> I'm taeching the beginning epsilon-delta course. Looking > ahead in the book (Ross, Elementary Analysis: the Theory > of Calculus) I see the following definition: Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for > every sequence (x_n) in A such that x -> a we have > f(x_n) -> L. At first I thought this must be a typo. But it turns out > he means it - later when he shows that this definition > is equivalent to the one in terms of epsilon and delta > the condition is |x - a| < delta, not 0 < |x-a| < delta. I'm shocked. Is this version of the definition actually > standard in some circles? My impression is that the David, What's the other definition? -k ==== >> I'm taeching the beginning epsilon-delta course. Looking >> ahead in the book (Ross, Elementary Analysis: the Theory >> of Calculus) I see the following definition: >Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for >> every sequence (x_n) in A such that x -> a we have >> f(x_n) -> L. >At first I thought this must be a typo. But it turns out >> he means it - later when he shows that this definition >> is equivalent to the one in terms of epsilon and delta >> the condition is |x - a| < delta, not 0 < |x-a| < delta. >I'm shocked. Is this version of the definition actually >> standard in some circles? My impression is that the > David, > What's the other definition? He told you. The other definition has 0 < |x-a| < delta, not just |x-a| < delta. But I think this has come up before, and the verdict was that yes, as unlikely as it may seem, this version of the definition is actually being taught in some places, notably in France. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== I'm taeching the beginning epsilon-delta course. Looking > ahead in the book (Ross, Elementary Analysis: the Theory > of Calculus) I see the following definition: Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for > every sequence (x_n) in A such that x -> a we have > f(x_n) -> L. At first I thought this must be a typo. But it turns out > he means it - later when he shows that this definition > is equivalent to the one in terms of epsilon and delta > the condition is |x - a| < delta, not 0 < |x-a| < delta. I'm shocked. Is this version of the definition actually > standard in some circles? My impression is that the David, What's the other definition? He told you. The other definition has 0 < |x-a| < delta, not just |x-a| > < delta. But I think this has come up before, and the verdict was that yes, as > unlikely as it may seem, this version of the definition is actually being > taught in some places, notably in France. For the definition of continuity of f in a it doesn't really matter of course, since 0 = |a-a| < d and |f(a)-f(a)| < e go together automatically. With limits there can be a problem since the second expression |f(a) - L| can obviously exceed epsilon if we have a function that is discontinuous *but* defined in a. But it not only happens in France. I have seen it in many places. It happens on Mathworld as well: http://mathworld.wolfram.com/Epsilon-DeltaDefinition.html | ... for all epsilon > 0 there is delta > 0 such that, | whenever |x-x0| < delta, then |f(x)-y0| < epsilon. | These two statements are equivalent formulations | of the definition of the limit | ( lim [ x->x0; f(x) ] = y0 ) OTOH on http://mathworld.wolfram.com/ContinuousFunction.html they say that continuity is sometimes defined in terms of epsilon-delta and limit, and then they give an example where they say (except possibly x0 itself) where they in fact should have said (and certainly at x0 itself), since they *are* defining continuity here ;-) I think it is a matter of mild sloppyness. At least in the case of these Mathworld pages... Dirk Vdm ==== > I'm taeching the beginning epsilon-delta course. Looking > ahead in the book (Ross, Elementary Analysis: the Theory > of Calculus) I see the following definition: Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for > every sequence (x_n) in A such that x -> a we have > f(x_n) -> L. At first I thought this must be a typo. But it turns out > he means it - later when he shows that this definition > is equivalent to the one in terms of epsilon and delta > the condition is |x - a| < delta, not 0 < |x-a| < delta. I'm shocked. Is this version of the definition actually > standard in some circles? My impression is that the David, What's the other definition? He told you. The other definition has 0 < |x-a| < delta, not just |x-a| > < delta. But I think this has come up before, and the verdict was that yes, as > unlikely as it may seem, this version of the definition is actually being > taught in some places, notably in France. How is the definition above not circular? I mean, how do you know what x->a (guess that must be x_n->a) means without knowing what a limit is? Wilbert ==== > I'm taeching the beginning epsilon-delta course. Looking > ahead in the book (Ross, Elementary Analysis: the Theory > of Calculus) I see the following definition: Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for > every sequence (x_n) in A such that x -> a we have > f(x_n) -> L. At first I thought this must be a typo. But it turns out > he means it - later when he shows that this definition > is equivalent to the one in terms of epsilon and delta > the condition is |x - a| < delta, not 0 < |x-a| < delta. I'm shocked. Is this version of the definition actually > standard in some circles? My impression is that the > > David, > > What's the other definition? He told you. The other definition has 0 < |x-a| < delta, not just |x-a| > < delta. But I think this has come up before, and the verdict was that yes, as > unlikely as it may seem, this version of the definition is actually being > taught in some places, notably in France. How is the definition above not circular? I mean, how do you know > what x->a (guess that must be x_n->a) means without knowing what > a limit is? I wondered the same thing at first, but the above is defining convergence of a function in terms of convergence of sequences of numbers. So sequential convergence must be defined elsewhere. - Randy ==== > How is the definition above not circular? I mean, how do you know > what x->a (guess that must be x_n->a) means without knowing what > a limit is? The book has a whole chapter on convergence of sequences. Now we are in the following chapter, where that is used to define convergence of functions. The idea is that learning to do these proofs, which is hard stuff (for many students), is best done first with sequences. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== >> I'm taeching the beginning epsilon-delta course. Looking >> ahead in the book (Ross, Elementary Analysis: the Theory >> of Calculus) I see the following definition: >> Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for >> every sequence (x_n) in A such that x -> a we have >> f(x_n) -> L. >> At first I thought this must be a typo. But it turns out >> he means it - later when he shows that this definition >> is equivalent to the one in terms of epsilon and delta >> the condition is |x - a| < delta, not 0 < |x-a| < delta. >> I'm shocked. Is this version of the definition actually >> standard in some circles? My impression is that the >> David, >> What's the other definition? >> He told you. The other definition has 0 < |x-a| < delta, not just |x-a| >> < delta. >> But I think this has come up before, and the verdict was that yes, as >> unlikely as it may seem, this version of the definition is actually being >> taught in some places, notably in France. > How is the definition above not circular? I mean, how do you know > what x->a (guess that must be x_n->a) means without knowing what > a limit is? The definition above? I'm guessing you mean the sequence definition, whereas the two definitions I was discussing have nothing to do with sequences. But it's certainly possible to define the limit of a sequence in a way that does not imply a particular definition for the limit of an arbitrary function at a point. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== I'm taeching the beginning epsilon-delta course. Looking > ahead in the book (Ross, Elementary Analysis: the Theory > of Calculus) I see the following definition: Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for > every sequence (x_n) in A such that x -> a we have > f(x_n) -> L. At first I thought this must be a typo. But it turns out > he means it - later when he shows that this definition > is equivalent to the one in terms of epsilon and delta > the condition is |x - a| < delta, not 0 < |x-a| < delta. I'm shocked. Is this version of the definition actually > standard in some circles? My impression is that the What's the other definition? For any epsilon>0 there exists delta>0 such that if 0 < |x-a| < delta then |f(x)-L| I'm taeching the beginning epsilon-delta course. Looking >> ahead in the book (Ross, Elementary Analysis: the Theory >> of Calculus) I see the following definition: >> Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for >> every sequence (x_n) in A such that x -> a we have >> f(x_n) -> L. >> At first I thought this must be a typo. But it turns out >> he means it - later when he shows that this definition >> is equivalent to the one in terms of epsilon and delta >> the condition is |x - a| < delta, not 0 < |x-a| < delta. >> I'm shocked. Is this version of the definition actually >> standard in some circles? My impression is that the >What's the other definition? >For any epsilon>0 there exists delta>0 such that if 0 < |x-a| < delta then >|f(x)-L|seen sequences used instead of neighbourhoods in the def'n of the limit of a >function. Surely the traditional def'n is better, because it generalizes >later to other kinds of spaces, just as Dr. U says. What bothered me was not sequences versus epsilons, what bothered me was that it was |x-a| < delta instead of 0 < |x-a| < delta. (But Edgar has pointed out I wasn't reading the notation carefully enough; the definition with |x-a| < delta was a definition of something other than lim_{x->a} f(x).) Ross seems to think that things will be easiest to follow if he first talks about convergent sequences and then bases everything else on them. I'm not sure that that's right but I don't have any big complaint with it. >LH >Times are bad. Children don't listen to their parents and everyone is >writing a book. >-- attributed to Cicero, 106-43 BC ************************ ==== Ross seems to think that things will be easiest to follow if he first > talks about convergent sequences and then bases everything > else on them. I'm not sure that that's right but I don't have any > big complaint with it. > That's the trouble with really understanding the subject. Equivalent definitions are, well, equivalent, so why not just go with the flow? We need people versed in pedagogy to teach courses, so they'll look at things from an educational point of view. Jon Miller ==== Ross seems to think that things will be easiest to follow if he first >> talks about convergent sequences and then bases everything >> else on them. I'm not sure that that's right but I don't have any >> big complaint with it. That's the trouble with really understanding the subject. Equivalent >definitions are, well, equivalent, so why not just go with the flow? We need people versed in pedagogy to teach courses, so they'll look at >things from an educational point of view. Fascinating. The fact that I'm not certain his opinion is correct but I'm also not certain it's wrong means I'm not looking at things from an educational point of view. Huh. Yes, it certainly is true that having people who really understand a subject teach it is a bad idea. (On what planet, exactly?) >Jon Miller > ************************ ==== > me was that it was |x-a| < delta instead of 0 < |x-a| < delta. (But wouldnt |x-a| < delta, delta > 0 (unless Ross didnt say that delta was > 0), be meaning the same as 0 < |x-a| < delta. I am not a mathematics professor but I am curious to know, why this notation would mean so much. -k ==== me was that it was |x-a| < delta instead of 0 < |x-a| < delta. (But wouldnt |x-a| < delta, delta > 0 (unless Ross didnt say that delta was > 0), >be meaning the same as 0 < |x-a| < delta. No, 0 < |x-a| < delta implies that x is not equal to a, while |x-a| < delta allows x = a. >I am not a mathematics professor but I am curious to know, why this notation >would mean so much. Define f(x) = 0 for all x except x = 0; let f(0) = 1. By one version of the definition lim_{x->0} f(x) = 0, while by the other version this limit does not exist. >-k > ************************ ==== No, 0 < |x-a| < delta implies that x is not equal to a, while |x-a| < > delta allows x = a. Silly of me to not have seen that. -k ==== >> me was that it was |x-a| < delta instead of 0 < |x-a| < delta. (But > wouldnt |x-a| < delta, delta > 0 (unless Ross didnt say that delta was > 0), > be meaning the same as 0 < |x-a| < delta. > I am not a mathematics professor but I am curious to know, why this notation > would mean so much. They don't mean the same, because 0 < |x-a| < delta specifically excludes the case x = a, while |x-a| < delta, delta > 0 does not. It makes a big difference when you are considering the limit of a discontinuous function. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== > I need some help in coming up with a method or equation to solve the > variables based on the pattern below. I've ran into a couple of > databases that store product attributes/elements in a matrix like > fashion and I'm trying to reverse engineer that matrix. Very much > kudos to anyone who could help me or point me in the right direction. The only information that would be available to me: > a1 + b1 + c1 = 7 > a1 + b1 + c2 = 11 > a1 + b2 + c1 = 12 > a1 + b2 + c2 = 16 a2 + b1 + c1 = 8 > a2 + b1 + c2 = 12 > a2 + b2 + c1 = 13 > a2 + b2 + c2 = 17 > a3 + b1 + c1 = 4 > a3 + b1 + c2 = 8 > a3 + b2 + c1 = 9 > a3 + b2 + c2 = 13 Goal: Need method/equation to solve a1, a2, a3, b1, b2, c1, c2 based > on the pattern. Note: Each variable a, b, c, can expand into unlimited elements beyond > this example, also what if there is 5 variables with many elements, > a?, b?, c?, d?. Cheat sheet of variables: > a1 = 5 > a2 = 6 > a3 = 2 > b1 = 0 > b2 = 5 > c1 = 2 First, your unknowns are not uniquely determined by the data. > In the solution you give (with, I assume, c2 = 6), add an arbitrary > constant to a1,a2,a3, and a possibly different arbitrary constant > to b1,b2; then subtract the sum of the constants from c1,c2. > The new values of a,b,c will still give the same data values. There are several steps to a solution. First we need some notation. > Let N = the number of data values. > Let L_a, L_b, ... = the number of unknown a's, b's, ... . > Let K = L_a + L_b + ... . > In your example, N = 12, L_a = 3, L_b = L_c = 2, and K = 7. Let y be the N-vector of data values, and let w be a K-vector of unknowns. > (It's traditional in these problems to call the vector of unknowns b, > but I'll use w, to avoid confusion with your b.) In your example, > y = (7,11,...,9,13), and w = (a1,a2,a3,b1,b2,c1,c2). Let X be an N x K matrix in which x_ij = 1 or 0 according as y_i does > or does not depend on w_j. (Actually, x_ij is a multiplier on w_j.) > For your example, X is a1 a2 a3 b1 b2 c1 c2 > 1 0 0 1 0 1 0 a1 + b1 + c1 = 7 > 1 0 0 1 0 0 1 a1 + b1 + c2 = 11 > 1 0 0 0 1 1 0 a1 + b2 + c1 = 12 > 1 0 0 0 1 0 1 a1 + b2 + c2 = 16 0 1 0 1 0 1 0 a2 + b1 + c1 = 8 > 0 1 0 1 0 0 1 a2 + b1 + c2 = 12 > 0 1 0 0 1 1 0 a2 + b2 + c1 = 13 > 0 1 0 0 1 0 1 a2 + b2 + c2 = 17 0 0 1 1 0 1 0 a3 + b1 + c1 = 4 > 0 0 1 1 0 0 1 a3 + b1 + c2 = 8 > 0 0 1 0 1 1 0 a3 + b2 + c1 = 9 > 0 0 1 0 1 0 1 a3 + b2 + c2 = 13 Then in matrix terms your model is y = X*w + e, where e is an > N-vector of random errors that happen to be zero in your example. The easiest way to handle the indeterminacy of the unknowns is to fix > the value of one of them (usually the last one, but it really doesn't > matter which one) in each set at zero, and to create a new unknown, say > m, that is present in every observation. In the matrices, the zeroed > unknowns are deleted from w and X, and new entries are created for m. > It is traditional to make m the first unknown (i.e., w_0), with a > corresponding column 0 in X. The general form of the model is still > y = X*w + e, but the new X has only K-G+1 columns, where G = the number > of original sets