mm-2049 > if the gamma function is restricted to the real interval [2,infinity] > than the inverse function exists. Does someone know a closed expression > of this inverse gammafunction or any other usable form, like an integral > function or power series,...? > dated 2001 Oct 25, in the thread Inverse gamma function at > . Yes, of course this helps al lot. However, I am wondering why I can find only little information about this function. I think it is an interesting topic and there should be some more results in the literature. But I do not find it or it is not as well examined as I assumed. === Subject: Re: inverse of the gamma function > if the gamma function is restricted to the real interval [2,infinity] > than the inverse function exists. Does someone know a closed > expression of this inverse gammafunction or any other usable form, > like an integral function or power series,...? > Would an approximation be adequate? If so, please see the second > . > Yes, of course this helps al lot. However, I am wondering why I can find > only little information about this function. I think it is an interesting > topic and there should be some more results in the literature. But I do > not find it or it is not as well examined as I assumed. of the literature. It certainly would not surprise me if there were some interesting published results of which I am unaware. Is anyone reading this aware of such results? Curiously, David Cantrell === Subject: isomorphic rings I'm studying for an exam and doing problems from the book Algebra: A Module-Based Approach. In one exercise it asks to show that the rings Z[X,Y]/(XY-1) and Z[Z] are isomorphic and that Z[X]/(X^n-1) and Z[Z_n] are isomorhic where Z[G] is the group ring for any group G. I haven't been able to come up with a suitable map... Any ideas? nojb === Subject: Re: Is there someone to help me PLEASE... 1-The first qustion is to find the largest set of natural numbers that if we choose two members a and b then ab+1 is square of a natural number.(is there any limit for the number of members of the set?) >suppose you had some a in such a set, then this would force also >a^2 + 1 to be a square, let's say a^2 + 1 = c^2 for some natural number c. > He's talking about two _distinct_ members. Show that if you have any natural number a, you don't have to look far from it to find a natural number b such that ab+1 is a square. === Subject: Re: Is there someone to help me PLEASE... 1-The first qustion is to find the largest set of natural numbers that if we choose two members a and b then ab+1 is > square of a natural number.(is there any limit for the number of members of > the set?) suppose you had some a in such a set, then this would force also >a^2 + 1 to be a square, let's say a^2 + 1 = c^2 for some natural > number c. > He's talking about two _distinct_ members. > Show that if you have any natural number a, you don't have to look far from > it to find a natural number b such that ab+1 is a square. So, a lot of sets have TWO members, but how about a large number of members..?? The problem is How large can the number of members be. Bob Pease === Subject: is this a harmonic oscillator equation? hi, i was looking at the differential equation f''(x)=( v + iwx^2)*f(x) where i is the imaginary unit. I believe this can be solved in terms of confluent hypergeometric functions. I was wondering if there is some simple way to express the result in terms of Hermite-Gauss functions, given that for w = -i it's pretty much the equation of a quantum harmonic oscillator (which makes me think a change of independent variable would help maybe...) yuric === Subject: Re: is this a harmonic oscillator equation? >hi, >i was looking at the differential equation >f''(x)=( v + iwx^2)*f(x) >where i is the imaginary unit. I believe this can be solved in terms >of confluent hypergeometric functions. I was wondering if there is >some simple way to express the result in terms of Hermite-Gauss >functions, given that for w = -i it's pretty much the equation of a >quantum harmonic oscillator (which makes me think a change of >independent variable would help maybe...) Maple 9 solves it in terms of WhittakerM and WhittakerW functions. The WhittakerW can be converted to a HermiteH, if that's a help. sol:=dsolve(de,f(x)); 1/2 (-1/8 + 1/8 I) v 2 sol := f(x) = _C1 WhittakerM(---------------------, 1/4, 1/2 w 1/2 1/2 2 / 1/2 (1/2 + 1/2 I) 2 w x ) / x + _C2 WhittakerW( / 1/2 (-1/8 + 1/8 I) v 2 1/2 1/2 2 / ---------------------, 1/4, (1/2 + 1/2 I) 2 w x ) / 1/2 / w 1/2 x > convert(sol,HermiteH); 1/2 (-1/8 + 1/8 I) v 2 f(x) = _C1 WhittakerM(---------------------, 1/4, 1/2 w 1/2 1/2 2 / 1/2 (1/2 + 1/2 I) 2 w x ) / x + 1/2 _C2 HermiteH( / 1/2 1/2 1/2 -6 w - v 2 + v 2 I --------------------------- + 1, 1/2 4 w 1/2 1/2 2 1/2 ((1/2 + 1/2 I) 2 w x ) ) / | | | 1/2 1/2 2 1/4 / | 1/2 ((1/2 + 1/2 I) 2 w x ) / x / / 1/2 1/2 1/2 |-6 w - v 2 + v 2 I| |---------------------------| | 1/2 | 4 w / 2 | | | 1/2 1/2 2 | exp((1/4 + 1/4 I) 2 w x )/ Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: James Harris Double In sci.math, G Frege : Sorry; my humor is a little off. :-) But my math is usually spot on, which is more than I can say for Mr. Harris's. (Not that his algebraic manipulation is all that bad, but the conclusions he jumps to!) > To be honest, I really don't understand why you guys STILL try to argue > with this ehrrr... you know. Good question. I'd say it wakes up the mind, but there are a lot of ways to do that, like writing algorithms to compute digits of pi.... :-) > F. -- #191, ewill3@earthlink.net -- nobody wants to compute the digits of 1/3, for some reason It's still legal to go .sigless. === Subject: Re: JSH: Resolution? > Harris's claim in Advanced Polynomial Factorization is > actually somewhat stronger than what I noted above. He > says that if P(x) is factored in the form > [*] P(x) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f), > where a1, a2, and a3 are algebraic integers, then > ... two of the a's have a factor that is f. [See p.3 > of APF.] > To see if this is true, let f = 5, u = 1, m = 1. > Then one readily sees that > P(x) = 25*(553*x^3 - 72*x + 5). > If this is factored in the form [*] above, clearly > r1 = -u*f/a1 = -5/a1 is a root. If a1 has a factor that is > f, then a1 = 5*g1, where g1 is an algebraic integer. > Therefore r1 = -1/g1. This implies > 553*(-1/g1^3) - 72*(-1/g1) + 5 = 0, or > 5*g1^3 + 72*g1 - 553 = 0. > Perhaps that should be: > 5*g1^3 + 72*g1^2 - 553 = 0. change my conclusion below: > The left side of this equation is a non-monic, > irreducible, primitive polynomial in g1. Therefore > g1 cannot be an algebraic integer, showing that > Harris's claim in APF is false. > I don't have doubts about this one. I am still maintaining my statement about doubts. Moreover: looking at Advanced Polynomial Factorization more closely, one sees that Harris is claiming that if [1] P(x) = (v^3 + 1)*x^3 - 3*v*(u*f)^2*x + (u*f)^3 is itself factored in the form [2] P(x) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f), where a1, a2, and a3 are algebraic integers, then one of a1, a2, and a3 is coprime to f (See: Adv Polyn. Fact., p. 3) : Therefore, one factor is coprime to f. This is false. Make the substition x = -u*f/y. Assume f = 5, m = 1, u = 1. Then equation [1] becomes P(x) = -13825*5^3/y^3 + 3*24*25*5/y + 125 = 125*(-25*553/y^3 + 72/y + 1). Setting this equal to 0 yields: [3] y^3 + 72*y^2 - 25*553 = 0. The left side is a monic irreducible primitive polynomial, which implies that (1) the roots are algebraic integers, and (2) NONE of the roots are coprime to 5 [by the basic Galois theory argument]. Note that a1, a2, and a3 happen to be roots of [3] because of the fact that -u*f/a1, etc., are the roots of [2], and y was specified in the substitution as x = -u*f/y. Therefore all of a1, a2, and a3 are NOT coprime to 5. Harris's conclusion in APF is therefore false. But it happens that he needs this claimed conclusion, that one of the a's is coprime to f, also in his proof of Fermat's Last Theorem. Conclusion: the FLT proof is also invalid. Nora B. > Nora B. > -- > There are two things you must never attempt to prove: the unprovable -- and the obvious. > -- > Democracy: The triumph of popularity over principle. === Subject: Re: JSH: Resolution? > Conclusion: [JSH's] FLT proof is also invalid. BIG surprise! ;-) F. === Subject: Re: JSH: Resolution? > Harris's claim in Advanced Polynomial Factorization is >actually somewhat stronger than what I noted above. He >says that if P(x) is factored in the form >[*] P(x) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f), >where a1, a2, and a3 are algebraic integers, then >... two of the a's have a factor that is f. [See p.3 >of APF.] > To see if this is true, let f = 5, u = 1, m = 1. >Then one readily sees that > P(x) = 25*(553*x^3 - 72*x + 5). > If this is factored in the form [*] above, clearly >r1 = -u*f/a1 = -5/a1 is a root. If a1 has a factor that is >f, then a1 = 5*g1, where g1 is an algebraic integer. >Therefore r1 = -1/g1. This implies > 553*(-1/g1^3) - 72*(-1/g1) + 5 = 0, or > 5*g1^3 + 72*g1 - 553 = 0. Perhaps that should be: 5*g1^3 + 72*g1^2 - 553 = 0. > change my conclusion below: >The left side of this equation is a non-monic, >irreducible, primitive polynomial in g1. Therefore >g1 cannot be an algebraic integer, showing that >Harris's claim in APF is false. >I don't have doubts about this one. > I am still maintaining my statement about doubts. > Moreover: looking at Advanced Polynomial Factorization > more closely, one sees that Harris is claiming that if > [1] P(x) = (v^3 + 1)*x^3 - 3*v*(u*f)^2*x + (u*f)^3 > is itself factored in the form > [2] P(x) = (a1*x + u*f)*(a2*x + u*f)*(a3*x + u*f), > where a1, a2, and a3 are algebraic integers, > then one of a1, a2, and a3 is coprime to f > (See: Adv Polyn. Fact., p. 3) : > Therefore, one factor is coprime to f. > This is false. Make the substition x = -u*f/y. > Assume f = 5, m = 1, u = 1. Then equation [1] > becomes > P(x) = -13825*5^3/y^3 + 3*24*25*5/y + 125 > = 125*(-25*553/y^3 + 72/y + 1). > Setting this equal to 0 yields: > [3] y^3 + 72*y^2 - 25*553 = 0. > The left side is a monic irreducible primitive polynomial, > which implies that (1) the roots are algebraic > integers, and (2) NONE of the roots are coprime > to 5 [by the basic Galois theory argument]. > Note that a1, a2, and a3 happen to be roots of [3] because > of the fact that -u*f/a1, etc., are the roots of [2], > and y was specified in the substitution as x = -u*f/y. > Therefore all of a1, a2, and a3 are NOT coprime to 5. > > Harris's conclusion in APF is therefore false. > But it happens that he needs this claimed conclusion, > that one of the a's is coprime to f, also in his proof > of Fermat's Last Theorem. Conclusion: the FLT proof > is also invalid. I find myself wondering if the only thing he will accept is explicitly computed numbers. Anything else doesn't seem to sink in. -- Will Twentyman email: wtwentyman at copper dot net === Subject: linear equation system with complex coefficients and boundary conditions hi everyone, i'd like to get some hints about the following problem i have a linear equations system like sum_{i=1^}^n c_ia_{ij}=0 for j=1..k and k < n (so, for example it looks like c1a11+c2a21+c3a31=0 c1a12+c2a22+c3a32=0 ) the a_ij are known and i want to compute the c_i both, a_ij and c_i are complex numbers and there is a special condition for the c_i, that is, the modulus of c_i (|c_i|) is 1, (so c_i=exp(I*d_i)) i have k=3 now and n not more than 20 any help is greatly appreciated :o) jan === Subject: Re: linear equation system with complex coefficients and boundary conditions >i have a linear equations system like >sum_{i=1^}^n c_ia_{ij}=0 for j=1..k and k < n >(so, for example it looks like >c1a11+c2a21+c3a31=0 >c1a12+c2a22+c3a32=0 ) >the a_ij are known and i want to compute the c_i >both, a_ij and c_i are complex numbers and there is a special condition >for the c_i, that is, the modulus of c_i (|c_i|) is 1, (so c_i=exp(I*d_i)) >i have k=3 now and n not more than 20 I might write the system in terms of u_i = Re(c_i) and v_i = Im(c_i), so you have a system of linear and quadratic equations. Groebner basis techniques might work if the problem is not too big. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: m/phi(m) average >Let phi(m) be >the Euler totient function, the number of positive integers <= m and >coprime to m. >Now, I am guessing that: >limit{n -> oo} >(1/n) *sum{m=1 to n} m/phi(m) >= zeta(2) zeta(3) /zeta(6), >where zeta(j) = sum{k=1 to oo} 1/k^j. >Is this well-known or even true? Yes, it's true. m/phi(m) = product_p (p/(p-1)) = product_p (1 + 1/(p-1)) the product being over primes dividing m. For any set S of primes, let f(S) = product_{p in S} 1/(p-1) and g(S) = product_{p in S} p. Sums over S in the following are always over sets of primes. 