g^11 = (dz|dz) = n^2 ==> det(g) = -1/(nf)^2 , and g = - uu + *u*u = - dT dT/f^2 + dz dz/n^2 in a CMCS. The energy-momentum tensor (EMT) is (the EMT T not to be confused with z^0 = T) T = Lg - dz dL/d(dz) = - rg + fdz dz/n = - rg + nf *u *u nf = r + p, so T = ruu + p*u*u = nfuu + pg. We have 2 equivalent eqns. from which to get the eq. of motion; The EL eq, d(fu) = df / u + f du = 0; take scalar product with u; P(df) + fa = 0 ; where P = g + uu = *u*u in 2D is the spatial projection operator, and a = du(u) = acceleration. Using the 1st law of thermo above gives a = - P(df/f) ; or nfa + P(dp) = 0 which is just the Euler eqn. of motion for a perfect ßuid. The EMT ==> div(T) = div(nfuu) + dp = nu df(u) + nfa + dp (u|div(T)) = dp(u) - n df(u) = T_e dS(u) = 0 ; this is conservation of energy for adiabiatic ßow with dS(u) = 0, i.e., entropy is const. along streamlines. The spatial part of div(T) = 0 is P(div(T) = nfa + P(dp) = 0 ; again, the Euler eq. of motion for a perfect ßuid. Van === Subject: Free maths source www.bossh.net -------China http://www.bossh.net is the best and largest maths web in china. We introduce math news,change our idears,and share our sources. Welcome you to BossH.net The forums :http://www.bossh.net/forums/index.php?act=idx === Subject: Re: Free maths source www.bossh.net -------China posting-account=997LIQ0AAAAYVSCeNZjd_pYD-JBOASeA Do provide us an english-translated link if exists so that we can also access the resources and participate in the forum. === Subject: Re: Infinity is not that big! Discussion, linux) >> oo is not a number >> oo has no magnitude > oo is a number greater than all the other positive numbers > oos magnitude is greater than all other numbers >> 0 is not a natural number >> 0 has no magnitude > 0 is a natural number, it was born way. > 0 is a magnitude, and has a magnitude of 0 >> if infinity was a number then oo = 0 > But you state that infinity is not a number, so how can it = 0. Because 0 is not a number. Duh[1]. On the other hand, *if* infinity is a number, then how can it equal the non-number 0? Footnotes: [1] If you take this comment seriously, then you *deserve* to be talking to Herc. -- The papers are currently at journals. [When published,] make no mistake, there will be no place on this planet where you can hide. Remember, Im not talking about something vague here. Im talking about publication in journals. James S. Harris. Wow. Journals. === Subject: Re: Infinity is not that big! Discussion, linux) >> oo is not a number >> if infinity was a number then oo = 0 > Also note that if infinity was a number lions would fall from the sky, > and moon pies would be movie critics. Start with a contradiction, and > you can prove anything. The claim that infinity is a number is contradictory only given his other claim that infinity is not a number. Without some hint of what one means by infinity and number, neither claim is meaningful. -- Meaningless movies on the screen behind the band thats blowing Waterboys, throwing shapes My Love is My Rock Half of the music is on tape in the Weary Land === Subject: Re: Infinity is not that big! >> oo is not a number >> if infinity was a number then oo = 0 > Also note that if infinity was a number lions would fall from the sky, > and moon pies would be movie critics. Start with a contradiction, and > you can prove anything. > The claim that infinity is a number is contradictory only given his > other claim that infinity is not a number. > Without some hint of what one means by infinity and number, neither > claim is meaningful. Exactly. Id been puzzled why AB claimed that considering infinity to be a number would lead to a contradiction. But then I noticed he said the following in an older thread: You seem to still be thinking of inifinity as a number like 4 or 7, but we know its different. For example whats 4*oo? If we assume our intuition is right, it would just be oo again. Now we then go what is 2*oo? oo again. So by this we could say that 4*oo/2*oo = oo/oo = 2. But, I could just as easily write 6*oo/2*oo = oo/oo = 3 = 2? This obvious contradiction means we made a mistake by assuming that oo is a number. This argument is not valid, of course. For example, one can replace every instance of oo above by 0 and thereby neatly prove that 0 cannot be a number either. David Cantrell, whos _still_ singing Aleph_nought bottles of beer on the wall... === Subject: Re: Infinity is not that big! _____________________ /| /| | | ||__|| | Do not feed the | / | --Mgt. | / |_____________________| / _ || / |____ || / | | | |____/ || / |_|_|/ | _|| / / |____| || / | | | --| | | | |____ --| * _ | |_|_|_| | -/ *-- _-- _ | || / _ | / ` * / _ /- | | | * ___ c_c_c_C/ C_c_c_c____________ === Subject: Re: Riemann Mapping Theorem Question >Basically Riemann said that you can take a surface in 3d and map it one >for one to >a plane. There are lots of ways that this is done in cartography and in >mathematics. >Using terms like biholomorphically doesnt convice anyone that > you know what a stereographic projection or a Gauss map is, does it? On the other hand using phrases like stereographic projection and Gauss map _does_ convince everyone that you have no idea what the question was about. >Try expaining what you want to know in universally understood English. >Try Google for: >Gauss Map >stereographic projection of the sphere >bielliptical Riemannian sphere mapping >Projective line mapping of a circle( 1d to 2d map) _You_ might try googling Riemann Mapping Theorem, since thats the topic of the thread. >>In the regular RMT for the complex plane, does the boundary get mapped >>biholomorphically to the other boundary? >> ************************ David C. Ullrich === Subject: Re: Riemann Mapping Theorem Question posting-account=2LgBuQ0AAAC2HZQp160dLwjzCEeNC7or Right, sorry I ment bicontinuously of course, ie homeomorphically. This was what I was searching for thank you! === Subject: Re: Riemann Mapping Theorem Question posting-account=2LgBuQ0AAAC2HZQp160dLwjzCEeNC7or that was what I was searching for! === Subject: Re: analysis........./ === Subject: analysis........./ > {x_n} : real-sequence > show that if > lim {x_(n+1) - x_n} = 1 > n->00 > then, lim (x_n) / n = 1 > n->00 > For each e > 0 there is an K such that > |x_(n+1) - x_n - 1| < e if n >= K. By telescoping, (k = K) |x_(n+k) - x_k - n| <= sum(=1,n) |x_(j+k) - x_k - 1| < ne <= (n+k)e |x_(n+k)/(n+k) - x_k/(n+k) - n/(n+k)| < e Thus for large enuf n, as lim(k->oo) (x_k/(n+k) + n/(n+k)) = 1 |x_(n+k)/(n+k) - 1| < 3e ---- === Subject: Re: analysis........./ > hello.....doctor~ > {x_n} : real-sequence > show that > if > lim {x_(n+1) - x_n} = 1 > n->00 > then, lim (x_n) / n = 1 > n->00 > --------------------------------------------- > this is a content of solution paper. > For each e > 0 there is an K such that > |x_(n+1) - x_n - 1| < e if n >= K. > if n >= K, > (x_K) + (n+1-K)(1-e) < x_(n+1) < (x_K) + (n+1-K)(1+e) > (*****) i cant understand left side[ (x_K) + (n+1-K)(1-e) < x_(n+1) ] > but i can understand right side. if n>=K, then x_(n+1) - x_n - 1 >-e => x_(n+1) > x_n + 1 -e. Therefore x_(K+1) > x_K + 1 -e. x_(K+2) > x_(K+1) + 1 -e . . . x_(n+1) > x_n + 1 -e. If we add these n-k+1 inequalities, we get x_(n+1) > x_K + (n-K+1) (1-e). Artur > because, > |x_(n+1) - x_n - 1| < e if n >= K. > => -e < x_(n+1) - x_n - 1 < e if n >= K. > => x_(n+1) < (x_K) + (1+e) > => x_(n+1) < (x_K) + (n+1-K)(1+e) (n-K >= 0 => n+1-K >= 1) > but i cant deduce left side > i need your advcie. > thank you very much. > so, apply limit to upper inequality. > lim [{(x_K)/(n+1)} + {(n+1-K)/(n+1)}*(1-e)] > n->00 > <= lim [{x_(n+1)} / (n+1)] > n->00 > <= lim [{(x_K)/(n+1)} + {(n+1-K)/(n+1)}*(1+e)] > n->00 > so, > 1-e <= lim [{x_(n+1)} / (n+1)] <= 1+e > n->00 > since e is any number, > lim (x_n) / n = 1 > n->00 === Subject: Q&A with Professor Didier Sornette Im running a moderated Q&A with Professor Didier Sornette, author of ÔWhy Stock Markets Crash: Critical Events in Complex Financial systems (http://www.ess.ucla.edu/faculty/sornette) Im looking for good questions which Ill put to him in a batch and publish his responses in my open-access webjournal and in the forum where you can post your questions: http://www.moneyscience.org/tiki/tiki-view_forum.php?forumId=9 Jacob === Subject: Re: JSH: Explaining the obvious by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i9UCG9014684; *Incapable*. Yes. They are *incapable*. JSH and [Hammond] should contribute more. Ad Hominem === Subject: area of circle before calculus Can the area of a circle be proven without using calculus? And was it used before it was really proven? - http://mysite.verizon.net/vze8adrh/news.html (profile) --Tim923 My email is valid. === Subject: Re: area of circle before calculus > Can the area of a circle be proven without using calculus? > And was it used before it was really proven? > .... Various ancient peoples had approximations, all assuming that the formula was (some constant)*(diameter)^2. The constant, which we now call pi/4, can easily grab our attention; but the first step needed was a proof that this formula really did have the correct form. The earliest surviving proof of that fact is Euclid XII.2: [Areas of] circles are to one another as the squares on the diameters. It used a rigorous limit argument stated in ancient Greek terms, possibly a bit hard for a modern reader to follow. In principle it was an integration (a limit of a sum), but of course it was long before the development of the calculus in its modern form from the 17th century onward. The only remaining problem then was to find the constant which we group in September 1999. > Some ancient peoples seem to have been quite content with the > approximation pi = 3. It was used in Mesopotamia (Iraq) and China, and > even occurs in the Bible (I Kings 7.23). The Egyptians calculated the > area of a circle by squaring (8/9 of the diameter), which amounts to > using 256/81 (about 3.1605) as an approximation to pi. Their reason > for this can only be guessed. The Greek mathematician Archimedes (3rd > century B.C.) used 96-gons inscribed and circumscribed to a circle, to > prove rigorously that pi is somewhere between 3 1/7 and 3 10/71. > Chinese mathematicians worked on improved approximations, culminating in > a lost book of Zu Chongzhi (5th century A.D.) which is believed to have > contained not only the well-known 22/7 (about 3.1429) but also the > excellent approximation 355/113 (about 3.1415929). However other > Chinese writers continued simply using 3. > Modern-style calculations of pi use infinite processes, such as > Gregorys series (already known in India by about 1400) > pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - ..... (inefficient for calculation), > or Walliss infinite product > pi/2 = (2/1)(2/3)(4/3)(4/5)(6/5)(6/7).... > or the integral > pi/4 = integral from 0 to 1 of 1/(1 + x^2) dx > or many others. These all use some calculus or other theory, rather than > dealing directly with circles. Ken Pledger. === Subject: Re: area of circle before calculus > Can the area of a circle be proven without using calculus? > And was it used before it was really proven? > http://mysite.verizon.net/vze8adrh/news.html (profile) --Tim923 My email is valid. It depends on what one accepts as a proof. More specifically, it depends on what system of axioms one begins with. Within the classical axiomatic framework of Greek (Euclidean) geometry, few would deny that Archimedes of Syracuse (287-212 BC) had successfully proven the formula for computing the area of a circle. He approximated the circle via his novel application of the method of exhaustion by inscribing and circumscribing regular polygons. The method of exhaustion was reportedly invented by Eudoxus of Cnidos (~408-355 BC). Archimedes techniques came so close to the modern methods of integral calculus that one generally reveres him as the father of the integral calculus. In other words, it would probably be fair to say that Archimedes invented the integral calculus to approximate the value of pi and to find the area of the circle. You asked whether or not the area formula was used before it was proven. Im no historian, but I think the likely answer is yes in view of reported evidence that the Babylonians had employed the Pythagorean Theorem long before it was officially established as a theorem in Greek geometry. To my knowledge--mathematical historians would correct me if Im wrong--the concept of a rigorous proof based on an axiomatic framework was singularly a Greek invention (first systematized by Euclid of Alexandria, ~325-265 BC). Perhaps someone can comment on whether the circle area formula was known to the Indians or the Chinese in ancient times before the Greeks. http://www.math.utah.edu/~alfeld/Archimedes/Archimedes.html