of unknowns. For your example, G = 3; the new X is m a1 a2 b1 c1 > 1 1 0 1 1 m + a1 + b1 + c1 = 7 > 1 1 0 1 0 m + a1 + b1 + 0 = 11 > 1 1 0 0 1 m + a1 + 0 + c1 = 12 > 1 1 0 0 0 m + a1 + 0 + 0 = 16 1 0 1 1 1 m + a2 + b1 + c1 = 8 > 1 0 1 1 0 m + a2 + b1 + 0 = 12 > 1 0 1 0 1 m + a2 + 0 + c1 = 13 > 1 0 1 0 0 m + a2 + 0 + 0 = 17 1 0 0 1 1 m + 0 + b1 + c1 = 4 > 1 0 0 1 0 m + 0 + b1 + 0 = 8 > 1 0 0 0 1 m + 0 + 0 + c1 = 9 > 1 0 0 0 0 m + 0 + 0 + 0 = 13 The new values of the unknowns are m = 2+5+6 = 13 > a1 = 5 - 2 = 3 > a2 = 6 - 2 = 4 > b1 = 0 - 5 = -5 > c1 = 2 - 6 = -4 In matrix terms, the solution is w = (X'X)^-1 * X'y. What you're doing > is called multiple regression with dummy-coded categorical predictors. > It should be covered in any good book on multiple regression. comprehensible until the very end. Could you please explain how you came up with the new values of the unknowns? m = 2+5+6 = 13 a1 = 5 - 2 = 3 a2 = 6 - 2 = 4 b1 = 0 - 5 = -5 c1 = 2 - 6 = -4 I don't understand where 2, 5, and 6 comes from in the answer for m. I don't understand where the 5, and 2 comes from in the answer for a1. I don't understand where the 6, and 2 comes from in the answer for a2. I don't understand where the 0, and 5 comes from in the answer for b1. I don't understand where the 2, and 6 comes from in the answer for c1. ==== > [...] > Could you please explain how you came up with the new values of the > unknowns? > m = 2+5+6 = 13 > a1 = 5 - 2 = 3 > a2 = 6 - 2 = 4 > b1 = 0 - 5 = -5 > c1 = 2 - 6 = -4 > [...] There are two ways to get the new values of the unknowns. The first is specific to your example. Changing the notation, we have y_ijk = a_i + b_j + c_k = (a_i - a_3) + (b_k - b_2) + (c_k - c_2) + (a_3 + b_2 + c_2) = ( new a_i ) + ( new b_j ) + ( new c_k ) + ( m ), The new a_3, b_2, c_2 are all zero. The other way is more general: evaluate w = (X'X)^-1 X'y, or (equivalently) solve the matrix equation X'Xw = X'y. ==== Where can I find the best (the simplest, elegant) proof of the prime number theorem (a) with complex variables allowed (b) without complex variables (the so called elementary proof)? ==== > Where can I find the best (the simplest, elegant) proof of the prime number > theorem > (a) with complex variables allowed > (b) without complex variables (the so called elementary proof)? a while back. He gives a very elegant and short proof (due to D.J. Newman), using barely more complex analysis than Cauchy's theorem: Zagier, D. Newman's short proof of the prime number theorem. Amer. Math. Monthly 104 (1997), no. 8, 705--708. Another excellent account of this proof can be found in Hlawka, Edmund; Schoissengeier, Johannes; Taschner, Rudolf Geometric and analytic number theory. Translated from the 1986 German edition by Charles Thomas. Universitext. Springer-Verlag, Berlin, 1991. x+238 pp. ISBN 3-540-52016-3 This book treats Newman's Tauberian theorem, uses it to deduce a weakened version of the Ikehara-Wiener theorem, and then applies the latter to give simple proofs of the (usual) prime number theorem, Hecke's prime number theorem for the Gaussian integers and the prime number theorem for arithmetic progressions. For (b), there is a readable and detailed presentation of Selberg's version of the elementary proof in Gioia, Anthony A. The theory of numbers. An introduction. Reprint of the 1970 original. Dover Publications, Inc., Mineola, NY, 2001. xii+207 pp. ISBN: 0-486-41449-3 Unfortunately I think Gioia isn't too careful about specifying the range over which his O-estimates are valid, which leads to some confusion near the end of the proof. But it can all be patched up by reading side by side with, e.g., Shapiro, Harold N. Introduction to the theory of numbers. Pure and Applied Mathematics. A Wiley-Interscience Publication. John Wiley & Sons, Inc., New York, 1983. xii+459 ISBN 0-471-86737-3 Shapiro presents both the Selberg and Erdos versions of the elementary proof. As far as I'm aware, this book represents the most thorough textbook-treatment of elementary prime number theory. Unfortunately it's been out of print for a while now. Hope this helps, Paul ==== > Where can I find the best (the simplest, elegant) proof of the prime > number theorem > (a) with complex variables allowed Apostol's Introduction to Analytic Number Theory runs through the proof in Chapter 4. I don't know if this is the best, but it is broken down in sections, and peppered with exercises. > (b) without complex variables (the so called elementary proof)? At the end of Chapter 4, there is a section that sketches the elementary proof, which may be helpful. Bart ==== >Where can I find the best (the simplest, elegant) proof of the prime number >theorem >(a) with complex variables allowed >(b) without complex variables (the so called elementary proof)? Actually there are non-complex proofs other than the elementary one, for example you can get it from Wiener's Tauberian Theorem (see Rudin Functional Analysis.) > ************************ ==== The subject of this post was the title of a paper my father gave me while I was growing up (more than 25 years ago). My fater was a clerk at the railroad and did not have a college degree but he did budget forecasting and hence worked with numbers a lot. My father brought home the paper The Complexity of Mere Simplification because he knew I had an interest in mathematics. The paper took the equation 1+1=2 and extrapolated into a complex mathematical equation that included many identities and well known formulas in mathematics. I kept this paper because I wanted to understand the formulas and before I graduated from Iowa State University with a bachelors degree in mathematics, the complexity of mere simplification was no wonder to me. post to this newsgroup a copy. I have long since lost this paper I would like to find a copy if I could. ==== Mike > > Does anyone know how to generate a random vector x = (x_1, ..., x_n) > uniformly from |x| = 1? Any recommendations on what books I need to ==== Mike Does anyone know how to generate a random vector x = (x_1, ..., x_n) > uniformly from |x| = 1? Any recommendations on what books I need to Mike What norm are you using? For the 1-norm, you want to generate from a > Dirichlet(1,1,...,1) distribution: Generate n iid exponentials > Y1,..., Y_n, set Y = > sum(Y_k) and X_k = I_k Y_k / Y, where the I_k are iid, > P{I1=1}=P{I1=-1}=1/2. For the 2-norm, generate n iid normals and normalize the vector > (though I wonder if there is a more efficient way). ==== To prove that the boundary of a manifold with boundary (what's the adjective? It can't be bounded manifold!) is well-defined, I need the following lemma: Let P^n = { x in R^n | x_1 >= 0 } and D^n = { x in R^n | x_1 == 0 } Suppose we have a homeomorphism f : P^n --> P^n . Prove that f(D^n)=D^n. I hope there's an anwser without algebraic topology :-! -- ==== In sci.math, AbsolutelyMagic : > hi, this problem is in the related rates section of the book Calculus > of a Single Variable Hostetler, Larson, etc. problem #48 in section > 2.6 I believe. An airplane is traveling in still air at 107 m/sec at an angle of 22 > degrees. Find the rate at which it is gaining altitude. You may need more data. Is the 22 degrees the attitude of the nose or the slope of the flight path? Assuming the latter, one can use simple trig. If you really meant the former one would need a lot of additional data such as the temperature and humidity of the air (hot air is thinner) and the precise construction of the wing of the plane. (The speed of sound is in there somewhere but can probably be derived from the temperature, humidity, and composition of the air. Fortunately, 107 m/s is not supersonic for most locales. :-) ) ok, i'm assuming 107 m/sec is the speed in the diagonal direction. so > you couldn't you just find the rate at which it is gaining altitude > (y-component of this) by using the tan fuction (no derivatives > needed). I can't even think of what equation to set up to > differentiate (Phythagorean theorem, tan A = y / x )??? any help > please..thanks > In your case it might be simpler to draw a diagram: //P 1 // | (hypotenuse) // | // | sin // | (side opposite) // | O------------G cos (side adjacent) (aint ASCII absolutely horrid for this sort of thing? :-) ) where O is the airport, P is the plane, and G is a point on the ground directly underneath P. (We assume here level ground.) We know the slope -- the angle GOP. (No puns of a certain elephantine nature, please. :-) ) We know the speed along the direction OP. We know what sin and cos are (at least, I hope so; you may want to review your mathbook). What you want: PG. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== In sci.math, M Rath : > I merely ponder this: 1D: Let y = 0 and then use the 2D formulas . > 2D: Distance = Sqrt(x^2 + y^2) > Angle = ArcTan(Sqrt(y^2) / x) > 3D: Distance = Sqrt(x^2 + y^2 + z^2) > Angle = ArcTan(Sqrt(y^2 + z^2) / x) This is the angle between a vector (x,y,z) and the plane (dimension N-1 subspace) where x = 0. A more general definition of angle might be represented ArcCos(dot(A,B) / (norm(A) * norm(B))) where A and B are arbitrary vectors. This works in any dimension N >= 2. Note that an angle could be measured in this manner > within a physical system. Also, the origin is zero for each > coordinate axis...as for the distance formula. An origin is nice, at that. As for the angle : you are measuring an angle for only a subset of all things for which one could measure angles. :-) > Beyond 3D: Distance = Sqrt(x^2 + y^2 + z^2 + a1^2 + ... + an^2) > Angle = ArcTan(Sqrt(y^2 + z^2 + a1^2 + ... + an^2) / x) > This extension does work, however. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== In sci.math, Larry Hammick : > This looks like a good way to get around in a rough neighborhood: > http://www.forceprotection.net/lion.html > There's also a family-size model: > http://www.forceprotection.net/gator.html > Dunno if they take Visa. > LH > But what's the gas mileage? :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== Probably a stupid question from a newbe, but here goes anywhy. Trying to determine when a constant becomes an irrational! At what level in this iteration of (e) does (e) become an irrational? e ~ 1/0! + 1/1! e ~ 1/0! + 1/1! + 1/2! e ~ 1/0! + 1/1! + 1/2! + 1/3! e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/8! e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/9! + 1/10! e = 1/0! + .. + 1/n! etc. Each of these iterations up to 1/10! has finite length CF's, thus all rows listed up to 1/10! are rationals. At some level of these iterations, will (e) become an irrational? If so, at what level? Dan ==== In sci.math, Dan <30pack@sbcglobal.net> : > Probably a stupid question from a newbe, but here goes anywhy. Trying to determine when a constant becomes an irrational! At what level in this iteration of (e) does (e) become an irrational? e ~ 1/0! + 1/1! > e ~ 1/0! + 1/1! + 1/2! > e ~ 1/0! + 1/1! + 1/2! + 1/3! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/8! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/9! + > 1/10! > e = 1/0! + .. + 1/n! etc. Actually, e ~ 1/0! + 1/1! + ... + 1/n! for any (positive integer) n. It's not equal unless one takes the limit. Each of these iterations up to 1/10! has finite length CF's, thus all > rows listed up to 1/10! are rationals. At some level of these iterations, will (e) become an irrational? > If so, at what level? The main problem is that e is irrational, and the equations above are rational approximations which get successively better as one increases n. However, they will never reach e -- although never is a funny term, as taking the limit essentially makes the jump, in a weird sort of twisted sense. A slightly more rigorous treatment might allow for the existence of e and successively better constructions by defining e_0 = 1/0! = 1 e_1 = 1/0! + 1/1! = 2 e_2 = 1/0! + 1/1! + 1/2! = 2.5 e_3 = 1/0! + 1/1! + 1/2! + 1/3! = 2.666... ... e_n = 1/0! + 1/1! + 1/2! + ... + 1/n! = e_{n-1} + 1/n! then defining e(n) = e_{floor(n)} and taking the limit as n -> +oo. It's not too difficult to prove that this will converge to something, namely e = 2.718281828459045235360287... The simplest method of proving this may be noting that term k -- namely, (1/k!) < (1/2^(k-1)) (for k > 1) -- and that therefore the series converges to a number e < 3. > > Dan -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > Probably a stupid question from a newbe, but here goes anywhy. Trying to determine when a constant becomes an irrational! At what level in this iteration of (e) does (e) become an irrational? e ~ 1/0! + 1/1! > e ~ 1/0! + 1/1! + 1/2! > e ~ 1/0! + 1/1! + 1/2! + 1/3! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/8! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/9! + > 1/10! > e = 1/0! + .. + 1/n! etc. Each of these iterations up to 1/10! has finite length CF's, thus all > rows listed up to 1/10! are rationals. At some level of these iterations, will (e) become an irrational? > If so, at what level? > It should be fairly obvious that every iteration is a sum of rationals and is therefore rational. It's the limit of the sums that is rational. This is an example of the well-known process of approximating an irrational by a sequence of rationals. ==== > Probably a stupid question from a newbe, but here goes anywhy. Trying to determine when a constant becomes an irrational! At what level in this iteration of (e) does (e) become an irrational? e ~ 1/0! + 1/1! > e ~ 1/0! + 1/1! + 1/2! > e ~ 1/0! + 1/1! + 1/2! + 1/3! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/8! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/9! + > 1/10! > e = 1/0! + .. + 1/n! etc. Each of these iterations up to 1/10! has finite length CF's, thus all > rows listed up to 1/10! are rationals. At some level of these iterations, will (e) become an irrational? > If so, at what level? > > Dan For all levels, the number e_n = 1/0! + ... + 1/n! is rational. You can use n! as denominator. This does not mean that e is rational; there are many sequences of rational numbers whose limit is irrational. Actually, one possibility to define the real numbers is by (inaccurately speaking) saying that they are all possible limits that sequences of rational numbers can have. So, for every real number, there are a lot of rational sequences converging to this particular (rational or irrational) number. -- ==== Probably a stupid question from a newbe, but here goes anywhy. Trying to determine when a constant becomes an irrational! At what level in this iteration of (e) does (e) become an irrational? e ~ 1/0! + 1/1! > e ~ 1/0! + 1/1! + 1/2! > e ~ 1/0! + 1/1! + 1/2! + 1/3! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/8! > e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/9! + > 1/10! > e = 1/0! + .. + 1/n! etc. Each of these iterations up to 1/10! has finite length CF's, thus all > rows listed up to 1/10! are rationals. At some level of these iterations, will (e) become an irrational? > If so, at what level? > > Dan > For all levels, the number e_n = 1/0! + ... + 1/n! is rational. You can use > n! as denominator. This does not mean that e is rational; there are many > sequences of rational numbers whose limit is irrational. > Actually, one possibility to define the real numbers is by (inaccurately > speaking) saying that they are all possible limits that sequences of > rational numbers can have. So, for every real number, there are a lot of > rational sequences converging to this particular (rational or irrational) > number. I had a suspicion that when +oo comes into play any logical thinking about whether a number is rational or irrational can not be deduced by any level of its iterations such as Exp(e). So basically, whatever the correct decimal length of (e) is calculated by summing the inverse of the factorials, it will always be rational unless expressed as a limit ---- e = 1/0! + 1/1! + 1/2! +.. 1/n!...