1/n sum_{m=1}^n m/phi(m) = 1/n sum_{m=1}^n sum_{S: g(S) divides m} f(S) = sum_S f(S) floor(n/g(S))/n since the number of m in [1..n] divisible by g(S) is floor(n/g(S)). Now as n -> infinity, floor(n/g(S))/n increases to 1/g(S) for each S, and thus lim_{n -> infinity} 1/n sum_{m=1}^n m/phi(m) = sum_S f(S)/g(S) = sum_S product_{p in S} 1/(p (p-1)) = product_{primes p} (1 + 1/(p (p-1))) = product_{primes p} (1-1/p^6)/((1-1/p^2)(1-1/p^3)) = zeta(2) zeta(3)/zeta(6) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Math expression *does* say it all In sci.math, James Harris realize it says it all rather quickly, and directly. > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) = > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >And what I've proven by checking the constant term P(0) is that Pedant point: If you're referring to the constant term (x^0) of P(m), which is u^3f^3, it is clear that that is *not* equal to P(0), which is 3u^2f^2x + u^3f^3 = f^2(3u^2x+u^3f), unless x=0. > That is incorrect. > The expression > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) > is NOT a polynomial, but is what I now call an uber-polynomial, which > contains an infinite number of polynomials, so it's more like a > family. I've got bad news for you: that expression is an element of the polynomial ring Z[x][f][m][u], and *is* a polynomial... with 4 indeterminates. It is *not* deserving of a special name. > By using P(m), I'm focusing on the variable m, so with P(m), m=0, > gives me the constant term with respect to m. That doesn't mean you can ignore u,f,x. > When considering a polynomial the constant term tells you which part > of it does not contain the key variable. You only have terms that are constant with respect to m, u, or x. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Math expression *does* say it all Visiting Assistant Professor at the University of Montana. [.snip.] The expression f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) is NOT a polynomial, but is what I now call an uber-polynomial, which contains an infinite number of polynomials, so it's more like a family. >I've got bad news for you: that expression is an element of the >polynomial ring Z[x][f][m][u], and *is* a polynomial... with 4 >indeterminates. It is *not* deserving of a special name. I agree it does not deserve a special name (and certainly not uber-polynomial!), but I think James is finally coming to understand that the roles that u, f, and m play in his expression are different from the role that x plays, and that what he really has is a family of polynomials in the [polynomial] variable x, indexed by the ->parameters<- m, f, and u. The fact taht he now is edging closer to realizing that he is dealing with a ->family<- of polynomials is certainly a step forward. Something that was explained to him some time ago, and he dismissed as anti-algebra. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Matrix Traversing Question -- I could really use a moment of help. I'm stuck. I have a task where I have to traverse a matrix in a way I cannot formulate into an equation easily. I suspect there is a concept to address this method already but I do not know how to begin looking for it. Could someone offer a bit of personal help? The concept is easy to demonstrait, but difficult to verbalize. === Subject: Re: more easy questions http://mygate.mailgate.org/mynews/sci/sci.math/3742d180305609327006111c978fa 4 6f.35661%40mygate.mailgate.org Folland book dealt with measure theory - I assumed it was only on analysis. But now I'll try and get hold of a copy. Is it written at a level an idiot can understand? David. -- === Subject: Re: more easy questions > Folland book dealt with measure theory - I assumed it was only on > analysis. Most graduate-level analysis books will also do some measure theory. Some undergraduate books do, too, usually toward the end of the book, e.g. Rudin's Principles of Mathematical Analysis and Strichartz' The Way of Analysis and Browder's Mathematical Analysis. > But now I'll try and get hold of a copy. Is it written at a > level an idiot can understand? It depends on your background, and what level of text you are comfortable reading. If you are comfortable reading Rudin's undergraduate analysis book, you should have no problem with Folland. If you prefer really talkative expositions like Strichartz' The Way of Analysis, then Folland may not be your style. Some other good choices would be Bruckner, Bruckner, and Thompson's Real Analysis, which does nearly 200 pages of measure theory before mentioning integrals (it's not as dense as Folland, but also covers more material); Wheeden and Zygmund's Measure and Integral, which concentrates mainly on Lebesgue measure and postpones the more abstract treatment until near the end of the book; and Bartle's The Elements of Integration and Lebesgue Measure, which I have only glanced through but it looked good. I personally like Folland's and Bruckner's books the best. If possible, you should go to a well-stocked university bookstore (or math library) and look at these and other books and choose one that suits you. There is far from universal agreement about what constitutes the best analysis book, and I say that with some trepidation as I know that Rudin's R&CA has some zealous adherents here in sci.math. Heck, some people even like Royden's book. === Subject: Re: more easy questions http://mygate.mailgate.org/mynews/sci/sci.math/242a26bb04dd999207a9ddcd28400 1 a1.35661%40mygate.mailgate.org > Most graduate-level analysis books will also do some measure theory. Some > undergraduate books do, too, usually toward the end of the book, e.g. Rudin's > Principles of Mathematical Analysis and Strichartz' The Way of Analysis and > Browder's Mathematical Analysis. > But now I'll try and get hold of a copy. Is it written at a > level an idiot can understand? > It depends on your background, and what level of text you are comfortable > reading. If you are comfortable reading Rudin's undergraduate analysis book, > you should have no problem with Folland. If you prefer really talkative > expositions like Strichartz' The Way of Analysis, then Folland may not be your > style. Some other good choices would be Bruckner, Bruckner, and Thompson's > Real Analysis, which does nearly 200 pages of measure theory before mentioning > integrals (it's not as dense as Folland, but also covers more material); Wheeden > and Zygmund's Measure and Integral, which concentrates mainly on Lebesgue > measure and postpones the more abstract treatment until near the end of the > book; and Bartle's The Elements of Integration and Lebesgue Measure, which I > have only glanced through but it looked good. I personally like Folland's and > Bruckner's books the best. > If possible, you should go to a well-stocked university bookstore (or math > library) and look at these and other books and choose one that suits you. There > is far from universal agreement about what constitutes the best analysis book, > and I say that with some trepidation as I know that Rudin's R&CA has some > zealous adherents here in sci.math. Heck, some people even like Royden's book. is organic chemistry. My (recreational) interest in quantum mechanics has brought me into contact with L^2 Hilbert spaces and this in turn has pointed me towards the mathematics, Lebesgue integration being the target. But, trying to be a little thorough, studying Lebesgue integration usually means you need to have some basics in set theory and measure theory and it is this I am trying to acquire. The problem, as ever, is that I'm trying to study something which is generally reserved for graduates in mathematics, so I'm on a hiding to nothing. So your suggestion of a talkative exposition such as Stricartz is appealing. I should be able to get it through my local library (as well as the other suggestions). I won't risk buying yet as I've had my fingers burnt one too many times - like when I bought Reed and Simon's book on functional analysis, and Sneddon's book on Fourier transforms! I guess I'm slowly picking up bits of set theory and measure theory but I'm a bit hit and with certain questions. I find the set theory stuff a bit abstract and have struggled with a lot of the concepts. I have photocopies from various books I've got form the library covering set theory/measure theory but I seem to flit from one to the other, so I guess my approach is good in that it draws on an eclectic mix of resources but perhaps a bit hit and miss as I never seem to stick with one of them for very long. I'll get there in the end, though. couple of the books you mention. David. -- === Subject: multivariable calculus I'm studying multivariable calculus and have been stuck for awhile on this proof. Any help/hints would be appreciated. Let T be a solid which has a piece-wise smooth boundary, S. Let f(x,y,z) and g(x,y,z) be scalar fields with continuous second partials everywhere. Set the vector field v = .84f x .84g. Show that the flux of v out of T is 0. Note: .84 means gradient. === Subject: Re: multivariable calculus > I'm studying multivariable calculus and have been stuck for awhile on this > proof. Any help/hints would be appreciated. > Let T be a solid which has a piece-wise smooth boundary, S. Let f(x,y,z) and > g(x,y,z) be scalar fields with continuous second partials everywhere. Set > the vector field v = .84f x .84g. Show that the flux of v out of T is 0. Note: .84 > means gradient. Hint: Stokes's Theorem / Divergence Theorem. Compute div(grad f x grad g) Dale. === Subject: Re: multivariable calculus > I'm studying multivariable calculus and have been stuck for awhile on this > proof. Any help/hints would be appreciated. > Let T be a solid which has a piece-wise smooth boundary, S. Let f(x,y,z) and > g(x,y,z) be scalar fields with continuous second partials everywhere. Set > the vector field v = .84f x .84g. Show that the flux of v out of T is 0. Note: .84 > means gradient. > Hint: Stokes's Theorem / Divergence Theorem. > Compute div(grad f x grad g) > Dale. Steve === Subject: Re: multivariable calculus > I'm studying multivariable calculus and have been stuck for awhile on > this > proof. Any help/hints would be appreciated. > Let T be a solid which has a piece-wise smooth boundary, S. Let f(x,y,z) > and > g(x,y,z) be scalar fields with continuous second partials everywhere. > Set > the vector field v = .84f x .84g. Show that the flux of v out of T is 0. > Note: .84 > means gradient. > Hint: Stokes's Theorem / Divergence Theorem. > Compute div(grad f x grad g) > Dale. > Steve Note: A x B = e_j epsilon_{jkm}A_k B_m with summation on repeated indices and e_j is the jth member of the set of three orthonormal unit vectors. In other words, each vector A can be written as A = A_i e_i, i = 1,2,3. Note: epsilon_{jkm} = +1 if jkm is an even permutation of 123, = -1 if it is an odd permutation of 123, and = zero otherwise. So, epsilon_{jkm} is antisymmetric for all pairs of its subscripts. For the problem to show that div [(grad f) x (grad g)] = 0 we need some other results. One of which is this e_i . e_j = delta_{ij}, the Kronecker delta, where delta_{ij} = 1 if i = j, and equal zero otherwise. Also, div Q = (e_i partial_i) . (Q_j e_j) where we used different subscripts because they are summed up independently of each other. OK, once more, div Q = (e_i partial_i) . (Q_j e_j) = (e_i . e_j) partial_i Q_j = delta_{ij} partial_i Q_j = partial_i Q_i The summed on subscripts are said to be dummy because they can be replaced by any other index so long as that index is not already in use in the same term. Just one more prior result to go: Let [M_{ij}] be any symmetric 3x3 matrix. Then epsilon_{ijk}M_{ij} = 0. Prove this to your satisfaction. Now we're ready. div [(grad f) x (grad g)] = e_ipartial_i . [e_jepsilon_{jkm}(partial_k f)(partial_m g)] = e_i . e_j partial_i [epsilon_{jkm}(partial_k f)(partial_m g)] = delta_{ij}partial_i [epsilon_{jkm}(partial_k f)(partial_m g)] = partial_j [epsilon_{jkm}(partial_k f)(partial_m g)] = epsilon_{jkm} partial_j [(partial_k f)(partial_m g) = epsilon_{jkm} { [( partial_jpartial_k f)(partial_m g)] + [(partial_k f)(partial_jpartial_m g)] } = [( epsilon_{jkm}partial_jpartial_k f)(partial_m g)] + [(partial_k f)(epsilon_{jkm}partial_jpartial_m g)] = 0 Now here's why we got zero. Because both terms on the right above are zero. I'll prove this for the first term only though. Because f and g have continuous second partials everywhere, the order of their partial derivatives is irrelevant. So, for the case of the derivative of f partial_jpartial_k f = partial_kpartial_j f This means that the 3x3 matrix of derivatives of f, [partial_jpartial_k f], is symmetric. So, by the lemma above epsilon_{jkm}(partial_jpartial_k f) = 0 (m = 1,2,3) Done. This proof employs just about every thing you'll need to know to use Cartesian tensors in advanced calculus. Patrick === Subject: multivariable MVT Howdy, My trusty old 2nd ed of Apostol's 'Mathematical Analysis' has a version of the mean value thm for functions from R^n to R^m (Thm 12.9). It says that if f is differentiable on an open set S in R^n and the straight line segment from x to y is contained in S, then for all a in R^m, there exists z in the line segment s.t. a . [f(y) - f(x)] = a . [f'(z) (y - x)] where . is dot product and f'(z) the Jacobian. z varies with a. In the exercises, Apostol points out that we cannot drop the dot product with a, giving the function f(t)=(cos t, sin t) from R to R^2 with y=2*pi and x=0 by way of example. All well and good. My question is: under what conditions on f can we drop the dot product bit and get a valid MVT of the form f(y) - f(x) = f'(z) (y - x) ? (excluding the case of real valued functions which I'm already aware of) I guess the periodic nature messes this up for Apostol's counterexample. Can we say anything if there is something like monotonicity in the components? Does anyone know of any references to this? TIA, Norm === Subject: Re: multivariable MVT > My question is: under what conditions on f can we drop the dot product > bit and get a valid MVT of