http://personal.bgsu.edu/~carother/pi/Pi3a.html http://www.maa.org/editorial/knot/GeometrySampling.html http://www.math.ubc.ca/~cass/archimedes/circle.html http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/ propXII2.html http://www.jimloy.com/geometry/pi.htm http://www.maths.uwa.edu.au/~schultz/3M3/L7Archimedes1.html Shedar === Subject: Re: area of circle before calculus ETAsAhQIIr4A/t/SPwcZMHKMjYcyjc+q4QIUbYdN0RY97UDMxSCyNxjJzL+6/ N0= There are really only two ways to evaluate the area of a general figure: dissection into a rectangle and taking limits involving objects taht can be so dissected. An example of the first method is the area of a triangle. Combine a given triangle with a like copy of itself to form a parallelogram, then transfer a triangular piece of the parallelogram from one side to the other to make a rectangle whose area is one side of the triangle times the corresponding altitude. Since two triangles were required to set p the rectangle, the area of one triangle is half that product. For a circle dissection into a rectangle cant be done directly because of the curved boundary. But we can consider a circle to be a limit of polygons, which are broken up into triangels whose areas are determined as above. Which is the method of exhaustion. Call it calculus if you will, even though the concept of exhaustion was in use long before the word calculus (in its modern sense) or the more general theory of limits. The point is, finding the area of a circle inevitably involves taking some sort of limit. --OL === Subject: Re: area of circle before calculus > There are really only two ways to evaluate the area of a general figure: > dissection into a rectangle and taking limits involving objects taht can > be so dissected. It would be amusing to consider a society of circles, living say in Flatland, which regard squares and rectangles as base and worthless, and wish to develop their theory of area using only circles--no squares or rectangles. Of course, it can be done, using some sort of swiss-cheese approach such as the Vitali covering lemma. But their notion of area would hardly be considered elementary, and would probably only be learned in graduate school, not third grade... --Ron Bruck === Subject: Re: area of circle before calculus > Can the area of a circle be proven without using calculus? > And was it used before it was really proven? By proving the area of a circle I suppose you mean prove that the area of a circle of radius r is pi times r squared. This was proved by Archimedes using his method of exhaustion. You can easily find presentations of his proof on the net. === Subject: Re: area of circle before calculus >Can the area of a circle be proven without using calculus? >And was it used before it was really proven? Has calculus been re-defined? pi*r^2 is simple trig. -- What is now proved was once only imagind. - William Blake, 1793. === Subject: Re: area of circle before calculus > Has calculus been re-defined? pi*r^2 is simple trig. pi*r^2 is just a formula. What proof that this yields the area of a circle do you have in mind? === Subject: Re: area of circle before calculus >Has calculus been re-defined? pi*r^2 is simple trig. The proof I saw that tried to avoid calculus still used limits. - http://mysite.verizon.net/vze8adrh/news.html (profile) --Tim923 My email is valid. === Subject: C^r functions in R^n Let G be an open set in R^n and f: G ->R a continuous function whose partial derivatives up to order r are uniformly continuous on G, i.e. f in C^r(overline{G}). Is it true that f can be extended to a C^r function on an open set containing $overline{G}$, the closure of G? Please give reference if the proof is too complicated. === Subject: Abstract Algebra Questions Is an operation a primative for functions. If this is the case is it no longer required that if the first elements of an ordered pair are equal, then the second elements of the same pairs are also equal: eg: for operation $: x $ 0 = a, x $ 0 = b, and a dne b how does this relate to existance and uniqueness? If the answers are too long to be posted to a message board, a link, or a book reference would be welcome. John === Subject: Re: Abstract Algebra Questions days. My association with the Department is that of an alumnus. >Is an operation a primative for functions. Is this a question? If so, it doesnt seem to make much sense to me. An operation on a set is a kind of function. More spefically, an n-ary operation (with n an ordinal) on the set A is a function from A^n (the set of maps from the ordinal n to A) to A. >If this is the case is it no longer >required that if the first elements of an ordered pair are equal, then the second >elements of the same pairs are also equal: Huh? >eg: for operation $: x $ 0 = a, x $ 0 = b, and a dne b >how does this relate to existance and uniqueness? Huh? I honestly have no idea what you are talking about. Existence and uniqueness of ->what<-? >If the answers are too long to be posted to a message board, a >link, or a book reference would be welcome. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Strong Induction clarification.... Before some time there was a post by Arturo Magidin saying: ------------------------------------------------------------- --------- Formally, you do not need to prove a base case. Strong mathematical induction involves proving that For all n in N, if for all k in N (kP(k)), then P(n). If you can really prove this proposition, then it takes care of the n=0 (the base case) automatically; because the antecedent of the implication (for all k in N, k P(k)) is true by vacuity, and so if you have really proven the implication, then P(n) must follow. ------------------------------------------------------------- --------- What i cant understand is the because he uses. Or to say it clearly how the base case is derived automatically, by proving the above proposition. Or to say it even more clearly: If we prove the aforementioned proposition, how can we prove that P(0) is true? Also another question. Is the following statement true: Something can be proved by Strong Induction IF AND ONLY IF it can be proven by Weak Induction. Or with other words: is Strong and Weak Induction equivalent? === Subject: Re: Strong Induction clarification.... > Before some time there was a post by Arturo Magidin saying: > ------------------------------------------------------------- --------- > Formally, you