---> oo. Is this also true for all the many different equations that calculate Pi? Probably not because some of the equations require the (sqrt) function in its calculation of Pi! ==== Tobias Fritz scribbled the following: >> Probably a stupid question from a newbe, but here goes anywhy. Trying to determine when a constant becomes an irrational! At what level in this iteration of (e) does (e) become an irrational? e ~ 1/0! + 1/1! >> e ~ 1/0! + 1/1! + 1/2! >> e ~ 1/0! + 1/1! + 1/2! + 1/3! >> e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! >> e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! >> e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! >> e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! >> e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/8! >> e ~ 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! + 1/9! + >> 1/10! >> e = 1/0! + .. + 1/n! etc. Each of these iterations up to 1/10! has finite length CF's, thus all >> rows listed up to 1/10! are rationals. At some level of these iterations, will (e) become an irrational? >> If so, at what level? >> >> Dan > For all levels, the number e_n = 1/0! + ... + 1/n! is rational. You can use > n! as denominator. This does not mean that e is rational; there are many > sequences of rational numbers whose limit is irrational. > Actually, one possibility to define the real numbers is by (inaccurately > speaking) saying that they are all possible limits that sequences of > rational numbers can have. So, for every real number, there are a lot of > rational sequences converging to this particular (rational or irrational) > number. For example consider this kind of series: n_0 = 1 n_1 = 0.4 n_2 = 0.01 n_3 = 0.004 ... n_i = 10^(-i) * (ith decimal of sqrt(2)) Being the product of two rationals, every n_i is rational. However, the sum of all n_i where i in N cup {0}, is equal to sqrt(2) which is irrational. This generalises to any other irrational. IOW the finite sums are approximations of the infinite sum. -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ No, Maggie, not Aztec, Olmec! Ol-mec! - Lisa Simpson ==== >there is a problem I find difficult. Let f:[0,1]->R be infinitely many >times differentiable such that for each x from [0,1] there exists n(x) >such that the >n(x) derivative of f in the point x is 0. Show that f is polynom. See the sci.math thread IT IS A POLYNOMIAL from May 1998 (and in Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >>there is a problem I find difficult. Let f:[0,1]->R be infinitely many >>times differentiable such that for each x from [0,1] there exists n(x) >>such that the >>n(x) derivative of f in the point x is 0. Show that f is polynom. See the sci.math thread IT IS A POLYNOMIAL from May 1998 (and in You mean this one, that someone reposted just to get it into the relevant folder in his newsreader? >Distribution: world > Can you give as clear and as simple as possible proof of the following >> in the Real Analysis? Let f be an infinitely differentiable function on [0, 1] such >> that at every x in [0, 1] an n-th order derivative of f is zero, >> where n depends on x. Prove that f is a polynomial. It brings back memories: this problem (the polynomial problem) was going >around U. of Chicago ca. 1970. Suppose f is not a polynomial. Let >C = {x: there is no neighbourhood of x on which f is a polynomial}. >A_n = {x: f^(n)(x) = 0}. Clearly C is closed and nonempty, and A_n closed with union_n A_n = [0,1]. >Applying the Baire Category Theorem to C, A_n intersect C has nonempty interior >in C for some n, i.e. there exist x0 in C and delta > 0 such that >C intersect (x0 - delta, x0 + delta) is contained in A_n. Suppose f^(k)(x0) <> 0 for some k > n. By Taylor's Theorem, we get f^(n)(x) <> 0 >for 0 < |x - x0| < eta (for some eta > 0). Taking eta < delta, this implies >C intersect (x0 - eta, x0 + eta) = { x0 }, but then f is a polynomial on >(x0 - eta, x0) and on (x0, x0 + eta) and hence on (x0 - eta, x0 + eta), >contradicting x0 in C. So we must conclude that f^(k)(x0) = 0 for all k >= n. >The same is true for all points in C intersect (x0 - delta, x0 + delta). Now suppose c in (x0 - delta, x0 + delta) C. Then f is a polynomial on some >neighbourhood of c. Let [a,b] be the maximal interval containing c on which >f is a polynomial. Then a or b (say b) is in >(x0 - delta, x0 + delta) intersect C, so that f^(k)(b) = 0 for all k >= n. >Now if f has degree p on [a,b], f^(p) <> 0 on [a,b] so p < n, and then >f^(k)(c) = 0 for all k >= n. We conclude that f^(k)(x) = 0 for all k >= n and all x in (x0 - delta, x0 + delta). But then f is a polynomial on (x0 - delta, x0 + delta), contradiction. Robert Israel israel@math.ubc.ca >Department of Mathematics (604) 822-3629 >University of British Columbia fax 822-6074 >Vancouver, BC, Canada V6T 1Z2 >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2 ************************ ==== >> there is a problem I find difficult. Let f:[0,1]->R be infinitely many >> times differentiable such that for each x from [0,1] there exists n(x) >> such that the >> n(x) derivative of f in the point x is 0. Show that f is polynom. The problem is quite easy for holomorphic functions in C or >> analytical in [0,1] but I cannot solve it in this case !!! I hope >> someone would help me with it. Mladen What characteristics does the set {x: f'(x)=0} have? (if it is not >everything, since then we would be done). Same for the set >{x:f''(x)=0}. And so on. >Do you know any theorem about countable unions of such sets? I imagine he does. I do. How does the result follow? ************************ ==== >there is a problem I find difficult. Let f:[0,1]->R be infinitely many >times differentiable such that for each x from [0,1] there exists n(x) >such that the >n(x) derivative of f in the point x is 0. Show that f is polynom. >>Are you certain this is true? (It's easy to show that there must >>be some interval on which f is a polynomial...) If I understood this correctly, couldn't you construct a >counterexample like this: n(x) = { ceil(1/x) when x in (0, 1] > { 1 when x = 0 If there _is_ a counterexample it seems to me that n(x) can certainly not be piecewise constant like that - if you patch together two polynomials what you get is not infinitely differentiable... >No matter which polynomial we choose (assume it is of degree k), we >can always pick x0 and epsilon so that in some neighborhood of x0 the >Taylor series expansion contains higher than degree k terms with >non-zero coefficients in it, and therefore the function cannot be a >polynomial. ************************ ==== >there is a problem I find difficult. Let f:[0,1]->R be infinitely many times differentiable such that for each x from [0,1] there exists n(x) such that the n(x) derivative of f in the point x is 0. Show that f is polynom. I suspect a proof could be based on the fact that some derivative must have uncountably many zeroes, posssibly dense in [0,1], and possibly that derivative might be zero on all of [0,1] except for a countable subset. If the latter were true, then continuity would show that that derivative is zero on all of [0,1], from which the polynomial nature of f(x) must follow. ==== there is a problem I find difficult. Let f:[0,1]->R be infinitely many > times differentiable such that for each x from [0,1] there exists n(x) > such that the > n(x) derivative of f in the point x is 0. Show that f is polynom. The problem is quite easy for holomorphic functions in C or > analytical in [0,1] but I cannot solve it in this case !!! I hope > someone would help me with it. Mladen Here is my attempt. There are some gaps, and a major one at the end, but I think it could work. Let U be an open interval of R and let f:U-->R be smooth. For each n in N, let Cn=((f^(n))^-1)(0) and suppose that UNION(n in N)Cn=U. I claim that there is an open interval V of U such that f|V is polynomial. Since U is uncountable, there is an n in N such that Cn is uncountable. Furthermore, Cn is closed. Since Cn is closed and uncountable, Cn contains an open interval V. Since f^(n)|V=0, f|V is polynomial. Again, let U be an open interval of R and let f:U-->R be smooth. For each n in N, let Cn=((f^(n))^-1)(0) and suppose that UNION(n in N)Cn=U. Let A be the family of all maximal open intervals V of U such that f|V is polynomial and let X=UNION(V in A)V. I claim that A is countable and X is dense in U. First of all, members of A must be disjoint. Since A is a family of disjoint open subsets of R, A must be countable. Suppose UX includes an open interval V. Then V includes a maximal open interval W such that f|W is polynomial and W is not in A, contradicting the definition of A. Therefore UX has empty interior so X is dense in U. I claim that UX has no isolated points. Let p in UX be isolated. Then there are open intervals (a,p) and (p,b) such that f|(a,p) and f|(p,b) are polynomial. Then all derivatives of f at p determine the coefficents of f|(a,p) and f|(p,b) so these coefficients must be identical. Hence f|(a,b) is polynomial so p is contains in some member of A, which contradicts the assumption that p is in UX. Therefore UX has no isolated points. To summarize, UX is closed, has empty interior, and contains no isolated points. I'm not sure how to prove this but I believe that UX must be empty in this case. This would imply that A={U} so f=f|U is polynomial. If f is actually defined on a closed interval instead of an open interval, then f is polynomial on the interior of the interval and it is easy to tack on the endpoints. Have a tolerable existence. Eli ==== >> there is a problem I find difficult. Let f:[0,1]->R be infinitely many >> times differentiable such that for each x from [0,1] there exists n(x) >> such that the >> n(x) derivative of f in the point x is 0. Show that f is polynom. The problem is quite easy for holomorphic functions in C or >> analytical in [0,1] but I cannot solve it in this case !!! I hope >> someone would help me with it. Mladen Here is my attempt. There are some gaps, and a major one at the end, but I >think it could work. Let U be an open interval of R and let f:U-->R be smooth. For each n in N, >let Cn=((f^(n))^-1)(0) and suppose that UNION(n in N)Cn=U. I claim that >there is an open interval V of U such that f|V is polynomial. Since U is >uncountable, there is an n in N such that Cn is uncountable. Furthermore, >Cn is closed. Since Cn is closed and uncountable, Cn contains an open >interval V. ??? An uncountable closed set need not contain an interval. But it's easy to see that there is an interval on which f is a polynomial... >Since f^(n)|V=0, f|V is polynomial. Again, let U be an open interval of R and let f:U-->R be smooth. For each n >in N, let Cn=((f^(n))^-1)(0) and suppose that UNION(n in N)Cn=U. Let A be >the family of all maximal open intervals V of U such that f|V is polynomial >and let X=UNION(V in A)V. I claim that A is countable and X is dense in U. >First of all, members of A must be disjoint. Since A is a family of >disjoint open subsets of R, A must be countable. Suppose UX includes an >open interval V. Then V includes a maximal open interval W such that f|W >is polynomial and W is not in A, contradicting the definition of A. >Therefore UX has empty interior so X is dense in U. I claim that UX has no isolated points. Let p in UX be isolated. Then >there are open intervals (a,p) and (p,b) such that f|(a,p) and f|(p,b) are >polynomial. Then all derivatives of f at p determine the coefficents of >f|(a,p) and f|(p,b) so these coefficients must be identical. Hence f|(a,b) >is polynomial so p is contains in some member of A, which contradicts the >assumption that p is in UX. Therefore UX has no isolated points. To summarize, UX is closed, has empty interior, and contains no isolated >points. I'm not sure how to prove this but I believe that UX must be >empty in this case. Imitate the traditional proof that the Cantor set is empty. > This would imply that A={U} so f=f|U is polynomial. >If f is actually defined on a closed interval instead of an open interval, >then f is polynomial on the interior of the interval and it is easy to tack >on the endpoints. Have a tolerable existence. Eli ************************ ==== > Are you certain this is true? (It's easy to show that there must > be some interval on which f is a polynomial...) I think it is true. The problem was given by a teacher from some book. I think that it can be shown that for each open set there exist open subset such that in it it is a polynomial. So this set is dense . But the polynomials can differ and I cannot show that it is impossible ! Greetings Mladen Savov ==== >> Are you certain this is true? (It's easy to show that there must >> be some interval on which f is a polynomial...) I think it is true. The problem was given by a teacher from some >book. I think that it can be shown that for each open set there exist >open subset such that in it it is a polynomial. So this set is dense . Yes, that much is easy. >But the polynomials can differ and I cannot show that it is impossible >! Greetings Mladen Savov ************************ ==== there is a problem I find difficult. Let f:[0,1]->R be infinitely many > times differentiable such that for each x from [0,1] there exists n(x) > such that the > n(x) derivative of f in the point x is 0. Show that f is polynom. The problem is quite easy for holomorphic functions in C or > analytical in [0,1] but I cannot solve it in this case !!! I hope > someone would help me with it. Mladen Let E_n={x:f^{n}(x)=0} Unions of E_n is [0,1] Look up the Baire category theorem. ==== there is a problem I find difficult. Let f:[0,1]->R be infinitely many >> times differentiable such that for each x from [0,1] there exists n(x) >> such that the >> n(x) derivative of f in the point x is 0. Show that f is polynom. The problem is quite easy for holomorphic functions in C or >> analytical in [0,1] but I cannot solve it in this case !!! I hope >> someone would help me with it. Mladen >Let E_n={x:f^{n}(x)=0} Unions of E_n is [0,1] Look up the Baire category theorem. Maybe it's just me being stupid, since several people seem to be giving hints like this. How does the result follow from the Baire category theorem? ************************ ==== > there is a problem I find difficult. Let f:[0,1]->R be infinitely many > times differentiable such that for each x from [0,1] there exists n(x) > such that the > n(x) derivative of f in the point x is 0. Show that f is polynom. The problem is quite easy for holomorphic functions in C or > analytical in [0,1] but I cannot solve it in this case !!! I hope > someone would help me with it. Mladen >Let E_n={x:f^{n}(x)=0} Unions of E_n is [0,1] Look up the Baire category theorem. Maybe it's just me being stupid, since several people > seem to be giving hints like this. How does the result > follow from the Baire category theorem? I'm not sure I follow gc's post but Baire category ideas seem to be the key. Define a bad point as a point x such that n is unbounded in every neighbourhood of x. The result follows easily if we can show there are no bad points. Suppose there exist bad points. The basic observation is that there is an open interval I with a bad point and an integer M such that every open subinterval I' of I with a bad point contains a bad point x with n(x) = M. This is easy to establish by contradiction - if it weren't true you could construct a bad point x (as the intersection of nested sets) with n(x) != M for all M, which is impossible. it follows that every point in I has n(x) <= M, which contradicts the existence of bad points in I. Michael <340b747b.0310280459.7de9f1dd@posting.google.com> ==== there is a problem I find difficult. Let f:[0,1]->R be infinitely many > times differentiable such that for each x from [0,1] there exists n(x) > such that the > n(x) derivative of f in the point x is 0. Show that f is polynom. >Let E_n={x:f^{n}(x)=0} >Unions of E_n is [0,1] >Look up the Baire category theorem. Maybe it's just me being stupid, since several people > seem to be giving hints like this. How does the result > follow from the Baire category theorem? Define a bad point as a point x such that n is unbounded in every > neighbourhood of x. The result follows easily if we can show there are > no bad points. Suppose there exist bad points. The basic observation > is that there is an open interval I with a bad point and an integer M > such that every open subinterval I' of I with a bad point contains a > bad point x with n(x) = M. This is easy to establish by contradiction > - if it weren't true you could construct a bad point x (as the > intersection of nested sets) with n(x) != M for all M, which is > impossible. > As [0,1] is compact and every f^(n)(x) is continuous as every f^(n+1)(x) exists, it follows that every f^(n)(x) is bounded. Further more if I is any interval, open, closed or half open, then I = Union { E_n | n in N }. Now as I