the form > f(y) - f(x) = f'(z) (y - x) ? (excluding the case of real valued > functions which I'm already aware of) > I guess the periodic nature messes this up for Apostol's counterexample. > Can we say anything if there is something like monotonicity in the > components? Does anyone know of any references to this? I don't think there's much hope here; it's in the nature of the beast. It's asking for too much serendipity. Different components will typically have different rates of change. For example, let f : R -> R^2 be defined by f(x) = (x^2,x^3). Then it's easy to see we can't have f(1) - f(0) = f'(z)(1 - 0) for any z in (0,1), even though both components are strictly increasing on the interval. However, there's always the mean value inequality, which is quite useful: |f(y) - f(x)| <= (sup_{z in [x,y]}) ||f'(z)||)*|y - x| Here | | denotes euclidean length, [x,y] is the line segment from x to y, and || || denotes the (L^2) matrix norm. === Subject: Re: multivariable MVT > I don't think there's much hope here; it's in the nature of the beast. It's > asking for too much serendipity. Different components will typically have > different rates of change. For example, let f : R -> R^2 be defined by f(x) > = (x^2,x^3). Then it's easy to see we can't have f(1) - f(0) = f'(z)(1 - 0) > for any z in (0,1), even though both components are strictly increasing on > the interval. > However, there's always the mean value inequality, which is quite useful: > |f(y) - f(x)| <= (sup_{z in [x,y]}) ||f'(z)||)*|y - x| > Here | | denotes euclidean length, [x,y] is the line segment from x to y, > and || || denotes the (L^2) matrix norm. Yes, I'm familiar with this inequality. Norm === Subject: My (Un)Originality (was Re: Reminder: Wages, Employment Not Determined By Supply, Demand) 1.0 INTRODUCTION I've recently demonstrated here that the determination of income distribution by the theory of supply and demand makes no sense. Rather than point out any errors in my exposition, certain irrationalists objected, incorrectly, that I was not echoing the literature. These people claimed that I was being original. 2.0 ANALYSIS AT LEVEL OF ECONOMY OR INTEGRATED INDUSTRY? For example, poor Chris Auld and JimT asserted, mistakenly, that the literature only applies so-called factor price frontiers reswitching, and capital-reversing to the economy as a whole, not to an individual vertially-integrated industry. Let's turn to the tape: To discuss this view of production we must refer to the notions of 'system of production' and 'integrated industry', applying them to the commodity A(i). [Footnote:] So far, we have used these notions only for the wage-commodity G, but they can be applied to any commodity A(i). A 'system of production' for A(i) is a set of methods of production; one method for A(i), and one for each of the commodities which enter directly or indirectly the production of A(i). Appropriate proportions of these industries will give the 'integrated industry' of A(i). In parallel with the assumptions of continuity inherent in the notions of 'marginal products', we must assume that the system for the production of A(i) 'changes continuously' with r. What we saw about the relations between r, w, and [net output per head], and their representation by means of the 'envelope' of wage-curves, can then be applied to the relations between r, the wage in terms of A(i), and the net physical product per worker in the production of A(i)... -- P. Garegnani, Heterogeneous Capital, the Production Function and the Theory of Distribution. Review of Economic Studies. 1970. 3.0 LABOR DEMAND OR CAPITAL THEORY ONLY? Some suggested, incorrectly, that I am being original in seeing that my favorite issues in capital theory have implications for labor economics. Once again, let's look at the literature: On the other hand, the assumption of a persisting equality between the employment of labour and its supply would be justified by a parallel mechanism at work in the labour market. A regular demand function for labour would exist for any given amount of capital employed in production, and competition could be thought of as bringing the wage to the level where wall labour finds employment... ...This elaborate theory of distribution therefore rests on the principle that a fall of r cheapens the more capital-intensive processes of production relative to the others. But this principle is no more valid than that of the equality between the rate of interest and the marginal product of capital: in just the same way, it is invalidated by the dependence of the value of capital goods on distribution. -- P. Garegnani, ibid. 4.0 NEOCLASSICAL THEORY AS AN EMPIRICAL GENERALIZATION Particular theoretical examples have forced the admission, in recent economic literature, that the switch of systems might operate in a direction contrary to the one traditionally assumed. The tendency however has been to label those cases as 'exceptions': as if the principle about capital-intensity had resulted from observed regularities, always liable to exception, and was not a pure deduction from postulates (like Bohm-Bawerk's 'average period of production') now generally admitted to be invalid. Instead, it must be recognized that the traditional principle, drawn from incorrect premises, is itself incorrect. Moreover, the examples of the Appendix do not seem to indicate that the conditions in which a fall of r results in a relative cheapening of the less capital-intensive processes are any less plausible than those in which the opposite would be true. This appears to undermine the ground on which rests the explanation of distribution in terms of demand and supply for capital and labour. -- P. Garegnani, ibid. 5.0 ONLY ABOUT AGGREGATION OF CAPITAL? the argument [is] over whether heterogenous capital goods can be treated as if they are one capital good. -- Poor Chris Auld Let's turn to some more tapes: Then Buddha said: ... But tell me, Subhuti, do you really believe that having only one homogeneous capital good will permit you to derive a rate of profit purely from the technical relationship between homogeneous capital and output? Subhuti replied: Thus it is said in some venerable books. Buddha said: Revere them, Subhuti, but trust them not. Suppose you do get the value of the marginal product of capital in terms of output of consumer goods. In what units will it be expressed? Physical units of additional consumer goods per unit of additional homogeneoues capital. But the rate of profit is a pure number. Surely you will need something more in going from the first to the second to reflect the relative price of the capital good vis-a-vis the consumer good. But the equilibrium price of capital in units of consumer goods depends on the rate of profit used for discounting, and a variation of the rate of profit can involve a variation of the value of the same physical capital in units of consumer goods. This difficulty is not eliminated by having one homogeneous good. -- Amartya Sen, On Some Debates in Capital Theory. Economica. V. 41, August 1974. Yet there finally came a third very curious phase, which should appear rather strange, given its weak theoretical and empirical underpinning, but which was greeted with relief by the theorists of mainstream economics. The essence of this third phase can be summed up with the following proposition: the criticisms of the traditional theory of capital raised by the phenomenon of re-switching (and consequent reverse capital-deepening) are valid, but only with reference to the neoclassical model conceived in aggregate terms. They do not apply to the neoclassical case of the general economic equilibrium model, conceived in disaggregated terms and based on the behaviour of individuals maximising inter-temporal functions of profits and of utility. This proposition actually has no objective foundation: phenomena of non-convexity, re-switchings of techniques and badly-behaved production functions, to take an expression that has been widely used ('behaving badly' meaning simply that they behave in a way as not to obey the assumptions of neoclassical economics), are not [CapitalEth] as has been amply demonstrated [CapitalEth] a consequence or a characteristic of any particular process of 'aggregation'. They may occur at any time and in any context, aggregated or disaggregated. Various authors have continued to demonstrate this point (e.g. Kurz 1987, Schefold 1997, Garegnani 1998, and others). But so things go. The contrary conviction had taken root and has continued to spread. Above all, the proposition cited above has been trundled out again and again, with no proof, but simply referring back to other sources, which in turn are either insufficient or inconsistent. -- Luigi L. Pasinetti, Critique of the Neoclassical Theory of Growth and Distribution, 2000? 6.0 CONCLUSION I don't expect poor Chris Auld or JimT to exhibit the intellectual integrity needed to acknowledge they were misrepresenting the peer- reviewed literature. -- Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html To solve Linear Programs: .../LPSolver.html r c A game: .../Keynes.html v s a Whether strength of body or of mind, or wisdom, or i m p virtue, are found in proportion to the power or wealth e a e of a man is a question fit perhaps to be discussed by n e . slaves in the hearing of their masters, but highly @ r c m unbecoming to reasonable and free men in search of d o the truth. -- Rousseau === Subject: Re: My FLT proof, *last* warning to critics Saying that my proof is wrong without checking it makes me angry, and > can be causing me financial harm. I reserve the right to seek redress > for such harm if it is provable that it occurred, and may do so in a > court of law. > Jimmy, > It's *been* checked. It's wrong. > You're not going to make it correct by indulging in that tedious tactic > of net.losers, of threatening litigation against anyone who doesn't > agree with you. > Now, sod off. We can't _make_ him sod off, how ever much we would like to. If he abandons math altogether, in favor of whinging and bluster, then he becomes tedious. Let us treat him accordingly - by ignoring him. Sci.math, attracts its share of cranks. I remember the old days, when our Resident Chief Crank was - oops - I almost pronounced his name - a Greek mathematician / a heavy radioactive metal. He still posts, but nobody responds. Now, can we be equally sane about JSH? -- Chris Henrich He had long ago come to the conclusion that there was nothing Man Was Not Meant To Know. He was willing to concede that there were things Man Was Too Dumb To figure Out. -- Michael Kurland === Subject: Near-integer values of polynomials on integers If f is a polynomial with real coefficients: How close can f(n) be to an integer, when n is an integer? How can I find the best values of n ? If f(n) = b n for some real number b, then the theory of continued fractions implies that f(n) is within O(1/n) of an integer infinitely often; moreover, the Euclidean algorithm gives us an efficient way to walk around GL(2,Z) looking for approximate solutions to b n - m = 0. For other polynomials f, I suppose it's true that the fractional parts of f(n) are uniformly distributed in [-1/2, 1/2), so that again f(n) ought to be within O(1/n) of an integer infinitely often, but I don't know how to find good values of n . (In fact, I don't even recall off-hand how to do this for f(n) = a + b n with nonzero real coefficients a,b.) I tested some small values of f(n) = a + b n and f(n) = a + b n + c n^2 where a=pi, b=e, c=sqrt(2). It certainly does appear from a plot that the points of the form ( log(n), n | { f(n) } | ) are uniformly distributed in the region (0, infty) x (0, 1) (where {x} = x - round(x) is the distance to the nearest integer). Here are the values of n < 3.10^6 where f(n) is within 1/n of an integer: {1, 2, 3, 4, 8, 11, 29, 36, 75, 107, 178, 501, 572, 1037, 2038, 3039, 4040, 11583, 20127, 29672, 47761, 65850, 83939, 256285, 446720, 655244, 1054203, 1453162, 1852121, 2251080} for the linear function and {1, 2, 3, 5, 24, 25, 28, 42, 79, 139, 354, 394, 467, 1357, 1933, 2173, 3905, 4097, 10218, 12310, 23629, 34644, 42266, 50277, 222996, 250375, 262688, 272302, 343133, 1226556, 1781633, 2107651} for the quadratic. My question is, how might I have found (some of) these values of n apart from an exhaustive search? dave === Subject: Re: non-Euclidean geometry at 05:07 PM, bhu@hotmail.com (Bhu Joshipura) said: >Every web-page I read on non-Euclidean geometry refered to problem >with Euclid's fifth postulate. That's because the term commonly refers specifically to Elliptic and Hyperbolic Geometry. It's strictly an issue of vocabulary. >Why does none talk about geometries with any other postulate >re-examined? Why do you assume that no one talks about them? Google for, e.g., Non-Archimedean Geometry. >Is it not possible or is it not interesting? No. It is both possible and interesting. There is a vast body of literature. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply === Subject: Re: non-Euclidean geometry >I am not a mathematician but I love mathematics. I love the warmth >with which this group responds to my questions. Here is one more: >Every web-page I read on non-Euclidean geometry refered to problem >with Euclid's fifth postulate. >Why does none talk about geometries with any other postulate >re-examined? >Is it not possible or is it not interesting? >-Bhu The postualtes are listed at http://mathworld.wolfram.com/EuclidsPostulates.html Notice the nature of the first four postulates: You can do _____ The fifth one, when phrased differently, is the only one that asserts the *unique* *existence* of an object. As a result, it is easier to ask what if when viewing this than the others. Also, if you look at the surface of a sphere and think of a line as a great circle, the fifth postulate is the one that is most obviously violated. > And another of Euclid's axioms is violated in this case too. In the > Hyperbolic case _only_ the fifth postulate is violated. This observation leads naturally to investigating the nature of parallel lines. True. Once you change a postulate, the next thing that must be checked is consistency. I'd have to look up which one is usually changed on the sphere. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Nora Baron is a palindrome, Not a person >In sci.math, michael >: > You know what happens after I make a post like this one? Sales of all products containing caffeine increase tenfold? >That reminds me. I need another cup of coffee/cocoa mixture. >Be right back... > Just remember, I've been looking at posts on this newsgroup for years, > and I've seen quite a few people come and go during that time. Then why are you posting this to alt.fiction.original, when it's of no interest to us whatsoever? >Most likely because JSH is a bit sloppy regarding newsgroup postings. :-) >Someone else is suggesting that JSH is sloppy in other areas >as well. :-) (Not me. I *know* JSH's math is a bit on the >careless side, not because of his equation manipulations but >because he leaps to conclusions that need to be carefully >tackled instead.) BTW, I see you're back to the Yahoo email address. I look forward to the next post from jstevh@msn.com telling us that this is a forgery. Go play in the traffic. >Now now, this *is* sci.math (among others); at least phrase >it as a creative math problem: >(1) A person decides to attempt to cross an 8-lane freeway, >for some reason which shall remain unspecified by the math >problem (why does Billy throw the ball to Jane anyway?). >The person can walk at 3 mi/hr = 4.4 ft/sec. Assuming a >uniform population of sedans of 16 feet in length and 6 >feet in width and exactly in the middle of each lane, that >the lanes are 12 feet wide, that every sedan is following >the speed limit and the 2-second rule [*], that the sedans >are otherwise randomly distributed, that the person, >once he starts to cross, blithely walks at a uniform >velocity straight across the highway, as opposed to doing >something more intelligent (like running zig-zags), and >that the sedans don't brake before hitting him, what is >the probability that he'll be struck? >(2) Same as (1), except the sedans are using a different rule, >the car-length every 10 mph rule. This rule is obviously >not quite as safe but I happen to live in a metro area so >know even this relaxed (?!) rule is broken routinely at speed. >As if the state Highway Patrol doesn't have enough to worry about... >(3) Same as (1) except the traffic is bumper-to-bumper stop-and-go. >(This one should be easy. Of course being struck at >2 mph isn't quite as deadly as 65 mph unless one's head >gets stuck under a wheel or something.) >(4) Same as (1), except we now assume a mix of cars: >8 parts sedan, 2 parts SUV (length 16 1/2 feet, width 6 1/2 feet). >(5) Same as (1) and (3), except we assume the mix 7 parts >sedan, 2 parts SUV, and 1 part semi-tractor trailer (length >64 feet (48 foot trailer, 16 foot cab), width 8 feet [+]). >Since truckers are (hopefully) more knowledgable we >increase their following distance to 4 seconds. >(6)-(10). Assume the sedans and SUV's can see a distance >of 1,000 feet and brake within a distance of 300 feet (of >which up to 50 feet can be reaction time). The SUVs take >350 feet since they're heavier. The trucks can brake >within 500 feet. Assume also that the cars and SUVs >don't skid out of control while braking and the trucks >don't jackknife. Assume perfect visibility to the vehicle's >right (e.g., no trees in the way or blind spots in the >vehicle). >(11)-(15) Assume in (6)-(10) a visibility of 200 feet >by either nighttime conditions, an obscuring hill, or fog. >(16)-(30). Now assume the person can only walk at 1.5 mph = 2.2 ft/sec. >(Good luck. He'll need it.) >[*] The 2-second rule is a common one, and basically stipulates >that the rear of the preceding car, assuming you and he are >traveling at the same speed, shall be 2 seconds in front of >your front bumper. Admittedly this rule has some interesting >ramifications with respect to highway capacity, as it indicates >that no matter how much one increases the speed the capacity of >the highway is largely constant, assuming everyone, erm, rigorously >follows this rule... >[+] a cargo container apparently has dimensions 40' >by 96; a trailer can be as big as 48' x 96 as mentioned -- >maybe even longer. I'm not sure how big a truck cab is but >I'm assuming it's similarly sized to a sedan or SUV except >that there's no area for the kids; it's mostly motor. :-) >Since 96 = 8' we're in fairly good shape here; also, the >truck cab partially occupies the trailer space because >of the hitch, making dimensions a little weird. But then, >this *is* a hypothetical math problem anyway... :-) Standard measurements for semis, is to measure overall length, from the front bumper to the rear of a trailer; thus the standard semi-rig with a 40' trailer will have an overall length of 55'; which is usually the limit for semis East of the Mississippi River. Going West of the 'Ole Miss, the length is uaually extended to 65', allowing the use of the 45' and 48' trailers. As for width, the standard used before 1990, was a 96 (8') max, although not the standard seems to have been slightly icreaded to 102 (8'6), of which the majority of this width is part of the side walls, since many trailers have more insulation in them than in previous years. Another point to consider, is whether or not the tractor unit ( pulling the trailer), is of the single driver axle type or the dual driver axle type, as truck-tractors with single driver axles frequently have less braking control that the dual power axle units do. Not that there is that much of a difference; but there is some, and it is measurable. Now given your 8-lane freeway (4-lanes in each direction; say an East/West positioning for the freeway -- and a north bound movement for the pedestrian); an average mix of vehicles, per mile of running length, would actually include 2.25 tractor-trailer rigs - 4.75 straight truck units (straight truck refering to vehicles like dump trucks, deliver trucks, tow trucks, etc) - 35.5 SUVs - 22.25 Vans (a van is anything not condsidered to be a delivery truck or an SUV), - with the bulk of the traffic being the compact auto. You would also find the occasional mid-size family vehicle and even rarer, the large family vehicle (concept based on observing mid-sized and bigger Fords and Chebbies moving along California Freeways). If the traffic were stop-n-go, bumper-to-bumper, type traffic, your pedestrain could fairly easily cross one half of the freeway, although crossing the other half, (if it were not bumper-to-bumper, stop-n-go traffic), might be more difficult. But then the laws of chance come into ply in this scene, and before your pedestrian could begin to cross the freeway, your local Highway Patrol officer, or County Sherriff deputy, or maybe a City cop, pulls up and demands to know why said pedestrian is trying to cross the freeway. Result: Pedestrian gets arrested for trying to hitchike, halued off to the hoosegow, and then cools his/her heels waiting to explain to the judge why he/she wanted to cross the freeway, and why cross at that particular point, when a walk of several hundred feet in any direction would have allowed the pesdestrian to cross safely. The above is only a response to a ? by the original poster on truck-trailer sizes, and not ment to be used to defame math people; unless one wished to defame such people. I use mat every day, from the time I get out of bed to the time I return to bed. At work, I can calculate the necessary size and numbers of cuttings to reduce a 55' long, 9' wide, 9'6 high RR boxcar, to proper mill size specs, and do so without regard to paper and pencil. I can also, aain without paper, calculate to within 80 pounds, how much an unchopped item will weigh after chopping. The only place I fall short in, is when writing a story, since even though mine may start out rated T , before long it has dropped to the XXX rating. The few times this has not happened, has been to working collaborations with others in turning out an accpetable (for publication) fiction work, which everyone can enjoy, (as opposed to a fiction work with only a small segment of the readers would enjoy). === Subject: Notation wanted for { 0, 1, ..., n } [Please excuse the crosspost. This issue comes up very frequently when I'm doing crypto, and may therefore be of interest to other cryppies; but the mathmos may have useful knowledge or suggestions.] Does anyone know of a decent, /terse/ notation for the set of natural numbers { i in N | i < n }, i.e., { 0, 1, ..., n - 1 }? A friend of mine suggested using $mathbb{N}_{[Please excuse the crosspost. This issue comes up very frequently when >I'm doing crypto, and may therefore be of interest to other cryppies; >but the mathmos may have useful knowledge or suggestions.] >Does anyone know of a decent, /terse/ notation for the set of natural >numbers { i in N | i < n }, i.e., { 0, 1, ..., n - 1 }? A friend of >mine suggested using $mathbb{N}_{looks a bit like > || | > || |and I'm quite fond of that. However, I'd rather use something standard. >I've seen [n] used to mean { 1, 2, ..., n }, but indexing things from 1 >confuses my little brain, and square brackets have way too many other >meanings. [0, n) implies real numbers to me. [0, n) intersect N is >just too heavyweight. I suspect you're going to be doing algebra with them, so block-Z-subscript-n (integers mod n) is what I'd suggest using. This is actually the standard notation for the ring of integers mod n, with addtion and multiplication, and not just the set, which is why I prefaced it as I did. The ASCII version is Z/nZ (the integers mod the ideal generated by n, which is just nZ). Block goes away in ASCII. Jon Miller === Subject: Re: Notation wanted for { 0, 1, ..., n } Supersedes: block-Z-subscript-n (integers mod n) is what I'd suggest using. Hmm... Z/nZ is the ring of equivalence classes mod n, which isn't strictly what I want. I'm sorry: I should have listed that, along with Z_n and Z/n (which I've also seen, but aren't as nice), as discarded candidates, but it slipped my mind. I tend to want to use N_{ Does anyone know of a decent, /terse/ notation for the set of natural > numbers { i in N | i < n }, i.e., { 0, 1, ..., n - 1 }? > I've seen [n] used to mean { 1, 2, ..., n }, but indexing things from 1 > confuses my little brain, and square brackets have way too many other > meanings. [0, n) implies real numbers to me. [0, n) intersect N is > just too heavyweight. Actually, in simplicial homology, the notation [n] is used for the set 0,1,...,n, but usually considered as an ordered set. Also, in the same field, some people use capital-delta-sub-n for the same set (Delta_n). I would suggest going with the [n]... > -- [mdw] Michael A. Van Opstall Padelford C-113 opstall@math.washington.edu http://www.math.washington.edu/~opstall/ === Subject: Re: Notation wanted for { 0, 1, ..., n } > [Please excuse the crosspost. This issue comes up very frequently when > I'm doing crypto, and may therefore be of interest to other cryppies; > but the mathmos may have useful knowledge or suggestions.] > Does anyone know of a decent, /terse/ notation for the set of natural > numbers { i in N | i < n }, i.e., { 0, 1, ..., n - 1 }? A friend of > mine suggested using $mathbb{N}_{ looks a bit like > || | > || | and I'm quite fond of that. However, I'd rather use something standard. > I've seen [n] used to mean { 1, 2, ..., n }, but indexing things from 1 > confuses my little brain, and square brackets have way too many other > meanings. N is sometimes used for {0, 1, 2, ..... } and sometimes used for {1, 2, 3, ....}. There is no standard that says whether or not to include 0. This is why Z^* is often a better notation for {0, 1, 2, ....