mm-2249 === Subject: Re: How to determine whether a point is contained inside an object/polygon? > Try computing the barycentric coordinates. > > The example there talks about barycentric coordinates in the plane, but > the same idea works in R^n, given n+1 points that do not all lie in the > same (n-1)-dimensional hyperplane. I will assume n = 3 in what follows. > Given the vertices at points p, q, r, s in R^3, and given a point x, > there is a unique representation of x in in the form > x = a * p + b * q + c * r + d * s (1) > where a, b, c, d are real numbers such that > a + b + c + d = 1. (2) > The condition (1) gives you three equations in four unknowns, and the > normalizing condition (2) provides the fourth equation. > If the resulting coordinates a, b, c, d are all positive, then x lies > inside the tetrahedron. If any coordinate is negative, then x lies > outside the tetrahedron. If one or more coordinates is zero and the rest > are positive, then x lies on a face of the tetrahedron or coincides with > a vertex. > -- > Dave Seaman > Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. > === Subject: Re: How to determine whether a point is contained inside an object/polygon? > Try computing the barycentric coordinates. > > The example there talks about barycentric coordinates in the plane, but > the same idea works in R^n, given n+1 points that do not all lie in the > same (n-1)-dimensional hyperplane. I will assume n = 3 in what follows. > Given the vertices at points p, q, r, s in R^3, and given a point x, > there is a unique representation of x in in the form > x = a * p + b * q + c * r + d * s (1) > where a, b, c, d are real numbers such that > a + b + c + d = 1. (2) > The condition (1) gives you three equations in four unknowns, and the > normalizing condition (2) provides the fourth equation. > If the resulting coordinates a, b, c, d are all positive, then x lies > inside the tetrahedron. If any coordinate is negative, then x lies > outside the tetrahedron. If one or more coordinates is zero and the rest > are positive, then x lies on a face of the tetrahedron or coincides with > a vertex. > -- > Dave Seaman > Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. > === Subject: Re: there is no such thing as infinity > Yes, we DO live in a FINITE universe. > Therefore, there is no such thing as infinity! > Otherwise we could have no > identity. > ?? > But if what you're trying futilely to prove is true, then > there is no such thing as calculus > I believe in Calculus but not the way it is taught in school. It > should be taught where dx=1/M, the good old fashioned way with > infinitesimals: > If f(x)=x^2, f'(x)=((x+h)^2-x^2)/h=(2xh+h^2)/h=2x+h approximately > equals 2x, > where h=1/M. See! I just revolutionized modern thinking and > mathematics into conforming to common sense and human decency! In your cosmology is the number pi rational or irrational? Same question goes for sqrt{2}. > We live in a finite universe and there is no room for infinity here. > We must blot out infinity from our world as it is causing the > breakdown of morals and standards in our fragile society - in an > infinite universe, anything is possible! Explain this breakdown of morals due to the concept of infinity. Patrick === Subject: Re: JSH: Pattern argument ... > Footnotes: > [1] Typically, newsservers have a finite capacity, so if you make > enough posts in a single newsgroup on any topic whatsoever, the server > may be forced to expunge some older posts. In this crude way, one > could try to suppress posts, but it wouldn't be very effective. Most > news servers have sufficient capacity that it would take at least > hundreds of posts for this to work. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Question for logarithm experts This is an interesting posting. I've never really thought about this before, but do all equations have solutions, (even though it may not be possible for us to find them in terms of simple functions) ? Graphing does not give us the whole story does it? What I am thinking about is that for example y = x^2 does not cross y = - 1, yet we know that x^2 = - 1 has solutions x = + or - i, (the sqrt of minus one). Does the topic you are studying suggest the use of complex numbers? If you replace x with z = a+b i, you could write 7^z = e^((ln7)z) = (e^(aln7)) (e^i(bln7))= (7^a)[cos(bln7) + i sin(bln7) ] = 4a + 4bi (7^a)cos(bln7) = 4a (7^a)sin(bln7) = 4b 16(a^2 +b^2) = 7^2a b = sqrt[((7^a)/4)^2 -a^2 ], but from above we have b = [arccos(4a/7^a)]/ln7 By graphing these like y equal to two different functions of x, I got a = 0.41756456 and b = 0.37824384 and so with z = a+bi, 7^z = 4z does have a solution of sorts. I hope you found this a helpful - Ian Hutcheson === Subject: Re: there is no such thing as infinity >> And does this zoned out doctor claim that all computers will stop at >> the same M? > That is an excellent point. That is why I am looking for collaborators > who will repeat the experiment on different computers. My theory is > this: > Even if different computers were to output different answers for M, we > could always take the average of these numbers and calculate the > variance too. Then we would at least have a good statistical estimate > of the range of M. > Dr. Ben Zona I just ran it on my computer. It went all the way up to 2147483647 and then went to to -2147483648 and started counting up again. I guess we've found our first candidate for M. - Tim Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University === Subject: Equation of Cyclotomic Polynomial Can someone please tell me how to find the equation of the 5th cyclotomic polynomial?? === Subject: Re: My Childhood and Infinity > This whole area of > mathematics has systematically been supressed by modern mathematicians > starting with Cantor. Aha! You and Harris should compare notes. === Subject: Re: JSH: Pattern argument, revisited > > > So I can have > > > > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > > where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x), or > > > > (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 2) and > > where the b's are roots of b^2 - (x - 2)b + 7(x^2 + x), or > > No. x - 2 should be x - 3 and 7 should be 17 in the line above. Typos. > > (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x + 2) > > where the c's are roots of c^2 - xc + 7(x^2 + x). > > No. 7 should be 2 in the line above. More typos. > > The more astute of you should see a pattern. Clearly the constant > factor on the right, which shows up as a coefficient on the left must > equal 2 mod 5 for this particular setup. > > But why? > > Clearly there is an *underlying* equation being multiplied in each > case, which I'll call > > (5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2 > > > No. You yourself see the difficulties with this approach below. There's no other approach Decker. Given the pattern there has to be some relevance of 2 mod 5 with prime multiples of the polynomial. > What's happening is that I'm multiplying that underlying equation by > integers that are 2 mod 5, which allows for *one* factorization if > you're to have integer coefficients. > > Don't believe me, then try to multiply by 13 instead of 7, like 13 > equals 3 mod 5, so it will not work. > Right. No more than your underlying equation will work when > you multiply by 1. This equals 1 mod 5, so in your own words, > it will not work. It seems that you're missing the point Decker. You emphasized the *single* example, whereas I've pointed out the pattern when you consider more. Now then, an enquiring mind should wonder why is 2 mod 5 important? > > The problem is that it's clear that y_1(x) and y_2(x) can't both be > algebraic integer functions. > Right again. Of course there's no reason why they should yield > algebraic integers. It's fairly easy to prove that with your own example (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x) that they'd have to exist, and have to be algebraic integers, if the exists algebraic integer functions A(x), and B(x) such that 7 A(x) B(x) = (5a_1(x) + 7)(5a_2(x) + 7) with the same restriction on the a's. > > Rick For other readers I want to point out that Rick Decker plays the nice guy role, when his behavior here is contemptible, and reflects badly not only upon him, but also upon Hamilton College. His *duty* as a professor is to tell the truth and promote elucidating the truth, not hiding it. like 2 and 17, it's clear that there's more here than his attempts at dismissal indicate. I hereby charge Rick Decker with academic fraud and note that his college is responsible for this rogue professor. I've deliberately involved an official at his college to take away plausible deniability for his school in a phone call I made months ago. His college has been made aware of his behavior. There may also be a civil matter involved at some point in the future. Some of you seem to think this is a game, or that Usenet doesn't matter. However, Usenet is a *public* forum read around the world. There is a burden upon professors to tell the truth, and upon the colleges and universities that promote them either directly or indirectly. Possibly an example needs to be made to convince you all that you have to be accountable for your public words. James Harris === Subject: Re: Limit Of Sum Over Some Rationals > First, a simple looking limit, where I have revealed the (somewhat > unexpected) closed-form at the post's bottom. > --- t > 1 r > limit ----- / ---- , > m -> oo m^2 --- e^r > r > which is, in linear-mode: > limit{m-> oo} (1/m^2) sum{r} r^t /e^r > And the r-sum is over all distinct positive rationals r > which have denominators, in their simplest forms, > which are each <= pi*m. > -- > Okay, the lesson for today: > Start with the finite double-sum identity: > sum{k=1 to n} sum{j=1 to m, GCD(k,j)=1} f(k,j) = > sum{i=1 to min(m,n)} mu(i) (sum{1<=k<=m/i} sum{1<=j<=n/i} f(ki,ji)) > - > Defining f(k,j) as > a(j/m) b(k/m) c(k/j), > and taking limits, after dividing both sides by m^2, we get: > For positive q and s, > --- > 1 > limit ----- / a(den(r)/m) b(num(r)/m) c(r) > m -> oo m^2 --- > r= distinct rationals, > 1<=den(r)<=q*m > 1<=num(r)<=s*m > 6 /q /s > = ---- | | > pi^2 | | a(x) b(y) c(y/x) dy dx > /0 /0 > (limit{m->oo} > (1/m^2) sum{r= distinct rationals,1<=den(r)<=q*m,1<=num(r)<=s*m} > a(den(r)/m) b(num(r)/m) c(r) = > (6/pi^2)* > integral{0 to q}integral{0 to s} a(x) b(y) c(y/x) dy dx) > where a(), b(),c() are such that the integral exists, is defined, and > the identity above is correct...) > ;) > (And num(r) is numerator of reduced r, den(r) is denomiator of > reduced r.) > -- I am probably making a too-broad statement, if it is correct at all, even when I write this more specific case of the previous result: /s 6 | --- | f(x) dx = pi^2 /0 1 --- limit --- m-> oo m^2 / f(num(r)/m) --- (linera-mode: (6/pi^2)*integral{0 to s} f(x) dx = limit{m -> oo} (1/m^2) *sum f(num(r)/m) ), where the sum is over those distinct positive rationals <= s where 1 <= num(r) <= m*s, 1 <= den(r) <= m. Under which restrictions on s and f() is this true? (Perhaps s needs to be positive, and f() needs to be Riemann-integrable in the range, at least.) > So, we might have, if I am right so far, > limit{m-> oo} (1/m^(t+u+2)) sum{r} (den(r))^u (num(r))^t /e^r > = t! q^(2+t+u) (6/pi^2)/(2+t+u), > where the r-sum is over all positive distinct rationals r, > where den(r) <= m*q, q = fixed positive real. > -- > So, in answer to the original question: > limit{m-> oo} (1/m^2) sum{r} r^t /e^r, > where the r-sum is over all distinct positive rationals r > which have denominators, in their simplest forms, > which are each <= pi*m, > is > 3 * t!. (?) > (I find this more interesting than otherwise because of the fact that > 3 is the integer closest to both pi and e, which are each part of the > original limit.) Leroy Quet === Subject: Re: Looking for primes of a particular form. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1A2UwB05274; >I am trying to find primes numbers of the following forms: > p = 6*(10^N)-1, q = 6*(10^N)+1 p is prime when n = {0,1,2,4,5,7,10,13,22,23,28,34,40,61,73,361,490,613,...} q is prime when n = {0,1,2,8,9,15,20,26,38,45,65,112,244,303,393,560,...} beyond this the number are rather large. As the others have said other metods are needed. === Subject: How to compute the minimal distanct between a point and curve in N-dim space In the N-dimensional space, give a data point A and a curve f, how to write the explicit expression for calculating the minimal distance between A and f? Or have to use some nonlinear optimization method to calcualte it? Fred === Subject: Re: Equation of Cyclotomic Polynomial > Can someone please tell me how to find the equation of the 5th > cyclotomic polynomial?? What it is: x^4 + x^3 + x^2 + x + 1. How to find them: van der Waerden, Modern Algebra I, pp.113-5. Ken Pledger. === Subject: simple integration problem hi i am trying to prove: If f(x)>=0 for all x in [a,b], f continuous on [a,b], and Int[f(x),{x, a, b}], then f(x)=0 for all x in [a,b]. here is my proof: Suppose there exists an x_0 in [a,b] such that f(x_0)>0. Since f is continuous we have that there exists a d>0 such that if |x-x_0|0. We immediately have that for any partition P of [a,b] L(P,f)=Sum[m_i dx_i] >= 2md > 0 where m=inf f(x) over all |x-x_0| 0. Similarly, Int[f(x), {x, a, x_0-d}] >= 0 and Int[f(x), {x, x_0+d, b}] >= 0, which implies that Int[f(x), {x, a, b}] > 0, contradicting that Int[f(x), {x, a, b}] = 0. Therefore f(x)=0 for all x in [a,b] Is that decent? === Subject: re:help! h(x) = cos^4x sinx [0, pi] I presume you mean value. To get the average, take 1/pi times the integral from 0 to pi. Make the substitution u=cosx. Then du=-sinxdx. Finally the average is 2/5pi, after integrating between 0 and 1 in u (2 times after splitting the x domain in half). http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: 3pi.com Pi3.com is the ultimate maths enthusiast web address, what about 3pi.com ? Its up for appraisal if anyone wants I'll make a bid for them, usually $xx. Herc -- QUADS give you STRENGTH BICEPS give you POWER PECS give you CONFIDENCE www.theBanner.net === Subject: polynomial degree in general polynomial Hello all, i am trying to prove that deg(fg)=deg(f)+deg(g) for f,g in F[x1,...,xn], where F is a field. I was able to prove that deg(fg)<=deg(f)+deg(g), but the full proof is still alluding me. Can anyone provide some help or hints? The notation is very cumbersome. === Subject: Re: I need a site about number theory . > I want to learn about number theory , If you know a good website about > that please let me know. > > Not a bad suggestion. A liitle more focussed is > http://www.numbertheory.org/ntw/web.html =-=-=-=- Have you evere gone to www.themathpages.com ?? It is a site where you learn precalculus,trigonometry,arithmetic .The texts there are so simple and learning there is so good. I need a site such as mathpages.com but about number theory and I === Subject: Re: Four Color Theorem > Robinson, Sanders, Seymour, & Thomas in their proof of the FCT, state I think you mean Robertson. > It has been known since 1913 that every minimal counterexample to the > Four Color Theorem is an internally 6-connected triangulation. > Kainen, 1986. pp 59, > If G is minimal [ie a minimal counterexample to the FCT], then G is > 5-connected. This isn't necessarily say that it is 5-connected and not 6-connected. If something is 6-connected, then it is connected, and 2-connected, 3-connected, and so on up to 5-connected, so there's no immediate conflict between these two. > If G is minimal [ie a minimal counterexample to the FCT], then G is > maximal planar [ie a triangulation]. > Is there a conflict between the two descriptions of an MCE or does > internally 6-conncted mean the same as 5-connected? What does internally connected mean? If you are reading the paper, it should define it there. J === Subject: Re: JSH: Pattern argument, revisited > > > So I can have > > > > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > > where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x), or > > > > (5b_1(x) + 17)(5b_2(x) + 17) = 17(25x^2 + 30x + 2) and > > where the b's are roots of b^2 - (x - 2)b + 7(x^2 + x), or > > No. x - 2 should be x - 3 and 7 should be 17 in the line above. > Typos. > > (5c_1(x) + 2)(5c_2(x) + 2) = 2(25x^2 + 30x + 2) > > where the c's are roots of c^2 - xc + 7(x^2 + x). > > No. 7 should be 2 in the line above. > More typos. > > The more astute of you should see a pattern. Clearly the constant > factor on the right, which shows up as a coefficient on the left must > equal 2 mod 5 for this particular setup. > > But why? > > Clearly there is an *underlying* equation being multiplied in each > case, which I'll call > > (5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2 > > > No. You yourself see the difficulties with this approach below. > There's no other approach Decker. Given the pattern there has to be > some relevance of 2 mod 5 with prime multiples of the polynomial. > What's happening is that I'm multiplying that underlying equation by > integers that are 2 mod 5, which allows for *one* factorization if > you're to have integer coefficients. > > Don't believe me, then try to multiply by 13 instead of 7, like 13 > equals 3 mod 5, so it will not work. > > Right. No more than your underlying equation will work when > you multiply by 1. This equals 1 mod 5, so in your own words, > it will not work. > It seems that you're missing the point Decker. > You emphasized the *single* example, whereas I've pointed out the > pattern when you consider more. > Now then, an enquiring mind should wonder why is 2 mod 5 important? > > The problem is that it's clear that y_1(x) and y_2(x) can't both be > algebraic integer functions. > > Right again. Of course there's no reason why they should yield > algebraic integers. > It's fairly easy to prove that with your own example > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x) > that they'd have to exist, and have to be algebraic integers, if the > exists algebraic integer functions A(x), and B(x) such that > 7 A(x) B(x) = (5a_1(x) + 7)(5a_2(x) + 7) > with the same restriction on the a's. > > Rick > For other readers I want to point out that Rick Decker plays the nice > guy role, when his behavior here is contemptible, and reflects badly > not only upon him, but also upon Hamilton College. > His *duty* as a professor is to tell the truth and promote elucidating > the truth, not hiding it. > like 2 and 17, it's clear that there's more here than his attempts at > dismissal indicate. > I hereby charge Rick Decker with academic fraud and note that his > college is responsible for this rogue professor. > I've deliberately involved an official at his college to take away > plausible deniability for his school in a phone call I made months > ago. > His college has been made aware of his behavior. > There may also be a civil matter involved at some point in the future. > Some of you seem to think this is a game, or that Usenet doesn't > matter. > However, Usenet is a *public* forum read around the world. There is a > burden upon professors to tell the truth, and upon the colleges and > universities that promote them either directly or indirectly. > Possibly an example needs to be made to convince you all that you have > to be accountable for your public words. > James Harris James, GET A LIFE. No college is going to fire a professor, especially tenured, for posting on Usenet. What do you expect them to do? You are a jerk who has no life whatsoever so you drag other's down with you. I use math all the time at work, why don't you contact the manager of the place I work for and tell her I'm using incorrect mathematics? I also teach others how to do the calculations I do so I guess I'm lying to them too, huh? If you e-mailed her, she'd likely tell you to get a life because she knows that I am well-educated in math