is Baire space, there's some n with interior E_n /= nulset. Hence some open interval J subset I for which f^(n)(x) = 0 for all x in J. Given both of those, show how easy to prove f is polynomial as you claim. > it follows that every point in I has n(x) <= M, which contradicts the > existence of bad points in I. > ==== > there is a problem I find difficult. Let f:[0,1]->R be infinitely many > times differentiable such that for each x from [0,1] there exists n(x) > such that the > n(x) derivative of f in the point x is 0. Show that f is polynom. > >Let E_n={x:f^{n}(x)=0} >Unions of E_n is [0,1] >Look up the Baire category theorem. >Maybe it's just me being stupid, since several people > seem to be giving hints like this. How does the result > follow from the Baire category theorem? Define a bad point as a point x such that n is unbounded in every > neighbourhood of x. The result follows easily if we can show there are > no bad points. Suppose there exist bad points. The basic observation > is that there is an open interval I with a bad point and an integer M > such that every open subinterval I' of I with a bad point contains a > bad point x with n(x) = M. This is easy to establish by contradiction > - if it weren't true you could construct a bad point x (as the > intersection of nested sets) with n(x) != M for all M, which is > impossible. As [0,1] is compact and every f^(n)(x) is continuous as every > f^(n+1)(x) exists, it follows that every f^(n)(x) is bounded. Further more if I is any interval, open, closed or half open, > then I = Union { E_n | n in N }. Now as I is Baire space, > there's some n with interior E_n /= nulset. Hence some open > interval J subset I for which f^(n)(x) = 0 for all x in J. Given both of those, show how easy to prove f is polynomial > as you claim. I'm not quite sure if you're filling in some details, or asking for clarification about my answer. In case it's the latter, I'll explain more precisely what I'm saying. We have assumed f is not a polynomial. It follows immediately that n(x) is unbounded in every neighbourhood of certain points. (For convenience I'm taking n(x) to be the _minimal_ n such that f^n(x) = 0. This wasn't actually said explcitly in the question, but hopefully it's clear what I mean.) I've called these points x, such that n is unbounded in every neighbourhood of x, bad points. Then I've claimed that there is an open interval I with a bad point and a positive integer M such that every open subinterval I' of I with a bad point contains a bad point x with n(x) = M. If this were false then that would mean that for every open interval I containing a bad point and every positive integer M you could find a subinterval I' with a bad point such that each bad point x of I' satisfies n(x) != M. OK so let I_0 be any open interval containing a bad point. (We've assumed bad points exist.) Now set M = 1 - by the hypothesis in the above paragraph we could find an open subinterval I_1 of I_0 containing a bad point such that every bad point of I_1 satisfies n != 1. Now set M = 2. Find an open subinterval I_2 of I_1 containing a bad point such that every bad point of I_2 satisfies n != 2. Etc. In general I_m are nested intervals each containing a bad point, and such that every bad point x of I_m has n(x) != m. Without loss of generality, the left and right endpoints of I_m are strictly increasing/decreasing respectively, and the lengths of I_m are tending to 0. Hence the intersection of I_m is a singleton. Must this singleton x be a bad point? Yes. Otherwise n would be bounded in some neighbourhood of x, so n would be bounded in some I_m which contradicts the fact I_m contains a bad point. So this singleton x is a bad point. But it is in I_m for each m so n(x) != m for each m. Contradiction. (I'm sure you can write this out much more compactly using Baire's theorem, but I personally find it more intuitive to write this sort of thing out from scratch. Hopefully that will change with time!) So after all that we've established the claim in the 5th paragraph: that there is an open interval I with a bad point and a positive integer m such that every open subinterval I' of I with a bad point contains a bad point x with n(x) = m. Now that means that the mth derivative of every bad point in I vanishes. Because given a bad point x in I you can find bad points in I arbitrarily close to x at which the mth derivative of f vanishes. Since f^m is continuous that means that the mth derivative at every bad point in I vanishes. But that means the m+1th derivative of every bad point in I vanishes. Any bad point x has other bad points in I arbitrarily close. Take a sequence x_n of bad points in I tending to x. Then: f^m+1(x) = lim(n->inf)(f^m(x_n)-f^m(x))/(x_n-x) = 0 since f^m(x_n) = f^m(x) = 0. By induction the m+2th, m+3th derivatives etc. all vanish at the bad points in I. Now take any point x in I. We can take a maximal subinterval I' of I containing x on which f is a polynomial of degree n(x). Pick an endpoint x' of I' which doesn't coincide with an endpoint of I. Then it is clear that x' is bad. Hence f^m(x') = f^m+1(x') = ... = 0 so it follows that the degree of the polynomial in I' is < m. Hence the mth,m+1th,m+2th,etc derivatives of f vanish at x. Therefore f is a polynomial on I so there are no bad points in I, which is a contradiction. Michael ==== > there is a problem I find difficult. Let f:[0,1]->R be infinitely many > times differentiable such that for each x from [0,1] there exists n(x) > such that the > n(x) derivative of f in the point x is 0. Show that f is polynom. The problem is quite easy for holomorphic functions in C or > analytical in [0,1] but I cannot solve it in this case !!! I hope > someone would help me with it. Mladen >Let E_n={x:f^{n}(x)=0} Unions of E_n is [0,1] Look up the Baire category theorem. Maybe it's just me being stupid, since several people > seem to be giving hints like this. How does the result > follow from the Baire category theorem? ************************ Whenever the union of countably many closed subsets of a Baire space has an interior point, then one of the closed subsets must have an interior point. The set of zeros of the nth derivative is closed, and [0,1] is a Baire space. So the function is a polynomial in a neighborhood of such an interior point. This gets things started anyway. ==== there is a problem I find difficult. Let f:[0,1]->R be infinitely many >> times differentiable such that for each x from [0,1] there exists n(x) >> such that the >> n(x) derivative of f in the point x is 0. Show that f is polynom. The problem is quite easy for holomorphic functions in C or >> analytical in [0,1] but I cannot solve it in this case !!! I hope >> someone would help me with it. Mladen >>Let E_n={x:f^{n}(x)=0} >>Unions of E_n is [0,1] >>Look up the Baire category theorem. Maybe it's just me being stupid, since several people >> seem to be giving hints like this. How does the result >> follow from the Baire category theorem? ************************ Whenever the union of countably many closed subsets of a Baire space >has an interior point, then one of the closed subsets must have an >interior point. The set of zeros of the nth derivative is closed, and [0,1] is a >Baire space. So the function is a polynomial in a neighborhood of such an >interior point. This gets things started anyway. Uh, thanks. I said it was easy to see the function must be a polynomial in some interval some time ago, in what appears to be the first reply to the OP. You say this gets things started. I don't see how: Now that we know f is a polynomial in some interval how does the result follow? (Or how _might_ the result follow? What's a possible approach to get from here to what we want to prove? I can't see any hint of a possible path from here to there - you do?) ************************ ==== > there is a problem I find difficult. Let f:[0,1]->R be infinitely many >> times differentiable such that for each x from [0,1] there exists n(x) >> such that the >> n(x) derivative of f in the point x is 0. Show that f is polynom. >> The problem is quite easy for holomorphic functions in C or >> analytical in [0,1] but I cannot solve it in this case !!! I hope >> someone would help me with it. >> Mladen >Let E_n={x:f^{n}(x)=0} Unions of E_n is [0,1] Look up the Baire category theorem. Maybe it's just me being stupid, since several people > seem to be giving hints like this. How does the result > follow from the Baire category theorem? ************************ Whenever the union of countably many closed subsets of a Baire space >has an interior point, then one of the closed subsets must have an >interior point. The set of zeros of the nth derivative is closed, and [0,1] is a >Baire space. So the function is a polynomial in a neighborhood of such an >interior point. This gets things started anyway. Uh, thanks. I said it was easy to see the function must be a > polynomial in some interval some time ago, in what appears to > be the first reply to the OP. You say this gets things started. I don't see how: Now that > we know f is a polynomial in some interval how does the > result follow? (Or how _might_ the result follow? What's > a possible approach to get from here to what we want to > prove? I can't see any hint of a possible path from here to > there - you do?) > Prove that f is polynomial in every component of V=Union(int(Cn)) and that UV has no isolated points. Then show that UV is empty (or else you'll get a contradiction). -gs- ==== > there is a problem I find difficult. Let f:[0,1]->R be infinitely >many > times differentiable such that for each x from [0,1] there exists >n(x) > such that the > n(x) derivative of f in the point x is 0. Show that f is polynom. >> The problem is quite easy for holomorphic functions in C or > analytical in [0,1] but I cannot solve it in this case !!! I hope > someone would help me with it. >> Mladen >>Let E_n={x:f^{n}(x)=0} >>Unions of E_n is [0,1] >>Look up the Baire category theorem. >Maybe it's just me being stupid, since several people >> seem to be giving hints like this. How does the result >> follow from the Baire category theorem? >************************ >Whenever the union of countably many closed subsets of a Baire space >>has an interior point, then one of the closed subsets must have an >>interior point. >>The set of zeros of the nth derivative is closed, and [0,1] is a >>Baire space. >>So the function is a polynomial in a neighborhood of such an >>interior point. >>This gets things started anyway. >Uh, thanks. I said it was easy to see the function must be a >> polynomial in some interval some time ago, in what appears to >> be the first reply to the OP. >You say this gets things started. I don't see how: Now that >> we know f is a polynomial in some interval how does the >> result follow? (Or how _might_ the result follow? What's >> a possible approach to get from here to what we want to >> prove? I can't see any hint of a possible path from here to >> there - you do?) Prove that f is polynomial in every component of V=Union(int(Cn)) and that >UV has no isolated points. I believe I can do that... >Then show that UV is empty (or else you'll get a contradiction). How does that follow? >-gs- > ************************ ==== >> there is a problem I find difficult. Let f:[0,1]->R be infinitely >many >> times differentiable such that for each x from [0,1] there exists >n(x) >> such that the >> n(x) derivative of f in the point x is 0. Show that f is polynom. >> The problem is quite easy for holomorphic functions in C or >> analytical in [0,1] but I cannot solve it in this case !!! I hope >> someone would help me with it. >> Mladen >>Let E_n={x:f^{n}(x)=0} >>Unions of E_n is [0,1] >>Look up the Baire category theorem. >> Maybe it's just me being stupid, since several people >> seem to be giving hints like this. How does the result >> follow from the Baire category theorem? >> ************************ >> Whenever the union of countably many closed subsets of a Baire space >has an interior point, then one of the closed subsets must have an >interior point. The set of zeros of the nth derivative is closed, and [0,1] is a >Baire space. So the function is a polynomial in a neighborhood of such an >interior point. This gets things started anyway. Uh, thanks. I said it was easy to see the function must be a > polynomial in some interval some time ago, in what appears to > be the first reply to the OP. You say this gets things started. I don't see how: Now that > we know f is a polynomial in some interval how does the > result follow? (Or how _might_ the result follow? What's > a possible approach to get from here to what we want to > prove? I can't see any hint of a possible path from here to > there - you do?) >Prove that f is polynomial in every component of V=Union(int(Cn)) and that >UV has no isolated points. I believe I can do that... Then show that UV is empty (or else you'll get a contradiction). How does that follow? > closed(=complete) Suppose that W=UV is non empty. By hypothesis U=Union(Cn) so W=Union(Dn) with Dn=Cn inter W. Dn is closed in W and W is complete (closed subset of U). By Baire's th. we know that there exist n0 and a,b in U with both I:=]a,b[ inter W non-empty and a subset of Cn0. I'll show that f^(n0)=0 on ]a,b[, hence ]a,b[ will be included in int(Cn0) which contradicts the non-emptiness of I. Hence W will be empty and f will be polynomial in U. First let x be in I, then I claim that f^(n)(x)=0 if n>=n0. It's easily shown by recurrence : Let (x_k) in I^N be an injective monotonous sequence with lim x_k=x (W has no isolated points) Assume that (x_k) is increasing. We can construct (x_1_k) with x_koo) so you find that f^(n)(x)=0 if n>n0. Now if x is in ]a,b[ inter V, then the component V_x of x has an extremity in x0 in I (coz I is non-empty) and we know that f = P in V_x. But x0 in I => f^(n)(x0)=P^(n)(x0)=0 if n>=n0, so P is a polynomial of deg> there is a problem I find difficult. Let f:[0,1]->R be infinitely >>many > times differentiable such that for each x from [0,1] there exists >>n(x) > such that the > n(x) derivative of f in the point x is 0. Show that f is polynom. >> The problem is quite easy for holomorphic functions in C or > analytical in [0,1] but I cannot solve it in this case !!! I hope > someone would help me with it. >> Mladen >Let E_n={x:f^{n}(x)=0} >>Unions of E_n is [0,1] >>Look up the Baire category theorem. >> Maybe it's just me being stupid, since several people > seem to be giving hints like this. How does the result > follow from the Baire category theorem? >> ************************ >> >>Whenever the union of countably many closed subsets of a Baire space >>has an interior point, then one of the closed subsets must have an >>interior point. >>The set of zeros of the nth derivative is closed, and [0,1] is a >>Baire space. >>So the function is a polynomial in a neighborhood of such an >>interior point. >>This gets things started anyway. >Uh, thanks. I said it was easy to see the function must be a >> polynomial in some interval some time ago, in what appears to >> be the first reply to the OP. >You say this gets things started. I don't see how: Now that >> we know f is a polynomial in some interval how does the >> result follow? (Or how _might_ the result follow? What's >> a possible approach to get from here to what we want to >> prove? I can't see any hint of a possible path from here to >> there - you do?) >>Prove that f is polynomial in every component of V=Union(int(Cn)) and >that >>UV has no isolated points. >I believe I can do that... >>Then show that UV is empty (or else you'll get a contradiction). >How does that follow? closed(=complete) >Suppose that W=UV is non empty. By hypothesis U=Union(Cn) so W=Union(Dn) >with Dn=Cn inter W. >Dn is closed in W and W is complete (closed subset of U). By Baire's th. we >know that there exist n0 and a,b in U with both >I:=]a,b[ inter W non-empty and a subset of Cn0. I'll show that f^(n0)=0 >on ]a,b[, hence ]a,b[ will be included in int(Cn0) which contradicts >the non-emptiness of I. Hence W will be empty and f will be polynomial in U. >First let x be in I, then I claim that f^(n)(x)=0 if n>=n0. It's easily >shown by recurrence : Let (x_k) in I^N be an injective monotonous sequence >with lim x_k=x (W has no isolated points) Assume that (x_k) is increasing. >We can construct (x_1_k) with x_kRolle's th. and we'll have lim x_1_k=x so we find that f^(n0+1)(x)=0 (using >continuity). Then you can construct by recurrence (x_n_k) with the property >: f^(n0+n)(x_n_k)=0 and lim x_n_k =0 (k->oo) so you find that f^(n)(x)=0 if >n>n0. >Now if x is in ]a,b[ inter V, then the component V_x of x has an extremity >in x0 in I (coz I is non-empty) and we know that f = P in V_x. >But x0 in I => f^(n)(x0)=P^(n)(x0)=0 if n>=n0, so P is a polynomial of >degAnd so we proved that f^(n0)(]a,b[)=0 !! (coool) >And it's easy to prove the result if U isn't closed now. >I hope there's no error (and sorry for the bad english ^o^) Took a few minutes, but that seems right. >-gs- > ************************ ==== suppose x=(x7,x6,...,x0) in R^8 space p(x) = 1/(sqrt(2pisigma^2))^n exp(-xx'/(2sigma^2)) if we wanna do integration on a 8 dimension sphere with radius r, what should we do? how to translate the p(x) into the express with only radius r? I know in 2 dimension, it is pretty easy but how about higher dimension? thanks a lot! ==== haha I got it it is pretty easy I just confused myself xx' = x7^2+x6^2+...