do not need to prove a base case. Strong mathematical > induction involves proving that > For all n in N, > if for all k in N (kP(k)), then P(n). > If you can really prove this proposition, then it takes care of the > n=0 (the base case) automatically; because the antecedent of the > implication (for all k in N, k P(k)) is true by vacuity, and so > if you have really proven the implication, then P(n) must follow. > ------------------------------------------------------------- --------- > What i cant understand is the because he uses. Or to say it > clearly how the base case is derived automatically, by proving the > above proposition. > Or to say it even more clearly: > If we prove the aforementioned proposition, how can we prove that P(0) > is true? The thing you need to prove is: For all n in N, ( (For all k in N, (k P(k))) => P(n) ) In particular you need to prove the case where n = 0, i.e. For all k in N (k<0 => P(k)) => P(0) k<0 is always false, so k<0 => P(k) is always true, so as part of the proof you show that true => P(0) i.e. that P(0) is true. === Subject: Re: Strong Induction clarification.... > Before some time there was a post by Arturo Magidin saying: > ------------------------------------------------------------- --------- > Formally, you do not need to prove a base case. Strong mathematical > induction involves proving that > > For all n in N, > > if for all k in N (kP(k)), then P(n). > > If you can really prove this proposition, then it takes care of the > n=0 (the base case) automatically; because the antecedent of the > implication (for all k in N, k P(k)) is true by vacuity, and so > if you have really proven the implication, then P(n) must follow. > ------------------------------------------------------------- --------- > > What i cant understand is the because he uses. Or to say it > clearly how the base case is derived automatically, by proving the > above proposition. > Or to say it even more clearly: > If we prove the aforementioned proposition, how can we prove that P(0) > is true? > The thing you need to prove is: > For all n in N, ( (For all k in N, (k P(k))) => P(n) ) > In particular you need to prove the case where n = 0, i.e. > For all k in N (k<0 => P(k)) => P(0) > k<0 is always false, so k<0 => P(k) is always true, so as part of the > proof you show that > true => P(0) > i.e. that P(0) is true. I couldnt have said it better, so I wont try. And yes, so-called strong and weak induction are equivalent. The easy way to see it is to see that they are each equivalent to the principle of well-ordering (for N): every non-empty subset of N has a least element. Now, assuming some proposition P(n) involving a free variable of type N were false, there would be a least n for which it fails. Go on from there. === Subject: Re: Strong Induction clarification.... days. My association with the Department is that of an alumnus. [.snip.] >> The thing you need to prove is: >> For all n in N, ( (For all k in N, (k P(k))) => P(n) ) >> In particular you need to prove the case where n = 0, i.e. >> For all k in N (k<0 => P(k)) => P(0) >> k<0 is always false, so k<0 => P(k) is always true, so as part of the >> proof you show that >> true => P(0) >> i.e. that P(0) is true. >I couldnt have said it better, so I wont try. And yes, so-called >strong and weak induction are equivalent. They are equivalent for the set of natural numbers (or for any set of order type omega). However, strong induction is, as might be expected, stronger when you have other order types. E.g., in w+w, weak induction only gets you half the elements, strong induction gets you every element. > The easy way to see it is >to see that they are each equivalent to the principle of well-ordering >(for N): every non-empty subset of N has a least element. Now, >assuming some proposition P(n) involving a free variable of type N >were false, there would be a least n for which it fails. Go on from >there. What you actually have to show is that (1) Weak induction in N. (2) Strong induction in N. (3) Well ordering of N. are equivalent. You are using Well ordering, but well ordering is in fact also equivalent to both induction properties. It is not hard to come up with a direct proof that any of the three imply any other. I usually do it (1)->(2)->(3)->(1). WEAK INDUCTION: If S is a subset of N such that: (i) 0 in S; and (ii) For every k in N, (k in S -> (k+1) in S) then S=N. STRONG INDUCTION: If S is a subset of N such that For every n in N (for all k in N (k < n -> k in S)) -> (n in S) then S = N. WELL ORDERING: If A is a nonempty subset of N, then A has a least element. (1)->(2): Let S be a set as in SI. Let S be the subset of S S = {n in S: {k in N: k(3). Let A be a nonempty subset, let S = N-A. Since S is not N, there exists n in N such that for all k in N, if k(1). Let S satisfy (i) and (ii); let A = N-S. If x_0 is the least element of A, then x_0 is not 0 by (i); so x_0 = k+1 for some k, and k is in S; then k+1=x_0 in S, a contradiction. Therefore, A is empty, so S=N. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Strong Induction clarification.... > [.snip.] > The thing you need to prove is: > > For all n in N, ( (For all k in N, (k P(k))) => P(n) ) > > In particular you need to prove the case where n = 0, i.e. > > For all k in N (k<0 => P(k)) => P(0) > > k<0 is always false, so k<0 => P(k) is always true, so as part of the > proof you show that > > true => P(0) > > i.e. that P(0) is true. >> I couldnt have said it better, so I wont try. And yes, so-called >> strong and weak induction are equivalent. > They are equivalent for the set of natural numbers (or for any set of > order type omega). However, strong induction is, as might be expected, > stronger when you have other order types. E.g., in w+w, weak induction > only gets you half the elements, strong induction gets you every > element. For much more on this see my post on 1998/05/07: --Bill Dubuque === Subject: Re: solution of equation (difficult) >6^(2x) - 2(6^x) + 2^x = 0. >So far, Ive been able to show that there are two solutions. One of >them x=0, the other is negative. But i havent been able to find the >exact value of the second solution. > Equations of this sort are very unlikely to have closed-form solutions. > According to Maple, the second solution is approximately > -0.345326898906577615347889861006. I mentioned a different method of solving