} See for example: http://mathworld.wolfram.com/N.html How about using something like {Z^n}_0 (Z with exponent n, and lower index 0)? --Anton === Subject: Re: Notation wanted for { 0, 1, ..., n } >[Please excuse the crosspost. This issue comes up very frequently when >I'm doing crypto, and may therefore be of interest to other cryppies; >but the mathmos may have useful knowledge or suggestions.] >Does anyone know of a decent, /terse/ notation for the set of natural >numbers { i in N | i < n }, i.e., { 0, 1, ..., n - 1 }? A friend of >mine suggested using $mathbb{N}_{looks a bit like > || | > || |and I'm quite fond of that. However, I'd rather use something standard. >I've seen [n] used to mean { 1, 2, ..., n }, but indexing things from 1 >confuses my little brain, and square brackets have way too many other >meanings. [0, n) implies real numbers to me. [0, n) intersect N is >just too heavyweight. > > I suspect you're going to be doing algebra with them, so > block-Z-subscript-n (integers mod n) is what I'd suggest using. This is > actually the standard notation for the ring of integers mod n, with > addtion and multiplication, and not just the set, which is why I > prefaced it as I did. The ASCII version is Z/nZ (the integers mod the > ideal generated by n, which is just nZ). Block goes away in ASCII. > Jon Miller But to get {0,1,2,...,n} as a complete set of representatives, you would need to use n+1 as your modulus! === Subject: Re: Notation wanted for { 0, 1, ..., n } > Does anyone know of a decent, /terse/ notation for the set of natural > numbers { i in N | i < n }, i.e., { 0, 1, ..., n - 1 }? The shortest possibility is to simply write n. The notation X^n for the set of functions from n to X is consistent with the notation X^n for the set of length-n sequences from X, if you number vectors starting from 0. In many set theories, this is part of the definition of natural numbers: each natural number n is equal to the set of smaller natural numbers. This idea is universally credited to von Neumann; does anyone know where it was first published? ---D. J. Bernstein, Associate Professor, Department of Mathematics, Statistics, and Computer Science, University of Illinois at Chicago === Subject: Re: Notation wanted for { 0, 1, ..., n } >[Please excuse the crosspost. This issue comes up very frequently when >I'm doing crypto, and may therefore be of interest to other cryppies; >but the mathmos may have useful knowledge or suggestions.] >Does anyone know of a decent, /terse/ notation for the set of natural >numbers { i in N | i < n }, i.e., { 0, 1, ..., n - 1 }? A friend of >mine suggested using $mathbb{N}_{looks a bit like > || | > || |and I'm quite fond of that. However, I'd rather use something standard. >I've seen [n] used to mean { 1, 2, ..., n }, but indexing things from 1 >confuses my little brain, and square brackets have way too many other >meanings. [0, n) implies real numbers to me. [0, n) intersect N is >just too heavyweight. I use [0..n], and more generally [a..b] for {a,a+1,...,b} for integers a <= b. There is some ambiguity here, because [a..b] can also be used to mean the ordered set [a,a+1,...,b], but usually that distinction is not critical. Derek Holt. === Subject: Re: Notation wanted for { 0, 1, ..., n } > Does anyone know of a decent, /terse/ notation for the set > {i in N | i < n} of natural numbers, i.e., {0, 1, ..., n - 1}? > The shortest possibility is to simply write n. Ah, right! Of course, in axiomatic set theory we (usually) have: n = {0, 1, ..., n-1} for any natural number n != 0. @Mark: You might consult P. R. Halmos' Naive Set Theory for a terse treatment of this stuff. > In many set theories, this is part of the definition of natural numbers: > each natural number n is equal to the set of smaller natural numbers. > This idea is universally credited to von Neumann; does anyone know where > it was first published? According to A. A. Fraenkel (in P. Bernays' Axiomatic Set Theory): J. von Neumann, Zur Einf.9fhrung der transfiniten Zahlen. Acta Lit. ac Sc. Univ. ... (Szeged), Sectio Sc. Math. 1, 199-208, 1923. In 1921 Fraenkel (Zu den Grundlagen der Cantor-Zermeloschen Mengenlehre, Mathematische Annalen, volume 86; pp. 230-237) still used Zermelo's definition(s) 1 = {0}, 2 = {1}, etc. F. === Subject: Re: Notation wanted for { 0, 1, ..., n } > I've seen [n] used to mean { 1, 2, ..., n }, but indexing things from 1 > confuses my little brain, and square brackets have way too many other > meanings. [0, n) implies real numbers to me. [0, n) intersect N is > just too heavyweight. {1 <= i <= n} Bob Kolker === Subject: Re: Notation wanted for { 0, 1, ..., n } > [Please excuse the crosspost. This issue comes up very frequently when > I'm doing crypto, and may therefore be of interest to other cryppies; > but the mathmos may have useful knowledge or suggestions.] > Does anyone know of a decent, /terse/ notation for the set of natural > numbers { i in N | i < n }, i.e., { 0, 1, ..., n - 1 }? A friend of > mine suggested using $mathbb{N}_{ looks a bit like > || | > || | and I'm quite fond of that. However, I'd rather use something standard. > I've seen [n] used to mean { 1, 2, ..., n }, but indexing things from 1 > confuses my little brain, and square brackets have way too many other > meanings. [0, n) implies real numbers to me. [0, n) intersect N is > just too heavyweight. > -- [mdw] The J notation for {0,1,...,n-1} is i. n http://www.jsoftware.com/books/help/primer/more_primitives.htm If you are going to do computations, you may just as well use executable notation. Nemo === Subject: Re: Notation wanted for { 0, 1, ..., n } > [Please excuse the crosspost. This issue comes up very frequently when > I'm doing crypto, and may therefore be of interest to other cryppies; > but the mathmos may have useful knowledge or suggestions.] > Does anyone know of a decent, /terse/ notation for the set of natural > numbers { i in N | i < n }, i.e., { 0, 1, ..., n - 1 }? A friend of > mine suggested using $mathbb{N}_{ looks a bit like > || | > || | and I'm quite fond of that. However, I'd rather use something standard. > I've seen [n] used to mean { 1, 2, ..., n }, but indexing things from 1 > confuses my little brain, and square brackets have way too many other > meanings. [0, n) implies real numbers to me. [0, n) intersect N is > just too heavyweight. > -- [mdw] Lots of good replies.What are you planning to publish? -- Russ Lyttle lyttlecatearthlink.net at = @ === Subject: Re: Notation wanted for { 0, 1, ..., n } As a notation for {0,1,2,...,n}, you might use 0..n, which indicates the range of integers from 0 to n, in a number of computer languages. === Subject: Re: Notation wanted for { 0, 1, ..., n } Does anyone know of a decent, /terse/ notation for the set of natural numbers { i in N | i < n }, i.e., { 0, 1, ..., n - 1 }? > The shortest possibility is to simply write n. The notation X^n for the > set of functions from n to X is consistent with the notation X^n for the > set of length-n sequences from X, if you number vectors starting from 0. > In many set theories, this is part of the definition of natural numbers: > each natural number n is equal to the set of smaller natural numbers. > This idea is universally credited to von Neumann; does anyone know where > it was first published? Miraminoff used a construction very similar to the von Neumann ordinals. This appears at a paper from 1917 I think. von Neumann's definition appears in his 1923 paper Zur Einfuhrung der transfiniten Zahlen, which -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Notation wanted for { 0, 1, ..., n } > Does anyone know of a decent, /terse/ notation for the set of natural > numbers { i in N | i < n }, i.e., { 0, 1, ..., n - 1 }? In france we use [[0;n-1]] It's almost the same notation that the segment's one with double bracket ( is it the right word for [ and ] ?) It can be used also more generaly : [[2;7]]={2,3,4,5,6,7}. === Subject: Re: Notation wanted for { 0, 1, ..., n - 1 } > The shortest possibility is to simply write n. [...] In many set > theories [...] each natural number n is equal to the set of smaller > natural numbers. > Yes! ;-) I'm not sure I'm going to get away with that -- I'll see how > confusing it is -- but its minimalism has definite appeal. Right. For example, instead of writing: n-1 SUM a_n i = 0 one might write SUM a_n i e n On the other hand, the latter notation seems to be rather uncommon, I guess. F. === Subject: Re: Notation wanted for { 0, 1, ..., n - 1 } -----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 | | |The shortest possibility is to simply write n. [...] In many set |theories [...] each natural number n is equal to the set of smaller |natural numbers. |>Yes! ;-) I'm not sure I'm going to get away with that -- I'll see how |>confusing it is -- but its minimalism has definite appeal. | | | Right. | | For example, instead of writing: | | n-1 | SUM a_n | i = 0 | | one might write | | SUM a_n | i e n | | On the other hand, the latter notation seems to be rather uncommon, I | guess. Um you mean the sum of a_i right? Otherwise it's just equal to n * a_n :-) In that case you can just say sum_{forall i in n} {a_i} Tom -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.2.2 (GNU/Linux) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org iD8DBQE/OtJfsP+tEsHHY0ARAoQZAJ0fVz3DldiL9yrASaAhJLf2Y+rBUgCgh+nI +X51LxOLm8xZruesuvVVSj0= =+4RD -----END PGP SIGNATURE----- === Subject: Re: Notation wanted for { 0, 1, ..., n - 1 } >The shortest possibility is to simply write n. [...] In many set >theories [...] each natural number n is equal to the set of smaller >natural numbers. Yes! ;-) I'm not sure I'm going to get away with that -- I'll see how confusing it is -- but its minimalism has definite appeal. > Right. > For example, instead of writing: > n-1 > SUM a_n > i = 0 > one might write > SUM a_n > i e n > On the other hand, the latter notation seems to be rather uncommon, I > guess. > F. Gottlieb, good to see you, you old fogy... :) === Subject: Re: Notation wanted for { 0, 1, ..., n - 1 } > Um you mean the sum of a_i right? Should read: [...] instead of writing: n-1 SUM a_i i = 0 one might write SUM a_i i e n If you prefer LaTeX, try this one: begin{displaymath} sum_{i in n} a_i end{displaymath} F. === Subject: Re: Notation wanted for { 0, 1, ..., n - 1 } > Gottlob, good to see you, you old fogy... :) Well, you know: 1000 years is like one day to the lord. Hallelujah! F. === Subject: Re: Notation wanted for { 0, 1, ..., n - 1 } Why not SUM a_i n === Subject: Re: Notation wanted for { 0, 1, ..., n - 1 } > Why not > SUM a_i > n Not a good idea, for 1.) the relation between i and n is not clear. This would lead to following difficulty in practical math: SUM PROD a_ij ??? n m could mean SUM PROD a_ij ??? i e n j e m as well as SUM PROD a_ij ??? j e n i e m If a_ij != a_ji you have a problem here. And 2.) there's the common (well established) notation SUM a_i i The sum over all i - hence your proposal would (imho) be a potential source of confusion (to say the least). -------------- Note: the proposed notation is related to a rather common notation; it's not just some artificial conception. :-) In set theory we have, for example: UNION A_i i e I and/or INTER A_i i e I (I guess you didn't realize this fact... ;-) F. P.S. Right... a more common notation certainly would be: SUM a_i i < n instead of SUM a_i i e n I guess. :-) === Subject: Re: Notation wanted for { 0, 1, ..., n - 1 } Sorry, two typos. :-( > For example, instead of writing: > n-1 > SUM a_n > i = 0 > one might write > SUM a_n > i e n > On the other hand, the latter notation seems to be rather uncommon, I > guess. Should read: [...] instead of writing: n-1 SUM a_i i = 0 one might write SUM a_i i e n F. === Subject: Re: Notation wanted for { 0, 1, ..., n - 1 } (was `..., n') > The shortest possibility is to simply write n. [...] In many set > theories [...] each natural number n is equal to the set of smaller > natural numbers. Yes! ;-) I'm not sure I'm going to get away with that -- I'll see how confusing it is -- but its minimalism has definite appeal. There is some ambiguity lurking here. Since 2 = { 0, 1 } is the set of bits, then 2^n is: * the nth power of 2, and hence the set { 0, 1, ..., 2^n - 1 }, and * the set of binary n-vectors, and I'm not /entirely/ convinced that they're the same. Oh, nuts. I've just noticed: I got the subject line /wrong/. Fixed here. -- [mdw] === Subject: Re: Notation wanted for { 0, 1, ..., n - 1 } (was `... n') Supersedes: would need to use n+1 as your modulus! I don't want in the set, just { 0, 1, ..., n - 1 }. That is, the n smallest natural numbers, starting at 0. So apart from the technicality of equivalence classes vs plain numbers, Z/nZ is a good suggestion. (My brain works best when it thinks about indexing things beginning with zero, and half-open intervals a <= i < b; otherwise there are overlap issues, and I make stupid off-by-one errors.) (And I got the subject line wrong. Now fixed; sorry about the confusion.) -- [mdw] === Subject: Re: One kg = w/g = 9.81 N/(9.81 m/sec.94) Cut< > No--Should we assume, Shead, you are blathering about standard > acceleration of gravity or the definition of a kilogram? > standard acceleration of gravity > http://physics.nist.gov/cgi-bin/cuu/Value?gn > 9.806 65 N Are you really telling me that the acceleration of gravity is 9.806 65 N? Sounds to me more like the force of gravity on a kilogram; at Sevres' France! === Subject: Re: One kg = w/g = 9.81 N/(9.81 =?iso-8859-1?Q?m=2Fsec=B2?