+x0^2 = r^2 right? > suppose x=(x7,x6,...,x0) in R^8 space p(x) = 1/(sqrt(2pisigma^2))^n exp(-xx'/(2sigma^2)) if we wanna do integration > on a 8 dimension sphere with radius r, what should we do? > how to translate the p(x) into the express with only radius r? I know in 2 dimension, it is pretty easy but how about higher dimension? thanks a lot! <3f9d2f6a$16$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 05:15 PM, magidin@math.berkeley.edu (Arturo Magidin) said: >I assume that by ordered set with certain properties you mean >something akin to defining a group as an ordered pair (G,*), where G >is a set, and * is a map GxG->G satisfying certain axioms. Yes. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 07:40 AM, kramsay@aol.com (KRamsay) said: >Mixing the levels, while claiming that a statement is true when it's >a theorem (or variations on that theme) runs afoul of Goedel's >construction. Not really. It just means that a statement may be true, false or neither, and that you may not know which it is. Of course, that greatly diminishes the utility of the term true. FWIW, I find both Formalism and Platonism to be unsatisfactory, but I don't see a viable alternative to the two of them :-( -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org <3f8fd096$5$fuzhry+tra$mr2ice@news.patriot.net> <3f957850$5$fuzhry+tra$mr2ice@news.patriot.net> <3f985227$11$fuzhry+tra$mr2ice@news.patriot.net> <3f9b2e04$2$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== >With the usual meaning of words, the observation or assumption that >an arithmetical sentence is true does not have to do with models. A >model is always a model /of a theory,/ but here we have no theory; >all we have are the natural numbers. Which leads right back to the question of what you mean by The Natural Numbers. How do you characterize statements about them outside of some formal system? How do you determine whether a formal system correctly captures some of their properties? -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > Which leads right back to the question of what you mean by The > Natural Numbers. How do you characterize statements about them > outside of some formal system? You formulation suggests that there is some inside formal systems to be invoked here, but it's unclear what you have in mind. Just what is the role of formal systems in exlaining what we mean in referring to the natural numbers? For example, is the conjecture that there are infinitely many twin primes to be explained in terms of formal sytems, and if so, how? <3f8c413c$18$fuzhry+tra$mr2ice@news.patriot.net> <3f8fd096$5$fuzhry+tra$mr2ice@news.patriot.net> <3f95780d$4$fuzhry+tra$mr2ice@news.patriot.net> <3f987867$18$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== >If you think the Peano Postulates do not follow from the definition I >gave, I don't think that you gave a definition. >The natural numbers do not form a finite collection, and we must be >careful about carrying over results about finite collections to >infinite collections. Which is precisely why your definition doesn't define anything. >I claim that if a particular statement cannot be decided from that >definition, then it is the statement that is problematic, not the >definition. That would certainly leave you a lot of wiggle room. Certainly there are a lot of open questions in number theory, and I certainly consider them to be reasonable questions. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > Despite some searching I have > not been able to find a definition of arithmetical property, Now I have. http://home.ddc.net/ygg/etext/godel/godel3.htm A relation (class) is called arithmetical, if it can be defined solely by means of the concepts +, . [addition and multiplication, applied to natural numbers]49 and the logical constants ò, ~, (x), =, where (x) and = are to relate only to natural numbers.50 The concept of arithmetical proposition is defined in a corresponding way. <3f9d06e2$0$69956$edfadb0f@dread12.news.tele.dk> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Terminate: SPA(GIS) ==== at 12:52 PM, Michael J¡rgensen said: >To me, understanding means to be able to explain to someone else. There have been cases of fallacious proofs that were accepted for a long time. People were able to explain the proof to each other, but the explanations were actually wrong. IMHO they did not understand the purported proof. To me, you understand the proof when you understand each step along the way. That's not as simple as it sounds, since sometimes a step can depend on very subtle distinctions or unstated assumptions. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== To me, you understand the proof when you understand each step along > the way. That's not as simple as it sounds, since sometimes a step can > depend on very subtle distinctions or unstated assumptions. > Well... A proof is always uncertain in philosophical sense and in a very real sense. All proof ultimately stands on a foundation of goodwill. I pursuade you with my arguments, but maybe both of us have missed something. As you say, this has happened before, and with a lot more than 2 people involved. We can construct very formal methods to lower the probability of errors. But they can still happen. And out formal method itself can be flawed. Ultimately this, again, stands on a foundation of goodwill. We can then construct computer programs that can prove theorems for us, but there can still be errors in this computer program. The computer program can itself be checked and proved to be correct. But again, ultimately the goodwill of humans are the last link in this chain. The proof or the checking program could be faulty. It is made by humans. In a philisophical sense, this is interesting. But it is also interesting for very practical reasons. A mathematical result can eventually influence the building of cars, space craft, medical equipment etc. How sure are we of the truth of the theorems that we've proved? This leads to another question: What is the most human-independant method of proof we can construct? Can we somehow prove the theorem of Pythagoras without human interaction (other than observing the result)? /David ==== the simplest proof taht i know of PT is the lunes one, which is just as simple as Einstien's, if you thinbk about it in terms of using compasses. the question is absurd, though! > What is the most human-independant method of proof we can construct? Can > we somehow prove the theorem of Pythagoras without human interaction > (other than observing the result)? --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac ==== > the simplest proof taht i know of PT is the lunes one, > which is just as simple as Einstien's, if you thinbk about it > in terms of using compasses. the question is absurd, > though! > Why is the question absurd? /David ==== you can have your cake & stuff it, two. now, it's always better to be able to read the proof, and decide on its correctness, than to have to go through the process of encoding it into a machine-language; that is a last resort, at this time. but, you make the question, Well, Jimmy Dean Harris, why do you not submit your proof to The Machine of your choice, and have it explain its correctness for all to see in the googolplex? you can't ever, these days, 9 years into the mission, get us to believe that you never make errors. have you tried acting and/or political science? > Checking proofs should be a mechanical process, best left to machines. > A math proof begins with a truth and proceeds by logical steps to a > conclusion which then must be true. --les ducs d'Enron! ==== > Depends on what you mean by understand the proof. What exactly does it mean to understand a proof? I mean, anyone can read a > proof and say, yeah, sure, I understand it. In fact, people may *think* > they understand a proof when in fact they do not. I thought that it might be interesting to quote G. H. Hardy here: There is strictly no such thing as mathematical proof; proofs are what > Littlewood and I call gas, rhetorical flourishes, devices to stimulate > the imaginations of pupils. > Jose Carlos Santos Bullshit. A proof begins with a truth and proceeds by logical steps to a conclusion which then must be true. It occurs to me that some of you think proofs are *created* when they have existed since long before any of you were born. Proofs are what you are not--permanent. James Harris ==== a proof has at least to be communicable to another party ... why, if you give the Republican Party a big contribution, they'll probably say taht it's true. the problem with your definition of proof, is that it's completely contigent upon the veracity (truth-value) and connection to-and-fro of each of your logical steps. now, if you've communicated this to any one, communicating by that Little Voice Within; eh? > A proof begins with a truth and proceeds by logical steps to a > conclusion which then must be true. --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac ==== > It occurs to me that some of you think proofs are *created* when they > have existed since long before any of you were born. Whether created or discovered or constructed, they seem not to be any part of JSH's world. ==== > Bullshit. A proof begins with a truth and proceeds by logical steps to a > conclusion which then must be true. Theorem: The square root of 2 is irrational. Proof: Assume that there are integers p and q such that (p/q)^2 = 2. ... I bet James Harris won't get it. ==== Bullshit. A proof begins with a truth and proceeds by logical steps to a >> conclusion which then must be true. Theorem: The square root of 2 is irrational. >Proof: Assume that there are integers p and q such that (p/q)^2 = 2. ... I bet James Harris won't get it. Actually, I think he's developed a new version of reductio ad absurdum. He starts from what's true and arrives at a contradiction so somewhere there must be a core error (and heaven forbid it should be in *his* algebra). ==== >[...] Checking proofs should be a mechanical process, best left to machines. Yet when I've brought that up before on this forum there has been an >*emotional* reaction from posters!!! It's as if there's some deep need to have people wasting their time >doing what machines could do far faster and better, and that's scary, >as it indicates that many people in math society don't understand what >a proof is. Excellent point. (Don't know if you noticed, but it follows from what you say here that you find your own posts scary...) >I think a lot of people in math society think of a proof as a bit of >mathematicians, where the truth is just a social construct, depending >on who believes. >James Harris ************************ ==== Checking proofs should be a mechanical process, best left to machines. Yet when I've brought that up before on this forum there has been an > *emotional* reaction from posters!!! It's as if there's some deep need to have people wasting their time > doing what machines could do far faster and better, and that's scary, > as it indicates that many people in math society don't understand what > a proof is. Or at least it indicates that James Harris doesn't know what a formal > proof is and just how difficult it is to formalize a standard > mathematical exposition. Formalizing mathematics so that standard proofs are machine checkable > is a long and arduous task. Creating a machine that can check the > correctness of proofs as they are presented in journals, say, is a far > more difficult task. There's considerable work on the former task and > also, I assume, on the latter, but neither are at a stage in which new > proofs may be feasibly checked by machine. How hard can it be? After all, James keeps saying he's going to produce his own machine checkable proof if no one does it for him. Of course, James has been told this before. <3f9d06e2$0$69956$edfadb0f@dread12.news.tele.dk> <3c65f87.0310270731.7ad47b8f@posting.google.com> <87wuaq65o3.fsf@phiwumbda.org> ==== >> Formalizing mathematics so that standard proofs are machine checkable >> is a long and arduous task. Creating a machine that can check the >> correctness of proofs as they are presented in journals, say, is a far >> more difficult task. There's considerable work on the former task and >> also, I assume, on the latter, but neither are at a stage in which new >> proofs may be feasibly checked by machine. How hard can it be? After all, James keeps saying he's going to produce > his own machine checkable proof if no one does it for him. That, and have congressional hearings and threaten the livelihood of every mathematician today. James does say a lot, doesn't he? -- However, you presuppose that certain numbers *are* prime ideals, ... when in fact ...* they are not... (Maybe I should look up 'prime ideals' but the effort doesn't seem to be worth it. I assume some poster will get excited ... if I messed up.) --James Harris ==== > Depends on what you mean by understand the proof. What exactly does it mean to understand a proof? I mean, anyone can read a > proof and say, yeah, sure, I understand it. In fact, people may *think* > they understand a proof when in fact they do not. People overrate understanding as, isn't it better to have correctness? How do you verify correctness without understanding? A math proof begins with a truth and proceeds by logical steps to a > conclusion which then must be true. People throw the word proof around, when usually it's a *claim* of > proof. You can understand a math argument that some people think is a proof, > believe it's a proof yourself, and it still be wrong because it has a > break in the logical chain, or it doesn't begin with a truth. Your irony meter is no-doubt broken. Mine is registering off-scale. This is pretty funny. > To me, understanding means to be able to explain to someone else. In this > particular situation, it would mean to be able to convince other > (mathematical) people that the theorem is true. Quite a daunting task, I my > eyes.... Do you care if your car mechanic can explain to you how your car > engine works, or do you want it to work? Knowledge is advanced by conveying it to others. Scientific results are verified when other people can reproduce them or build on them. Mathematics is advanced when other people accept a proof, understand it, and build on it. Despite it's esoteric nature, Wiles proof has been understood by a significant and increasing number of people. The proof of that is that his result has been generalized. Got that? There is another paper out there with a STRONGER result. > Mathematicians have created a *social* environment, where people > debate math proofs when a correct math argument is perfect, and not > open to debate. A proof is always open to examination. Nobody's going to accept a proof with a big black cover nailed over the middle of it labelled don't look in here. > For instance, consider Wiles's work, don't you think it might be a > problem for some people to admit that it's not a proof? You conveniently forget that the first version of the paper was withdrawn when it became clear that it wasn't a valid proof. You think that happened because people accepted it on face value and didn't understand what he was doing? > Checking proofs should be a mechanical process, best left to machines. So do it. You've