such equations in the thread Solving A e^(ax) + B e^(bx) = 1 at . To use that method on our 6^(2x) - 2(6^x) + 2^x = 0, we must rewrite it in the form A e^(ax) + B e^(bx) = 1. There are different ways to do that. One of the most obvious gives 1/2 e^(log(6) x) + 1/2 e^(-log(3) x) = 1. But then using series (*) at the above link gives us merely the trivial solution x = 0 [since we have A = B = 1/2 and the series was in positive powers of (A + B - 1)]. In order to get the nontrivial solution by this method, it is helpful to graph 36^x - 2(6^x) + 2^x and note that its other zero is roughly at x = -1/3. With that in mind, lets substitute y - 1/3 for x in our equation, giving 36^(y - 1/3) - 2*6^(y - 1/3) + 2^(y - 1/3) = 0 36^(y - 1/3) + 2^(y - 1/3) = 2*6^(y - 1/3) 1/2*6^(y - 1/3) + 1/2*(1/3)^(y - 1/3) = 1 1/(2*6^(1/3)) * 6^y + 3^(1/3)/2 * (1/3)^y = 1 We may then use series (*) with A = 1/(2*6^(1/3)), a = log(6), B = 3^(1/3)/2 and b = -log(3). Using only the first ten terms of (*) to approximate the solution, we get y = -0.0119935655723 and so x = y - 1/3 = -0.3453268989056 approximately. Comparing this with the value given earlier by Robert Israel, we see that our approximation is correct through the eleventh decimal place. David Cantrell === Subject: Best Math discussion forums I am trying to track down the busiest Math discussion forums on the net. Anyone care to point out a few? Also ones dealing with statistics. Jon === Subject: C^1 function problem im new here, what do you say about the following problem? f in C^1([0,2pi]) ==> the maximum a such that |k|^a hat{f}(k) -> 0 (k->infty) is a=1. What do you suggest to prove this problem? Alex Semioli === Subject: Re: C^1 function problem > im new here, > what do you say about the following problem? > f in C^1([0,2pi]) ==> the maximum a such that |k|^a hat{f}(k) -> 0 > (k->infty) is a=1. What do you suggest to prove this problem? Thats not quite correct as stated. For example if f is C^2, then |k|^2 hat{f}(k) -> 0. And of course if f is C^2, then f is C^1. What I believe you mean to say is: the maximum a such that |k|^a*hat{f}(k) -> 0 for all f in C^1 is a = 1. Hints for proof: We know from integration by parts that a = 1 works. If a > 1, choose b, 1 < b < a, and consider f(t) = sum (k=1,oo) exp(ikt)/k^b. === Subject: Re: C^1 function problem > im new here, > what do you say about the following problem? > f in C^1([0,2pi]) ==> the maximum a such that |k|^a hat{f}(k) -> 0 > (k->infty) is a=1. What do you suggest to prove this problem? > Thats not quite correct as stated. For example if f is C^2, then |k|^2 > hat{f}(k) -> 0. And of course if f is C^2, then f is C^1. > What I believe you mean to say is: the maximum a such that |k|^a*hat{f}(k) > -> 0 for all f in C^1 is a = 1. Hints for proof: We know from integration > by parts that a = 1 works. If a > 1, choose b, 1 < b < a, and consider f(t) > = sum (k=1,oo) exp(ikt)/k^b. Oops, forget the last sentence. Consider instead f(t) = sum_(n=1,oo) exp(i2^n*t)/(n^2*2^n), which is C^1 and fails |k|^a*hat{f}(k) -> 0 for every a > 1. === Subject: Re: 4 In sci.math, kalikinkar it is a concept. > it is a feeling of void. > never aim at it. A difficult task since the traditional aiming-point is a series of red concentric zeroes... [rest snipped] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: analysis....! hello.....doctor~ f : (0, 00) -> R : differentiable. lim f(x) = L => for any h > 0, lim {f(x+h) - f(x)} / h = L ----------------------------------------------- um........ lim f(x) = lim lim {f(x+h) - f(x)} / h = L um....... i cant progress more. i need your advice. thank you very much for your advice. === Subject: Re: analysis....! > hello.....doctor~ > f : (0, 00) -> R : differentiable. > lim f(x) = L > x->00 > => for any h > 0, > lim {f(x+h) - f(x)} / h = L > x->00 > ----------------------------------------------- > um........ > lim f(x) = lim lim {f(x+h) - f(x)} / h = L > x->00 x->00 h->0 > um....... i cant progress more. > i need your advice. > thank you very much for your advice. This isnt really the best way to approach it. The trick is to use the mean value theorem. (f(x+h)-f(x))/h=f(c) for some c between x+h and x. Since as x->oo , f(x)->L if we choose x large enough then f will be close to L on the interval from x to x+h. This needs to be all formalized of course, but that is the general idea. Hope this helps. -Ron === Subject: Re: Mathematical Proof for FFT using Sparse Matrices Multiplication > If it is so trivial indeed, I would welcome a Rigorous Mathematical > Proof pl. > Take, for example, the simple radix-2 decimation in time FFT algorithm > (see e.g. Numerical Recipes). This computes the DFT by log2(n) passes > over the array, where each pass is a linear operation with O(n) > operations. A linear operation on n elements with O(n) operations is > a sparse matrix, therefore this is the product of log2(n) sparse > matrices. Q.E.D. Given Sundar Krishnans apparent level of knowledge and request for a rigorous proof, do you think that is likely to help him? It is not Borowski and Borwein: rigorous adj. (of a proof) making completely explicit the validity of the successive steps, usually with reference to some underpinning formal system. Your paragraph just has general descriptions, not explicit steps. Maybe you can see immediately that a simple summation is a linear operation and can be written as a sparse matrix -- but the fact that you can see it does not make it explicit. The summation is not the matrix -- a linear operation on n elements with O(n) operations is NOT a sparse matrix. The first is a function with certain properties. The second is a collection of data. They may be very closely tied in your mind because multiplication of a sparse matrix by a vector of n elements IS a linear operation with O(n) operations (provided we define sparse to be O(n) out of n^2), but they are NOT the same thing, any more than 3 is the same thing as multiplying by 3. Sometime in the past, you were explicitly told that one could be used for the other, or explicitly told something closely related to this. Dont forget that other people have to learn too. If you dont have the patience for this, it would be better not to respond. Sundar Krishnan didnt do anything rude that I can see and does not deserve your scorn. To put my money where my mouth is, here is a rigorous demonstration. Let the length of the vector to be transformed