=) > Are you really telling me that the acceleration of gravity is 9.806 65 N? > Sounds to me more like the force of gravity on a kilogram; at Sevres' > France! Don't be stooopid Shead, acceleration has units of m/s^2 (in the metric system). See: http://physics.nist.gov/cgi-bin/cuu/Value?gn === Subject: Re: online/self-study 4 year math degrees? > Can anyone provide me a list of some universities which offer such a > program? I am in the military and so attending classes like normal > people is pretty much nixed. But I very much desire to take classes > in math. Im already self teaching myself alot of maths (heck, of all > the maths i HAVE managed to take here and there at various colleges, > the only one i hadnt already taught myself- thus the only one that > wasn't just a matter of warming a chair to get a credit- was linear > algebra) so i am well disciplined and prepared to survive such a > program. > PS please post answers to sci.math since this email account doesn't > exist > I do not know any online-programs. > However, if it is only a matter of attendance, tradidional programs of > distance learning (i.e. via correspondence) might help too. > Look at > Can anyone provide me a list of some universities which offer such a > program? I am in the military and so attending classes like normal > people is pretty much nixed. But I very much desire to take classes > in math. Im already self teaching myself alot of maths (heck, of all > the maths i HAVE managed to take here and there at various colleges, > the only one i hadnt already taught myself- thus the only one that > wasn't just a matter of warming a chair to get a credit- was linear > algebra) so i am well disciplined and prepared to survive such a > program. > PS please post answers to sci.math since this email account doesn't > exist University of Wisconsin used to have a pretty decent selection of correspondence courses in math, and you could ask if you were interested in courses not specifically offered in the correspondence catalog. University of Waterloo (Ontario, Canada) also had some range of math courses. There exist various external degree programs to which such credits could be applied. David Ames === Subject: OT: variation on Traveller problem, or is it about IMO... I thought of this problem after seeing a thread with disscusion about Reid Barton. Then I looked on IMO pages and found that problems rather easy, too much from one group and somehow mechanistical. (Lots of them were too often beaten to the death) I somehow think that if something is International and mathemathic it shouldn't be flooded by problems that were solved in 1950 or sooner. There are books filled with solution of such problems. Person that enters such contest should be required to think, at least little. So here is something simple from person that was kicked out from Highschool, or was it College, just for beeing born in with poor health and in low income family. It's a kind of traveller problem so KPD could be happy (he might add it to his program if he wants). BTW Xantian do you wanna some pictures to your program? The problem: Lets say you have a lots of points say a amount. And you must visit each of them. You would like to have shortest route possible. Now something a little more difficult. You need to do it in 3D. Yes they are somewhere in space. And more, you might like to visit them more than once. You might like to have from one to three visits. You would receive information about what point would you like visit more than once after you had travelled at least 1/3 minimal distance for single visits. (If you found a way how visit each point exactly once and divided that distance by three. Yes, thats the distance after you'd receive that information). (my explanations are sometimes as clear as explanations of Bernhard from Yes minister) We might modify this problem to: after travelling 1/3 TOTAL distance you'd receive this information. BTW how many points you'd be able to visit if you'd neet to visit some of them five times? Say points are in the cube. Try to solve it for a=1000 (for 400MHz Celerons) to 10000(If you have something faster). I don't think coding should be longer than 2 weeks, (If you'd like it to be challenge) and your programs shouldn't for sure run for years. I consider it as a simple problem, I have already encountered more complicated, but It would suffice as and example of something more difficult. You could call it Raghar problem. ( it's very simple problem so, throught I believe in calling problems in propper way, I wont be offended by this). === Subject: Other proof? (was: Re: Relationship between Undecidability, Incompleteness, and NP Completeness??) > an easy diagonalization shows that r.e. is different from > co-r.e. Which reminds me: Does anyone know of a second proof of this fact, other than diagonalization? A concrete algorithm, for example, with some 'direct' proof that it's domain is not decidable? Or something else? Herman Jurjus === Subject: Phasetype-Distribution / Negative Binomial The Negative Binomial distribution can be represented as a Phasetype Distribution. Thats what i know for sure but, is there anyone who can tell me how the representation alpha,T (in Phasetype notation) of the Negative Binomial Distribution looks like? Mathew === Subject: Physical repn in noneuclidean spaces? When we make a construction in 2-D euclidean geometry, we use a compass, straightedge, pencil, and paper. The paper (or chalkboard) represents a plane; the straightedge represents a line segment; and the pencil tip represents a (moving) point. If we move to a sphere, we can still use the compass and pencil -- (assume the student has a steady hand) -- but now it makes better sense to use a string (held taut) instead of a straightedge. What then do we use when we move to more complicated surfaces? Neither the straightedge nor the taut string is guaranteed to work well on a surface with negative curvature. (Try it: you'll sometimes get the string leaving the surface altogether, or else (in theory) hugging alternate sides of the surface, or other kinds of trouble.) What do we use for constructions on nonstandard spaces or noneuclidean geometries? Please don't tell me we just give up on constructions. Ted Shoemaker shoemakerted@yahoo.com === Subject: Re: Physical repn in noneuclidean spaces? 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > When we make a construction in 2-D euclidean geometry, > we use a compass, straightedge, pencil, and paper. > The paper (or chalkboard) represents a plane; > the straightedge represents a line segment; > and the pencil tip represents a (moving) point. > If we move to a sphere, we can still use the > compass and pencil -- > (assume the student has a steady hand) -- > but now it makes better sense to use a string (held taut) > instead of a straightedge. > What then do we use when we move to more complicated surfaces? > Neither the straightedge nor the taut string is > guaranteed to work well on a surface with negative curvature. > (Try it: you'll sometimes get the string leaving the surface > altogether, or else (in theory) hugging alternate sides of > the surface, or other kinds of trouble.) > What do we use for constructions on nonstandard spaces or > noneuclidean geometries? Please don't tell me we just > give up on constructions. What do you mean when you talk about nonstandard spaces or noneuclidean geometries? The most natural, to me, would be to do the geometry entirely within the space, not referring to any embedding of that space in higher dimensional Euclidean space. So a string stretched tight in hyperbolic space can not fall off, because there is nowhere for it to fall off to. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: Physical repn in noneuclidean spaces? Ted Shoemaker > When we make a construction in 2-D euclidean geometry, > we use a compass, straightedge, pencil, and paper. ... > What do we use for constructions on nonstandard spaces or > noneuclidean geometries? Please don't tell me we just > give up on constructions. Informally speaking, we can discard the ruler and compass, and speak instead of the set of zeros of a linear or (a certain kind) of quadratic function in the coordinates. Such notions still make sense in some other settings, such as a circular torus or some other algebraically defined surface. But of course, what specifically can be said is dependant on the structure of the noneuclidean (nonstandard?) space. Using synthetic axioms and methods, instead of coordinates, we can still get circles in e.g. an inversive plane, and conics in a projective plane. Roughly speaking, these appear as fixed points of an automorphism (i.e. a symmetry) of the space, or as orbits of a point under a group of such symmetries. LH === Subject: Re: Physical repn in noneuclidean spaces? > Ted Shoemaker > What do we use for constructions on nonstandard spaces or > noneuclidean geometries? Please don't tell me we just > give up on constructions. > Informally speaking, we can discard the ruler and compass, and speak instead > of the set of zeros of a linear or (a certain kind) of quadratic function in > the coordinates. That makes good sense, in terms of the *mental* exercise involved. I was hoping for a discussion of the *hands-on* aspect. (1) How, in a concrete sense, do kinesthetic learners get involved with noneuclidean geometry? (2) Hands-on work isn't to be disdained. It was good enough for Euclid and his contemporaries. Can we not still learn from hands-on work, now with noneuclidean questions? (Getting your hands on a 19-dimensional hyperbolic space might be a little tricky, though.) Ted Shoemaker shoemakerted@yahoo.com === Subject: Re: Physical repn in noneuclidean spaces? 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > That makes good sense, in terms of the *mental* exercise involved. > I was hoping for a discussion of the *hands-on* aspect. > (1) How, in a concrete sense, do kinesthetic learners get > involved with noneuclidean geometry? > (2) Hands-on work isn't to be disdained. It was good enough for > Euclid and his contemporaries. Can we not still learn from > hands-on work, now with noneuclidean questions? I have in my office a big set of geoshapes -- rigid plastic polygons (mostly equilateral triangles) that can be clicked together edge-to-edge. One useful hands-on exercise is to construct a discrete model of the hyperbolic plane: connect 7 (or some other number larger than 6) equilateral triangles surrounding a common vertex, then add more triangles around each of the outer vertices of this complex so that it is again surrounded by 7 triangles, etc. You can also do this with paper cutouts. You will quickly discover that the hyperbolic plane does not embed very cleanly into 3d space... -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: Physical repn in noneuclidean spaces? |> Ted Shoemaker |> |> What do we use for constructions on nonstandard spaces or |> noneuclidean geometries? Please don't tell me we just |> give up on constructions. |> Informally speaking, we can discard the ruler and compass, and speak instead |> of the set of zeros of a linear or (a certain kind) of quadratic function in |> the coordinates. | |That makes good sense, in terms of the *mental* exercise involved. |I was hoping for a discussion of the *hands-on* aspect. |(1) How, in a concrete sense, do kinesthetic learners get |involved with noneuclidean geometry? it's not that hard to create 2d and 3d hyperbolic geometry virtual realities for a person to walk/swim/fly/navigate in. a primitive low-tech approximation of this can be done pretty easily with ordinary computer graphics on a pc. -- [e-mail address jdolan@math.ucr.edu] === Subject: please help me b^2 - 2i(1-f)b + P0 = 0 I want to sovle this equation, where f is a defined time-depenedent function and P0 is an operator which is 1/i * .8d.94 / .8d.94 t Please help me === Subject: prime ideals in Z[x] Today I came accross the following problem, which has resisted my attemps to solve it: Show that every prime ideal of Z[X] are of one of the following two forms: 1. with f in Z[x] irreducible 2. with p in Z prime and f in Z[x] such that f* in Z_p[x] irreducible where f* is f with the coefficients reduced mod p. Any ideas? nojb. === Subject: Re: prime ideals in Z[x] Visiting Assistant Professor at the University of Montana. >Today I came accross the following problem, which has resisted my >attemps to solve it: >Show that every prime ideal of Z[X] are of one of the following two >forms: >1. with f in Z[x] irreducible >2. with p in Z prime and f in Z[x] such that f* in Z_p[x] >irreducible where f* is f with the coefficients reduced mod p. >Any ideas? Assuming I made no mistake, the only caveat is that I have a reputation for doing things the hard way, so there may be a much simpler way of doing this... Take a prime ideal I of Z[x], and let J={a in Z: a in I}. Note that J is an ideal of Z. Therefore, J=(r) for some r in Z. Moreover, J is a ->prime<- ideal of Z, since I is a prime ideal of Z[x]. So either J=(0), or else J=(p) for some prime p. Abusing language, let's write J=(p), with p=0 or p a prime. If I-(p) is empty, then I=(p) and we are done. Otherwise, of the polynomials in I-(p) of minimal positive degree, let f(x) be one with the smallest positive leading coefficient. Suppose that g(x) is another polynomial in I-(p) of minimal positive degree and with the same leading coefficient as f(x). Let f*(x) and g*(x) be the polynomials in Z/pZ[x] obtained by reducing the coefficients of f and g modulo p (if p=0, that just means doing nothing). Since f(x)-g(x) must be a polynomial in (p), and therefore f*(x)-g*(x)=0, so f*(x)=g*(x). Therefore, the ideals (f(x),p) and (g(x),p) are equal in Z[x]. We will prove that I=(f(x),p); the above discussion shows that although there may be many choices for f(x), the choice is immaterial. Assume first that p>0: Since there is some coefficient of f(x) which is not a multiple of p, coefficient to obtain a polynomial of smaller degree; so writing f(x) = a_n x^n + ... + a_1 x + a_0, with a_n not divisible by p, we may find r and s integers such that ra_n + sp = 1, and so taking r f(x)+sp x^n we obtain a monic polynomial of degree equal to f(x). Thus we conclude that f(x) is monic. Note also that f(x) is such that f*(x) is irreducible in Z/pZ[x]. For otherwise, let g(x) and h(x) be nonconstant polynomials, with leading coefficient not divisible by p, such that