and know what I'm doing. David Moran === Subject: Re: I need a site about number theory . > I want to learn about number theory , If you know a good website about > that please let me know. > > Not a bad suggestion. A liitle more focussed is > http://www.numbertheory.org/ntw/web.html > =-=-=-=- > Have you evere gone to > www.themathpages.com ?? > It is a site where you learn precalculus,trigonometry,arithmetic .The > texts there are so simple and learning there is so good. > I need a site such as mathpages.com but about number theory and I You want to learn number theory, but you haven't got time to search with Google? === Subject: Re: Rationals are Uncountable > I agree that if X' is the set of all rationals in (-1-r,1-r) with > r irrational and |r|<1 then 0 is in X'. > What I am proving is that if x_n is a sequence that > contains every rational in X=(-1,1) then there is no > way to convert x_n to x'_n such that x'_n contains 0. > I just realized that there is a way to do this which is piecewise > linear, and therefore order-preserving. I am assuming that x'_n preserves the order of x_n. > For any pair of intervals (a,b) and (c,d), with a, b, c, and d rational, > there is a linear transformation taking the first interval to the > second: f(x) = c+(d-c)*(x-a)/(b-a) But, r is not rational. > For example, consider r = (sqrt(5)-1)/2 I don't understand how you are choosing the intervals (a,b) and (c,d). Could you please explain? I am also curious what happens when we construct the sequences a'_n and b'_n from x'_n. Is the LUB of a'_n irrational? If x'_n preserves the order of x_n then the members of x'_n that make up a'_n must be the same members of x_n that make up a_n. For example, if: a_1 = x_1 a_2 = x_8 a_3 = x_21 ... then a'_1 = x'_1 a'_2 = x'_8 a'_3 = x'_21 ... Russell - 2 many 2 count === Subject: misunderstanding my fault by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1A4B3m13015; . Divisors of n not divisible by 9 are 54717 - 9333 = 45384, not divisible by 9. Divisors of n not divisible by 169 are 54717 - 43433 = 11284, not divisible by 169. Divisors of n not divisible by 61 are 54717 - 52338 = 2379, which _is_ divisible by 61. I must apologise for my clumsy terseness in using an example with just three primes each dividing once into the sample number, but my claim about symmetry between zero and the LCM of the factors means that there is no reason for sums of factors of 3, 13, and 61 to _not_ be symmetrically positioned between zero and a multiple of LCM3x13x61. But the relevant LCM in the example you give is 3.3.13.13.61, so it will be partition sums and complements of two composites [9, 169] and one prime [61] which must all divide by their appropriate factors for 3.3.13.13.61 to be a perfect number. Since the factor sum, 54717, is 23 x 2379 [2379 = 3.13.61], the sum of factors not dividing by 61 also divides by 61, but this fails for 9 and 169, whose complements are nowhere near their LCM with 61, which is n = 92781. Does this make any sense? Best wishes, Mark G. . === Subject: Unitary operator basics, lebesgue measure Let H = L^2(R,dx). For each a in R (the reals), defined T_a : H ----> H by (T_a f)(x) = f(x-a), where f is in H (the hilbert space) Firstly, why is f(x-a) necessarily in H if f is in H? Why is T_a a unitary operator (unitary operator I believe means that for an operator A, A* = A^{-1}) ? I suppose this might be easy if I could represent T_a as a matrix. But then how would this matrix multiply by f? Could I represent the function f by a matrix(with respect to a basis of L^2(R,dx)?)? What is a basis for L^2(R,dx)? Moshe === Subject: Web Sites for Complexity Theory Are there any good books in this field, or good web sites that provide an introduction or lecture notes with definitions and examples? I am really trying to teach myself this stuff before I take the class next semester. Blake Manner === Subject: Qabala & Physics The Others as The Ecclesia (see also Herman Hesse's Magister Ludi (AKA The Glass Bead Game'). Before I forget to Paul Zielinski - the Cartan spin connection valued in the local Lorentz group Lie algebra o(3,1) is the tensor part of the Levi-Civita connection. The spin connection can take an additional torsion field that is not in Einstein's theory - but it need not have it. The gravity field is really the non-trivial part of the tetrad field in the sense of Rovelli's formalism. The tetrad field is the spin 1 gauge force compensating local field from breaking global translational symmetry, i.e. introducing local tidal curvature. The translations and Lorentz rotations intetwine as seen in the Sagnac and Thomas precession effects which do not depend on locally gauging the Lorentz rotations, but if we do that we get the additional torsion field of Shipov, Akimov et-al in Russia. The spin 2 field is the strain tensor of the tetrads. Einstein's metric guv is quadratic in the tetrads and in quantum theory 1 + 1 = 2 or 0. The 0 is from locally gauging the dilations of the conformal group. There is a second set of tetrads from locally gauging the 4 remaining special conformal transformations that give Tony's conformal gravitons, so now I am beginning to understand intuitively what Tony is talking about at least partly. My intuition is powerful, I mean I have a kind of analog computer simulation in my mind's I and I can tweak it this way and that without having to do the formal symbol manipulations, then later put it into symbols. I think this is what Einstein meant when he said he thought kinesthetically? With regard to the 22 Hebrew letters mentioned by Suares: A useful way of looking at the 22 letters of the Hebrew alphabet is: 3 mothers - quaternion imaginaries, 3 spatial dimensions, 3 QCD colors 7 doubles - octonion imaginaries, 3 spatial dimensions + 4 internal symmetry space dimensions 7 charge-carrying first-generation fermions (electron; red,blue,green down quarks; red,blue,green up quarks) 12 simples - 12 root vectors of the SU(2,2) conformal group, 12 gauge bosons of SU(3)xSU(2)xU(1) When 5 finals are added to the 22, you get the 27 dimensions of the exceptional Jordan algebra J3(O), which is the dimensionality of bosonic string M-theory (which I think is related to the quantum interactions of world-lines in the many-worlds). Its complexified automorphism group is E6. All this (and a lot more) is inherent in my physics model Tony Too bad Carlo Suares is not alive to see this. :- === Subject: Re: Rationals are Uncountable > But, r is not rational. Which is why I have to do it in pieces: For example, consider r = (sqrt(5)-1)/2 x_n x'_n 0 0 (0,1/2) (0,1/3) 1/2 1/3 (1/2,3/4) (1/3,3/8) 3/4 3/8 (3/4,7/8) (3/8,8/21) etc. (-1/2,0) (-3/2,0) -1/2 -3/2 (-3/4,-1/2) (-8/5,-3/2) -3/4 -8/5 (-7/8,-3/4) (-21/13,-8/5) etc. (For some other irrational r, the sequence 1/3,3/8,8/21,... would be replaced with some other increasing sequence of positive rationals that converges to 1-r, and the sequence -3/2,-8/5,-21/13,... would be replaced with some other decreasing sequence of negative rationsl that converges to -1-r.) > I don't understand how you are choosing the intervals (a,b) and (c,d). > Could you please explain? For example, if 0 < x_n < 1/2, the intervals (a,b) and (c,d) are (0,1/2) and (0,1/3); we have x'_n = 2*x_n/3. If -7/8 < x_n < -3/4, the intervals (a,b) and (c,d) are (-7/8,-3/4) and (-21/13,-8/5); we have x'_n = (8*x_n-98)/65. > I am also curious what happens when we construct the sequences > a'_n and b'_n from x'_n. Is the LUB of a'_n irrational? The reason that the LUB of a_n had to be irrational applies just as well to the LUB of a'_n. If the LUB of a_n was (sqrt(5)-1)/2 and the transform is as I described above, the LUB of a'_n ends up as (sqrt(5)+2)/12 . -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: misunderstanding my fault > think that your example should read > Divisors of n not divisible by 9 are > 54717 - 9333 = 45384, not divisible by 9. > Divisors of n not divisible by 169 are > 54717 - 43433 = 11284, not divisible by 169. > Divisors of n not divisible by 61 are > 54717 - 52338 = 2379, which _is_ divisible by 61. If the divisors of n sum to a number divisible by all the prime powers dividing n (hence, divisible by n), the number is said to be multiperfect. You're trying to show there are no odd perfect numbers by showing there are no odd multiperfect numbers. It is widely believed that there are no odd multiperfect numbers, but not proved. It seems safe to say that elementary methods have been exhausted by now, and if anyone is going to prove that there are no odd multiperfects she's going to have to use a lot stuff way beyond anything you've shown us. There's a lot of literature on odd multiperfects. If you're really interested in the topic, I suggest you have a good long look at what's already been done. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Unitary operator basics, lebesgue measure > Let H = L^2(R,dx). For each a in R (the reals), defined T_a : H ----> H by > (T_a f)(x) = f(x-a), where f is in H (the hilbert space) > Firstly, why is f(x-a) necessarily in H if f is in H? Why is T_a a unitary > operator (unitary operator I believe means that for an operator A, A* = > A^{-1}) ? I suppose this might be easy if I could represent T_a as a > matrix. But then how would this matrix multiply by f? Could I represent > the function f by a matrix(with respect to a basis of L^2(R,dx)?)? What is > a basis for L^2(R,dx)? The graph of f(x-a) is just the graph of f shifted a units to the right. That answers the first question. Verify that (T_a)* = T_(-a) and you'll have the answer to the second question. === Subject: confused on substitution hello, been working on a problem too hard and now have started to become unsure as to the correct nature of a substitution: the problem is this, solve the heat diffusion equation, c*u(x,t)_xx = u(x,t)_t [1] where underscores are partial derivitives and c=constant, such that u(x,t) satisfies the condition u(x,t)_x = u(x,t)*F(u(x,t),t) [2] for all x and t where F is an arbitrary function of u and t. first of all is this problem well defined, and does it have a solution (i think it might not)? one would assume that from [2] we can find u_xx and substitute this into the LHS of [1]. however do we also have to solve for u from [2] and substitute that into the RHS of [1] as well to obtain a function that satisfies both [1] and [2]? or is the first substitution enough to guarantee this. i assume that we probably need to substitute for u from [2] in both sides of [1] in order to be consistent and keep the equality....but my brain is fried and i have a mental block on this issue at present and it doesnt seem obvious to me at the moment. If we do substitute for both sides of [1] using [2] then we get... c*u_x*(F+u*F_u)^2 = u_xt - u*F_t [3] So to reiterate more succinctly will [3] provide me a solution that satisfies both [1] and [2]. Is there a better way to solve this problem. i have thought about using greens functions for [1] and forcing it to satisfy [2] but i cant seem to construct them. need help with the analysis of this system. any help appreciated. cheers moth === Subject: Re: Unitary operator basics, lebesgue measure > Let H = L^2(R,dx). For each a in R (the reals), defined T_a : H ----> H by > (T_a f)(x) = f(x-a), where f is in H (the hilbert space) > Firstly, why is f(x-a) necessarily in H if f is in H? Why is T_a a unitary > operator (unitary operator I believe means that for an operator A, A* = > A^{-1}) ? I suppose this might be easy if I could represent T_a as a > matrix. But then how would this matrix multiply by f? Could I represent > the function f by a matrix(with respect to a basis of L^2(R,dx)?)? What is > a basis for L^2(R,dx)? > The graph of f(x-a) is just the graph of f shifted a units to the right. > That answers the first question. Verify that (T_a)* = T_(-a) and you'll > have the answer to the second question. How can I think about (T_a)* ?? First I need to represent T_a as a matrix, but what matrix is it? Then I would transpose the matrix and conjugate...(still confused) probably silly for not seeing the second) === Subject: Re: existence of bounded linear functional Ah, of course. I still have one lingering question, though. Given {x_k} a sequence in X and {a_k} a sequence of scalars, define a bounded linear functional on the subspace generated by x_1,...,x_n by f(b_1x_1 + ... + b_nx_n) = b_1f(x_1) + ... + b_nf(x_n) , f(x_k)=a_k. How does the requirement that | b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| guarantee that ||f|| = M? It's obvious that ||f|| <= M, but how do you assert equality? >Let X normed linear space, X' it's dual. a_1,...,a_n some scalars, >x_1,...,x_n some elements of X. How do you show that >| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| >for some M for each n and for scalars b_1,...,b_n implies that there exists >x' in X' satisfying x'(x_k)=a_k and ||x'|| = M. >i don't know how to show this without explicitly constructing the bounded >linear functional on X, and of course i'm stuck on constructing it (hence >the problem in the first place).. >if X is finite dimensional, then each x in X can be written x = b_1 x_1 + >... + b_n x_n on some basis {x_k}..then define x'(x)=b_1 x'(x_1) + ... + b_n >x'(x_n) and defining x'(x_k)=a_k. This is bounded by the given constraint. > Careful: the x_n's were given, and might not be a basis. >But what to do if X is infinite-dimensional??? > Use the Hahn-Banach Theorem. > 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: JSH: Tautological spaces [posted and mailed] >> Oh yeah, I don't think the human being has been born yet that can >> handle complex solutions in a complex topological space. I call a >> person who can handle such a challenge a third generation >> mathematician. >> Today's mathematicians are first generation. >> James Harris > Very eloquently stated. ..... .....I need someone to confirm my findings. > Could you possibly help me? I believe there is much profit if I can > find M. > Dr. Ben Zona Lagest Number = 0 Zero does not exist Zero is not a number M is 0 M does not Exist QED Cheeres Carl === Subject: easy complex analysis problem...... hello........ complex function f(z) = 1 / [2*{x^(1/2)}] int f(z) dz = ?? |z| = 1 ------------------------- i think.....f(z) have isolated singular point on z=0 but i can't use residue theorem to this problem. thus let z = e^i@ (@= theta) dz = i(e^i@) d@ thus int [1/{2*e^(i@/2)}]*ie^i@ d@ 0~2pi = (i/2)*int e^(i@/2) d@ = -2 it's right?? let me check up, please.....thank you teacher.... === Subject: Re: My Childhood and Infinity > When I was a young three and a half year old lad, I used to write > numbers on paper and line them across the room. My goal was to get to > the largest number possible which I now call M. Then my mother told > me that the largest number is infinity. I asked her what things are > infinite. She could not answer me. Then I asked her what do you mean > goes on for ever? How can something go on for ever? She could not > answer me and told me to ask my teacher, as that is her job. > So the next day I went to nursery school and was playing on the swing > and I bumped my nose and started crying. The nursery school teacher > saw me crying and tried to make me feel better. I decided at that > point that it would be a good opportunity to ask her what infinity is. > So I did. But she didn't understand me because I was still crying. > For years, I blocked this out of my memory, this disturbing incident. > And I went to college and eventually got a PhD from the University of > San Moritz, a non-accredited but well-respected university, writing my > thesis on the role of life experience in scientific thought. Then I > differential equations and started to shake. What I was doing was > meaningless! What my teachers told me was a lie! Partial differential > equations presuppose the existence of infinity! And how do I know that > there is an infinity? No one knows in fact. Then I remembered myself > as a child lining up numbers trying to find M, the largest number in > the universe. I decided that I would convert my partial differential > equation paper about difference equations instead. Then my life > changed. I decided that I would pick up the pieces of where I left off > as a child and renew my search for M. I am currently using my computer > to find this number, as I have concluded that this is not a job for a > human being. > I am looking for collaborators for this big project. Anyone who wants > to participate in this revolution, literally, in mathematics and > science should please contact me. I can be found at my brother-in-law > Craig Feinstein's email address. Just do not make it known to him that > I am using it, as he thinks it is not healthy for me to continue this > project, and that I should just do what my doctor says. > Dr. Ben Zona Infinity has been misunderstood as a number or not understood properly in the history of science and mathematics while it is the reality behind this universe. It is this unobservable infinity which appears as the observers' universe which forms the foundation for the observer entangled quantum theory and the observer related relativity theory, the modern theories of physics. S S Shastry === Subject: Ellipse? Hipothesis: Let p and q be two positive real numbers. Let Ox and Oy be two cartesian axis. Let l be the variable line that intersects Ox and Oy in A and B such that AB measures p. Let C be *one* of the two points in l which distance to B is q. Thesis: the locus of C is an ellipse I'm looking for an euclidean proof. Any hint? === Subject: Re: JSH: Pattern argument <8765eg19vc.fsf@phiwumbda.org> Discussion, linux) > ... > > Footnotes: > > [1] Typically, newsservers have a finite capacity, so if you make > > enough posts in a single newsgroup on any topic whatsoever, the server > > may be forced to expunge some older posts. In this crude way, one > > could try to suppress posts, but it wouldn't be very effective. Most > > news servers have sufficient capacity that it would take at least > > hundreds of posts for this to work. Google's not a typical newsserver. I did suggest that one could hide a post on Google by posting in *other* threads, so that the thread containing that post one wants to bury no longer appears on the first page for that group. This is, of course, not at all what James accuses folks of doing. Instead, he claims that a group of evil mathematicians have cunningly concluded that, to hide a thread started by JSH, one should post many, many messages in that thread. Clever, clever. Of course, even my suggestion of hiding posts on Google isn't particularly effective. But I'd wager that, if one could measure how many times a particular post is read via Google[1], my strategy would be a lot more effective than the one James is on about. Footnotes: [1] Even Google can't measure this very well, since if I choose to view the thread, they don't know which other posts in a block of ten that I read. -- Jesse F. Hughes That's the base tautological space where by tautological space I mean a region of truth. -- James S. Harris does philosophy of mathematics. JSH is a renaissance man. === Subject: Hamilton College announces new course A new course, Introduction to Harristotelian Logic, will be presented by new Hamilton College faculty member, James S. Harris. The foundations of the method will first be established by defining several innovative proof techniques: Proof by Assertion Proof by Repetition Proof by Exhaustion Proof by Vituperation Next, the topic of Object Algebras will be introduced, starting with definitions and basic axioms, and leading into some fundamental theorems. Based on the preceding, the new concept of the Tautological Space will be developed, in conjunction with prime number counting using partial difference equations, and certain results will be proved, in particular Harris's Error in Core Mathematics Theorem. If there is time a new proof of Fermat's Last Theorem will be presented. === Subject: Re: JSH: Pattern argument, revisited Discussion, linux) > I hereby charge Rick Decker with academic fraud and note that his > college is responsible for this rogue professor. I have nothing to add. I just want to compliment your consistently inventive use of the word rogue. It makes one wonder whether Decker wears an eyepatch, or perhaps an ammunition belt strapped across his chest. Anyway, good luck with the threatened civil action. I bet there are just dozens of lawyers aching to take this case on a contingency basis. -- Jesse Hughes Besides, discoverers are too proud to kiss butt. Indiana Jones would never kiss some academic's ass to get published, and neither will I. --James Harris === Subject: Re: construct a circle of unit area? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1A6lWv24980; >> how to construct a circle with area = exactly one unit? >> any applications for it outside of probability? > 1) Define an arbitrary circle to have unit area. > 2) Pi is irrational and transcendental. There is no way to >construct a unit area circle by Greek rules (straightedge and >compass, linear and/or quadratic functions). How about Pi = 4/[sqrt(Golden Section)] Ref:http://www.stefanides.gr/piquad.htm http://www.stefanides.gr/quadcirc.htm The impossible of solution by compass and straight edge was base on the use of : e^[iPi]=-1 However the complete form for THETA=Pi is: e^[iPi]=-1+i[0] , so Only the real pert was considered ,and the imaginary part i.e. : e^[iPi]=i[0] , or e^[iPi]= 0 LINDEMAN , HERMITE , ETC, ETC IGNORED THIS SOLUTION Panagiotis Stefanides http://www.stefanides.gr > 3) Construct a unit square. Convert to a unit circle using a >quadratrix curve. There are lots! Alas, no quadratrix curve can be >constructed by Greek rules. >-- >Uncle Al >http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) >Quis custodiet ipsos custodes? The Net! === Subject: Re: construct a circle of unit area? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1A6lWP24984; >>how to construct a circle with area = exactly one unit? >Depends on (a) you mean by construct; and (b) what you mean by >unit. >from that, again through well known methods, you would be able to >construct two points whose distance is the square of sqrt(pi), which >is pi. >However, pi is a transcendental number It would be nice somebody here to reproduce this TRANSCENDANCE ! How about Pi = 4/[sqrt(Golden Section)] Ref:http://www.stefanides.gr/quadrature.htm http://www.stefanides.gr/theo_circle.htm http://www.stefanides.gr/piquad.htm http://www.stefanides.gr/quadcirc.htm The impossible of solution by compass and straight edge was base on the use of : e^[iPi]=-1 However the complete form for THETA=Pi is: e^[iPi]=-1+i[0] , so Only the real pert was considered ,and the imaginary part i.e. : e^[iPi]=i[0] , or e^[iPi]= 0 LINDEMAN , HERMITE , ETC, ETC IGNORED THIS SOLUTION Panagiotis Stefanides http://www.stefanides.gr === Subject: Re: Unitary operator basics, lebesgue measure >> Let H = L^2(R,dx). For each a in R (the reals), defined T_a : H ----> H >> (T_a f)(x) = f(x-a), where f is in H (the hilbert space) >> The graph of f(x-a) is just the graph of f shifted a units to the right. >> That answers the first question. Verify that (T_a)* = T_(-a) and you'll >> have the answer to the second question. >How can I think about (T_a)* ?? First I need to represent T_a as a matrix, >but what matrix is it? Then I would transpose the matrix and >conjugate...(still confused) No, a matrix representation is neither needed nor desirable: it would just obscure matters. (T_a)* is defined by <(T_a)* f, g> = where <,> is the inner product. Write out what <(T_{-a} f, g> = means in terms of integrals and you should see it immediately. 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: How to determine whether a point is contained inside an object/polygon? > Hello all, > I am trying to determine whether a given point is contained inside a > tetrahedron, and I believe I am following the correct method in order > to do so but need it double-checked just in case! > The tetrahedron is specified by the points: > (10,20,5),(5,10,5),(10,10,5) and (10,10,10). > What I did was obtain the inner surface normal for each face and then > determine whether the point lies inside or outside the face. I > discovered that it lies outside three faces but inside one face. I > assume for it to lie inside the tetrahedron it must lie inside all the > faces, which it doesn't in this case, so the point must be outside the > tetrahedron? I have used the vector cross product. Going around the polygon, calculate side x ( point - begin of side ) If the product does not change its sign, then point is inside of the polygon. arto.huttunen at estlab.com === Subject: Re: existence of bounded linear functional >Ah, of course. I still have one lingering question, though. Given {x_k} a >sequence in X and {a_k} a sequence of scalars, define a bounded linear >functional on the subspace generated by x_1,...,x_n by f(b_1x_1 + ... + >b_nx_n) = b_1f(x_1) + ... + b_nf(x_n) , f(x_k)=a_k. >How does the requirement that >| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| >guarantee that ||f|| = M? It's obvious that ||f|| <= M, but how do you >assert equality? Oh. It doesn't, of course, and I don't. If x_1,...,x_n already span X, you're clearly out of luck, and your statement is false. Hahn-Banach gives you x' with ||x'|| <= M. If x_1,...,x_n don't span X (e.g. if X is infinite-dimensional), it also gives you y' <> 0 such that y'(x_j) = 0 for j=1..n. For suitable t, x' + t y' will be what you want. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 >>Let X normed linear space, X' it's dual. a_1,...,a_n some scalars, >>x_1,...,x_n some elements of X. How do you show that >>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| >>for some M for each n and for scalars b_1,...,b_n implies that there >exists >>x' in X' satisfying x'(x_k)=a_k and ||x'|| = M. >>i don't know how to show this without explicitly constructing the bounded >>linear functional on X, and of course i'm stuck on constructing it (hence >>the problem in the first place).. >>if X is finite dimensional, then each x in X can be written x = b_1 x_1 + >>... + b_n x_n on some basis {x_k}..then define x'(x)=b_1 x'(x_1) + ... + >b_n >>x'(x_n) and defining x'(x_k)=a_k. This is bounded by the given >constraint. >> Careful: the x_n's were given, and might not be a basis. >>But what to do if X is infinite-dimensional??? >> Use the Hahn-Banach Theorem. === Subject: Re: :: towards a constructive education :: (news server friendly) > Hi Mitchie, still not back from kiddie league? I'm still waiting for > an answer more worthy of yourself. > Mostly, it isn't worth the effort. Oh. > I spent the last year dealing with someone exhibiting the same crude behavior when I was > trying to talk about logic and set theory. Why don't you give me some sense of what you > know about Boolean-valued forcing, partition characteristics of large cardinals, or even > descriptive set theory? Because that is not what we are discussing. You posted some crap quotation of Jeremy Bentham. I gave you my answer to it, quoted some became the one showing crude behaviour; hey that's my part in this dialogue! As far as I am concerned, this is about the objective existence of right and wrong. Is there anything else you want to say about that? > I do not have Galathaea's wide background. So, it is unlikely I would be for now it looks as if Galathaea is all width and no depth, all references and no meat. Like someone out of a Borgean nightmare. We'll see. > able to mount a > decent argument against someone as esteemed as yourself. > My suspicion, however, is that you do not even understand the numbers they keep score with > in my league. But, then maybe you do. No, I guess you're right. I'm also happy to have provided you with an opportunity to state what a big shot you are, you must sorely need everyone you get. However on a more pedestrian level the score is for now against you. The ball is on your side of the court. === Subject: Re: Rationals are Uncountable > The reason that the LUB of a_n had to be irrational applies just as well > to the LUB of a'_n. If the LUB of a_n was (sqrt(5)-1)/2 and the > transform is as I described above, the LUB of a'_n ends up as > (sqrt(5)+2)/12 . I will have to think about this. For example, I wonder if your method actually gets all of the rationals close to r. For now, I will agree that your x'_n does maintain the order of x_n and it does contain 0. Russell - 2 many 2 count === Subject: Graph Theory: Number of maximal Matchings let G = (V, E) be a bipartite graph with matching number m(G) = n. Assume that V is the disjoint union of the subset U1 and U2 both with cardinality |U1| = |U2| = n. Edges connect each vertex from U1 with one or more vertices of U2. Is there any result about the number of maximal matchings in G? === Subject: Re: :: towards a constructive education :: (news server friendly) > Hi Mitchie, still not back from kiddie league? I'm still waiting for > an answer more worthy of yourself. > > Mostly, it isn't worth the effort. > Oh. > I spent the last year dealing with someone exhibiting the same crude behavior when I was > trying to talk about logic and set theory. Why don't you give me some sense of what you > know about Boolean-valued forcing, partition characteristics of large cardinals, or even > descriptive set theory? > Because that is not what we are discussing. You posted some crap > quotation of Jeremy Bentham. I gave you my answer to it, quoted some > became the one showing crude behaviour; hey that's my part in this > dialogue! As far as I am concerned, this is about the objective > existence of right and wrong. Is there anything else you want to say > about that? No. But I thank you for this insight. A while back I would have been much more patient about someone choosing to be argumentative toward a post from the outset. I became angered from the moment I opened your first response and set myself up for this. As I said, it is not worth the effort. mitch === Subject: Re: My Childhood and Infinity >I ask you what is 2 times infinity? Infinity, you might say, but that >doesn't make any sense and you know it. I am claiming that there is no >evidence for the Peano axioms that say that every number has a >successor! It seems clear to me that no one has come close to proving >this. On the other hand, there is abundant evidence for the number M >existing, i.e., Asimov's number. And 2M cannot exist by definition. Mr. Feinstein, you seem to contradict the proof of Mr. Feinstein on the impossibility of proving the Collatz 3n+1 -conjecture. It relied on the necessity of calculating infinitely many values of a function in order to prove the conjecture. Surely if M exists then the Collatz conjecture can be proven/disproven through exhaustion alone. -- I'm not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: A little help is needed In sci.math, Arturo Magidin : >>f: Q x Q -> Q >>f(r1, r2) = (a+c)/(b+d) where r1 = a/b, r2 = c/d, >> a,b,c,d integer, b > 0, d > 0, >> gcd(a,b)=1, gcd(c,d)=1 [*] > [.snip.] >>[*] probably should also state that 0 = 0/1 is the unique >> representation of 0 as well. > That follows from the definition: gcd(0,b) = |b| for all integers > b. For gcd(x,y) is the nonnegative integer d that satisfies: > (1) d|x and d|y; > and (2) if c is an integer such that c|x and x|y, then x|d. > Clearly, |b| divides b and divides 0. And if x is any integer that > divides both b and 0, then it divides |b|. Thus, gcd(0,b)=|b|. So if > a=0, for gcd(a,b) to equal 1, with b>0, you must have b=1. OK, then maybe we *don't* have to specify 0 = 0/1 as the unique representation for this function. As long as it works, and from the looks of it, it'll work over the domain Q^2; there's no horribly obvious ambiguity, if done right. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Rationals are Uncountable >> Not needed to disprove your claim. One only needs -1-r < 0 < 1-r, which >> follows immediately from |r| < 1. Though it is qite possible that you, >> Russell, cannot follow that simple argument without help. > I agree that if X' is the set of all rationals in (-1-r,1-r) with > r irrational and |r|<1 then 0 is in X'. > What I am proving is that if x_n is a sequence that > contains every rational in X=(-1,1) then there is no > way to convert x_n to x'_n such that x'_n contains 0. ? Define convert. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: e is transcendental (was: classes of transcendental numbers ? >> hey do you guys have a copy of the proof of e is transcendental? ..id really >> appreciate if somebody could help me. ive been searching and found nothing. >Spivack, CALCULUS, 2nd edition, Chapter 21. Are you sure that the proof there is for e being transcendental and not just irrational? I know that the proof of the irrationality of e is simple enough for a Calculus book, but that the transcendence is quite a bit more complicated. Rob Johnson take out the trash before replying === Subject: Re: confused on substitution >solve the heat diffusion equation, >c*u(x,t)_xx = u(x,t)_t [1] >where underscores are partial derivitives and c=constant, such that u(x,t) >satisfies the condition >u(x,t)_x = u(x,t)*F(u(x,t),t) [2] >for all x and t where F is an arbitrary function of u and t. first of all >is this problem well defined One of the requirements is that the problem must have a unique solution given the boundary conditions. Consider: Let F(u,t) = 1. Then u(x,t) = C exp(x) exp(ct) is a solution to [1] for all C in R and fulfills [2]. Not good. You need more boundary conditions. -- I'm not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: e is transcendental (was: classes of transcendental numbers ? > hey do you guys have a copy of the proof of e is transcendental? ..id > really appreciate if somebody could help me. ive been searching and > found nothing. >>Spivack, CALCULUS, 2nd edition, Chapter 21. > Are you sure that the proof there is for e being transcendental and not > just irrational? I know that the proof of the irrationality of e is > simple enough for a Calculus book, but that the transcendence is quite a > bit more complicated. And Spivak actually does it! -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Hamilton College announces new course >A new course, Introduction to Harristotelian Logic, will be presented by > new Hamilton College faculty member, James S. Harris. >The foundations of the method will first be established by defining >several innovative proof techniques: >Proof by Assertion >Proof by Repetition >Proof by Exhaustion >Proof by Vituperation There are also other techniques he has successfully employed: Proof by Contradiction 1. State your theorem. 2. Wait for someone to disagree. 3. Contradict them. Fool Proof 1. State your theorem. 2. Invite colleagues to comment. 3. If they don't agree, exclaim loudly: You Fools! Proof by Stubborness Even when presented with a counterexample, the author insists his proof is correct and demands his readers find the logical flaw for him. Proof by Appeal to Infants Even a child can see it's true. Proof by Elimination of Intermediate Steps Any intermediate steps in the reasoning are removed to obscure holes in the proof. Proof by Conspiracy Theory You are right but the establishment is trying to silence you from publishing your result and disturbing the status quo. Proof by Leap-of-logic Make general statements that have little to do with the proposition, then make a wild jump to completely unrelated conclusions. >Based on the preceding, the new concept of the Tautological Space will >be developed, in conjunction with prime number counting using partial >difference equations, and certain results will be proved, in particular >Harris's Error in Core Mathematics Theorem. Is that 'Harris's Error in Core Mathematics Theorem' or 'Harris's Error in Core Mathematics Theorem' or 'Harris's Error in Core Mathematics Theorem.'? >If there is time a new proof of Fermat's Last Theorem will be presented. Good thing too it was his last, the Internet could not handle any more kooks. -- I'm not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: one quetion Today, I read the proof of Every real # is a limit point of rational #s and I understand that proof. Now, I am curious about correctness of Every real # is a limit point of irrational #s. Is it true ? I think this is true. Because, For any real # p, Let a_n=(p -1/n,p +1/n). By density of irrational #, a_n contains irrational #s. Let b_n=irrational # in a_n, then b_n is a sequence of irrational #s and b_n converges to p Is my thinking correct ? === Subject: Re: Repunits prime factors: a result. I suppose that anyone discovered this result (the proof is so simply). It's curious!: it's not known about prime repunits and we have that result!. Well, Xan. > escribi.97: >> Is the result [point 7] in somehow interesting? Just say: I will know >> if the found result is original at least. > 7. Theorem: For all b>=1 such that 2,3,5 do not divide b, then there > exist n>=0 such that b divides R(n) > Yes, it is known, and the exception of 3 is unnecessary. Actually, if > gcd(b, 10) = 1 exist n such taht b divides R(n). > Dem.: > gcd(9b, 10) = 1 ==> 10^(phi(9b)) = 1 (mod 9b) (Th. Euler-Fermat) ==> > k*b = R(phi(9b)) > If gcd(b, 3) = 1, is enough with n = phi(b). > By example, it is a problem in > Divulgaciones Matem.87ticas (Universidad del Zulia, Venezuela), Vol. 7, > n.bc 1, (p. 103) > Ignacio Larrosa Ca.96estro > A Coru.96a (Espa.96a) > ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: existence of bounded linear functional Duh, the norm is continuous. I'm officially an idiot. > Hmmm...another sticking point (apart from this norm stuff). Suppose > | b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| > for some M for each n and for scalars b_1,...,b_n. We want to construct a > functional x' in X' such that x'(x_j)=a_j for ALL j = 1,2,.... > Consider the subspace spanned by ALL the x_k, and call it S. Then any x in S > can be written > x = b_1x_1 + b_2x_2 + .... > Define f(x) = b_1a_1 + b_2a_2 + .... > Then f(x_j)=a_j for all j = 1,2... > But is f bounded? The given inequality only works for finite n, and > |f(x)| = |b_1a_1 + ... | = |lim b_1a_1 + ... + b_na_n| = lim |b_1a_1 + ... > + b_na_n| < = lim M ||b_1x_1 + ... b_nx_n||, but > lim ||b_1x_1 + ... + b_nx_n|| = ||lim b_1x_1 + ... b_nx_n||, so how do you > show f is bounded??? >Ah, of course. I still have one lingering question, though. Given {x_k} a >sequence in X and {a_k} a sequence of scalars, define a bounded linear >functional on the subspace generated by x_1,...,x_n by f(b_1x_1 + ... + >b_nx_n) = b_1f(x_1) + ... + b_nf(x_n) , f(x_k)=a_k. >How does the requirement that >| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| >guarantee that ||f|| = M? It's obvious that ||f|| <= M, but how do you >assert equality? > Oh. It doesn't, of course, and I don't. If x_1,...,x_n already > span X, you're clearly out of luck, and your statement is false. > Hahn-Banach gives you x' with ||x'|| <= M. If x_1,...,x_n don't > span X (e.g. if X is infinite-dimensional), it also gives you > y' <> 0 such that y'(x_j) = 0 for j=1..n. For suitable t, > x' + t y' will be what you want. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 >>Let X normed linear space, X' it's dual. a_1,...,a_n some scalars, >>x_1,...,x_n some elements of X. How do you show that >> >>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| >> >>for some M for each n and for scalars b_1,...,b_n implies that there >exists >>x' in X' satisfying x'(x_k)=a_k and ||x'|| = M. >> >>i don't know how to show this without explicitly constructing the > bounded >>linear functional on X, and of course i'm stuck on constructing it > (hence >>the problem in the first place).. >> >>if X is finite dimensional, then each x in X can be written x = b_1 > x_1 + >>... + b_n x_n on some basis {x_k}..then define x'(x)=b_1 x'(x_1) + ... >b_n >>x'(x_n) and defining x'(x_k)=a_k. This is bounded by the given >constraint. >> >> Careful: the x_n's were given, and might not be a basis. >> >>But what to do if X is infinite-dimensional??? >> >> Use the Hahn-Banach Theorem. === Subject: Re: existence of bounded linear functional Hmmm...another sticking point (apart from this norm stuff). Suppose | b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| for some M for each n and for scalars b_1,...,b_n. We want to construct a functional x' in X' such that x'(x_j)=a_j for ALL j = 1,2,.... Consider the subspace spanned by ALL the x_k, and call it S. Then any x in S can be written x = b_1x_1 + b_2x_2 + .... Define f(x) = b_1a_1 + b_2a_2 + .... Then f(x_j)=a_j for all j = 1,2... But is f bounded? The given inequality only works for finite n, and |f(x)| = |b_1a_1 + ... | = |lim b_1a_1 + ... + b_na_n| = lim |b_1a_1 + ... + b_na_n| < = lim M ||b_1x_1 + ... b_nx_n||, but lim ||b_1x_1 + ... + b_nx_n|| = ||lim b_1x_1 + ... b_nx_n||, so how do you show f is bounded??? >Ah, of course. I still have one lingering question, though. Given {x_k} a >sequence in X and {a_k} a sequence of scalars, define a bounded linear >functional on the subspace generated by x_1,...,x_n by f(b_1x_1 + ... + >b_nx_n) = b_1f(x_1) + ... + b_nf(x_n) , f(x_k)=a_k. >How does the requirement that >| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| >guarantee that ||f|| = M? It's obvious that ||f|| <= M, but how do you >assert equality? > Oh. It doesn't, of course, and I don't. If x_1,...,x_n already > span X, you're clearly out of luck, and your statement is false. > Hahn-Banach gives you x' with ||x'|| <= M. If x_1,...,x_n don't > span X (e.g. if X is infinite-dimensional), it also gives you > y' <> 0 such that y'(x_j) = 0 for j=1..n. For suitable t, > x' + t y' will be what you want. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 >>Let X normed linear space, X' it's dual. a_1,...,a_n some scalars, >>x_1,...,x_n some elements of X. How do you show that >> >>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| >> >>for some M for each n and for scalars b_1,...,b_n implies that there >exists >>x' in X' satisfying x'(x_k)=a_k and ||x'|| = M. >> >>i don't know how to show this without explicitly constructing the bounded >>linear functional on X, and of course i'm stuck on constructing it (hence >>the problem in the first place).. >> >>if X is finite dimensional, then each x in X can be written x = b_1 x_1 + >>... + b_n x_n on some basis {x_k}..then define x'(x)=b_1 x'(x_1) + ... + >b_n >>x'(x_n) and defining x'(x_k)=a_k. This is bounded by the given >constraint. >> >> Careful: the x_n's were given, and might not be a basis. >> >>But what to do if X is infinite-dimensional??? >> >> Use the Hahn-Banach Theorem. === Subject: Re: JSH: Howard Aiken quote, my situation >> Not quote Howard Aiken, but anybody know Jonathan Aitkin? He was a >> politician, an MP for the Tories, can't remember his constituency >> (Hatton?) who was revealed in the satirical press to be a liar and a >> bit of a bastard. He promised to fight his detractors with the 'shield > The MP for *T*atton was Neil Hamilton, one of the other lying bastard > Tory MPs. Aitken was MP for South Thanet. >> of truth' and the 'sword of justice'... he ended up in prison for >> perjury after he sued a newspaper for libel. As a liberal I wnjoy that >> story anyway, but just thought I'd share it, and I can't quite put my >> finger on why it's passably relevant.... > Also sparked an excellent Guardian (the paper he had sued) front page the > day after the trial. Huge photo of him, with the headline 'A liar and a > cheat'. And of course it wasn't libel, as it has been proved in court:) No idea why this is on topic -- apart from Jonathan Aitken having a different surname from Howard Aiken --- but Aitken recently published -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: easy topology problem..... when a, b in Q and a < b , show that collection of closed interval [a,b] is not basis for a topology on R. ----------------------- i can't find contradiction. help me....please.... === Subject: Re: confused on substitution > hello, > been working on a problem too hard and now have started to become unsure as > to the correct nature of a substitution: > the problem is this, > solve the heat diffusion equation, > c*u(x,t)_xx = u(x,t)_t [1] > where underscores are partial derivitives and c=constant, such that u(x,t) > satisfies the condition > u(x,t)_x = u(x,t)*F(u(x,t),t) [2] > for all x and t where F is an arbitrary function of u and t. first of all > is this problem well defined, and does it have a solution (i think it might > not)? I am inclined to agree. > one would assume that from [2] we can find u_xx and substitute this > into the LHS of [1]. Yes. > however do we also have to solve for u from [2] Given F, [2] is a non-linear PDE for u, so no. > and > substitute that into the RHS of [1] as well to obtain a function that > satisfies both [1] and [2]? or is the first substitution enough to guarantee > this. The first substitution is enough, assuming such a u exists. > i assume that we probably need to substitute for u from [2] in both > sides of [1] in order to be consistent and keep the equality....but my brain > is fried and i have a mental block on this issue at present and it doesnt > seem obvious to me at the moment. Largely because that would be wrong. > If we do substitute for both sides of > [1] using [2] then we get... > c*u_x*(F+u*F_u)^2 = u_xt - u*F_t [3] We actually get u_xx = u*F_x + u_x*F F_x = F_u*u_x u_xx = u_x*(F_u*u + u_x*F) c*u_x*(F_u*u + u_x*F) = u_t [3a] > So to reiterate more succinctly will [3] provide me a solution that > satisfies both [1] and [2]. No, but [3a] will. > Is there a better way to solve this problem. i have thought about using > greens functions for [1] and forcing it to satisfy [2] but i cant seem to > construct them. need help with the analysis of this system. [1] and [2] are two simultaneous PDE's in one unknown (u). This system is in general overdetermined. However, [3a] may be soluble; whether it is well-posed when combined with the boundary and initial data for [1] and [2] is another matter. -- P.A.C. Smith The vast majority of Iraqis want to live in a peaceful, free world. And we will find these people and we will bring them to justice. === Subject: Re: Probability Question >Rob Johnson wibbled: >{OrigPoster was talking abut the UK national lottery - pick 6 to match 6 >from 49} I didn't know the rules of the lottery, so I answered as best as I could without this information, using only what was said in the original post. With this additional information, I can deduce that the £10 ticket would be for choosing 3 out of 6 correct (246820/13983816 ~= .01765040387). Choosing 100 different tickets would give a probability of C(13736996,100-m)C(246820,m)/C(13983816,100) of getting m tickets that would win £10. The chances against getting a £10 winning ticket would be C(13736996,100)/C(13983816,100). Note that this is 246820 246820 246820 ( 1 - -------- )( 1 - -------- ) ... ( 1 - -------- ) 13983816 13983815 13983717 which is between (1-246820/13983816)^100 and (1-246820/13983717)^100; both are 16.850% to 5 places. Choosing 1398382 different tickets would give a probability of C(13983815,1398382-m)C(1,m)/C(13983816,1398382) of getting m tickets that would win the jackpot. The chances against winning the jackpot would be C(13983815,1398382)/C(13983816,1398382). Note that this is 13983815 13983814 12585434 -------- -------- ... -------- 13983816 13983815 12585435 which is exactly 12585434/13983816 = 1 - 1398382/13983816 = 90.000% to 5 places. Therefore, choosing 1398382 different tickets would have a 10.000% chance of winning the jackpot. Choosing 1380000 different tickets would give a 9.86855% chance of winning the jackpot. >Only one set of six numbers can win the jackpot. It's relaively common >in the UK for the jackpot to be split between holders of identical >tickets: there is no all tickets different restriction in the total >sale distribution. What I meant was that only one combination of numbers wins the jackpot, whereas for smaller prizes there are many combinations of numbers which can win. I did not mean that only one person could win the jackpot. If someone else had submitted the same ticket, i.e. the same combination of numbers, they would have shared in the jackpot. What I meant by different tickets was that you didn't choose duplicate tickets, that is, more than one ticket with the same numbers. >> the odds you give, there are 14 million different lottery tickets >Roughly. Whatever 49C6 is - just under 14million. As I said above, I was going simply on the information given in the original post. Since the probability of winning was 7.15x10^-8, it follows that the count of winning numbers must be the reciprocal of that, or between 13976240 and 13995801. C(49,6) = 13983816 falls in that range. As the probability was given to 3 places, I took 3 places in the reciprocal; 14 million. I should have said, approximately. >> possible. Again assuming that you are asking only about a single >> lottery, you would need to buy 1.4 million _different_ tickets to have >> a 10% chance of winning the jackpot. >It'll be N, where the Nth root of .9 is 1-p where p is the probability >he gave. He might get away with 1.38 million No, if you have 13983816 different tickets (combinations of numbers), then if you place 1398382 of them, you have as close to 10% chance of winning the jackpot as possible. Again, I had rounded to 3 places. Rob Johnson take out the trash before replying === Subject: Re: JSH: Howard Aiken quote, my situation > Not quote Howard Aiken, but anybody know Jonathan Aitkin? He was a > politician, an MP for the Tories, can't remember his constituency > (Hatton?) who was revealed in the satirical press to be a liar and a > bit of a bastard. He promised to fight his detractors with the 'shield >> The MP for *T*atton was Neil Hamilton, one of the other lying bastard >> Tory MPs. Aitken was MP for South Thanet. > of truth' and the 'sword of justice'... he ended up in prison for > perjury after he sued a newspaper for libel. As a liberal I wnjoy that > story anyway, but just thought I'd share it, and I can't quite put my > finger on why it's passably relevant.... >> Also sparked an excellent Guardian (the paper he had sued) front page the >> day after the trial. Huge photo of him, with the headline 'A liar and a >> cheat'. And of course it wasn't libel, as it has been proved in court:) > No idea why this is on topic -- apart from Jonathan Aitken having > a different surname from Howard Aiken --- but Aitken recently published Hey I did say it was not quite Aiken, well, ok I said not quote Aiken, but never mind. It now turns out JSH wants to start a civil suit against Rick Decker for misleading the public. You start something as a mildly amusing joke (to me and perhaps no one else, fine) and look what happens... if* JSH loses, can we expect a post by his evil cabal of Nora, Dik, Arturo, and Rick posted under A liar and cheat. * 'if' meaning should he ever get it to court, where he will lose. Still, moving further OT: hasn't Aitken now turned to religion? I've yet to see JSH cite divine right as a proof, but it can't be far off. OK, enough, I will not mention this again. === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational > Hello > > How can we prove [arccos((sqrt(5)-1)/2 )] / pi [is irrational]? > The real part of a root of unity cannot be > an algebraic integer unless it is 0, 1, or -1. How can we prove this? Could you please outline a proof? Amanda > If u=[arccos((sqrt(5)-1)/2 )]/pi, then e^(i pi u) has real part > (sqrt(5)-1)/2, which is an algebraic integer. So e^(i pi u) > is not a root of unity, that is: u is irrational. === Subject: Re: e is transcendental (was: classes of transcendental numbers ? > hey do you guys have a copy of the proof of e is transcendental? ..id really > appreciate if somebody could help me. ive been searching and found nothing. >>Spivack, CALCULUS, 2nd edition, Chapter 21. >Are you sure that the proof there is for e being transcendental and not >just irrational? I know that the proof of the irrationality of e is >simple enough for a Calculus book, but that the transcendence is quite a >bit more complicated. In my version of Spivak, the transcendence of e is proven in Chapter 20, and the irrationality of pi (or actually, the stronger result, the irrationality of pi^2) is proven in Chapter 16. Both chapters are marked with an asterisk, presumably denoting that the chapters are intended for the more advanced student. I note that Spivak describes these chapters as optional. As others have noted, another good book is Irrational Numbers by Ivan Niven, in which a proof of the transcendence of e, pi, and many others, is provided in the first chapter. Specifically, Niven gives the proof of the result that if a_1, ..., a_n are distinct algebraic numbers, then exp(a_1), ..., exp(a_n) are linearly independent over the algebraic numbers. The transcendence of e is then proven by taking a_1 = 0, a_2 = 1, and the transcendence of pi is proven by taking a_1 = 0, a_2 = i pi. David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. === Subject: Countable perfect set It is known that every perfect (every point is an accumulation point) closed subset of a complete metric space is uncountable. But, what about a non-complete metric space ? The proof of the previous case cannot be adapted, and I don't manage to find a counter example. Does anyone have any idea ? Nobody knows a perfect, closed and countable subset of a metric space ? Vinect G. === Subject: Re: e is transcendental (was: classes of transcendental numbers ? >> hey do you guys have a copy of the proof of e is transcendental? ..id >> really >> appreciate if somebody could help me. ive been searching and found nothing. >Spivack, CALCULUS, 2nd edition, Chapter 21. > Are you sure that the proof there is for e being transcendental and not > just irrational? I know that the proof of the irrationality of e is > simple enough for a Calculus book, but that the transcendence is quite a > bit more complicated. Yep. Irrationality is just a half-page in Chapter 20, using the Taylor polynomial. But then Spivack adds Chapter 21. It includes a 3-page proof that e is transcendental. But easy enough for calculus students (at least those reading Spivack's book... much more than the usual calculus book) to understand. The proof is due to Hermite, as simplified by Hilbert. It involves integrals related to the Gamma function. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Countable perfect set > It is known that every perfect (every point is an accumulation point) > closed subset of a complete metric space is uncountable. > But, what about a non-complete metric space ? The proof of the > previous case cannot be adapted, and I don't manage to find a counter > example. Take the metric space of rational numbers, with the usual metric. > Does anyone have any idea ? Nobody knows a perfect, closed and > countable subset of a metric space ? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: easy topology problem..... > when a, b in Q and a < b , > show that > collection of closed interval [a,b] is not basis for a topology on R. > ----------------------- > i can't find contradiction. help me....please.... The intersection of two such sets may be nonempty without containing such a set. So it is not a base for a topology. === Subject: Re: Modular Arithmetic > I was wondering if anyone knows a way to reduce 2^100 mod 5. I know > the answer is 1 (2^100 mod 5 = 1), but I had to use Maple to find what > lot. Quick method note 2^2 = 4 ==> -1 MOD 5 2^100 = 4^50 { 50 is even} -1^{even}= 1 So [2^100] MOD 5 = 1 A^N MOD B ==> [A MOD B]^N MOD B === Subject: Re: How far can one go with self-study nowadays? In , on >Just thinking about all the neat things available online, compared to >when I was in school 20ish years ago, I'm wondering how far one can >go nowadays in mathematics with self-study. Quite far, but it will be a lot easier if you have access to a good library. Access to a good university is better. If a good local university allows you to audit classes, that's even better, but at that point it's no longer self study. >Compare that to nowadays. You can find textbooks easily online, and >find customer reviews of them, I don't have much confidence in those customer reviews. I'd much rather talk to a few professors in a good Mathematics department and pay careful attention to what they have to say. >If you get stuck on something, there is a fair chance that a >Google search will find a solution, and if not, there is usenet--a >post to sci.math will probably get an answer. The WWW and Usenet are certainly valuable resources, but they can't take the place of personal contact, attendance at colloquia, etc. >I think one should be able to reasonably get to at least a fair way >into graduate level. Of course, but it will be a lo0t harder than if you had local resources. -- Shmuel (Seymour J.) Metz, SysProg and JOAT 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 === Subject: One question for calculus experts please help me solving this problem.. Let Omega be an open, bounded subset of R^n , let u : Omega .9a> R^n be a function: Then, we can identify the curl of u as vector in R^( n(n-1)/2 ). Now, let T: R^( n(n-1)/2) -> R^( n(n-1)/2) be a linear isomorphism. Then, is it true that the composition map Tcurlu: Omega .9a>R^( n(n-1)/2 ) is the curl of some function v defined on Omega ???? === Subject: Re: :: towards a constructive education :: (news server friendly) > While you are reading this, your retinal rods and cones are [etc. appreciated constructivism promotion campaign snipped.] Would you care to give references for each of the applications that you mention? Because in most cases, i have no idea what you are talking about. And for the rest, i had no idea they had anything to do with Heyting algebras perse... Or perhaps there already is a book Applications of Heyting Algebras ? If not, you may consider writing one yourself. Herman Jurjus (posting from the planet sci.math) === Subject: Re: one quetion > Today, I read the proof of Every real # is a limit point of rational #s > and I understand that proof. Which proof? Based on which definition of the real numbers? > Now, I am curious about correctness of > Every real # is a limit point of irrational #s. > Is it true ? > I think this is true. Yes. > Because, For any real # p, > Let a_n=(p -1/n,p +1/n). > By density of irrational #, a_n contains irrational #s. Are you taking the density of the irrationals as an axiom, or did you prove it? In fact, that assertion is the whole heart of the matter. > Let b_n=irrational # in a_n, then b_n is a sequence of irrational #s and > b_n converges to p > Is my thinking correct ? Yes, basically. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: JSH: Howard Aiken quote, my situation >[...] >It now turns out JSH wants to start a civil suit against Rick Decker for >misleading the public. Really? Keen - I haven't seen that post yet. Every once in a while he says something about lawsuits, but we're all always _so_ disappointed when he never actually gets around to hiring a lawyer. I mean it would be like the trial of the century... > You start something as a mildly amusing joke (to me >and perhaps no one else, fine) and look what happens... if* JSH loses, can >we expect a post by his evil cabal of Nora, Dik, Arturo, and Rick posted >under A liar and cheat. >* 'if' meaning should he ever get it to court, where he will lose. >Still, moving further OT: hasn't Aitken now turned to religion? I've yet >to see JSH cite divine right as a proof, but it can't be far off. >OK, enough, I will not mention this again. ************************ David C. Ullrich === Subject: Re: e is transcendental (was: classes of transcendental numbers ? > hey do you guys have a copy of the proof of e is transcendental? ..id really > appreciate if somebody could help me. ive been searching and found nothing. >>Spivack, CALCULUS, 2nd edition, Chapter 21. >Are you sure that the proof there is for e being transcendental and not >just irrational? I know that the proof of the irrationality of e is >simple enough for a Calculus book, but that the transcendence is quite a >bit more complicated. Evidently you've never seen the book. It's not at all like the typical book titled Calculus. >Rob Johnson >take out the trash before replying ************************ David C. Ullrich === Subject: Re: JSH: Pattern argument, revisited >[...] >For other readers I want to point out that Rick Decker plays the nice >guy role, when his behavior here is contemptible, and reflects badly >not only upon him, but also upon Hamilton College. >His *duty* as a professor is to tell the truth and promote elucidating >the truth, not hiding it. >like 2 and 17, it's clear that there's more here than his attempts at >dismissal indicate. >I hereby charge Rick Decker with academic fraud and note that his >college is responsible for this rogue professor. >I've deliberately involved an official at his college to take away >plausible deniability for his school in a phone call I made months >ago. >His college has been made aware of his behavior. >There may also be a civil matter involved at some point in the future. Whee!!!! Wandering off the deep end again, are we? Excellent. >Some of you seem to think this is a game, or that Usenet doesn't >matter. >However, Usenet is a *public* forum read around the world. There is a >burden upon professors to tell the truth, and upon the colleges and >universities that promote them either directly or indirectly. >Possibly an example needs to be made to convince you all that you have >to be accountable for your public words. Absolutely. Actually _you're_ the one who seems to think this is all a game. For one thing, you've _said_ it's all just a game many times. But more to the point, every once in while you mumble something about lawsuits, but you never carry through. A lot of us are really looking forward to reading the transcript. It's cruel of you to tease us this way. >James Harris ************************ David C. Ullrich === Subject: Re: Ellipse? Haroldo Stenger a .8ecrit: > Hipothesis: > Let p and q be two positive real numbers. > Let Ox and Oy be two cartesian axis. > Let l be the variable line that intersects Ox and Oy in A and B such > that AB measures p. > Let C be *one* of the two points in l which distance to B is q. > Thesis: the locus of C is an ellipse > I'm looking for an euclidean proof. > Any hint? Just a hint then... (in case of homework) Let I middle of AB on the OI line : OM=AP and OC=AB PM intersects OB at H Prove that HP/HM=PB/AP (repeated Thales on various triangles) Locus of P is the affine (stretch ?) transformed of locus of M. -- philippe (chephip at free dot fr) === Subject: Re: JSH: Pattern argument, revisited >> I hereby charge Rick Decker with academic fraud and note that his >> college is responsible for this rogue professor. >I have nothing to add. I just want to compliment your consistently >inventive use of the word rogue. It makes one wonder whether Decker >wears an eyepatch, or perhaps an ammunition belt strapped across his >chest. >Anyway, good luck with the threatened civil action. I bet there are >just dozens of lawyers aching to take this case on a contingency >basis. No doubt. And of course Decker would be totally screwed - there's no way he could possibly find an expert witness to testify on his behalf. (I mean sure, more or less every mathematician on the planet has explained to James that he's all wet. But we know they're just saying that because they're afraid of the Truth - no way they'd lie like that under oath.) ************************ David C. Ullrich === Subject: Re: Unitary operator basics, lebesgue measure >> Let H = L^2(R,dx). For each a in R (the reals), defined T_a : H ----> H >> >> (T_a f)(x) = f(x-a), where f is in H (the hilbert space) >> >> Firstly, why is f(x-a) necessarily in H if f is in H? Why is T_a a >unitary >> operator (unitary operator I believe means that for an operator A, A* = >> A^{-1}) ? I suppose this might be easy if I could represent T_a as a >> matrix. But then how would this matrix multiply by f? Could I >represent >> the function f by a matrix(with respect to a basis of L^2(R,dx)?)? What >> a basis for L^2(R,dx)? >> The graph of f(x-a) is just the graph of f shifted a units to the right. >> That answers the first question. Verify that (T_a)* = T_(-a) and you'll >> have the answer to the second question. >How can I think about (T_a)* ?? First I need to represent T_a as a matrix, No, you don't. Let f and g be elements of L^2(R,dx), then, using the mathematician's inner product (linear in the first factor and antilinear in the second), ((T-a)*f,g) = (f,T_a g) = int_R f(x) g(x-a)^* dx = int_R f(x+a) g(x)^* dx by substitution, and so [(T_a)* f](x) = f(x+a) = [T_(-a) f](x), i.e. (T_a)* = T_(-a). David McAnally Despite anything you may have heard to the contrary, the rain in Spain stays almost invariably in the hills. === Subject: Are these two polynomials equivalent? Can I claim that f(x)=a+b(x-e)+c(x-e)^2+d(x-e)^3 , g(x)=m+n(x-e)+p1(x-e)^3+...+pn(x-e)^3 are equivalent? a,b,c,d,e,m,n,p1,p2,...,pn are constants, n is arbitrary The key is, is it that there always exists p1,...,pn such that we can drop the quadratic term c(x-e)^2 in f(x) ? (my problem arises from the representation of a spline, because the book seems to say that a cubic spline CAN always be expressed in the second form) === Subject: Re: Equation of Cyclotomic Polynomial boundary=----=_NextPart_000_0039_01C3EFDE.49659C80 --------------------------------------------------------------------- The answer to your Q is below but I hope you want to know why. .93.bc(5) = (x-e^i(2 pi/5)) (x-e^i(4 pi/5)) (x-e^i(6 pi/5)) (x-e^i(8 pi/5)) = x^4 +x^3 +x^2 +x +1 THE PRIMITIVE ROOTS OF UNITY LEAD TO CPs Solutions to x^n - 1 = 0, known as the n-th roots of unity, are e^i(2kp/n) for k=1 to n, (equivalent to k=0 to n-1) The primitive roots of unity are e^(2pi)ik/n where k and n coprime, (no common factors). Therefore, there are .93.bc(n) different primitive n-th roots of unity, where .93.bc(n) is defined to be the number of positive integers less than or equal to n and coprime to n. For example, .93.bc(3) = 2, since the two numbers 1, and 2, (but not 3), have no common factors with 3 .93.bc(5) = 4 since the two numbers 1,2,3 and 4 are coprime to 5. .93.bc(8) = 4 since the four numbers 1, 3, 5 and 7 are coprime to 8. The primitive n-th roots of unity can be used to define and generate the nth cyclotomic polynomial .93.bcn(x), by stating that they are the zeros of this polynomial. Examples will make this clearer. .93.bc3(x) = (x-e^i(2 pi/3))(x-e^i(4 pi/3)) = x^2 +x +1 and .93.bc5(x) = (x-e^i(2 pi/5)) (x-e^i(4 pi/5)) (x-e^i(6 pi/5)) (x-e^i(8 pi/5)) = x^4 +x^3 +x^2 +x +1 If we only select the roots where k and n are co-prime, (have no common factors), known as the primitive n-th roots of unity, these may be used to recompose a special type of polynomial, which has the primitive n-th roots as its solution. Polynomials constructed in this way are known as the n-th cyclotomic polynomials. An Considering a symmetrical example, the 8th roots of unity are e^i(2k pi/8) = e^i(k pi/ 4), for k = 1 to 8 so, the primitive 8th roots of unity are e^i(k pi/ 4) for k = 1, 3, 5 and 7, but due to symmetry, e^i(7 pi/ 4) = e^-i( pi/4) , (etc), so we may pair off complex conjugates to write the factors as .93.bc8(x) =(x - e^i( pi/4))(x - e^-i( pi/4))(x - e^i(3 pi/4))(x - e^-i(3 pi/4)) = (x^2 - (sqrt2)x +1) (x^2 + (sqrt2)x +1) = x^4 + 1 CYCLOTOMIC POLYNOMIALS - SHORTCUTS FOR FINDING THEM a) Whenever n is an odd prime, p, the CP becomes (x^p-1)/(x-1) which from geometric progressions we recognise as a simple sum of descending powers of x, starting with x^(p-1) e.g. .93.bc5(x) = x^4 +x^3 +x^2 +x +1 I give no more examples of these here as the pattern is clear. b) Whenever n is a composite n = rs with r and s different, and assuming you have a technical calculator, that has 'expand' as one of its algebraic functions, you can use the fact that, Phi rs(x) = (x-1)(x ^(rs) - 1)/[(x ^(r) - 1)(x^(s) - 1)] e.g. Phi 6(x) = (x-1)(x ^(6) - 1)/[(x ^(2) - 1)( x^(3)- 1)] Here I list a few examples of rs CPs Phi 6(x) = x^2 - x +1 Phi 10(x) = x^4 - x^3 + x^2 - x +1 Phi 14(x) = x^6 - x^5+ x^4 - x^3 + x^2 - x +1 Phi 15(x) = x^8 - x^7 + x^5 - x^4 +x^3 - x +1 Phi 21(x) = x^12 - x^11 +x^9 - x^8 + x^6 - x^4 + x^3 - x +1 Note that with composite n = rs, certain powers can be missing and signs alternate. c) Whenever n is a square composite n = r^2, again using 'expand', you can use the fact that, Phi r^2(x) = (x^(r^2)-1)/(x^(r) - 1) here are the first few examples Phi 4(x) = (x ^(4) - 1)/(x ^(2) - 1) = x^2 + 1 Phi 9(x) = (x ^(9) - 1)/(x ^(3) - 1) = x^6 + x^3 + 1 Phi 16(x) = (x ^(16) - 1)/(x ^(4) - 1) = x^12 + x^8 + x^4 +1 Phi 25(x) = (x ^(25) - 1)/(x ^(5) - 1) = x^20 + x^15 + x^10 + x^5 +1 Here there are r positive terms, starting x^(r(r-1)) with powers descending in steps of r. I hope you find this interesting and useful - Ian Hutcheson === Subject: Re: JSH: Pattern argument, revisited >[...] >For other readers I want to point out that Rick Decker plays the nice >guy role, when his behavior here is contemptible, and reflects badly >not only upon him, but also upon Hamilton College. >His *duty* as a professor is to tell the truth and promote elucidating >the truth, not hiding it. >like 2 and 17, it's clear that there's more here than his attempts at >dismissal indicate. >I hereby charge Rick Decker with academic fraud and note that his >college is responsible for this rogue professor. >I've deliberately involved an official at his college to take away >plausible deniability for his school in a phone call I made months >ago. >His college has been made aware of his behavior. >There may also be a civil matter involved at some point in the future. > Whee!!!! Wandering off the deep end again, are we? Excellent. >Some of you seem to think this is a game, or that Usenet doesn't >matter. >However, Usenet is a *public* forum read around the world. There is a >burden upon professors to tell the truth, and upon the colleges and >universities that promote them either directly or indirectly. >Possibly an example needs to be made to convince you all that you have >to be accountable for your public words. > Absolutely. Actually _you're_ the one who seems to think this is all > a game. For one thing, you've _said_ it's all just a game many times. > But more to the point, every once in while you mumble something > about lawsuits, but you never carry through. > A lot of us are really looking forward to reading the transcript. > It's cruel of you to tease us this way. That'd be a pretty short transcript. Something like: Judge: James, You are a crank. Case dismissed. James: Everyone is against me! Everyone's lying!! >James Harris > ************************ > David C. Ullrich David Moran === Subject: block matrix, determinant zero I know that upper or lower triangular matrices have determinant zero, which can be directly seen from the cofactor formula. How does this extend to the fact that det(A,0; 0,D) = det(A)det(D) ? Or is this proven in other ways ? === Subject: Re: The Real Da Vinci Code Broken > My Meetings with Carlo Suares in Paris 1973 that resulted in the book > ñSpace-Time and Beyondî > With Bob Toben and Fred Alan Wolf > The Real Da Vinci Code I'll counter with this gem, taken from the Rapture Ready BBS, a site for fundementalist Christians: >Please do not quote me out of context. I said the THEORY [of Relativity] had >nothing to do with physics. And you can whip out any old definition you want, >but its still a fact that its a theory and nothing more. And by the way, I've >never heard of gravity being called a theory. Everyone else I know of >considers it to be proven true. Why, just the other day I fell down. - Dorko, RaptureReady === Subject: triangulation Is there anyone in this group, who can help me with the triangulation of a given polygon P under these two restrictions: 1. Each triangle shall be as equilateral as possible 2. No triangle has an area that exceeds a given constant O Hans van Duijnhoven (Randers, Danmark). === Subject: Re: Question for logarithm experts by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1AE2qL25626; >This is an interesting posting. I've never really thought about this before, >but do all equations have solutions, (even though it may not be possible for >us to find them in terms of simple functions) ? Look at the following: 7^X=12*X has no solutuion Try: 7^X=[12.34683945]*X >Graphing does not give us the whole story does it? Try a relevant graph topic: http://www.stefanides.gr/why_logarithm.htm What I am thinking about >is that for example >y = x^2 does not cross y = - 1, yet we know that x^2 = - 1 has solutions >x = + or - i, (the sqrt of minus one). >Does the topic you are studying suggest the use of complex numbers? >If you replace x with z = a+b i, you could write >7^z = e^((ln7)z) = (e^(aln7)) (e^i(bln7))= (7^a)[cos(bln7) + i sin (bln7) ] = >4a + 4bi 7^Z may be expanded : 7^Z=1+Z*[ln[7]+[{Z*[ln[7])^2]/2!+...... , according to Ref: http://www.stefanides.gr/why_logarithm.htm Panagiotis Stefanides http://www.stefanides.gr === Subject: Re: Silly question for someone with a big calculator. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1AE2qg25622; >>4/3 = 1.333333... >>24/17 = 1.411764... >>816/577 = 1.414211... >>941664/665857 = 1.414213... >> >>If we let k represent that ratio for one of the exponents, the ratio for >>the next in this list is (4k)/(k^2+2); it can be shown that the ratios >>will converge to sqrt(2). >> Sorry, I think I must be overlooking something (I do a lot). It >> certainly *looks* to me as if the above sequence will converge to >> sqrt(2); however, I am neither sure what your k is in (4k)/(k^2+2) >> nor how one can show that that the sequence of ratios converges, >> of which no closed form exists- unless I`m overlooking something, >> of course. >If k = 4/3, (4k)/(k^2+2) = 24/17. >If k = 24/17, (4k)/(k^2+2) = 816/577. >etc. >If we let delta = 2-k^2, the delta for the following k will be >2*delta^2/(k^2+2)^2 . For example, for k=4/3, we have delta = 2/9. For >k=24/17, we have delta = 2/289 = 2*(2/9)^2/((4/3)^2+2)^2 . Since >repeating this often enough will get you a delta as close to 0 as you >want, k will get as close to sqrt(2) as you want. are, even though you explained them correctly the first time round. However, forgive me for being quite stubborn, aren't we effectively saying - instead of the usual: let (k_n) be the sequence of ratios, then for every epsilon > 0 exists N in naturals: |k_n - sqrt(2)| < epsilon for all n > N - for any epsilon > 0, we can find you a bunch of examples such that... ? (Indirectly of course, since we are trying to show delta -> 0 through substitution instead of k_n -> sqrt(2) ). Sincerly, C. Dement >Daniel W. Johnson >panoptes@iquest.net >http://members.iquest.net/~pano ptes/ >039 53 36 N / 086 11 55 W === Subject: Re: block matrix, determinant zero > I know that upper or lower triangular matrices have determinant zero, > which can be directly seen from the cofactor formula. Not so. The determinant of a triangular matrix is the product of the diagonal entries, which can be directly seen from the cofactor formula. === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational >How can we prove [arccos((sqrt(5)-1)/2 )] / pi? Any idea? I suspect >this has something to with the golden ratio, and maybe with Fibonacci >numbers or continued fractions? But I could't start. > There are several possible methods. > What do you know of either Chebyshev polynomials or cyclotomic > polynomials? Or what can you say about the polynomial > 1+2 t-2 t^2+2 t^3+t^4? Well, I know what cyclotomic polynomials are, but have never studies them. The polynomial 1+2 t-2 t^2+2 t^3+t^4 is reciprocal and 2 of its roots satisfy x + 1/x = (sqrt(5)-1)/2. I'm a bit lost, could you outline a prrof, please? Amanda === Subject: Re: Web Sites for Complexity Theory check out this book (pdf - free): http://www.cis.upenn.edu/~wilf/AlgComp2.html good luck, Jurgen > Are there any good books in this field, or good web sites that provide > an introduction or lecture notes with definitions and examples? I am > really trying to teach myself this stuff before I take the class next > semester. > Blake Manner === Subject: Re: Equation of Cyclotomic Polynomial REPEATED WITH SYMBOLS CLARIFIED The answer to your Q is below but I hope you want to know why. Phi5 (x)=(x-e^i(2 pi/5))(x-e^i(4 pi/5))(x-e^i(6 pi/5)) (x-e^i(8 pi/5)) = x^4 +x^3 +x^2 +x +1 THE PRIMITIVE ROOTS OF UNITY LEAD TO CPs Solutions to x^n - 1 = 0, known as the n-th roots of unity, are e^i(2kpi/n) for k=1 to n, (equivalent to k=0 to n-1) If we only select the roots where k and n are co-prime, (have no common factors), known as the primitive n-th roots of unity, these may be used to recompose a special type of polynomial, which has the primitive n-th roots as its solution. So the primitive roots of unity are e^(2pi)ik/n where k and n have no common factors and there will be Phi (n) of them, where Phi (n) is defined to be the number of positive integers less than or equal to n and coprime to n. For example, Phi (3) = 2, since the two numbers 1, and 2, (but not 3), have no common factors with 3 Phi (5) = 4 since the two numbers 1,2,3 and 4 are coprime to 5. Phi (8) = 4 since the four numbers 1, 3, 5 and 7 are coprime to 8. Thus the primitive n-th roots of unity can be used to define and generate the n-th cyclotomic polynomials, Phi n (x) , by stating that they are the zeros of this polynomial. Examples will make this clearer. Phi3 (x) = (x-e^i(2 pi/3))(x-e^i(4 pi/3)) = x^2 +x +1 and Phi5 (x) = (x-e^i(2 pi/5)) (x-e^i(4 pi/5))(x-e^i(6 pi/5)) (x-e^i(8 pi/5))=x^4 +x^3 +x^2 +x +1 Considering a symmetrical example, the 8th roots of unity are e^i(2kpi/8) = e^i(kpi/4), for k = 1 to 8 so, the primitive 8th roots of unity are e^i(kpi/4) for k = 1, 3, 5 and 7, but due to symmetry, e^i(7pi/4) = e^-i(pi/4) , (etc), so we may pair off complex conjugates to write the factors as Phi8 (x) =(x - e^i(pi/4))(x - e^-i(pi/4))(x - e^i(3pi/4))(x - e^-i(3pi/4)) = (x^2 - (sqrt2)x +1) (x^2 + (sqrt2)x +1) = x^4 + 1 CYCLOTOMIC POLYNOMIALS - SHORTCUTS FOR FINDING THEM a) Whenever n is an odd prime, p, the CP becomes (x^p-1)/(x-1) which from geometric progressions we recognise as a simple sum of descending powers of x, starting with x^(p-1) e.g. Phi5 (x) = x^4 +x^3 +x^2 +x +1 I give no more examples of these here as the pattern is clear. b) Whenever n is a composite n = rs with r and s different, and assuming you have a technical calculator, that has 'expand' as one of its algebraic functions, you can use the fact that, Phi rs(x) = (x-1)(x ^(rs) - 1)/[(x ^(r) - 1)(x^(s) - 1)] e.g. Phi 6(x) = (x-1)(x ^(6) - 1)/[(x ^(2) - 1)( x^(3)- 1)] Here I list a few examples of rs CPs Phi 6(x) = x^2 - x +1 Phi 10(x) = x^4 - x^3 + x^2 - x +1 Phi 14(x) = x^6 - x^5+ x^4 - x^3 + x^2 - x +1 Phi 15(x) = x^8 - x^7 + x^5 - x^4 +x^3 - x +1 Phi 21(x) = x^12 - x^11 +x^9 - x^8 + x^6 - x^4 + x^3 - x +1 Note that with composite n = rs, certain powers can be missing and signs alternate. c) Whenever n is a square composite n = r^2, again using 'expand', you can use the fact that, Phi r^2(x) = (x^(r^2)-1)/(x^(r) - 1) here are the first few examples Phi 4(x) = (x ^(4) - 1)/(x ^(2) - 1) = x^2 + 1 Phi 9(x) = (x ^(9) - 1)/(x ^(3) - 1) = x^6 + x^3 + 1 Phi 16(x) = (x ^(16) - 1)/(x ^(4) - 1) = x^12 + x^8 + x^4 +1 Phi 25(x) = (x ^(25) - 1)/(x ^(5) - 1) = x^20 + x^15 + x^10 + x^5 +1 Here there are r positive terms, starting x^(r(r-1)) with powers descending in steps of r. I hope you find this interesting and useful - Ian Hutcheson === Subject: Re: e is transcendental (was: classes of transcendental numbers ? >> hey do you guys have a copy of the proof of e is transcendental? ..id >> really appreciate if somebody could help me. ive been searching and >> found nothing. >Spivack, CALCULUS, 2nd edition, Chapter 21. >>Are you sure that the proof there is for e being transcendental and not >>just irrational? I know that the proof of the irrationality of e is >>simple enough for a Calculus book, but that the transcendence is quite a >>bit more complicated. > Evidently you've never seen the book. It's not at all like the typical > book titled Calculus. Quite right, it's actually good. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: there is no such thing as infinity > I just ran it on my computer. It went all the way up to 2147483647 > and then went to to -2147483648 and started counting up again. I > guess we've found our first candidate for M. Very inefficient. I modified the program to count in increments of M/10, and it was done very quickly. It turns out that M= 11(M/10), which is embarrassing mostly because if we'd have thought of this sooner, we could have worked it out by hand. Well, I'm glad that's finally settled. (Hmmm...I wonder what journal I should send this off to....) Bart === Subject: Partition question Re:my Catalan question - if commutativity holds, the number of different bracketings is still lower and can be though of as ways in a partition graph (example n=4): 4 / 22 311 / 211 | 1111 One path would be ((a*a)*(a*a)) and the other (a*(a*(a*a))). Unluckily, #ways != #bracketings, as n=5 already shows: 2111->221->23 or 2111->311->32 are different ways to the same bracketing. But surely the sequence is known? -- Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de als man ankam wollte man werden, die geschichte schreiben, die doofen sollen sterben, der plan als man damals nach hamburg kam (Kettcar) === Subject: Re: Countable perfect set by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1AEUFu27848; I don't know if you will consider this a satisfying example, but how about the set Q of rational numbers. Q is closed in itself, is countable, and every point is an accumulation point. Or if you want a proper subset of Q, I suppose the rationals in [0,1] would work. Mark Motley >It is known that every perfect (every point is an accumulation point) >closed subset of a complete metric space is uncountable. >But, what about a non-complete metric space ? The proof of the >previous case cannot be adapted, and I don't manage to find a counter >example. >Does anyone have any idea ? Nobody knows a perfect, closed and >countable subset of a metric space ? >Vinect G. === Subject: Re: polynomial degree in general polynomial > Hello all, i am trying to prove that deg(fg)=deg(f)+deg(g) for f,g in > F[x1,...,xn], where F is a field. > I was able to prove that deg(fg)<=deg(f)+deg(g), but the full proof is > still alluding me. Can anyone provide some help or hints? The notation > is very cumbersome. If you refine the order on the exponents in a suitable way, every polynomial has a unique leading monomial i.e., an monomial that has the highest exponent w.r.t. that order with nonvanishing coefficient. Then you can imitate the proof for the case n=1. For example, you can compare two exponents by total degree first and break ties by comparing them lexicographic. For a = (a1, ... ,an) in N^n let |a| := sum( ai | i+1..n ) Then order the exponents via: a < b <==> |a| < |b| or ( |a| = |b| and am < bm ) where m = min { i | ai <> bi }. You can convince yourself, that this gives a total order on the exponents, which is also compatible with addition of exponents i.e., a < b ==> a+c < b+c. In particular, the product of the leading monomials of f and g is indeed the leading monomial of fg. Marc === Subject: Re: polynomial degree in general polynomial Adjunct Assistant Professor at the University of Montana. >Hello all, i am trying to prove that deg(fg)=deg(f)+deg(g) for f,g in >F[x1,...,xn], where F is a field. >I was able to prove that deg(fg)<=deg(f)+deg(g), but the full proof is >still alluding me. Can anyone provide some help or hints? The notation >is very cumbersome. Prove it for monomials. Then show that if deg(f) is different from deg(g), then deg(f+g) = max(deg(f),deg(g)), and use this to prove it for the product of a monomial with a polynomial, and then a polynomial with a polynomial. Of course, you need to be careful, depending on how you define degree. The easiest thing to do is to refine the degree by letting a monomial be of higher degree than another monomial if either the first has larger total degree, or else they have equal total degree and the first is larger lexicographically (compare the powers of x1, and take the largest; if they are equal, compare the powers of x2, etc.) -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Pattern argument, revisited >The more astute of you should see a pattern. Clearly the constant >factor on the right, which shows up as a coefficient on the left must >equal 2 mod 5 for this particular setup. >But why? >Clearly there is an *underlying* equation being multiplied in each >case, which I'll call >(5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2 >> >>No. You yourself see the difficulties with this approach below. > There's no other approach Decker. Given the pattern there has to be > some relevance of 2 mod 5 with prime multiples of the polynomial. >What's happening is that I'm multiplying that underlying equation by >integers that are 2 mod 5, which allows for *one* factorization if >you're to have integer coefficients. >Don't believe me, then try to multiply by 13 instead of 7, like 13 >equals 3 mod 5, so it will not work. >>Right. No more than your underlying equation will work when >>you multiply by 1. This equals 1 mod 5, so in your own words, >>it will not work. > It seems that you're missing the point Decker. > You emphasized the *single* example, whereas I've pointed out the > pattern when you consider more. > Now then, an enquiring mind should wonder why is 2 mod 5 important? Ummm. Might it have anything to do with that 2 that shows up in 25x^2 + 30x + 2? Let's see. Suppose we take k to be an arbitrary integer and consider (5a_1(x) + (2+5k))(5a_2(x) + (2+5k)) = (2+5k)(25x^2 + 30x + 2) The RHS can be rearranged to yield 25[(2+5k)(x^2+x)] + 5[(2+5k)(x-k)] + [5(2+5k)k+(2+5k)2] = 25[(2+5k)(x^2+x)] + 5[(2+5k)(x-k)] + [(2+5k)^2] So now if we expand the LHS we have 25a_1(x)a_2(x) + 5(2+5k)(a_1(x) + a_2(x)) + (2+5k)^2 and if we take a_1(x)a_2(x) = (2+5k)(x^2+x) a_1(x) + a_2(x) = x - k then we see that the a's satisfy a^2 - (x - k)a + (2+5k)(x^2+x) So it's obvious that the reason we can do this is that the constant terms on both sides are equal to (2+5k)^2, which is a consequence of the facts that (1) you decompose the RHS into a linear combination of powers of 5 and (2) the constant term of 25x^2+30x+2 is 2. This is so trivial that I presume your question why is 2 mod 5 important? is merely a rhetorical device. >The problem is that it's clear that y_1(x) and y_2(x) can't both be >algebraic integer functions. This is only a problem if you are somehow convinced that in (5y_1(x) + 1)(5y_2(x) + 2) = 25x^2 + 30x + 2 the y's have to be algebraic integer functions. There's no reason they should be, as you note above. > It's fairly easy to prove that with your own example > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x) > that they'd have to exist, and have to be algebraic integers, if the > exists algebraic integer functions A(x), and B(x) such that > 7 A(x) B(x) = (5a_1(x) + 7)(5a_2(x) + 7) > with the same restriction on the a's. Well, sure. We'd need two functions A, B: Z -> Algebraic integers such that A(x)B(x) = 25x^2 + 30x + 2 There are lots of functions A and B that satisfy this: A(x) = 1, B(x) = 25x^2 + 30x + 2 or A(x) = B(x) = sqrt(25x^2 + 30x + 2) But that's immaterial here. The point is that if you have (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where the a's are defined as above, then for almost all values of x, 7 DOES NOT divide either of a_1(x) or a_2(x). This is (1) an undisputable fact that (2) has been demonstrated here many times. > For other readers I want to point out that Rick Decker plays the nice > guy role, when his behavior here is contemptible, and reflects badly > not only upon him, but also upon Hamilton College. Oh, I think my reputation and the College's will survive. > His *duty* as a professor is to tell the truth and promote elucidating > the truth, not hiding it. > like 2 and 17, it's clear that there's more here than his attempts at > dismissal indicate. Better would be to replace more with less in the sentence above. > I hereby charge Rick Decker with academic fraud and note that his > college is responsible for this rogue professor. Charge noted and ignored with a wry chuckle. I did like the sobriquet rogue professor; it's on my office door now. > I've deliberately involved an official at his college to take away > plausible deniability for his school in a phone call I made months > ago. I LOVE Usenet. You couldn't pay for entertainment like this. BTW, you should probably direct any further calls to our Dean of Faculty. If you don't want to involve him, you could always notify my department Chair. See the from line of this post for his contact information. We do have a new President, but she'd probably just pass on any complaints to the Dean. > His college has been made aware of his behavior. > There may also be a civil matter involved at some point in the future. I'd love that. You'd better get a job first, though. I understand these things can be expensive. > Some of you seem to think this is a game, or that Usenet doesn't > matter. Right on both counts. > However, Usenet is a *public* forum read around the world. There is a > burden upon professors to tell the truth, and upon the colleges and > universities that promote them either directly or indirectly. > Possibly an example needs to be made to convince you all that you have > to be accountable for your public words. Bring it on, James. I urge you to take your best shot. Cordially, Rick === Subject: Re: there is no such thing as infinity >> I just ran it on my computer. It went all the way up to 2147483647 >> and then went to to -2147483648 and started counting up again. I >> guess we've found our first candidate for M. > Very inefficient. I modified the program to count in increments > of M/10, and it was done very quickly. It turns out that > M= 11(M/10), which is embarrassing mostly because if we'd have > thought of this sooner, we could have worked it out by hand. > Well, I'm glad that's finally settled. (Hmmm...I wonder what > journal I should send this off to....) The two-line program that I posted previously, namely print *, huge(1) end prints 2147483647 on my machine. Huge is a standard intrinsic function. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: CHALLENGE Hi I have a beautiful puzzle, and i also have a general method for finding the answer but what i'm looking for is a proof for why the method works. Here's the puzzle: Consider n number of poeple standing in a circle. All of them are numbered from 1 to n. Now, Man no. 1 is given a sword. He kills the person next to him (i.e Man No. 2) and passes the sword on to Man no.3. Man 3 in turn kills the man next to him (No. 4) and passes the sword on to the next man (i.e Man No. 5). This sequence continues until only one man is alive. Find which number is it. For eg, take n=5 then: |1| 2 3 4 5 (the number in | | has the sword) 1 |3| 4 5 1 3 |5| |3| 5 |3| Thus in the end man no. 3 remains. Try the puzzle if you haven't seen it before, its a nice exercise for the brain. Here's my conjecture for a general method to find the solution. 1) Find the binary representation for the number of people. eg for n=5 it is 101 2) take the first 1 of the representation and put it at the end. so, here it becomes 011. 3) Now convert this new binary number back to decimal and that is the answer. The decimal of 11 is 3, which is the answer. I have checked this method for many many values using a java program that I made , and it does work. But, I could not think of a proof for this. I will greatly appreciate it if you could post the proof should you find it. === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational > But is an algebraic number divided by a transcendental number always > irrational? > It's always transcendental, 0 is algebraic, right? === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational > The real part of a root of unity cannot be > an algebraic integer unless it is 0, 1, or -1. > How can we prove this? Could you please outline a proof? > Amanda > > If u=[arccos((sqrt(5)-1)/2 )]/pi, then e^(i pi u) has real part > (sqrt(5)-1)/2, which is an algebraic integer. So e^(i pi u) > is not a root of unity, that is: u is irrational. G.A.Edgar: What is an algebraic integer? I don't see (sqrt(5)-1)/2 being integral, and if this just means an algebraic number, then roots of unity can have much different real parts from 0, 1, -1. J === Subject: Re: My Childhood and Infinity > Mr. Feinstein, you seem to contradict the proof of Mr. Feinstein > on the impossibility of proving the Collatz 3n+1 -conjecture. It > relied on the necessity of calculating infinitely many values of a > function in order to prove the conjecture. Surely if M exists then the > Collatz conjecture can be proven/disproven through exhaustion alone. Don't tell my brother-in-law Craig Feinstein I said this, as he doesn't know I'm using his email address: His proof as you mentioned it is a complete fraud. I tried to tell him, but he threatened to have me confined. Dr. Ben Zona === Subject: Re: I need a site about number theory . > > I want to learn about number theory , If you know a good website > about > that please let me know. > > > Not a bad suggestion. A liitle more focussed is > > http://www.numbertheory.org/ntw/web.html > =-=-=-=- > Have you evere gone to > www.themathpages.com ?? > It is a site where you learn precalculus,trigonometry,arithmetic .The > texts there are so simple and learning there is so good. > I need a site such as mathpages.com but about number theory and I > You want to learn number theory, but you haven't got time to search with > Google? Why when somebody else has done that before I have to waste my time ? When you search with Google you will finde thousands of site that many of them are so boring that .... === Subject: How to compute the minimal distanct between a point and curve in N-dim space In the N-dimensional space, give a data point A and a curve f, how to write the explicit expression for calculating the minimal distance between A and f? Or have to use some nonlinear optimization method to calcualte it? Fred === Subject: Integro-differential equation I need to solve the following partial integro-differential equation for P(x,t) P_t(x,t) = a*P_x(x,t) + b*P_{xx}(x,t) - c*P(x,t) + c*{ int_0^{x} P(x-v,t)*e^{-v/d}/d ,dv + e^{-x/d} } subject to P(0,t) = 1 P(x,0) = 0 and a,b,c, and d are constants. My approach so far has been to Laplace transform first for t (t->s1, get ordinary integro-differential equation for P(x,s1)) and then for x (x->s2, get algebraic equation for P(s2,s1)). Then I can solve for P(s2,s1), do the inverse transform to get P(x,s1). Unfortunately the expression for P(x,s1) is to complex to find the inverse transform so I'm left with the solution in the form of a Bromwich integral (which I presumably could evaluate numerically for all relevant values). Does anyone know of a better way to attack this problem? Stefan Nilsson Marine ecology G.9ateborg University === Subject: Re: My Childhood and Infinity >> >> When I was a young three and a half year old lad, I used to write >> numbers on paper and line them across the room. My goal was to get to >> the largest number possible which I now call M. > 2M=M, obviously. The computer program to search for M is trivial: > Set N=0. Loop, incrementing N, until N+1=N. When the loop exits, N=M. > It's easy and requires no expensive equipment. Somebody ought to look. > Lead, follow, or get out of the way, Al. > Assume that the largest number is an integer. Assume > further that it's a positive integer. This means you > can use an unsigned integer format for the search > variable. It also means that you can use the following > shortcut method: Initialize the variable to zero, then > rather than repeatedly adding one until you reach the > highest value, subtract one instead. This is the short > way 'round via negative infinity! Why waste cycles? I get M = 4,294,967,295. Anyone else? -- KCS === Subject: Re: Qabala & Physics > The Others as The Ecclesia (see also Herman Hesse's Magister Ludi (AKA > The Glass Bead Game'). The Smashing of the Vessels is the Kabalistic version of the Big Bang. Bob Kolker === Subject: Mod operator Most mathematicians would say that x mod 7 is always a value between 0 and 6, regardless of the sign of x. But many computer languages do this the wrong way. For example, most mathematicians would insist that -15 mod 7 = 6 but many (but not all) programming languages give the wrong result, -1. In fact, the programming language Ada has two mod operators. One which returns 6 and one which returns -1 in the above example. Insightful comments welcome. -- Mail sent to this email address is automatically deleted (unread) on the server. Send replies to the newsgroup. === Subject: Re: JSH: Pattern argument, revisited >> [...] >> For other readers I want to point out that Rick Decker plays the nice >> guy role, when his behavior here is contemptible, and reflects badly >> not only upon him, but also upon Hamilton College. >Oh, I think my reputation and the College's will survive. >> His *duty* as a professor is to tell the truth and promote elucidating >> the truth, not hiding it. >> like 2 and 17, it's clear that there's more here than his attempts at >> dismissal indicate. >Better would be to replace more with less in the sentence above. >> I hereby charge Rick Decker with academic fraud and note that his >> college is responsible for this rogue professor. >Charge noted and ignored with a wry chuckle. I did like the >sobriquet rogue professor; it's on my office door now. >> I've deliberately involved an official at his college to take away >> plausible deniability for his school in a phone call I made months >> ago. >I LOVE Usenet. You couldn't pay for entertainment like this. BTW, >you should probably direct any further calls to our Dean of >Faculty. If you don't want to involve him, you could always >notify my department Chair. See the from line of this >post for his contact information. We do have a new President, >but she'd probably just pass on any complaints to the Dean. >> His college has been made aware of his behavior. >> There may also be a civil matter involved at some point in the future. >I'd love that. You'd better get a job first, though. I understand >these things can be expensive. >> Some of you seem to think this is a game, or that Usenet doesn't >> matter. >Right on both counts. >> However, Usenet is a *public* forum read around the world. There is a >> burden upon professors to tell the truth, and upon the colleges and >> universities that promote them either directly or indirectly. >> Possibly an example needs to be made to convince you all that you have >> to be accountable for your public words. >Bring it on, James. I urge you to take your best shot. Good to see you putting a brave face on things. You must be terrified he's going to actually go through with it or you wouldn't feel the need to pretend this way. (Just curious, who did he actually call? And is that the same person as the official that he's deliberately