be 2^N. Define N matrices, each a 2^N * 2^N matrix, named R[n], where the subscript n runs from 0 to N-1. Set all elements of R[n] to zero except: For integers k, 0 <= k < 2^N, write k as 2^(N-n)*k0 + 2^(N-n-1)*k1 + k2, where integers k0, k1, and k2 satisfy 0 <= k1 < 2 and 0 <= k2 < 2^(N-n-1). This rewriting of k is equivalent to writing k as an N-bit binary numeral and setting k0 equal to the first n bits, k1 to the next bit, and k2 to the last N-n-1 bits. For each such k, for j1 equal to 0 and 1, let R[n][k, 2^(N-n)*k0 + 2^(N-n-1)*j1 + k2] = e^(2*pi*i * j1 * (r(k1) + r(2*k0))). That is, let: R[n][k, 2^(N-n)*k0 + k2] = 1. R[n][k, 2^(N-n)*k0 + 2^(N-n-1) + k2] = e^(2*pi*i * (r(k1) + r(2*k0))). Here, R[n][a, b] refers to the element in row a, column b, of the matrix R[n]. r(x) is my notation; it stands for the number you get by writing x as a binary numeral, reversing the digits, and putting a binary point in front of them. E.g., r(13) = r(binary 1101) = binary .1011 = 11/16. A more rigorous definition of r is in the paper referenced below. I have assigned 2*2^N non-zero elements to R[n] (one for each of two values of j1 and each of 2^N values of k), and R[n] has (2^N)^2 elements, so it is sparse, if N is more than 1. Consider multiplying R[n] by some column vector v, and call the result u. Observe that: u[k] = sum for 0 <= j1 < 2 of exp(2*pi*i * j1 * (r(k1) + r(2*k0))) * v[2^(N-n)*k0 + 2^(N-n-1)*j1 + k2]. This of course follows immediately from the definition of matrix-vector multiplication and the sparseness of R[n]. (The definition of multiplying a matrix A by a column-vector v to get a vector result u is u[i] = sum of A[i, j] * v[j], where j runs over As columns.) I include this oberservation just in case Ive made a mistake like transposing dimensions or something, so readers can figure out my intent and correct the mistake. I claim that R[N-1] * R[N-2] * ... * R[1] * R[0] * v is the DFT of v, with the outputs in bit-reversed order. It is tedious to write the proof for this, so I will omit it here, but you can find one at http://edp.org/work/Construction.zip. -- edp (Eric Postpischil) http://edp.org That probably would have sounded more commanding if I wasnt wearing my yummy sushi pajamas. -- Buffy the Vampire Slayer === Subject: Re: Mathematical Proof for FFT using Sparse Matrices Multiplication > It is not > rigorous. ... (more rigorous proof omitted). Any time anyone posts please prove this theorem I think it is incumbent upon readers/ potential responders to ask (a) Is this question relevant to the newsgroup on which it is posted? (b) Is it just a request to do someones math homework? [in real life proofs of well-known though specialized results are rarely needed!] (c) Will posting the result be relevant to the newsgroup (as opposed to just answering the question in private mail)? A lengthy post which is also mostly or entirely off topic should be avoided. By answering off-topic questions, we encourage them, and further dilute the newsgroup. I have encountered several people in the last month who used to read sci.math.symbolic [where I read this thread], who said they stopped reading it. Lack of focus would be a euphemistic way of portraying their comments. There is another place, the Dr.Math website, especially for homework questions. mathforum.org/dr.math RJF === Subject: Generalized Eigenvalue Problem for Three matrices Hello Everyone, In my work I have encountered a problem of the following form (A - lambda B - mu C) x = 0 where x is an unknown 3x1 vector and A,B,C are nx3 matrices. The scalars mu and lambda are also unknown. The case when lambda = 0 or mu = 0 is an instance of the standard generalized eigenvalue problem, and is discussed in the standard texts, but I do not seem to be able to find anything on the above problem. The problem is a slight simplification of the general bilinear system x^T P_i y = 0 where P_i are a collection of 3x3 matrices. I can make some assumptions about the components of x not being zero, and hence work with the ratios instead, which leads to the above mentioned eigenvalue problem. About the only reference I could find on this is the Technical Report by Cohen and Tomasi on Systems of Bilinear Equations, which says that the generalized eigenvalue problem for 3 or more matrices is unsolved, but it was written in 1994 and I am wondering if there have been any developments since, or if there are particular simplifications that occur as a result of the size of the problem. Any insights into the solution of the above problems or pointers to the literature will be much appreciated. Sameer === Subject: Re: Generalized Eigenvalue Problem for Three matrices > Hello Everyone, > In my work I have encountered a problem of the following form > (A - lambda B - mu C) x = 0 > where x is an unknown 3x1 vector and A,B,C are nx3 matrices. The > scalars mu and lambda are also unknown. > The case when lambda = 0 or mu = 0 is an instance of the standard > generalized eigenvalue problem, and is discussed in the standard > texts, but I do not seem to be able to find anything on the above > problem. The problem is a slight simplification of the general > bilinear system > x^T P_i y = 0 > where P_i are a collection of 3x3 matrices. I can make some > assumptions about the components of x not being zero, and hence work > with the ratios instead, which leads to the above mentioned eigenvalue > problem. About the only reference I could find on this is the > Technical Report by Cohen and Tomasi on Systems of Bilinear > Equations, which says that the generalized eigenvalue problem for 3 > or more matrices is unsolved, but it was written in 1994 and I am > wondering if there have been any developments since, or if there are > particular simplifications that occur as a result of the size of the > problem. > Any insights into the solution of the above problems or pointers to > the literature will be much appreciated. > Sameer It has been studied, although not as much as lambda matrices. Googling two parameter eigenvalue problems returns 435 hits, although many seem to be just paper abstracts. === Subject: Re: Generalized Eigenvalue Problem for Three matrices >In my work I have encountered a problem of the following form >(A - lambda B - mu C) x = 0 >where x is an unknown 3x1 vector and A,B,C are nx3 matrices. The >scalars mu and lambda are also unknown. You want lambda and mu so that A - lambda B - mu C has rank < 3. Now this is equivalent to saying that every 3 x 3 submatrix has determinant 0. For any choice of the three rows, the determinant of this submatrix is a polynomial in lambda and mu. In the easy case, two of these polynomials will have a nonzero resultant with respect to one of the variables, and then you have a finite number of possibilities to check. In general, I think you can use Groebner basis methods to find the solutions. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Generalized Eigenvalue Problem for Three matrices >>In my work I have encountered a problem of the following form >>(A - lambda B - mu C) x = 0 >>where x is an unknown 3x1 vector and A,B,C are nx3 matrices. The >>scalars mu and lambda are also unknown. >You want lambda and mu so that A - lambda B - mu C has rank >< 3. Now this is equivalent to saying that every 3 x 3 submatrix >has determinant 0. For any choice of the three rows, >the determinant of this submatrix is a polynomial in lambda and mu. >In the easy case, two of these polynomials will have a nonzero >resultant with respect to one of the variables, and then you have >a finite number of possibilities to check. In general, I think >you can use Groebner basis methods to find the solutions. For an instructive example, try [use fixed-width font] [ 1 -1 1 ] [ 1 0 0 ] [-1 1 0 ] [ 0 0 0 ] [-1 -1 2 ] [ 1 1 2 ] A = [-1 2 1 ], B = [ 1 1 0 ], C = [ 0 1 0 ] [ 1 0 2 ] [ 1 0 0 ] [ 0 1 -1 ] p_1(lambda, mu) = 7*lambda-3*mu-9*lambda^2-6*lambda*mu-mu^2+2*lambda^3-4*lambda *mu^2-2*mu^3 p_2(lambda, mu) = 11*lambda-7*lambda^2+2*lambda^2*mu-7*lambda*mu-3*mu+mu^3-6*mu ^2-lambda*mu^2 +2*lambda^3 The resultant of these wrt lambda is 16*mu*(25*mu^6+100*mu^5+180*mu^4+215*mu^3+101*mu^2-35*mu+22)* (-3+mu)^2 For mu = 0, gcd(p_1(lambda,0), p_2(lambda,0)) = lambda. However, for lambda = mu = 0, A has rank 3. For mu = 3, gcd(p_1(lambda,3), p_2(lambda,3)) = 2*lambda^2+7*lambda+9. If lambda is one of the roots of this, A - lambda B - 3 C again has rank 3 (the second row turns out to be a multiple of the third). Let r be a root of 25*mu^6+100*mu^5+180*mu^4+215*mu^3+101*mu^2-35*mu+22. Then gcd(p_1(lambda,r), p_2(lambda,r)) = lambda - s where s = 119/914 + 25/1828 r + 630/457 r^2 + 3105/1828 r^3 + 2425/1828 r^4 + 525/1828 r^5 This time A - s B - r C has rank 2, with nullspace spanned by [ 425*r^5+875*r^4-4015*r^3-20810*r^2-34233*r-15650 ] [ -825*r^5-7075*r^4-24465*r^3-35950*r^2-26023*r-11342 ] [ 10968 ] So these are the solutions of your problem in this example. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: what kind of distribution is it? Let X be a generalized gaussian distribution random variable i.e. p(X=x) = v*c(v,sigma)*exp(-|c(v,sigma)*x|^v) / (2 gamma(1/v)) gamma is the gamma function basically can write as p(X=x) = a*exp(-|bx|^v) Now I am wondering what the distribution of z = |bx|^v is I can get p(Z = z) = 1/gamma(1/v) x^{1/v - 1} exp(-x) is it some famous distribution? === Subject: Re: Q: properties of covariance matrices > It may not be precisely the covariance matrix, but it is the > correlation matrix, and there is a strong connection to the covariance > = autocorrelation matrix. All you have to do is to consider a set of > random variables, say Z, which is the union of X and Y: > Z_i = X_i if 0 = Y_{i-N} if N > Then your matrix is just an off-diagonal submatrix of the covariance > matrix of Z > This is getting a bit complicated. Is it involved in the solution of a > natural problem? Zigoteau, Im interested in the following measurement problem. We want to measure A = B + C, where B and C are normally distributed n-tuplets. However, we do not observe A, B, or C directly. Instead, we observe D = B + F, where F is another normally distributed n-tuplet. So, I want to construct a definition of how good a measure that D is of A. Obviously, this is related to the covariance between D and A. Second, I want to get a sense of whether the various matrix elements of this covariance matrix must be positive or negative. -- Ricky === Subject: Re: Q: properties of covariance matrices > This is getting a bit complicated. Is it involved in the solution of a > natural problem? > Im interested in the following measurement problem. > We want to measure > A = B + C, where B and C are normally distributed n-tuplets. > However, we do not observe A, B, or C directly. > Instead, we observe > D = B + F, where F is another normally distributed n-tuplet. Why cant you say A = D + G where G=C-F. Is there a need to consider F and C separately? > So, I want to construct a definition of how good a measure > that D is of A. Obviously, this is related to the covariance > between D and A. Yes, I would have thought that the correlation matrix of the measurement error was what you needed, with the possible addition of and for comparison. If is nonzero, then the measurement is biased and you can get a better answer by combining measurements. Any attempt to produce a single goodness of measurement parameter would eliminate useful distinctions. > Second, I want to get a sense of whether the various > matrix elements of this covariance matrix must be positive or negative. Well I guess that both and must be positive definite, but apart from that there are no restrictions. Zigoteau. === Subject: A weird classification of compact subset of R of zero measure posting-account=2LgBuQ0AAAC2HZQp160dLwjzCEeNC7or If K subset mathbb{R} is compact how does one see that the following condictions are equivalent: (1) For all x in K we can assign an uncountable set F_x in mathbb{R} such that |x - y| leq d(F_x,F_y) for all x,y in K. (2) K is of 0 measure. I have no idea of even where to start. === Subject: A weird classification of compact subsets of R of zero measure posting-account=2LgBuQ0AAAC2HZQp160dLwjzCEeNC7or Assume that C subset mathbb{R} is compact, how does one show that the following conditions are equivalent? (1) For all x in C there is a set F_x such that |x - y| leq d(F_x,F_y) for all x,y in C. (2) C is of 0 measure. === Subject: Re: A weird classification of compact subsets of R of zero measure For easier reading, Ive removed the TeX. > Assume that C subset R is compact, how does one show that the > following conditions are equivalent? > (1) For all x in C there is a set F_x such that |x - y| <= > d(F_x,F_y) for all x,y in C. Do you intend For all x,y in C there are sets F_x, F_y such that |x - y| <= d(F_x,F_y) for all x,y in C. If not, then whats F_y? Im also puzzled by for all x in C included twice. Are you defining d(A,B) = inf{ d(x,y) | x in A, y in B } ? Then can I not have F_x = { x } and F_y = { y } ? If perchance Ive fathomed (1) correctly, then its true of all sets. > (2) C is of 0 measure. === Subject: Re: A weird classification of compact subsets of R of zero measure posting-account=2LgBuQ0AAAC2HZQp160dLwjzCEeNC7or Ive found the source. http://www.cs.elte.hu/~schw02/feladatok/engproblems.dvi So, it is what you say in the first paragraph with the condition that F_x,F_y are uncountable. I think by dist they mean the Hausdorff metric, no? Hmm. === Subject: Re: A weird classification of compact subsets of R of zero measure > Ive found the source. > http://www.cs.elte.hu/~schw02/feladatok/engproblems.dvi > So, it is what you say in the first paragraph with the condition that > F_x,F_y are uncountable. > I think by dist they mean the Hausdorff metric, no? Hmm. This is getting to hard to follow, with patches, removed context, references to which I dont have access and now maybe, just maybe, the Hausdorff metric. I suggest you abandon this thread, take the time to produce a clear readable description of the problem and present it in a new thread, in hopes of getting the attention of those more knowledgeable than I. === Subject: Re: A weird classification of compact subsets of R of zero measure posting-account=xHEctg0AAACtCPlnsA9MK6ZUwuCulk6O I doubt perhaps, (1) is true for all subsets of $mathbb{R}. Indeed, $C$ be any subset of $mathbb{R}$ (for example let $C = [0,1]$. For each $x in C$, define $F_x = {x}$. Then $|x - y| = d(F_x,F_y)$. But the measure of $C$ is not zero in our particular case. Dinesh Karia === Subject: Re: A weird classification of compact subsets of R of zero measure posting-account=2LgBuQ0AAAC2HZQp160dLwjzCEeNC7or I forgot something very vital, of course that is even true for F_x countable. F_x must be uncountable. === Subject: Re: 0.9999... = 1? > Hi! > Ulf! > Anyhow, to me, this whole discussion about whether 0.999... =/= 1 > seems just like a matter of notation. > I think a/b, a and b being integers is the real notation for > rationals. I dont think you can confuse the name of something and the thing itself. Essentially 1/3, 3/15, 0.333..., a third, ein Drittel demote the same thing. The easiset notation to use in a particular set of circumstances is another matter. > We can add numbers together using decimal expansions: > 1/3 + 1/3 = 2/3 > 0.333... + 0.3333... = 0.6666... > Since 2/3 really has the decimal expansion 0.666..., it works in this > case. Yes, you are showing that even though objects are given different names, they obey the same laws. Changing the name of an object does not chnage its properties. > Some people says that it should not work for some cases, such as 2/3 + > 1/3. I think that the main problem with the notation 0.999... or 0.444... is that it is taken by some as an instruction. In other words, take 0.9 then add 0.09 then add 0.009 then add .... and so on - a never-ending process performed over an infinite time. But numbers are constants and do not chnage in value over time. The assumption is that symbols have an intrinsic meaning irrespective of the context in which they are used. We know what + what ... means and we know what 1,2,3,... are therefore, say, 1+2 + 3 + 4 +.... must in and of itself be meaningful. These symbols are being used in a new context and may be quite meanuingless. Most sci.math debates about 1/0 =00, .999.. <> 1, infinite integers and the James Harris dictum the algebnra cannot lie or some such drivel arise because it is thought that symbols magically remain meaningful no matter how they are used. Although Euler and Heaviside seem to have been symbol abusers who did turn up interseting maths > Those people are WEIRD! Do not talk to them! ;) === Subject: Re: 0.9999... = 1? _____________________ /| /| | | ||__|| | Do not feed the | / | --Mgt. | / |_____________________| / _ || / |____ || / | | | |____/ || / |_|_|/ | _|| / / |____| || / | | | --| | | | |____ --| * _ | |_|_|_| | -/ *-- _-- _ | || / _ | / ` * / _ /- | | | * ___ c_c_c_C/ C_c_c_c____________ === Subject: Re: another nim game Originator: joshp@xoxy.net (joshp) > Two players; > Just one pile of 22 fruits; > I can only take one, two or three fruits at a time; > Whoever takes the last fruit, looses; > I know the solution for the game but what I would like to know is the math > model behind it. Combinatorial game theory deals with these issues. Berlekamp, Conway, and Guys _ Winning Ways _ takes a lighthearted, informal, yet comprehensive approach to the subject. The specific game you describe above (with the exception that the last player to remove an item *wins*) is discussed in Volume 1, Chapter 4, page 83: We might modify the game of Nim by requiring that in any move the number of beans taken is at most three. This will mean that for the nim-values we have G(n) = mex(G(n-1),g(n-2),G(n-3)) so that the nim-sequence is n = 0 1 2 3 4 5 6 7 8 9 ... G(n) = 0.1 2 3 0 1 2 3 0 1 ... and a single heap is a P-position (previous player winning) just if its size is a multiple of 4. We could instead allow a heap to be reduced by any number up to k, when the nim-sequence would be n = 0 1 2 ... k-1 k k+1 k+2 ... 2k 2k+1 2k+2 2k+3 ... G(n) = 0.1 2 ... k-1 k 0 1 ... k-1 k 0 1 1 and a single heap would be a P-position just if its size were a multiple of k+1. Here, G(x) refers to the combinatorial game-theoretic value of a game consisting of a single heap of size x. For our purposes, in the case of the game above, this value will be a non-negative integer, and the second player to move will have a normally have winning strategy just when G(x) = 0. But when the last player to move *loses*, you should play exactly as you would play Normal Nim unless your move would leave an even number of singleton heaps [a singleton heap is a heap of size 1] and no other heap. Then leave an odd number instead. (page 393). Ônim-sequence is defined on page 82: [I]f heaps of sizes 0,1,2,3... have values [a,b,c,d,...] we shall say that the game has the nim-sequence a.bcd... Ômex(a,b,c,...) is defined on page 82: [W]e shall call the least number (from 0,1,2,3,...) which is missing from a set {x,y,z,..} the mex (minimal excluded number) of that set. mex(0,1,3,7) = 2, mex(2,4,5) = 0 and the mex of the empty set is 0.