g*(x)h*(x) = f*(x). Again, we may assume that g(x) and h(x) are monic, and so we have that g(x)h(x) = f(x) + p*m(x) for some polynomial m(x) of degree strictly smaller than f(x). Since this lies in I, either g(x) or h(x) lie in I, contradicting the minimality of deg(f). This shows that f*(x) is irreducible If I-(f(x),p) is empty, then we are done. Otherwise, let h(x) be a polynomial of minimal degree and smallest positive leading coefficient in I-(f(x),p). Note that deg(h) is no smaller than deg(f). Write f(x) = x^n + ... + a_1 x + a_0 h(x) = x^m + ... + b_1 x + b_0. Then x^{m-n}f(x) - h(x) lies in I, and is of strictly smaller degree than h(x). So it must lie in (f(x),p). But that means that h(x) itself lies in (f(x),p), contradicting the choice of h(x). So I=(f(x),p) and we are done. Now assume that p=0; this time we have that f(x) is uniquely determined as the polynomial of smallest degree (I assume the zero polynomial has no degree) in I with smallest positive leading coefficient, though we do not know if f(x) is monic. We also know that f(x) is irreducible in Z[x], because if g(x)h(x)=f(x), then either g(x) or h(x) would like in I; and therefore (f) is a prime ideal since Z[x] is a UFD. The claim is that I=(f). If not, let g(x) be a polynomial in I-(f) of minimal degree; it must be of degree at least equal to deg(f). Write f(x) = a_n x^n + ... + a_0 g(x) = b_m x^m + ... + b_0. Note that m>=n. Consider h(x)=b_m x^{m-n}f(x) - a_n g(x). This element lies in I; if it were in (f), then a_n g(x) would be in (f), and since (f) is a prime either, either a_n lies in (f), which is impossible, or else g(x) lies in (f), which is also impossible. So h(x) is in I-(f), and has degree strictly smaller than g(x), which contradicts the choice of g(x). Therefore, I=(f), f irreducible, as needed. Of course, you should also prove that these two types of ideals do indeed yield prime ideals... It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: prime ideals in Z[x] I had tried to do something like this ... but couldn't take it all the way to the end ... nojb. >Today I came accross the following problem, which has resisted my >attemps to solve it: >Show that every prime ideal of Z[X] are of one of the following two >forms: >1. with f in Z[x] irreducible >2. with p in Z prime and f in Z[x] such that f* in Z_p[x] >irreducible where f* is f with the coefficients reduced mod p. >Any ideas? > Assuming I made no mistake, the only caveat is that I have a > reputation for doing things the hard way, so there may be a much > simpler way of doing this... > Take a prime ideal I of Z[x], and let J={a in Z: a in I}. > Note that J is an ideal of Z. Therefore, J=(r) for some r in > Z. Moreover, J is a ->prime<- ideal of Z, since I is a prime ideal of > Z[x]. So either J=(0), or else J=(p) for some prime p. Abusing > language, let's write J=(p), with p=0 or p a prime. > If I-(p) is empty, then I=(p) and we are done. > Otherwise, of the polynomials in I-(p) of minimal positive degree, let > f(x) be one with the smallest positive leading coefficient. > Suppose that g(x) is another polynomial in I-(p) of minimal positive > degree and with the same leading coefficient as f(x). Let f*(x) and > g*(x) be the polynomials in Z/pZ[x] obtained by reducing the > coefficients of f and g modulo p (if p=0, that just means doing > nothing). > Since f(x)-g(x) must be a polynomial in (p), and therefore f*(x)-g*(x)=0, so > f*(x)=g*(x). Therefore, the ideals (f(x),p) and (g(x),p) are equal in > Z[x]. > We will prove that I=(f(x),p); the above discussion shows that > although there may be many choices for f(x), the choice is immaterial. > Assume first that p>0: > Since there is some coefficient of f(x) which is not a multiple of p, > coefficient to obtain a polynomial of smaller degree; so writing > f(x) = a_n x^n + ... + a_1 x + a_0, > with a_n not divisible by p, we may find r and s integers such that > ra_n + sp = 1, and so taking r f(x)+sp x^n we obtain a monic > polynomial of degree equal to f(x). Thus we conclude that f(x) is > monic. > Note also that f(x) is such that f*(x) is irreducible in Z/pZ[x]. For > otherwise, let g(x) and h(x) be nonconstant polynomials, with leading > coefficient not divisible by p, such that g*(x)h*(x) = f*(x). Again, > we may assume that g(x) and h(x) are monic, and so we have that > g(x)h(x) = f(x) + p*m(x) > for some polynomial m(x) of degree strictly smaller than f(x). Since > this lies in I, either g(x) or h(x) lie in I, contradicting the > minimality of deg(f). This shows that f*(x) is irreducible > If I-(f(x),p) is empty, then we are done. Otherwise, let h(x) be a > polynomial of minimal degree and smallest positive leading coefficient > in I-(f(x),p). Note that deg(h) is no smaller than deg(f). > Write f(x) = x^n + ... + a_1 x + a_0 > h(x) = x^m + ... + b_1 x + b_0. > Then x^{m-n}f(x) - h(x) lies in I, and is of strictly smaller degree > than h(x). So it must lie in (f(x),p). But that means that h(x) itself > lies in (f(x),p), contradicting the choice of h(x). So I=(f(x),p) and > we are done. > Now assume that p=0; this time we have that f(x) is uniquely > determined as the polynomial of smallest degree (I assume the zero > polynomial has no degree) in I with smallest positive leading > coefficient, though we do not know if f(x) is monic. We also know that > f(x) is irreducible in Z[x], because if g(x)h(x)=f(x), then either > g(x) or h(x) would like in I; and therefore (f) is a prime ideal since > Z[x] is a UFD. > The claim is that I=(f). If not, let g(x) be a polynomial in I-(f) of > minimal degree; it must be of degree at least equal to deg(f). > Write f(x) = a_n x^n + ... + a_0 > g(x) = b_m x^m + ... + b_0. > Note that m>=n. Consider h(x)=b_m x^{m-n}f(x) - a_n g(x). This element > lies in I; if it were in (f), then a_n g(x) would be in (f), and since > (f) is a prime either, either a_n lies in (f), which is impossible, or > else g(x) lies in (f), which is also impossible. So h(x) is in I-(f), > and has degree strictly smaller than g(x), which contradicts the > choice of g(x). Therefore, I=(f), f irreducible, as needed. > Of course, you should also prove that these two types of ideals do > indeed yield prime ideals... > It's not denial. I'm just very selective about > what I accept as reality. > --- Calvin (Calvin and Hobbes) > Arturo Magidin > magidin@math.berkeley.edu === Subject: Re: prime ideals in Z[x] > Show every prime ideal of Z[X] has one of the following two forms: > 1. with f in Z[x] irreducible > 2. with p in Z prime and f in Z[x] such that f* in Z_p[x] > irreducible where f* is f with the coefficients reduced mod p. It is very easy. Suppose that P is a prime ideal of Z[x]. 1. Primes P with P / Z = (0) biject with primes in Q[x], 2. Primes P with P / Z = (p) biject with primes in F[x], F = Z/(p), by basic properties of fraction and residue rings, resp. So Arturo's hunch is correct -- there is a much simpler way, which I mentioned to him in an earlier post (deja vu hunch?) The above technique of reducing complex problems to simpler problems in homomorphic images is a fundamental algebraic problem solving method. One should be sure to master it before proceeding to hairier problems. For another simple prototypical example consider proof of Gauss' Lemma: -Bill Dubuque : : Assuming I made no mistake, the only caveat is that I have a : reputation for doing things the hard way, so there may be a much : simpler way of doing this... : : Take a prime ideal I of Z[x], and let J={a in Z: a in I}. : : Note that J is an ideal of Z. Therefore, J=(r) for some r in : Z. Moreover, J is a ->prime<- ideal of Z, since I is a prime ideal of : Z[x]. So either J=(0), or else J=(p) for some prime p. Abusing : language, let's write J=(p), with p=0 or p a prime. : : If I-(p) is empty, then I=(p) and we are done. : : Otherwise, of the polynomials in I-(p) of minimal positive degree, let : f(x) be one with the smallest positive leading coefficient. : : Suppose that g(x) is another polynomial in I-(p) of minimal positive : degree and with the same leading coefficient as f(x). Let f*(x) and : g*(x) be the polynomials in Z/pZ[x] obtained by reducing the : coefficients of f and g modulo p (if p=0, that just means doing : nothing). : : Since f(x)-g(x) must be a polynomial in (p), and therefore f*(x)-g*(x)=0, so : f*(x)=g*(x). Therefore, the ideals (f(x),p) and (g(x),p) are equal in : Z[x]. : : We will prove that I=(f(x),p); the above discussion shows that : although there may be many choices for f(x), the choice is immaterial. : : Assume first that p>0: : : Since there is some coefficient of f(x) which is not a multiple of p, : coefficient to obtain a polynomial of smaller degree; so writing : : f(x) = a_n x^n + ... + a_1 x + a_0, : : with a_n not divisible by p, we may find r and s integers such that : ra_n + sp = 1, and so taking r f(x)+sp x^n we obtain a monic : polynomial of degree equal to f(x). Thus we conclude that f(x) is : monic. : : Note also that f(x) is such that f*(x) is irreducible in Z/pZ[x]. For : otherwise, let g(x) and h(x) be nonconstant polynomials, with leading : coefficient not divisible by p, such that g*(x)h*(x) = f*(x). Again, : we may assume that g(x) and h(x) are monic, and so we have that : : g(x)h(x) = f(x) + p*m(x) : : for some polynomial m(x) of degree strictly smaller than f(x). Since : this lies in I, either g(x) or h(x) lie in I, contradicting the : minimality of deg(f). This shows that f*(x) is irreducible : : : If I-(f(x),p) is empty, then we are done. Otherwise, let h(x) be a : polynomial of minimal degree and smallest positive leading coefficient : in I-(f(x),p). Note that deg(h) is no smaller than deg(f). : : Write f(x) = x^n + ... + a_1 x + a_0 : h(x) = x^m + ... + b_1 x + b_0. : : Then x^{m-n}f(x) - h(x) lies in I, and is of strictly smaller degree : than h(x). So it must lie in (f(x),p). But that means that h(x) itself : lies in (f(x),p), contradicting the choice of h(x). So I=(f(x),p) and : we are done. : : Now assume that p=0; this time we have that f(x) is uniquely : determined as the polynomial of smallest degree (I assume the zero : polynomial has no degree) in I with smallest positive leading : coefficient, though we do not know if f(x) is monic. We also know that : f(x) is irreducible in Z[x], because if g(x)h(x)=f(x), then either : g(x) or h(x) would like in I; and therefore (f) is a prime ideal since : Z[x] is a UFD. : : The claim is that I=(f). If not, let g(x) be a polynomial in I-(f) of : minimal degree; it must be of degree at least equal to deg(f). : : Write f(x) = a_n x^n + ... + a_0 : g(x) = b_m x^m + ... + b_0. : : Note that m>=n. Consider h(x)=b_m x^{m-n}f(x) - a_n g(x). This element : lies in I; if it were in (f), then a_n g(x) would be in (f), and since : (f) is a prime either, either a_n lies in (f), which is impossible, or : else g(x) lies in (f), which is also impossible. So h(x) is in I-(f), : and has degree strictly smaller than g(x), which contradicts the : choice of g(x). Therefore, I=(f), f irreducible, as needed. : : Of course, you should also prove that these two types of ideals do : indeed yield prime ideals... === Subject: Prime Numbers Is this the right place to answer something about prime number generating function? THanx === Subject: Re: Prime Numbers >Is this the right place to answer something about prime number generating >function? Sure, why not? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Prime Numbers I think I have found a very efficient algoritmh for generating prime numbers. I'd like to understand which is its complexity but I'm stuck. Ok... first of all... I have a program that works whit a main cicle like this: void main(){ for(int x = 0; x < MAX; x++){ } } where MAX is the number of iterations to be done... (prime numbers to be generated) into the cicle I have to do some comparisons... these comparisos are much smaller in number than the main cicle number of iterations: very low grow rate: for instance if I'm computing the 3000th prime number this is the snapshot of the computation: void main(){ for(int x = 0; x < MAX; x++){ .... for(int y = 0; y < NUMOFCOMPARISONS; y++){ .... } ... } } I've just to do 5 comparisons to obtain it, the 3001th prime number is obtained directly from the previous (the numbers between them don't take part in the computation). as the number of primes to be generated grows also the number of comparisons grow but its rate is really small... as I said at 1000th iteration 10 comparison, at 20000th iteration about 45 comparisons per cicle... at 30000th about 50 ... at which is the complexity? Thnx! === Subject: Re: Prime Numbers 2.6 _ Oh... I forgot... I think the complexity can be x * / x where x is the number of prime to generate. 