involved? Can't tell from the above...) >Cordially, >Rick ************************ David C. Ullrich === Subject: Re: Math Too Advanced For Mainstream Economists > I have written up a demonstration that wages and employment need not > be determined by the intersection of well-behaved supply and demand > curves for labor: > The poster claims to have constructed a model that has something to do with the intersection of well-behaved supply and demand curves for labor, but he has not. Rather, he comes close to presenting a model of the factor demand for labor, but he confuses movements along a demand curve with shifts of that curve. Further, he has confusion about the competition assumptions that go with the basic supply-demand setting, which set the rate of profits independently from the wage rate. As such, he may have a model of something, but it's nothing related to the workings of a competitive labor market. Even if this were a model of labor market behavior that contradicted the usual results of the standard supply and demand model, it would not be an interesting contribution unless it did a better job than the supply and demand (or other established labor models) of describing something interesting about the observed world. There are plenty of labor market models that use tools other than standard supply and demand analysis, I've cited some of them in this thread, but their value comes from their empirical relevance. Does this model show something of interest to any serious researcher? What do we observe that it explains better than competing approaches? > I think this thread would amusing to those who find JSH threads > amusing, but with a twist. > I was delighted to find in a dictionary the word MUMPSIMUS, > which means stubborn persistence in an error that has been > exposed. > -- Joan Robinson > > ...the original poster was unable to find a single piece of > empirical support for the labor market model presented... > The above, of course, is untrue, and it is hard to see how poor > Mark Witte cannot know it is untrue, if he had any clue about > what he is talking about. > A > demand curve is a relationship between price and quantity demanded. > If something else changes, such as prices of related goods in demand > (here the discount rate would be an example), this shifts the curve. > Poor Mark Witte is confused. A discount rate is not a price. > To be charitable to the original poster, his confusion probably stems > from his notion that a change by the firm in question in the number of > workers it hires would change the discount rate for the economy. > The above is a strawperson. No such claim is to be found at the above > URL. > Notice no claim is made about the discount rate for the economy > being changed by a change in how many workers firm hires. In fact, > an economy-wide market for financial capital is not described at > my URL. > Poor Mark Witte just doesn't understand accounting. If the best > return a firm can obtain is 10%, then that's what the firm gives > up in investing elsewhere. If costs are different for the best > investment (e.g., because the level of wages is different), then the > firm gives up some other rate of return than 10%. > To help poor Mark Witte out, here's some explanations of a concept > important for understanding Step 2 of the algorithm given at my > URL: > > My URL is more a tutorial to results well-established in the literature. > No claim is made to novelty there. For that matter, no claim is made > there that the relationship shown is a demand curve for labor. But that > claim is made in the 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 Childhood and Infinity Wow, shouldn't the Subject line read My Childhood at Infinity? > Mr. Feinstein, you seem to contradict the proof of Mr. Feinstein > on the impossibility of proving the Collatz 3n+1 -conjecture. It > relied on the necessity of calculating infinitely many values of a > function in order to prove the conjecture. Surely if M exists then the > Collatz conjecture can be proven/disproven through exhaustion alone. > Don't tell my brother-in-law Craig Feinstein I said this, as he > doesn't know I'm using his email address: His proof as you mentioned > it is a complete fraud. I tried to tell him, but he threatened to have > me confined. > Dr. Ben Zona === Subject: Re: Hamilton College announces new course > A new course, Introduction to Harristotelian Logic, will be presented by > new Hamilton College faculty member, James S. Harris. > The foundations of the method will first be established by defining > several innovative proof techniques: > Proof by Assertion > Proof by Repetition > Proof by Exhaustion > Proof by Vituperation > Next, the topic of Object Algebras will be introduced, starting with > definitions and basic axioms, and leading into some fundamental theorems. > Based on the preceding, the new concept of the Tautological Space will > be developed, in conjunction with prime number counting using partial > difference equations, and certain results will be proved, in particular > Harris's Error in Core Mathematics Theorem. > If there is time a new proof of Fermat's Last Theorem will be presented. of scheduling courses for next year. Should this course be cross-listed with any other departments? Should I alert my colleagues in, say, Philosophy, Psychology, or Sociology? Perhaps I should contact the folks in Creative Writing? Rick === Subject: Re: JSH: Pattern argument, revisited > (Just curious, who did he actually call? And is that the same > person as the official that he's deliberately involved? > Can't tell from the above...) I have no evidence that any such call was ever made. Of course the administration might be keeping mum while they prepare their case. Guess we'll just have to wait and see. Rick === Subject: Re: Mod operator > In fact, the programming language Ada has two mod operators. One > which returns 6 and one which returns -1 in the above example. Are they both called mod? -- === Subject: Re: Rings by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1AG2E803159; >> A Gaussian Ring is an integral domain verifying >> (i) the intersection of two principal ideals is a principal ideal >> Equivalent conditions are >> (ii) each couple of elements has a GCD >> (iii) each couple of elements has a LCM >> (iv) any couple of elements is elements is contained in a minimum principal ideal >> (v) any ideal generated by two elements is contained in a minimum principal ideal. >> I. Consider G=Z[X][[X1,X2,...]]. >> a) Is G a Gaussian Ring ? >yes, because every factor ring verify (ii) I am not so sure that it is a factor ring. The ring to be considered is a formal power series ring on the ring Z[X]. If a ring A is a principal ring, then the formal power series ring A[[X]] is a factor ring. But the condition A factor ring does not imply A[[X]] factor ring. And Z[X] is a non principal factor ring, even if Z is principal. >> b) Is G a Factor Ring ? >Yes >It's a well known problem : if A is a factor ring, so is A[X] See above under a) >> c) Does G verify the Bezout condition ? >GCD(X1,X2)=1 >But PX1+QX2 = 1 is false. Assuming PX1+QX2 =1, you substitute 0 to X1 >ans X2 you find 0=1. >> II. Any (other?) example of a Gaussian non Factor non Bezout Ring ? >Now you should find an example of a gaussian ring that is not a factor ring. That remains to be done. === Subject: Re: Request for comments on antiquated algebraic topology online-book Originator: israel@math.ubc.ca (Robert Israel) > Algebraic Topology > by Solomon Lefschetz > http://www.ams.org/online_bks/coll27/ > --- > I was wondering if anyone familiar with algebraic topology (assuming you > have the time and/or inclination) could skim through this text and let me > know what they think of it. I'm particularly interested in knowing of any > differences in terminology or definitions that may be present in this book > when compared to more modern treatments. Of course, I would be grateful for > any comments at all that people may have. > At first glance it seems quite good (starts from the very bottom), and is > easily an order of magnitude easier to understand than Hatcher's 'Algebraic > Topology' - also a free ebook, but much more modern. I only had time to take a quick glance at it, but it looks pretty out of date. I think that at the time this book was written, not all of the material that constitutes a standard first-year algebraic topology course today had been worked out, and much of that which had been worked out was still being polished. The book emphasizes some topics which are not emphasized so much anymore, and uses some obsolete terminology, or modern terminology in slightly different ways. For example, if you look at the definition of complex at the beginning of Chapter 3, it is (almost) equivalent to the modern definition of a chain complex (over the integers), but looks completely different. I didn't see singular homology theory in there (but I didn't have time to download and scan the whole book) while this is an essential part of most modern treatments. Exact sequences, which simplify and unify so many formulas in algebraic topology, are apparently not in the book (and maybe hadn't been invented yet). One could probably say much more about how this book differs from modern approaches but I don't want to do a historical study. Suffice it to say that when flipping through it, not too much of what I saw was immediately recognizable. Anyway, while reading old-fashioned texts won't do you any harm, I think that for learning well-established material, in general it is better to get a modern introduction first, because things get simplified and improved so much over time (and of course you need to know the modern terminology and approaches if you want to understand current developments and if you don't want people to look at you funny when you talk to them). After you have done that, then it can be fun and enlightening to look at older literature to get some idea of how people originally thought of things (because some intuition can get lost in the polishing process). For an introduction to algebraic topology, I would look for something written after 1960 or so. Hatcher's book is great, but quite sophisticated. The book by Greenberg and Harper gives a pretty concise introduction. I also like the book by Bredon a lot, if you can find a copy. === Subject: Re: How to compute the minimal distanct between a point and curve in N-dim space Fred escribi.97: > In the N-dimensional space, give a data point A and a curve f, > how to write the explicit expression for calculating the > minimal distance between A and f? > Or have to use some nonlinear optimization method to calcualte it? Do you has parmetric equations of the curve? If so, the distance from A to a generic point of the curve corresponding a value t of the parameter, is only function of t. Then minimize that one variable function. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: My Childhood and Infinity Discussion, linux) >> Assume that the largest number is an integer. Assume >> further that it's a positive integer. This means you >> can use an unsigned integer format for the search >> variable. It also means that you can use the following >> shortcut method: Initialize the variable to zero, then >> rather than repeatedly adding one until you reach the >> highest value, subtract one instead. This is the short >> way 'round via negative infinity! Why waste cycles? > I get M = 4,294,967,295. Anyone else? I got 13, but I did the search manually. -- All intelligent men are cowards. The Chinese are the world's worst fighters because they are an intelligent race[...] An average Chinese child knows what the European gray-haired statesmen do not know, that by fighting one gets killed or maimed. -- Lin Yutang === Subject: Riemann Integration let f:[0,1]->R be continious. then set of all such f's such that inegration with limits 0 to 1 of t^n*f9t) dt=0 is a) single element b)contably infinite C) infinite http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups ---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: Re: there is no such thing as infinity > Dr. Ben Zona Is your brother called Ari? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Mod operator > Most mathematicians would say that x mod 7 is always a No Real Mathematicians talk about x mod 7. They recognize mod as part of the ternary relation of congruence: a = b (mod n) means that (a-b)/n is an integer. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Silly question for someone with a big calculator. > are, even though you explained them correctly the first time round. > However, forgive me for being quite stubborn, aren't we > effectively saying - instead of the usual: let (k_n) be the sequence > of ratios, then for every epsilon > 0 exists N in naturals: > |k_n - sqrt(2)| < epsilon for all n > N - for any epsilon > 0, > we can find you a bunch of examples such that... ? > (Indirectly of course, since we are trying to show delta -> 0 > through substitution instead of k_n -> sqrt(2) ). Since k_{n+1} would fall between k_n and sqrt(2), we would only need to find a first example k_N, and then all k_n with n>N will be closer to sqrt(2). -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Rationals are Uncountable > For example, I wonder if your method actually gets all of the > rationals close to r. In X (-1,1) or in X' (-1-r,1-r)? Because if r > 1/2, it will be nowhere near X'. If you mean something else, the question is How could it not?. In X, if x_n is positive and less than 1, the sequence 1/2, 3/4, 7/8, ... has an eventual term larger than x_n, and x_n will be in the corresponding interval (or be an endpoint). In X', if a positive rational is less than 1-r, the sequence approaching 1_r will have an eventual term larger than that rational, and again it is in the corresponding interval (or is an endpoint). And much the same applies to negative rationals in the intervals. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: Riemann Integration > let f:[0,1]->R be continious. then set of all such f's such that > inegration with limits 0 to 1 of t^n*f9t) dt=0 is Please re-read your posts before posting them; it should be t^n*f(t). > a) single element > b)contably infinite > C) infinite The first possibility is the correct one. Jose Carlos Santos === Subject: Re: there is no such thing as infinity > In your cosmology is the number pi rational or irrational? Same > question goes for sqrt{2}. That's a good question. Pi an sqrt(2) are rational, obviously, because there is no such thing as irrational numbers. It's just an illusion invented by the Evil Empire, Ancient Greece. The way to calculate pi and sqrt(2) is just compute the decimals until you get to the Mth decimal place and you won't be able to go any further after that. > Explain this breakdown of morals due to the concept of infinity. Another good question. Morality is based on truth and common sense. The concept of infinity is the antithesis to both. Modern society is brainwashed into not only thinking that there is infinity but that there are lots of different kinds of infinity. This is the extent that common sense has been eliminated from society by Cantor and his gang! The Evil Empire, Ancient Greece, accepted to some extent, the concept of infinity. And they were one of the most immoral societies around. They destroyed common sense with their philosophy - and with its destruction, morality was also destroyed. Dr. Ben Zona === Subject: Re: JSH: Pattern argument > I've been watching the one and only James Harris Show for some time > now, but I'm still at a loss what James is actually trying to prove > (if anything). Is this a crucial part of some purported proof of FLT > or what? > Just curious. Do you have to ask? Of course it's FLT. What else is there? Nathan === Subject: Re: Are these two polynomials equivalent? > Can I claim that f(x)=a+b(x-e)+c(x-e)^2+d(x-e)^3 , > g(x)=m+n(x-e)+p1(x-e)^3+...+pn(x-e)^3 > are equivalent? > a,b,c,d,e,m,n,p1,p2,...,pn are constants, n is arbitrary > The key is, is it that there always exists p1,...,pn such that we can > drop the quadratic term c(x-e)^2 in f(x) ? > (my problem arises from the representation of a spline, because the > book seems to say that a cubic spline CAN always be expressed in the > second form) g(x) can simply be written as m+n(x-e)+p(x-e)^3, where p = p1+p2+...+pn. So the question becomes: given a,b,c,d,e, can you find m,n,p such that: a+b(x-e)+c(x-e)^2+d(x-e)^3 = m+n(x-e)+p(x-e)^3 ? Let's look at the expansions: f(x) = a + bx-be + cx^2-2cxe+ce^2 + dx^3-3dex^2+3de^2x-de^3 = (d)x^3 + (c-3de)x^2 + (b-2ce+3de^2)x + (a-be+ce^2-de^3) g(x) = m + nx-ne + px^3-3pex^2+3pe^2x-pe^3 = (p)x^3 + (-3pe)x^2 + (n+3pe^2)x + (m-ne-pe^3) So if f(x) = g(x), d = p c-3de = -3pe b-2ce+3de^2 = n+3pe^2 a-be+ce^2-de^3 = m-ne-pe^ Looking at the first 2 lines, f(x) = g(x) means c-3de = -3de means c=0. Having said all that, I haven't worked with cubic splines, but I suspect that the definition of the cubic spline is what causes the (x-e)^2 term to not exist. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: easy complex analysis problem...... > complex function f(z) = 1 / [2*{x^(1/2)}] I do not understand the definition of f. Try to explain it in a better way. Jose Carlos Santos === Subject: Re: Mod operator > Most mathematicians would say that x mod 7 is always a > value between 0 and 6, regardless of the sign of x. But > many computer languages do this the wrong way. For > example, most mathematicians would insist that > -15 mod 7 = 6 > but many (but not all) programming languages give the wrong > result, -1. In fact, the programming language Ada has two > mod operators. One which returns 6 and one which returns -1 > in the above example. > Insightful comments welcome. -1 mod 7 = 6 The issue is more one of convention than anything, IMO. What the mod operator actually does is define sets of numbers that are considered equivalent under a particular relation. So, for instance, -1 mod 7 = 6 really says that -1 is in the same set as 6 as defined by the (mod 7) relation. You could as easily say 6 mod 7 = -15 though most people would think it looks strange. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: triangulation 3QLpj-NoP*NzsIC,boYU]bQ]H'y<#4ga3$21: > Is there anyone in this group, who can help me with the triangulation > of a given polygon P under these two restrictions: > 1. Each triangle shall be as equilateral as possible > 2. No triangle has an area that exceeds a given constant O > Hans van Duijnhoven (Randers, Danmark). Most problems of optimal triangulation of polygons (without holes) can be solved by dynamic programming. -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science === Subject: Re: Mod operator by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1AHJhQ09552; >Most mathematicians would say that x mod 7 is always a >value between 0 and 6, regardless of the sign of x. No they would not. To all mathematicians x mod 7 is an equivalence class. It is not a single number. Furthermore, there are alternative ways of representing the reduced equivalence class. One can, as you suggest, represent it as [0,...6]. However, it can also be represented as [-3,....3]. There is no unique way of doing it. Further, in a computer application, which way is best depends on the application. >But >many computer languages do this the wrong way. There is no right way or wrong way. >For >example, most mathematicians would insist that > -15 mod 7 = 6 >but many (but not all) programming languages give the wrong >result, -1. -1 and 6 are equivalent mod 7. There is no wrong answer. It is a matter of convention. === Subject: Re: Countable perfect set > It is known that every perfect (every point is an accumulation point) > closed subset of a complete metric space is uncountable. > But, what about a non-complete metric space ? The proof of the > previous case cannot be adapted, and I don't manage to find a counter > example. > Does anyone have any idea ? Nobody knows a perfect, closed and > countable subset of a metric space ? Maybe I'm not seeing this properly (perhaps have a definition wrong). What if you only have one point x in your space X={x}? Give it the obvious topology ({} and {x}=X are open). Make your metric be d(x,y) = 0 since x must be equal to y. I think x is the only possible accumulation point. -- === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational > Hello > > How can we prove [arccos((sqrt(5)-1)/2 )] / pi [is irrational]? > The real part of a root of unity cannot be > an algebraic integer unless it is 0, 1, or -1. Isn't there a mistake here? If z^n =1, then z is an algebraic integer and so is it's conjugate z'. Since Re(z) = (z+z')/2, it follows Re(z) is half the sum of 2 algebraic integers, therefore an algebraic integer. That is, the real part of a root of unity is always an algebraic integer. Amanda > If u=[arccos((sqrt(5)-1)/2 )]/pi, then e^(i pi u) has real part > (sqrt(5)-1)/2, which is an algebraic integer. So e^(i pi u) > is not a root of unity, that is: u is irrational. === Subject: Re: Request for comments on antiquated algebraic topology online-book Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow) Originator: israel@math.ubc.ca (Robert Israel) >For an introduction to algebraic topology, I would look for something >written after 1960 or so. Hatcher's book is great, but quite >sophisticated. The book by Greenberg and Harper gives a pretty >concise introduction. I also like the book by Bredon a lot, if you >can find a copy. One advantage of Greenberg and Harper is that they introduce homotopy groups early on, which many elementary texts don't. Apart from that, though, I think it's too terse to serve as the primary text for self- study. I rather like Munkres's Elements of Algebraic Topology, partly because it spends quite a lot of time on simplicial homology. Some people might dislike it for the same reason, feeling that singular homology is more relevant for the budding topologist and that simplicial homology is just a distraction. But I think that simplicial homology is easier for the beginner to grasp, and it is still an important tool today in (for example) combinatorics. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Proofs - Please help quickly! > I'm really having trouble with 4 trigonometric functions. Here they are: Use the formulas you were taught in class > 1) Find an exact expression for: sin(pi/12) And again use the formulas > 2) Solve to 4 decimal places (0_ sin2x + cosx = 0 Use the formulas... might need more than just trig ones here. > 3) Solve sin^2 x - 2sinx - 1 = 0 and find the general solution. yes, this one is hard, I don't think I can do it > 4) Prove the following (this is a real toughy): tanx/secx = secx > Please help! Soon! === Subject: Re: Request for comments on antiquated algebraic topology online-book Originator: israel@math.ubc.ca (Robert Israel) > For an introduction to algebraic topology, I would look for something > written after 1960 or so. Hatcher's book is great, but quite > sophisticated. The book by Greenberg and Harper gives a pretty > concise introduction. I also like the book by Bredon a lot, if you > can find a copy. I also found J. Rotman's An Introduction to Algebraic Topology, (Springer Verlag, 1988) pretty good; as readable as it gets, IMHO. Mario. === Subject: de moivre's formula + polar representation z^n = w where z=x+iy and w=a+ib. w = r(cos(x) + i sin(x)) z = p(cos(y) + i sin(y)) z^n = p^n (cos(nqy + i sin(ny)). it follows p^n = r = |w| by uniqueness of the polar representation. can anyone detailed explain what is the uniqueness of the polar representation and why p^n = r = |w|? === Subject: branch of log z What exactly is to choose a branch of log z for complex z? I'd appreciate if someone help me picture this. === Subject: Re: simple integration problem > hi i am trying to prove: > If f(x)>=0 for all x in [a,b], f continuous on [a,b], and Int[f(x),{x, > a, b}], then f(x)=0 for all x in [a,b]. You forgot to add =0 after the integral. > here is my proof: > Suppose there exists an x_0 in [a,b] such that f(x_0)>0. Since f is > continuous we have that there exists a d>0 such that if |x-x_0| then f(x)>0. We immediately have that for any partition P of [a,b] > L(P,f)=Sum[m_i dx_i] > >= 2md > 0 > where m=inf f(x) over all |x-x_0| 0? If you take, for instance, f(x) = x, x_0 = 1, and d = 1, then m = 0. > Is that decent? No, but it easy to correct it. Or look for a recent thread called easy....analysis problem......... Jose Carlos Santos === Subject: Re: simple integration problem > hi i am trying to prove: > If f(x)>=0 for all x in [a,b], f continuous on [a,b], and Int[f(x),{x, > a, b}], then f(x)=0 for all x in [a,b]. > here is my proof: > Suppose there exists an x_0 in [a,b] such that f(x_0)>0. Since f is > continuous we have that there exists a d>0 such that if |x-x_0| then f(x)>0. We immediately have that for any partition P of [a,b] > L(P,f)=Sum[m_i dx_i] > >= 2md > 0 > where m=inf f(x) over all |x-x_0| L(P,f) used in Rudin, where L(P,f) is the lower sum for some partition > P of [a,b], m_i is smallest f-value on [x_(i-1), x_i], and dx_i = > x_(i-1) - x_i. The above inequality implies that Int[f(x), {x, x-d, > x+d}] > 0. Similarly, Int[f(x), {x, a, x_0-d}] >= 0 and Int[f(x), {x, > x_0+d, b}] >= 0, which implies that Int[f(x), {x, a, b}] > 0, > contradicting that Int[f(x), {x, a, b}] = 0. Therefore f(x)=0 for all > x in [a,b] > Is that decent? ************************************************ You have a small flaw in your proof: m could be zero even if f(x)>0 for all x with |x-x_0| f(x_0)/2 whenever |x-x_0|= f(x_0)/2. _________________________________________________________ Eric J. Wingler (wingler@math.ysu.edu) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 44555-0001 330-941-1817 === Subject: Re: branch of log z > What exactly is to choose a branch of log z for complex z? I'd > appreciate if someone help me picture this. See http://mathworld.wolfram.com/PrincipalBranch.html Jose Carlos Santos === Subject: Re: Unitary operator basics, lebesgue measure >How can I think about (T_a)* ?? First I need to represent T_a as a >matrix, Why? That would only confuse matters. -- Shmuel (Seymour J.) Metz, SysProg and JOAT 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 === Subject: Re: de moivre's formula + polar representation > z^n = w where z=x+iy and w=a+ib. > w = r(cos(x) + i sin(x)) > z = p(cos(y) + i sin(y)) > z^n = p^n (cos(nqy + i sin(ny)). it follows p^n = r = |w| by > uniqueness of the polar representation. can anyone detailed explain > what is the uniqueness of the polar representation If r and s are real numbers greater than zero and if a and b are real numbers, then we have r.(cos(a) + i.sin(a)) = s.(cos(b) + i.sin(b)) if and only if r = s and (a - b)/(2.pi) is an integer. > and why p^n = r = |w|? Jose Carlos Santos === Subject: Re: My Childhood and Infinity >I ask you what is 2 times infinity? Infinity, you might say, but that >doesn't make any sense and you know it. I am claiming that there is no >evidence for the Peano axioms that say that every number has a >successor! It seems clear to me that no one has come close to proving >this. On the other hand, there is abundant evidence for the number M >existing, i.e., Asimov's number. And 2M cannot exist by definition. > Mr. Feinstein, you seem to contradict the proof of Mr. Feinstein > on the impossibility of proving the Collatz 3n+1 -conjecture. It > relied on the necessity of calculating infinitely many values of a > function in order to prove the conjecture. Surely if M exists then the > Collatz conjecture can be proven/disproven through exhaustion alone. Hey, I just got an idea! This number M must be on the Collatz tree somewhere. And since there aren't any numbers larger than M, M itself must be the excursion for the Collatz sequence starting from M. Therefore, we have Mensanator's Theorem: M is even Proof: Assume M is odd. The next number in the Collatz sequence would then be 3*M+1 which contradicts the fact that M is the largest number, so the assumption that M was odd must be false, therefore, M is even. QED This is an important development in determining what M must be, since we can alter the FORTRAN program to only search for even numbers. It is so important, that we should start referring to M as Mensanator's Number === Subject: Re: Mod operator Adjunct Assistant Professor at the University of Montana. >Most mathematicians would say that x mod 7 is always a >value between 0 and 6, regardless of the sign of x. No. That would be the remainder of dividing x by 7 (assuming that x is an integer, anyway). Most mathematicians would say that x is congruent modulo 7 to one and only one of 0, 1, 2, 3, 4, 5, and 6; but it is also congruent to one and only one of -3, -2, -1, 0, 1, 2, and 3, and one and only one of 15, 16, 17, 18, 19, 20, and 21. In fact, they will know that the relation congruent modulo 7 partitions the integers into seven equivalence classes, of which 0, 1, 2, 3, 4, 5, 6 are a set of representatives, but that there are an infinite number of such sets of representatives. > But >many computer languages do this the wrong way. For >example, most mathematicians would insist that > -15 mod 7 = 6 >but many (but not all) programming languages give the wrong >result, -1. No; mathematicians would agree that -15 = 6 (mod 7), and they would also agree that -15 = -1 (mod 7). What they may not agree on is whether -1 is the remainder of dividing -15 by 7, since remainders are defined to be nonnegative. >In fact, the programming language Ada has two >mod operators. One which returns 6 and one which returns -1 >in the above example. While it is usually convenient to pick the standard set of representatives for the congruent modulo k relation to be 0, 1, 2, 3, 4, ..., k-1 (for instance, that makes the representative of the class of n modulo k to be exactly equal to the remainder of dividing n by k), it is also often very useful to pick as a set of representatives the ones with smallest absolute value, which would be, depending on the parity of k, either (-k/2)+1,...,0,1, 2, ..., (k/2) or (-k/2), (-k/2)+1,...,0,1,2,...,(k/2)-1 if k is even; and (1-k)/2, (1-k)/2+1,...,-1,0,1,2,...,(k-1)/2 if k is odd. (For some applications, other choices may even be preferable). Which one the computer program chooses is immaterial, so long as you know which one it is choosing and does so in a consistent manner. -- 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: there is no such thing as infinity @mozo.cc.purdue.edu: >> Very inefficient. I modified the program to count in increments >> of M/10, and it was done very quickly. It turns out that >> M= 11(M/10), which is embarrassing mostly because if we'd have >> thought of this sooner, we could have worked it out by hand. >> Well, I'm glad that's finally settled. (Hmmm...I wonder what >> journal I should send this off to....) > The two-line program that I posted previously, namely > print *, huge(1) > end > prints 2147483647 on my machine. Huge is a standard intrinsic > function. Must be a glitch, since 11/10 * Huge isn't even an integer. Bart === Subject: Re: existence of bounded linear functional O.K., so you've got f(x)=b_1*a_1 + ... for x = b_1*x_1 + ... in the subspace generated by all of the x_i. Then |f(x)| <= M ||x|| for all x in the subspace. Before we even bother extending this functional to the rest of X, how do we know ||f|| = M? The OP must have misstated his problem... (maybe he means ||x|| <= M ) > Duh, the norm is continuous. I'm officially an idiot. > Hmmm...another sticking point (apart from this norm stuff). Suppose > | b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| > for some M for each n and for scalars b_1,...,b_n. We want to construct a > functional x' in X' such that x'(x_j)=a_j for ALL j = 1,2,.... > Consider the subspace spanned by ALL the x_k, and call it S. Then any x in > S > can be written > x = b_1x_1 + b_2x_2 + .... > Define f(x) = b_1a_1 + b_2a_2 + .... > Then f(x_j)=a_j for all j = 1,2... > But is f bounded? The given inequality only works for finite n, and > |f(x)| = |b_1a_1 + ... | = |lim b_1a_1 + ... + b_na_n| = lim |b_1a_1 + > ... > + b_na_n| < = lim M ||b_1x_1 + ... b_nx_n||, but > lim ||b_1x_1 + ... + b_nx_n|| = ||lim b_1x_1 + ... b_nx_n||, so how do > you > show f is bounded??? >Ah, of course. I still have one lingering question, though. Given {x_k} > a >sequence in X and {a_k} a sequence of scalars, define a bounded linear >functional on the subspace generated by x_1,...,x_n by f(b_1x_1 + ... + >b_nx_n) = b_1f(x_1) + ... + b_nf(x_n) , f(x_k)=a_k. >How does the requirement that >| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| >guarantee that ||f|| = M? It's obvious that ||f|| <= M, but how do you >assert equality? > > Oh. It doesn't, of course, and I don't. If x_1,...,x_n already > span X, you're clearly out of luck, and your statement is false. > Hahn-Banach gives you x' with ||x'|| <= M. If x_1,...,x_n don't > span X (e.g. if X is infinite-dimensional), it also gives you > y' <> 0 such that y'(x_j) = 0 for j=1..n. For suitable t, > x' + t y' will be what you want. > > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 > > >>Let X normed linear space, X' it's dual. a_1,...,a_n some scalars, >>x_1,...,x_n some elements of X. How do you show that >>| b_1a_1 + ... + b_na_n | < = M || b_1x_1 + ... + b_nx_n|| >>for some M for each n and for scalars b_1,...,b_n implies that there > exists >>x' in X' satisfying x'(x_k)=a_k and ||x'|| = M. >>i don't know how to show this without explicitly constructing the > bounded >>linear functional on X, and of course i'm stuck on constructing it > (hence >>the problem in the first place).. >>if X is finite dimensional, then each x in X can be written x = b_1 > x_1 + >>... + b_n x_n on some basis {x_k}..then define x'(x)=b_1 x'(x_1) + > ... > + > b_n >>x'(x_n) and defining x'(x_k)=a_k. This is bounded by the given > constraint. >> >> Careful: the x_n's were given, and might not be a basis. >> >>But what to do if X is infinite-dimensional??? >> >> Use the Hahn-Banach Theorem. > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > === Subject: Re: triangulation >> Is there anyone in this group, who can help me with the triangulation >> of a given polygon P under these two restrictions: >> 1. Each triangle shall be as equilateral as possible >> 2. No triangle has an area that exceeds a given constant O >> Hans van Duijnhoven (Randers, Danmark). >Most problems of optimal triangulation of polygons (without holes) can >be solved by dynamic programming. > Hans === Subject: Re: Mod operator > In fact, the programming language Ada has two mod operators. One > which returns 6 and one which returns -1 in the above example. > Are they both called mod? IIRC, it depends upon the types being operated on. In Ada 95 the integer types are subdivided into signed integer types and modular types. The signed integer types are Integer and so on. The modular types are unsigned integer types which exhibit cyclic arithmetic. Hopefully someone will correct me if my memory fails me. -- Mail sent to this email address is automatically deleted (unread) on the server. Send replies to the newsgroup. === Subject: Re: My Childhood and Infinity >Don't tell my brother-in-law Craig Feinstein I said this, as he >doesn't know I'm using his email address: His proof as you mentioned >it is a complete fraud. I tried to tell him, but he threatened to have >me confined. And what does Social Security Administration have to say about you trolling sci.math from their computers? -- I'm not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: My Childhood and Infinity by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1AIMqS15158; >> You were ingorant as a child. You are stupid as an adult. How big is >> 2M, git? >You missed the point of my argument entirely. 2M cannot exist. I feel this is a good point; as, if M exists, M is the largest number, then 2M is indeed nonsense. Hisanobu Shinya === Subject: Re: [arccos((sqrt(5)-1)/2 )] / pi is irrational > Hello > How can we prove [arccos((sqrt(5)-1)/2 )] / pi [is irrational]? >> The real part of a root of unity cannot be >> an algebraic integer unless it is 0, 1, or -1. > Isn't there a mistake here? If z^n =1, then z is an algebraic integer > and so is it's conjugate z'. Since Re(z) = (z+z')/2, it follows Re(z) > is half the sum of 2 algebraic integers, therefore an algebraic > integer. Mmm. Methinks algebraic integers don't always work like ordinary ones. For instance, is (1+sqrt(2))/2 an algebraic integer? Its minimal polynomial is 4x^2-4x-1, so... That is, the real part of a root of unity is always an > algebraic integer. > Amanda >> If u=[arccos((sqrt(5)-1)/2 )]/pi, then e^(i pi u) has real part >> (sqrt(5)-1)/2, which is an algebraic integer. So e^(i pi u) >> is not a root of unity, that is: u is irrational. === Subject: Re: there is no such thing as infinity >The way to calculate pi and sqrt(2) is just compute the decimals until >you get to the Mth decimal place and you won't be able to go any >further after that. >Dr. Ben Zona I just calculated sqrt(2) that way. Then I multiplied them back together with my M&M simulator. The machine got very warm doing this because it was having trouble with the 2M decimal places in the answer. Two unfortunate things occurred: 1. The answer didn't come out 2. 2. It melted the chocolate, exposing the nut. --Lynn === Subject: Newbie: Integration? y = x^4 - 3x^2 +5 that we can integrate. I am reading some stuff where: For 10,000 values of x we have corresponding observed values for y. Now, this can be plotted on a scatter plot but it is not really available an equation. Then they integrate it. They use matrix notation - which probably is the confusing part as there is no equation like above. Can someone explain how they do this? Point to some place where I can read basic stuff on it. Of course, if the question is not clear, I will repost. === Subject: Re: CHALLENGE .91Ä siddharth jain .91[CapitalEth].91ñ.93»[EDouble Dot]±.93Ì.91[Micro] .93¡.93.b3.91Ë .91.b9.91¬.91ü.93á[EDo ubleDot].b9.91± > Here's the puzzle: > Consider n number of poeple standing in a circle. All of them are > numbered from 1 to n. Now, Man no. 1 is given a sword. He kills the > person next to him (i.e Man No. 2) and passes the sword on to Man > no.3. Man 3 in turn kills the man next to him (No. 4) and passes the > sword on to the next man (i.e Man No. 5). This sequence continues > until only one man is alive. Find which number is it. > For eg, take n=5 > then: > |1| 2 3 4 5 (the number in | | has the sword) > 1 |3| 4 5 > 1 3 |5| > |3| 5 > |3| > Thus in the end man no. 3 remains. It's ill defined. You didn't specify that the person with the sword always kills a person with a higher number. And even if this is the case, at some point someone is bound to start killing people with lower numbers, since its a modulo circle. So: At step 2 above, it could easily be: 1 |3| 4 5 4 |5| |5| > Try the puzzle if you haven't seen it before, its a nice exercise for > the brain. > Here's my conjecture for a general method to find the solution. > 1) Find the binary representation for the number of people. eg for n=5 > it is 101 > 2) take the first 1 of the representation and put it at the end. so, > here it becomes 011. > 3) Now convert this new binary number back to decimal and that is the > answer. The decimal of 11 is 3, which is the answer. > I have checked this method for many many values using a java program > that I made , and it does work. But, I could not think of a proof for > this. I will greatly appreciate it if you could post the proof should > you find it. -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ------------------------------------------ Eventually, _everything_ is understandable === Subject: Re: My Childhood and Infinity