2.6 _ Is it a likely complexity if it is related to the algorith I show and if / x is the grow rate of the comparisons done the inner cicle? === Subject: Re: Prime Numbers > I think I have found a very efficient algoritmh for generating prime > numbers. > I'd like to understand which is its complexity but I'm stuck. Could you please provide the complete algorithm? I'm afraid we cannot calculate the complexity if we don't have the actual comparisons you are making. Two for loops are not enough to tell, it depends on what's inside these loops. Sam -- Fear is the path to the dark side. Fear leads to anger, anger leads to hatred, hatred leads to suffering. I sense much fear in you. === Subject: Re: Prime Numbers > I think I have found a very efficient algoritmh for generating prime > numbers. > I'd like to understand which is its complexity but I'm stuck. > Ok... first of all... I have a program that works whit a main cicle > like this: > void main(){ > for(int x = 0; x < MAX; x++){ > } > where MAX is the number of iterations to be done... (prime numbers to be > generated) > into the cicle I have to do some comparisons... > these comparisos are much smaller in number than the main cicle number of > iterations: very low grow rate: > for instance if I'm computing the 3000th prime number this is the snapshot > of the computation: > void main(){ > for(int x = 0; x < MAX; x++){ > .... > for(int y = 0; y < NUMOFCOMPARISONS; y++){ > .... > } > ... > } > I've just to do 5 comparisons to obtain it, the 3001th prime number is > obtained directly from the previous (the numbers between them don't take > part in the computation). > as the number of primes to be generated grows also the number of comparisons > grow but its rate is really small... > as I said at 1000th iteration 10 comparison, at 20000th iteration about 45 > comparisons per cicle... at 30000th about 50 > ... at > which is the complexity? It depends if those comparisons are growing proportionally to the iteration, 1, 10 2, 45 3, 50 ?? Assuming the inner loop is linear, the complexity is O(X^2) Because they are so small its effectively O(X) though. Which is basically what your are describing, the millionth prime takes 1000 times longer to calculate than the 1000th prime, plus a small factor. If NUMOFCOMPARISONS is linear to x then > for(int y = 0; y < NUMOFCOMPARISONS; y++){ is equivalent to > for(int y = 0; y < x; y++){ which is a near standard nested loop with MAX^2 / 2 iterations. so its O(X^2) but with a very small constant. i.e. at moderate large primes: 100,000,000 the NUMOFCOMPARISONS will be say 1,000,000 and the square nature will kick in. Herc === Subject: Probability of elements from one set will beat elements from another I have a few sets / distributions of values, and sample sets of these. Call the real sets A,B,C,D and the sample sets a,b,c,d (I do not know if there is a convension in notation). I'm now trying to determine the probability that a random single value taking from the distribution A will be higher than samples taken from B, C and D, using only my sample sets. Can this be done without knowledge of what type of distribution A,B,C and D have? (How?) Even if so, can it be done better with assumptions about the type of distributions? (How?) / Freed === Subject: Re: Problem with Algebraic Integers: Detailed Exposition |What's important here for readers is the *assertion* that the w's are |functions. | |Here there is the assertion that they are functions of f and m. First, what you mean by saying they are functions is usually described by saying they are _non-constant_ functions. w(m,f)=5 is a function of m and f; it just doesn't vary with m and f. Second, what's important here is not the claim that they vary with m; the claim that they _do not_ vary with m is what bears the burden of proof. |Remember w_1 w_2 w_3 = f. | |Also remember that when f=3, w_1, w_2, and w_3 are provably |*constants*, and in fact in that case w_1 = w_2 = 1, and w_3 = 3. | | P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 - | | (-1+mf^2 )x u^2 + u^3 = | | (b_1/w_1 x + u)(b_2/w_2 x + u)(b_3/w_3 x + u). | | |So why does the poster assert that the w's are functions of m? Doesn't matter! Why does the poster assert that w's for different values of m have to be the same? Desire for glory perhaps? Keith Ramsay === Subject: Re: Problem with Algebraic Integers: Detailed Exposition >A mathematical proof begins with a truth, and proceeds by logical steps to a conclusion which then must be true. That's important which is why I keep emphasizing it. I've pulled a detailed exposition of a short argument that quickly shows a problem with algebraic integers. It starts after the reference. Now here's a math proof. Those who doubt that fact can believe it's a claim of proof, but it's verified to be a proof by tracing the argument out. In this case, I begin with an expression. The expression exists, so that is the truth from which you start. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). That is, I have the identity which defines P(m) in terms of various symbols, and it's all in the ring of algebraic integers, which means that the symbols can only represent numbers that are algebraic integers. Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) > James, You sometimes get into an argument as to whether something is a > function of something else. It might help if you made it clear at the > beginning that P is a function of f, m, u and x. And that the b_i and > w_i are functions of f and m. You might then be able to show, > subsequently, that something that looked at though it depended on > something else, in fact does not. That's not a useful view and it can lead people to false conclusions, > like the ones this poster made. See below... > > For instance: Let f, m, u and x be algebraic integers. Consider the function: P(f,m,u,x)=f^2((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3(-1+mf^2)x u^2 + u^3 f) Since the set of algebraic integers is an integral domain, P(f,m,u,x) > is necessarily an algebraic integer. Aside: Is P(f,m,u,x)/f^2 necessarily an algebraic integer? It is possible to find functions (of f and m) b_1, b_2, b_3, w_1, w_2, > and w_3 such that P(f,m,u,x)/f^2 = > (b_1(f,m) x + u w_1(f,m))* > (b_2(f,m) x + u w_2(f,m))* > (b_3(f,m) x + u w_3(f,m)) Clearly these functions are not uniquely determined. What's important here for readers is the *assertion* that the w's are > functions. Here there is the assertion that they are functions of f and m. If we define (as I did earlier): > P(f,m,u,x)=f^2((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3(-1+mf^2)x u^2 + u^3 f) we can give values to f,m and x and get a cubic polynomial in u. We > can then find the zeros of that polynomial. Specifically, the zeros of P(4,9,u,1) are (roughly) -19.02218070771525900968938 > 23.33415993093670440580763 > 102.93802077677855460388174 the zeros os P(4,10,u,1) are (roughly) -21.15053567454956316464579 > 25.94497362387374822705549 > 114.45556205067581493759030 > So? > Do you consider the b_i and w_i to be functions of m? > The b's can be considered to be functions of m, but the w's cannot > reasonably be considered to be functions or m, or even dependent on > it. James, I think I must be missing something really obvious here. The coefficient of x u^2 in P(f,m,u,x)/f^2 is - 3(-1+mf^2) Is that right? So, b1 w2 w3 + b2 w3 w1 + b3 w1 w2 = - 3(-1+mf^2) Is that right? So, b1 w2 w3 + b2 w3 w1 + b3 w1 w2 is a function of m. Is that right? BUT w1, w2, and w3 are not functions of m? Is that what you are saying? I am sorry if I am making some stupid mistake here. > In fact, assertions otherwise just don't make sense. Consider the > numbers you used above for your second example, which are f=4, m=10, > u=u, x=1, with P(m). > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f), so > > P(10) = 16((1000(256) - 3(100)(16) + 30)1^3 - > 3(-1+10(16) )(1)u^2 + 4u^3), so > P(10) = 4019680 - 7632 u^2 + 64 u^3. > Now why would you ever believe that 16 divides off from that > expression as a function of m? > Why don't you do the calculation for P(9) yourself and ask yourself > why you thought the approximations you give above are significant. > What I'll suggest to you is that it's not about mathematical logic. > James Harris Math Fan === Subject: Re: Problem with Algebraic Integers: Detailed Exposition >This does two things which James does not seem to like. >1) It makes the formulas messier looking. >2) It forces him to clearly define what he is talking about. >I made similar suggestions to him a month or two ago and was informed >that he had no interest in issues of style. After several posts back >and forth it became clear that he doesn't understand the difference >between style and clarity, nor does he seem to appreciate that clarity >often comes at the expense of having simple looking expressions. Notice the insulting tone, and I want to emphasize to the newsgroups that these posters have a mission, which is to *convince* others that there isn't a problem in mathematics. Did I distort the facts here? I did state an oppinion, but I have yet to see a reason to change it. Rather than get to the truth, they continually send up information meant to convince other readers, and often engage in personal attacks. It probably IS personal to them as they work to protect math society by trying to keep the world from knowing about this strange, esoteric error with algebraic integers. Remember above you saw an *assertion* that the w's are functions of m, that I destroyed simply by letting f=3. Consider the following: If I define g as: g(f,m) = (f-3)*m + 2, then g is clearly a function of f and m. However, g(3,m) = (3-3)*m + 2, so g(3,m) = 2, which is NOT a function of m. Does the behavior of g when f=3 accurately reflect the behavior of g in general? What you are saying here is that it does. Based on this example, I have to disagree. If you want to show that the w's are *not* functions of m, you will have to do so in general, not with a special case. >The fact that this may be part of why his papers don't get accepted >doesn't seem to be something he understands. These posters are working a campaign to convince readers that my work isn't important, and there are errors, when in actuality I keep catching them in errors or making assertions that are false. And you keep refusing to address the errors that are pointed out, and have at least once changed your paper without notifying us in sci.math that you have changed your paper. As a result, we at times find ourselves talking about completely different things. Mathematicians have your trust already, remember? James Harris I don't want your trust, I want you to address the math. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Problem with Algebraic Integers: Detailed Exposition >A mathematical proof begins with a truth, and proceeds by logical steps to a conclusion which then must be true. That's important which is why I keep emphasizing it. I've pulled a detailed exposition of a short argument that quickly shows a problem with algebraic integers. It starts after the reference. Now here's a math proof. Those who doubt that fact can believe it's a claim of proof, but it's verified to be a proof by tracing the argument out. In this case, I begin with an expression. The expression exists, so that is the truth from which you start. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). That is, I have the identity which defines P(m) in terms of various symbols, and it's all in the ring of algebraic integers, which means that the symbols can only represent numbers that are algebraic integers. Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) > James, You sometimes get into an argument as to whether something is a > function of something else. It might help if you made it clear at the > beginning that P is a function of f, m, u and x. And that the b_i and > w_i are functions of f and m. You might then be able to show, > subsequently, that something that looked at though it depended on > something else, in fact does not. That's not a useful view and it can lead people to false conclusions, > like the ones this poster made. See below... > > For instance: Let f, m, u and x be algebraic integers. Consider the function: P(f,m,u,x)=f^2((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3(-1+mf^2)x u^2 + u^3 f) Since the set of algebraic integers is an integral domain, P(f,m,u,x) > is necessarily an algebraic integer. Aside: Is P(f,m,u,x)/f^2 necessarily an algebraic integer? It is possible to find functions (of f and m) b_1, b_2, b_3, w_1, w_2, > and w_3 such that P(f,m,u,x)/f^2 = > (b_1(f,m) x + u w_1(f,m))* > (b_2(f,m) x + u w_2(f,m))* > (b_3(f,m) x + u w_3(f,m)) Clearly these functions are not uniquely determined. What's important here for readers is the *assertion* that the w's are > functions. Here there is the assertion that they are functions of f and m. If we define (as I did earlier): > P(f,m,u,x)=f^2((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3(-1+mf^2)x u^2 + u^3 f) we can give values to f,m and x and get a cubic polynomial in u. We > can then find the zeros of that polynomial. Specifically, the zeros of P(4,9,u,1) are (roughly) -19.02218070771525900968938 > 23.33415993093670440580763 > 102.93802077677855460388174 the zeros os P(4,10,u,1) are (roughly) -21.15053567454956316464579 > 25.94497362387374822705549 > 114.45556205067581493759030 So? Do you consider the b_i and w_i to be functions of m? The b's can be considered to be functions of m, but the w's cannot > reasonably be considered to be functions or m, or even dependent on > it. > James, > I think I must be missing something really obvious here. > The coefficient of x u^2 in P(f,m,u,x)/f^2 is - 3(-1+mf^2) > Is that right? Yes. > So, b1 w2 w3 + b2 w3 w1 + b3 w1 w2 = - 3(-1+mf^2) > Is that right? Yes. > So, b1 w2 w3 + b2 w3 w1 + b3 w1 w2 is a function of m. > Is that right? Yes. > BUT w1, w2, and w3 are not functions of m? > Is that what you are saying? The b's are functions of m, while the w's provably are not, and in fact, in general you have b_1 f + b_2 f + b_3 = -3(-1+mf^2) as in general w_1 w_2 = 1, w_3 = f. Now you may wish that the w's be functions of m, but following the math shows otherwise. James Harris