been invited before. Give us a machine validation of your proof. Yet when I've brought that up before on this forum there has been an > *emotional* reaction from posters!!! I'm not sure what reaction you're talking about, except the one from people when you suggested somebody else should invest the months of effort to build the machine implementation because you couldn't be bothered. You don't see why that should get an emotional reaction? It's as if there's some deep need to have people wasting their time > doing what machines could do far faster and better, Do you have some data to back up your statement that any current math paper could be done faster and better in a machine? > and that's scary, > as it indicates that many people in math society don't understand what > a proof is. Well, since you're the only one who does, I guess it's up to you to lead the way with the machine validation of yours. - Randy ==== There's a certain extra je ne sais quoi to understanding a proof. > It's not enough to know not only the sequence of intermediate results which, > with the rules of inference, proves the theorem. One must also understand > why these intermediate points are chosen, that is, why the proof > proceeds in a certain direction rather than another one. Here's an > example which once appeared here: > http://google.com/groups?selm=301jo2%24a0g%40agate.berkeley.edu > Not intending to be mean to that thread's OP, it's clear that he > didn't understand the proof, even though he was convinced that he did, > and even though, superficially, he understood the steps and could > convey them to someone else. Here is the referenced proof (posted by a high-school teacher) >> QUESTION: Given n red points and n blue points on the plane >> (no three points collinear), is there always a way to pairwise >> connect the points (red to blue) with segments such that no >> segments intersect? Yes. [Why? Think a bit before you go to >> the next paragraph.] ANSWER: Connect them pairwise (red to blue) in any manner, and >> then for any intersection, reconnect the four points the other >> way and thereby reduce the length of segments. Q.E.D. There must also be a similar elusive quality to understanding the criticism of a proof, perhaps having to do with your innate bias towards the competence of high-school teachers? If you didn't know that the OP is a high-school teacher would you have been so critical of the proof? Perhaps not, esp. if OP is a distinguished professor. Please explain why it's clear that he didn't understand the proof. -Bill Dubuque ==== [I had commented that understanding a proof including more than understanding the individual steps.] >Here is the referenced proof (posted by a high-school teacher) >> QUESTION: Given n red points and n blue points on the plane > (no three points collinear), is there always a way to pairwise > connect the points (red to blue) with segments such that no > segments intersect? Yes. [Why? Think a bit before you go to > the next paragraph.] > > ANSWER: Connect them pairwise (red to blue) in any manner, and > then for any intersection, reconnect the four points the other > way and thereby reduce the length of segments. Q.E.D. There must also be a similar elusive quality to understanding the >criticism of a proof, perhaps having to do with your innate bias >towards the competence of high-school teachers? If you didn't know >that the OP is a high-school teacher would you have been so critical >of the proof? Perhaps not, esp. if OP is a distinguished professor. >Please explain why it's clear that he didn't understand the proof. not as good example as I remembered it to be. That is, I concede it's not clear that he didn't understand the proof. If the basis of the proof is color however you like and then fix the problems, then there needs to be a recognition that there might be a circularity in the repair procedure. There had been a prior thread in which someone used the notion of distance (which is not a priori relevant here) to circumvent the problem. How, exactly, the introduction of distance saves the day takes a bit of reflection. Did this poster see why the introduction of lengths helps? My memory of the thread at the time was that the answer was not really; now on second thought I'll just say I can't tell. But yes, I do think it's fair to take into account a person's background when assessing whether or not they understand a proof. I don't know this teacher but I do teach others who are about to become high school math teachers, and if they turned in this proof I would be much more critical than I would if you (B.D.) or another person of trusted mathematical strength were giving this proof. Is that really surprising? To give another example, consider this gem: is there an equilateral triangle whose vertices lie on the integer lattice? The answer is no and my proof is: Pick's Theorem. Cool, huh? A proof which is certainly short enough to remember! Now, does a person who hears that understand the proof? Frankly this sort of thing happens to me all the time with students (including, but not limited to, some who are about to become teachers): I present this proof in class, and then ask for a synopsis later, and get a mention of Pick's theorem and integrality, but nary a mention of sqrt(3). Sorry, but I would not be convinced that the student understands the proof in that case. dave ==== [snip] >> QUESTION: Given n red points and n blue points on the plane >> (no three points collinear), is there always a way to pairwise >> connect the points (red to blue) with segments such that no >> segments intersect? Yes. [Why? Think a bit before you go to >> the next paragraph.] ANSWER: Connect them pairwise (red to blue) in any manner, and >> then for any intersection, reconnect the four points the other >> way and thereby reduce the length of segments. Q.E.D. > [snip] > not as good example as I remembered it to be. That is, I concede it's > not clear that he didn't understand the proof. I missed the fact that the author is a high-school teacher, and I completely agreed that he didn't understand the proof. Perhaps it would be more accurate to say that he may have understood the proof, but he certainly did a very poor job at presenting it. To me, the latter pretty much implies the negation of the former. FWIW, I think this is a *great* example. ==== [snip] >> QUESTION: Given n red points and n blue points on the plane >> (no three points collinear), is there always a way to pairwise >> connect the points (red to blue) with segments such that no >> segments intersect? Yes. [Why? Think a bit before you go to >> the next paragraph.] ANSWER: Connect them pairwise (red to blue) in any manner, and >> then for any intersection, reconnect the four points the other >> way and thereby reduce the length of segments. Q.E.D. >[snip] not as good example as I remembered it to be. That is, I concede it's > not clear that he didn't understand the proof. I missed the fact that the author is a high-school teacher, and I > completely agreed that he didn't understand the proof. Perhaps it > would be more accurate to say that he may have understood the proof, > but he certainly did a very poor job at presenting it. To me, the > latter pretty much implies the negation of the former. Apart from the fact that I would have to prove that if the line from A to B intersects the line from A' to B' then the line from A to B' doesn't intersect the line from A' to B, and the sum of the lengths of these two lines is shorter, it seems quite obvious to me that this would then prove the theorem. ==== Regarding this proof (posted by a high-school teacher), from http://google.com/groups?selm=301jo2%24a0g%40agate.berkeley.edu | QUESTION: Given n red points and n blue points on the plane | (no three points collinear), is there always a way to pairwise | connect the points (red to blue) with segments such that no | segments intersect? Yes. [Why? Think a bit before you go to | the next paragraph.] | | ANSWER: Connect them pairwise (red to blue) in any manner, and | then for any intersection, reconnect the four points the other | way and thereby reduce the length of segments. Q.E.D. >not as good example as I remembered it to be. That is, I concede >> it's not clear that [the author=OP] didn't understand the proof. I missed the fact that the author is a high-school teacher, and I > completely agreed that he didn't understand the proof. Perhaps it > would be more accurate to say that he may have understood the proof, > but he certainly did a very poor job at presenting it. To me, the > latter pretty much implies the negation of the former. FWIW, I think this is a *great* example. Then please do justify how you reach the conclusion that the author didn't understand the proof, esp. without bias towards H.S. teachers. Surely a novice might make the same objections regarding a proof by an expert, only because the novice regarded some missing step as essential. But the missing step might be obvious to all experts, and hence should be omitted on grounds of obscuring the main idea (assuming that the proof was presented in a forum for experts). The OPs proof omitted some details about the descent (or induction). I think he should be given that liberty given the intended mature audience of sci.math (versus a less mature forum, say k12.ed.math). In summary, I don't think it is fair to criticize the correctness of the proof, since it would be deemed correct by a sufficiently mature audience. Instead, one can only argue that the author might be assuming too much knowledge from the generic sci.math reader, which is just an opinion about the knowledge of sci.math readers, not a counterexample to a proof. -Bill Dubuque ==== To give another example, consider this gem: is there an equilateral > triangle whose vertices lie on the integer lattice? The answer is no > and my proof is: Pick's Theorem. Cool, huh? A proof which is > certainly short enough to remember! Now, does a person who hears that > understand the proof? Frankly this sort of thing happens to me all > the time with students (including, but not limited to, some who are > about to become teachers): I present this proof in class, and then ask > for a synopsis later, and get a mention of Pick's theorem and integrality, > but nary a mention of sqrt(3). Sorry, but I would not be convinced > that the student understands the proof in that case. Yes, that is a well-known gem. In this case, I agree with you that irrationality is a crucial component of the proof and needs mention, as opposed to the rather obvious descent in the original problem. It's interesting that you mention this particular problem because it too has a very simple descent proof similar to that for the original problem. For if such a triangle T existed then one easily shows (hint below) that all vertices have equal parity, therefore the midpoints of the sides are lattice points. So connecting the midpoints dissects T into 4 smaller such triangles. Here's the parity hint: choose the origin as one vertex, equate the squares of the side lengths, then draw conclusions (modulo 4). I.e. if the vertices are denoted (0,0), (X,Y), (x,y) then show that X^2 + Y^2 = x^2 + y^2 = (X-x)^2 + (Y-y)^2 => X,Y,x,y all even An almost identical proof is attributed to Lucas (1878) in [1, p.250]. This paper contains much more of interest on this and related problems. For higher dimensional generalizations see Robin Chapman's posts [2]. -Bill Dubuque [1] Michael J. Beeson: Triangles with vertices on lattice points, Amer. Math. Monthly, 99 (1992) 243-252. http://links.jstor.org/sici?sici=0002-9890(199903)99:3%3C243%3E [2] Robin Chapman, n-dim generalization http://google.com/groups?selm=6loo52%24foo%241%40nnrp1.dejanews.com http://google.com/groups?selm=bht5lg%24affov%241%40athena.ex.ac.uk ==== : :If T is an equilateral lattice triangle then one easily shows (hint below) :that all vertices have equal parity, hence the midpoints of the sides are :lattice points. So connecting the midpoints dissects T into 4 smaller :such triangles. Here's the parity hint: choose the origin as one vertex, :equate the squares of the side lengths, then draw conclusions (modulo 4). :I.e. if the vertices are denoted (0,0), (x,y), (X,Y) then prove that :x,y,X,Y are all even if they satisfy xx+yy = XX+YY = (x-X)^2 + (y-Y)^2 [*] i.e. v.v = V.V = v.v + V.V - 2 v.V [= (v-V).(v-V) ] <=> v.v = V.V = 2 v.V [a = A = a+A-A' <=> a=A=A'] => v.v + V.V = 4 v.V i.e. xx+yy + XX+YY = 0 (mod 4). But ZZ = 0 or 1 (mod 4) => xx,yy,XX,YY all 0 or all 1 (mod 4), else 0 < sum < 4 => x,y,X,Y all even or all odd, but not all odd by [*]. QED I thought I'd better present a proof lest I stand accused of not understanding it, per previous posts in this thread :-) Hopefully the descent itself is clear to sci.math readers. If this isn't so please do speak up. This would help us all better comprehend the distribution of knowledge levels here. -Bill Dubuque ==== >> Depends on what you mean by understand the proof. >What exactly does it mean to understand a proof? ... To be able to sit down and rewrite the entire proof BUT including one more level of detail for every step. Leslie Lamport once said that he had found he needed to write down the proof so that it was completely obvious how to get from each step to the next AND THEN write out TWO MORE levels of detail than that! Only then he found his mistakes. And he is a very bright guy. <3f9d06e2$0$69956$edfadb0f@dread12.news.tele.dk> <3f9d8023_5@127.0.0.1> ==== > Depends on what you mean by understand the proof. >What exactly does it mean to understand a proof? > ... To be able to sit down and rewrite the entire proof > BUT including one more level of detail for every step. That makes it a bit difficult for logic students. After all, suppose you're studying a purely formal proof of, say, one of the De Morgan laws. If it's already purely formal, you'll never understand it, since you *can't* right down any more detailed proof. (Another problem logicians face, I suppose, is a tendency to take everything at its most literal meaning -- not that I have any examples at hand.) -- [N]ow for once I might actually have an audience that realizes that [my proof of Fermat's Last Theorem is correct], because you see, they'll finally know what's in it for them--cold, hard cash. --James Harris embarks on a new mathematical strategy. ==== >> Depends on what you mean by understand the proof. >What exactly does it mean to understand a proof? >> ... >> To be able to sit down and rewrite the entire proof >> BUT including one more level of detail for every step. >That makes it a bit difficult for logic students. After all, suppose >you're studying a purely formal proof of, say, one of the De Morgan >laws. If it's already purely formal, you'll never understand it, >since you *can't* right down any more detailed proof. I suppose that my suggestion must then be constrained to only apply in such cases that there is at least one more level of detail. Examples of things like this would be the proof of FLT given by Mr. Harris or the proof of the orbifold theorem given by Dr. Thurston. disrespect by my statement, both those individuals have more energy than I ever will, those were ONLY two examples that came to mind. X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 03:11 AM, Steven Margolin said: >I have heard that with Choice, the reals are, while still >uncountable, well-ordered. Zorn's Lemma states that any set can be well ordered. It is equivalent to the Axiom of Choice. >This does not make sense to me. If they are well-ordered, then they >can be listed, What do you mean by listed? The index set is uncountable, so the diagonal argument is inapplicable. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > Zorn's Lemma states that any set can be well ordered. It is equivalent > to the Axiom of Choice. > That's not Zorn's Lemma, that's the Well-Ordering Principle (also equivalent to the Axiom of Choice). -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== > Something that I've been wondering is: Is it known whether there is a > well-ordering of the real numbers that can actually be defined explicitly? Have a tolerable existence. Eli Incredibly naive question coming up, since the answer is obviously no, but for the life of me I can't figure out what's wrong with it... why can't we define the ordering a < b in the seemingly obvious way? Since real numbers have decimal expansions, choose for each real number the terminating expansion if possible and compare digits until you find a difference and order that way? ==== > Incredibly naive question coming up, since the answer is obviously no, > but for the life of me I can't figure out what's wrong with it... why > can't we define the ordering a < b in the seemingly obvious way? Since > real numbers have decimal expansions, choose for each real number the > terminating expansion if possible and compare digits until you find a > difference and order that way? The trouble is that your ordering is not a well-ordering. For example, the set of positive reals fails to have a least element. Consider the set Q of rational numbers. Q is not well-ordered with the standard ordering, for the same reason. There are subsets (such as the set of positive rationals) that have no least element. However, it is known that Q is countable. Choose a bijection f: N -> Q, and define a new ordering <' on Q by x <' y iff f^(-1)(x) < f^(-1)(y) where the < on the RHS is the one for natural numbers, which therefore defines a well ordering. The question is, is there a way to apply a similar trick for the reals? Since R is uncountable, it's not possible to reduce the problem to an ordering on the naturals. But it's conceivable that there might be some uncountable ordinal alpha and a bijection f: alpha -> R. That's where the question lies. In the absence of the axiom of choice, it's possible that no such ordinal exists. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== Out of curiosity, how does one define an 'ordering' without formulating a > bijection with a countable set like the naturals? Syntactically, I mean.. > I'm clueless, which is why I ask Others have given you good partial answers, so I'll just comment on the syntactically part. Working in ZFC+V=L you can construct a formula phi with two free variables, such that ZFC+V=L |- AxAy( x and y are reals --> phi(x,y) or phi(y,x) ) ZFC+V=L |- Ax(x is a real --> ~phi(x,x) ) ZFC+V=L |- AxAyAz( x, y and z are reals --> phi(x,y) & phi(y,z) --> phi(x,z) ) ZFC+V=L |- Ax(x is a subset of reals --> Ey in x Az in x( phi(y,x)) This gives also something stronger. Since any ordering of the reals is a subset of the cartesian product of the set of reals with itself, by separation the set {(x,y) | phi(x,y)} exists. Usually, we need not concern ourselves with formulas, but merely with sets. Formally, an ordering is a special binary relation on a set. An n-ary relation R on a set A is subset of A^n (the n-fold cartesian product of A with itself, i.e. the set of all tuples (x_1,...,x_n) with x_i in A). To be an ordering, the relation needs to satisfy additional requirements, namely the ordering axioms: aRb & bRc --> aRc (transitivity) aRb or bRa (connectedness) ~(aRb & bRa) (antisymmetry) There is no need to specify any bijections here, but if the relation is a well-ordering, i.e. it's an ordering and in addition Ax(x is a subset of A --> Ey in x Az in x(yRz)) (every subset of A has a unique R minimal element) then there is an unique ordinal alpha, s.t. the less than-relation for ordinals (which for von Neumann ordinals is just the membership relation or the subset relation (these produce the same results)) on alpha is isomprhic to the ordered set (A,R). -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus ==== > Out of curiosity, how does one define an 'ordering' without formulating a > bijection with a countable set like the naturals? Syntactically, I mean.. > I'm clueless, which is why I ask -- > Quaternion The usual ordering of the reals does not require a bijection with any countable set. For example, using upper Dedekind cuts to define the reals, order among reals can be based on the subset relation among sets of rationals. ==== I have heard that with Choice, the reals are, while still uncountable, >well-ordered. This does not make sense to me. If they are well-ordered, >then they can be listed, so what about Cantor's Diaganolization Argument? Somehow you got the idea that well-ordered implies countable. Not so. I sort of understand that, but the reals, with unique ternary representations, are uncountable by a number of proofs, and one of them is CDA. However, with WOP, a roster notation of the reals is possible, so take that roster and list it. Now, we have a (perhaps uncountable) listing of the reals, but Cantor says we cannot have that. For example, with the set of ordinals, less than w_2, CDA would be meaningless. Using CDA on a well-ordering of the reals would likely produce a number that, while in the list, is uncountably far away. I have also heard that you need to able to give a finite number to >correspond to any number I name, but this can be solved by putting the >definable numbers first. >Me Please Explain. Hard to explain, since I have no idea what the previous paragraph > means. > ************************ ==== >I have heard that with Choice, the reals are, while still uncountable, >>well-ordered. This does not make sense to me. If they are well-ordered, >>then they can be listed, so what about Cantor's Diaganolization Argument? >Somehow you got the idea that well-ordered implies countable. Not so. >I sort of understand that, but the reals, with unique ternary >representations, are uncountable by a number of proofs, and one of them is >CDA. However, with WOP, a roster notation of the reals is possible, so take >that roster and list it. Now, we have a (perhaps uncountable) listing of >the reals, but Cantor says we cannot have that. Huh????????? > For example, with the set >of ordinals, less than w_2, CDA would be meaningless. Using CDA on a >well-ordering of the reals would likely produce a number that, while in the >list, is uncountably far away. You need to explain this much more carefully, otherwise people will assume you're just babbling nonsense when they realize they have no idea what you're talking about. >>I have also heard that you need to able to give a finite number to >>correspond to any number I name, but this can be solved by putting the >>definable numbers first. >>Me Please Explain. >Hard to explain, since I have no idea what the previous paragraph >> means. >> ************************ ************************ ==== I have heard that with Choice, the reals are, while still uncountable, >well-ordered. This does not make sense to me. If they are well-ordered, >then they can be listed, so what about Cantor's Diaganolization Argument? Somehow you got the idea that well-ordered implies countable. Not so. >I sort of understand that, but the reals, with unique ternary >representations, are uncountable by a number of proofs, and one of them is >CDA. However, with WOP, a roster notation of the reals is possible, so take >that roster and list it. Now, we have a (perhaps uncountable) listing of >the reals, but Cantor says we cannot have that. Huh????????? For example, with the set >of ordinals, less than w_2, CDA would be meaningless. Using CDA on a >well-ordering of the reals would likely produce a number that, while in the >list, is uncountably far away. You need to explain this much more carefully, otherwise people will > assume you're just babbling nonsense when they realize they have > no idea what you're talking about. You are right, I do I believe that I am talking out my arse, and I think that my question was answered about 10 posts ago when someone said that you cannot create real number with an uncountable number of digits. I have also heard that you need to able to give a finite number to >correspond to any number I name, but this can be solved by putting the >definable numbers first. >Me Please Explain. Hard to explain, since I have no idea what the previous paragraph > means. > ************************ > ************************ ==== >I have heard that with Choice, the reals are, while still uncountable, >>well-ordered. This does not make sense to me. If they are >well-ordered, >>then they can be listed, so what about Cantor's Diaganolization >Argument? >Somehow you got the idea that well-ordered implies countable. Not so. >>I sort of understand that, but the reals, with unique ternary >>representations, are uncountable by a number of proofs, and one of them >is >>CDA. However, with WOP, a roster notation of the reals is possible, so >take >>that roster and list it. Now, we have a (perhaps uncountable) listing of >>the reals, but Cantor says we cannot have that. >Huh????????? >> For example, with the set >>of ordinals, less than w_2, CDA would be meaningless. Using CDA on a >>well-ordering of the reals would likely produce a number that, while in >the >>list, is uncountably far away. >You need to explain this much more carefully, otherwise people will >> assume you're just babbling nonsense when they realize they have >> no idea what you're talking about. >You are right, I do I believe that I am talking out my arse, and I think >that my question was answered about 10 posts ago when someone said that you >cannot create real number with an uncountable number of digits. Well I have to say I was surprised to read that. I was expecting this thread to start veering towards the dark side... congratulations. >>I have also heard that you need to able to give a finite number to >>correspond to any number I name, but this can be solved by putting the >>definable numbers first. >>Me Please Explain. >Hard to explain, since I have no idea what the previous paragraph >> means. >> ************************ ************************ ************************ ==== Steven Margolin says... >...the reals, with unique ternary >representations, are uncountable by a number of proofs, and one of them is >CDA. However, with WOP, a roster notation of the reals is possible, so take >that roster and list it. Now, we have a (perhaps uncountable) listing of >the reals, but Cantor says we cannot have that. For example, with the set >of ordinals, less than w_2, CDA would be meaningless. Using CDA on a >well-ordering of the reals would likely produce a number that, while in the >list, is uncountably far away. I'm not sure why you think Cantor proves that we can't have an uncountable listing of reals, or why you think that Cantor's diagonalization argument applies to such a listing. If there is a well-ordering of the reals, then (according to ZFC, anyway) there is an association between reals and the ordinals less than some specific ordinal, the cardinality of the reals. So let C be the cardinality of the reals, and for every ordinal alpha less than C, let r_alpha be the real corresponding to ordinal alpha. How do you think that you are going to diagonalize to come up with a real r that is unequal to r_alpha, for every ordinal alpha? You can certainly diagonalize out of the first omega-many reals: You can define a real r by its decimal expansion as follows: the nth ternary digit of r = some number different from the nth decimal place of r_n But that only proves that r is unequal to r_alpha when alpha is finite. It doesn't eliminate the possibility that r = r_alpha for some infinite ordinal alpha. You can't diagonalize using more than the first omega many reals in the listing r_alpha, because real numbers only have omega many decimal places. -- Daryl McCullough Ithaca, NY ==== > Steven Margolin says... >>...the reals, with unique ternary >>representations, are uncountable by a number of proofs, and one of them is >>CDA. However, with WOP, a roster notation of the reals is possible, so take >>that roster and list it. Now, we have a (perhaps uncountable) listing of >>the reals, but Cantor says we cannot have that. For example, with the set >>of ordinals, less than w_2, CDA would be meaningless. Using CDA on a >>well-ordering of the reals would likely produce a number that, while in the >>list, is uncountably far away. > I'm not sure why you think Cantor proves that we can't have an uncountable > listing of reals, or why you think that Cantor's diagonalization argument > applies to such a listing. > If there is a well-ordering of the reals, then (according to ZFC, anyway) > there is an association between reals and the ordinals less than some > specific ordinal, the cardinality of the reals. So let C be the cardinality > of the reals, and for every ordinal alpha less than C, let r_alpha be the > real corresponding to ordinal alpha. There need not be one. Just as a countable set can be enumerated so that each member corresponds to ordinal less than w, a set of cardinality alpha can be enumerated so that each member corresponds to an ordinal less than alpha. > How do you think that you are going to diagonalize to come up with > a real r that is unequal to r_alpha, for every ordinal alpha? > You can certainly diagonalize out of the first omega-many reals: You > can define a real r by its decimal expansion as follows: Any well-ordering of an uncountable set necessarily has a member that maps to w, and that is enough to defeat the diagonal argument. > the nth ternary digit of r = some number different from the nth > decimal place of r_n > But that only proves that r is unequal to r_alpha when alpha is finite. > It doesn't eliminate the possibility that r = r_alpha for some infinite > ordinal alpha. > You can't diagonalize using more than the first omega many reals in > the listing r_alpha, because real numbers only have omega many decimal > places. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== Dave Seaman says... If there is a well-ordering of the reals, then (according to ZFC, anyway) >> there is an association between reals and the ordinals less than some >> specific ordinal, the cardinality of the reals. So let C be the cardinality >> of the reals, and for every ordinal alpha less than C, let r_alpha be the >> real corresponding to ordinal alpha. There need not be one. Just as a countable set can be enumerated so that >each member corresponds to ordinal less than w, a set of cardinality >alpha can be enumerated so that each member corresponds to an ordinal >less than alpha. I think that's what I just said: If the reals have cardinality C, then the set of reals can be enumerated so that each real corresponds to some ordinal alpha less than C. -- Daryl McCullough Ithaca, NY ==== > Dave Seaman says... > If there is a well-ordering of the reals, then (according to ZFC, anyway) > there is an association between reals and the ordinals less than some > specific ordinal, the cardinality of the reals. So let C be the cardinality > of the reals, and for every ordinal alpha less than C, let r_alpha be the > real corresponding to ordinal alpha. >>There need not be one. Just as a countable set can be enumerated so that >>each member corresponds to ordinal less than w, a set of cardinality >>alpha can be enumerated so that each member corresponds to an ordinal >>less than alpha. > I think that's what I just said: If the reals have cardinality C, then > the set of reals can be enumerated so that each real corresponds to > some ordinal alpha less than C. So you did. My apologies. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== >I have heard that with Choice, the reals are, while still uncountable, >well-ordered. This does not make sense to me. If they are well-ordered, >then they can be listed, so what about Cantor's Diaganolization Argument? Somehow you got the idea that well-ordered implies countable. Not so. > I sort of understand that, but the reals, with unique ternary > representations, are uncountable by a number of proofs, and one of them is > CDA. However, with WOP, a roster notation of the reals is possible, so take > that roster and list it. Now, we have a (perhaps uncountable) listing of > the reals, but Cantor says we cannot have that. No, we *can* have that. In Cantor's diagonal argument, we cannot have a listing labeled by the natural numbers, because the labels for the digits are also the natural numbers, so we get a diagonal where the label of the number equals the label of the digit. But if we have an uncountable labelling for the real numbers, we cannot define a diagonal anymore. For example, with the set > of ordinals, less than w_2, CDA would be meaningless. Using CDA on a > well-ordering of the reals would likely produce a number that, while in the > list, is uncountably far away. This just doesn't make sense. >I have also heard that you need to able to give a finite number to >correspond to any number I name, but this can be solved by putting the >definable numbers first. Nor does this. >Me Please Explain. Hard to explain, since I have no idea what the previous paragraph > means. > ************************ > > ==== > Out of curiosity, how does one define an 'ordering' without formulating a > bijection with a countable set like the naturals? Syntactically, I mean.. > I'm clueless, which is why I ask Perhaps by formulating a bijection with an uncountable ordinal? Every ordinal is well-ordered, but only a few of them (the ones below w_1) are countable. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 01:15 AM, creighde@yahoo.de (C. Dement) said: Why do you assume that there is no topology? More precisely, why do you assume that you can't define a topology that yields the convergence that you have defined? Isn't this analogous to the situation in a metric space, where there is no topology, yet the metric lets you define a unique topology compatible with it? -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > I propose that physicists should no longer be allowed > to teach relativity and that the subject be taught by > mathematicians. AHAhahaha.......ahahahaha.... > Have you ever looked at sci.math? The ones you hope to get rescue > from do quarrel at great length over whether 1+1 is really 2. > Good luck to you, > ahahahaha......ahahanson Hanson, how long has it been since you had your meds adjusted? Perhaps you should get a blood test soon. ==== .....and the polynomial is x^3 - 147 x + 286. > Nice. And to take it one step farther put x -> x-1 to get f(x) = x^3 - 3x^2- 144x + 432 which has all the terms and all nonzero integer factors. --Lynn ==== >>What is the first number that can be divided byt he frist three prime >>numbers? >Go through all the numbers from one to twenty and ask >> Is this number divisable by 2 AND 3 AND 4? >If you find one that is, let us know. Seems like a little bit of attitude from one who thinks that 4 is prime! >adam > True. adam ==== Now what I alluded to in my previous post is whether a batter such as B. Bonds can hit a homerun from simply just tossing the ball in the air from homeplate and then seeing whether a homerun can be hit. Those who have played baseball know this test in that they do fielding practice in many instances where a batter with bat in one hand and with ball in another tosses the ball into the air and tries to hit it. One can say the ball has 0 speed from the pitcher's mound direction. And I really do not know what the slowest speed attainable from the pitcher in major league sports to carry the ball over the homeplate??? pitcher pitches a 90 mph ball. This test that I speak of is designed to see whether a genuine slugger of baseball can hit a homerun from a 0 speed ball from homeplate. I heard from the announcers that the field park in Florida for the Marlins is vastly deeper than the park in New York. But I am wondering of a Barry Bonds test of a 0 speed ball and whether Bonds can hit that ball a homerun?? If he can hit a 0 speed ball from Yankee stadium as a homerun then my opinion is that baseball stadiums are grossly underdistanced and that they need to be altered such that no hitter in baseball is able to hit a homerun from a 0 speed ball. In fact, my opinion is that the slowest speed pitch possible, what is it??? Is it 10 mph that a pitcher in the major leagues can get across the plate?? My opinion is that all pitches of this slowest speed should not be able to be hit as a homerun in any major league baseball park and that these parks should be re landscaped such that their distances make it impossible for any slugger to hit a 10 mph pitch as a homerun. Now I understand that aluminum bats are not allowed in major leagues. I feel they should be. And do the above testing with aluminum bats. I would love to hear the sound of a pitch hit from a aluminum bat. But if these bats can cause death and injury to players then keep the ban on aluminum bats. If a Barry Bonds is unable to hit a 0 speed ball as a homerun then that indicates to me that the Optimal Strategy Pitching in Baseball is not in the direction of fast pitching but in the direction of slow pitching. I need a test of the average slow speed of a baseball that gets it across the plate as a strike. I am simply guessing of 10 mph but need an accurate number. And the reason that no pitchers in modern baseball do a slow pitch is because of social antagonism that they demand and expect all pitchers to deliver fast pitches. But if the rules in Baseball allow slow pitches then that will be the future trend. And the distance of all baseball parks outfield should be a physics-science determination. Obviously a Barry Bonds is able to hit homeruns from 90 mph pitches in virtually every baseball park. But is he able to hit homeruns in every park from a 30 mph pitch?? So the distance of every major league baseball park should be determined by the speed of pitch and those parks that are inadequate should be relandscaped and designed better. So how can you justify to any baseball fan that a Barry Bonds can hit a homerun in their stadium from a simple hand tossed 0 pitch speed?? You cannot justify that and that the field distance needs to be fixed. Almost as if a Grandmother can go out to that field and hit a homerun. So that in the future, when a Barry Bonds comes to the plate and scared of him hitting a homerun, then a stream of slow pitches is rendered instead of a forced walk to the base. In fact, if a pitcher in the future perfects a slow pitch with a spin on the ball that almost every batter will have an impossible time of hitting a homerun. But whether the scores in baseball will become larger because so many singles or doubles are hit remains a question for slow pitching. The trouble with baseball at the moment is that science and physics have never rally given the sport any inputs and that the rules and agenda of baseball as a sport has been governed more by the tillation effect on fans. Most every fan of baseball is unaware at the moment of these writings that the pitcher is the dominate feature of baseball where most fans immediately think of the batting slugger. And what if a machine, a pitching machine that is anchored on the pitching mound designed to deliver a fast ball at a specified speed in every pitch. The fans would balk and baseball lose much of its appeal. And in that circumstance we can truly say that the slugger in baseball is of all importance and not the pitcher. So I have raised many important questions above. (1) what is the slowest speed to deliver in major league pitches (2) can the present day sluggers hit a homerun in the stadiums from a 0 speed toss up (3) are most stadiums inadequate in distance of homerun (4) link and connect speed of pitch to outfield distance (5)research the advantages of slow speed pitching in baseball Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ==== >I have set of two equations on functions x(t) y(t) (derivstion on t): >x' = a*y + b*x*y +c*x + d >y' = a*y + b*x*y + d Maple 9 says: / d 2 [{y(t) = _a &where [{|--- _b(_a)| _b(_a) + (_b(_a) d + a _a c d_a / 2 2 - _b(_a) - _b(_a) b _a - _a _b(_a) c + _a d c)/_a = 0}, d {_b(_a) = -- y(t), _a = y(t)}, dt / | 1 {t = | ------ d_a + _C1, y(t) = _a}]}, | _b(_a) / /d |-- y(t)| - a y(t) - d dt / {x(t) = ----------------------}] b y(t) Basically what it's doing is eliminating x and t and reducing the system to a first-order differential equation for v = dy/dt as a function of y. Start with b x y = v - a y - d; from the equation for x' you then get an equation for v' and y' in terms of y and v. And then dv/dy = v'/y' = b y + c + (v-d)/y - (a y + d) c/v. But there doesn't seem to be any closed-form solution for that differential equation. Maple calls it an Abel equation of the second type, class B. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== Lately I found some simplification coming out from model. To give better overview about my problem I start from begining by describing problem that leads me to those equations. I am analysing chemical reactions: X + Y -> Z (ratio v1) Z -> X + Y (ratio v2) -> X (ratio k1) X -> (ratio k2) My goal is to find out average of Z molecule, in function of time. (In truth this is not my original problem, which is larger, but only sub problem. In my original problem ratio k1 is not constant but vary depending on time (there are more reaction that produce X), I can calculate this function, but assume that this is constant. At the end I am looking for variance not average) How I get equations: P_{x}(t) = probability that in time t there is x molecules type X. [similar P_{y}, P_{z}] Ex(t) = average of x molecules in time t. [similar Ey, Ez] P_{x}(t+dh)=P_{x-1}(t)*(k1 + Ez(t)*v2)*dt + P_{x+1}(t)*[k2*(x+1) + Ey(t)*(x+1)*v1]*dt + P_{x}(t)*[1 - (k1 + Ez(t)*v2 + k2*x + Ey(t)*x*v1)*dt]. After multiply by x and summing it by x we got: Ex'(t) = k1 + v2*Ez(t) - k2*Ex(t) - v1*Ex(t)*Ey(t) Repeating same steps for y we got: Ey'(t) = v2*Ez(t) - v1*Ex(t)*Ey(t) To get ride off Ez(t) we can observe that at each time t number of molecules Y plus number of molecules Z are constant, so we got: Ez(t) = k3 - Ey(t) so: Ex'(t) = v2*k3 - v2*Ey(t) - v1*Ex(t)*Ey(t) - k2*Ex(t) + k1 Ey'(t) = v2*k3 - v2*Ey(t) - v1*Ex(t)*Ey(t) If we assume that k1 = 0, and change variable names, we got my equations: x' = a*y + b*x*y + c*x + d y' = a*y + b*x*y + d Yesterday I observe one more thing in my model, that we can calculate average number of molecule x+z: E_{x+z}(t)=f(t) and we have: Ex = Ey - k3 + f(t) so in new notation x = y + f(t), which leads us to: y' = b*y*y + [b*f(t)+a]*y + d Which is a bit less complex. However I steel need help on solving this one. By solving I understand exacly symbolic formula for y(t). Dargen ==== > y' = b*y*y + [b*f(t)+a]*y + d But a = -v2 b = -v1 d = v2*k3 and f(t) >=0 as it is E_{x+z} so (b*f(t)+a)^2 + 4v2*v2*k3 > 0 and I cane split this equation into two linear equations. Is this correct? Dargen ==== > But > a = -v2 > b = -v1 > d = v2*k3 > and f(t) >=0 as it is E_{x+z} > so (b*f(t)+a)^2 + 4v2*v2*k3 > 0 > and I cane split this equation into two linear equations. Is this correct? I cancel this question :) I can't split it, I don't know why, I thought that I have zero on left side:). If I'm correct this is Riccati equation. Dargen ==== >I have set of two equations on functions x(t) y(t) (derivstion on t): x' = a*y + b*x*y +c*x + d >y' = a*y + b*x*y + d I will be very thankful for any help on solving it. It seems to be very >regular, despite its nonlinear, but I have no idea how to solve it. > > These seem similar to to equations arising in the population dynamics of infectious diseases. The latter do not have a closed form solution, though one can describe qualitative properties. Bailey is a standard reference. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== >I have set of two equations on functions x(t) y(t) (derivstion on t): x' = a*y + b*x*y +c*x + d >y' = a*y + b*x*y + d I will be very thankful for any help on solving it. It seems to be very >regular, despite its nonlinear, but I have no idea how to solve it. Define solve. Since the equations are autonomous (there is no t) you can solve the problems graphically -- not in terms of t but at least showing how x and y will vary together. Simply make little marks in the plane whose slope at a point (x,y) is dy/dx = y'/x' = (a*y + b*x*y + d)/(a*y + b*x*y +c*x + d) and connect the dots. The resulting curves traces out the combinations of (x,y) which can possibly be attained as t varies. (Of course, you'll only be interested in the one curve passing through your initial point (x(0),y(0)). ) You can also solve numerically. A simple method estimates the values (x(n),y(n)) of the functions for t = n using the recurrence x(n+1) = x(n) + a*y(n) + b*x(n)*y(n) +c*x(n) + d y(n+1) = y(n) + a*y(n) + b*x(n)*y(n) + d but of course numerical analysts can suggest much more robust approximations. What you probably can't do, except for very fortunate combinations of a,b,c,d, is to give symbolic formulas for the functions x(t), y(t). dave ==== > I have set of two equations on functions x(t) y(t) > (derivstion on t): x' = a*y + b*x*y +c*x + d > y' = a*y + b*x*y + d I will be very thankful for any help on solving it. > It seems to be very regular, despite its nonlinear, > but I have no idea how to solve it. (1) y' = ay + bxy + d (2) x' = ay + bxy + cx + d (3) y = (x' - cx - d)(a + bx)^-1 (4) y' = (x'-cx-d)'(a+bx)^-1 + (x'-cx-d)(a+bx)^-1' = (x-cx')(a+bx)^-1 - (x'-cx-d)(a+bx)^-2(bx') = (a+bx)^-2[(x-cx')(a+bx) - (x'-cx-d)(bx')] (a+bx)^-2[(x-cx')(a+bx) - (x'-cx-d)(bx')] = a(x'-cx-d)(a+bx)^-1 + bx(x'-cx-d)(a+bx)^-1 + d does not seem that simple to me. is that a textbook problem? > thanks, > Dargen ==== > The first eqn can be written as: > x' = y' + cx well, yeah, and the second can be written as y' - (a + bx)y = d but we were hoping for something that would actually make progress... > MB I have set of two equations on functions x(t) y(t) (derivstion on t): x' = a*y + b*x*y +c*x + d > y' = a*y + b*x*y + d I will be very thankful for any help on solving it. It seems to be very > regular, despite its nonlinear, but I have no idea how to solve it. thanks, > Dargen ==== >.... If you can focus on any particular parts of Coxeter that you'd > like to follow up, then no doubt various people will have suggestions. Logical foundations. > Sorry I've been a bit slow coming back to this thread. I almost forgot that I'd started it! There seem to have been three main 20th-century approaches to the > foundations of geometry. The book by Borsuk & Szmielew which I mentioned follows up the > already have. Another good book in that style is Henry George Forder, > The Foundations of Euclidean Geometry, which was reprinted by Dover. In 1959 Tarski proposed studying geometry as a first-order theory > using point variables, so for example you never say For every line .... > Most of the subsequent literature is in Polish; but if you know something > about logic and first order theories you might appreciate Wanda Szmielew, A German school centred on Friedrich Bachmann produced a version of > geometry based firmly upon group theory. The standard reference is in > German: Bachmann's Aufbau der Geometrie aus dem Spiegelungsbegriff. > However, you can get a good introduction from H Behnke, F. Bachmann, K. > Fladt & H Kunle (editors), Fundamentals of Mathematics, Volume II, > Geometry, translated by S.H. Gould, M.I.T. Press, 1974. I think this is > out of print, but you can probably find second-hand copies. I looked into the three English volumes some years ago and I recall that chapter. Yes, it is out of print. Come to think of it, that last book Fundamentals.... may well be > the very best thing to read after Coxeter. As well as introducing the > group-theoretic approach (e.g. in chapters 5 and 2), it covers a > fascinating range of different topics within geometry. Yes _ I shoyld > have advised you to find a copy of Fundamentals.... in the first place. I know that Coventry Public Library (in the UK) has it. I shall obediently return to it. Ken Pledger. -- G.C. ==== A German school centred on Friedrich Bachmann produced a version of >geometry based firmly upon group theory. The standard reference is in > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > When this view is pursued too far, it leads to 'hot potato' reactions like > those generated by http://www.oswego.edu/~baloglou/103/seventeen.html, a > purely geometrical/elementary classification of planar crystallographic > groups (wallpaper patterns) Yes, it's a nice site which I recommend to my group theory class. > since we can study and classify these beasts > using group theory, since any two of them belong to the same type if and > only if their isometry groups are isomorphic, why look at other approaches? > .... Such very valuable use of symmetry groups is not at all the same thing as the Bachmann school's approach to foundations. They use a group-theoretic equation to express the fact that a point lies on a line, another equation to say that two lines are perpendicular, etc. It really is an alternative _foundation_ for geometry, Euclidean and non-Euclidean. Ken Pledger. ==== > A German school centred on Friedrich Bachmann produced a version of >geometry based firmly upon group theory. The standard reference is in > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > When this view is pursued too far, it leads to 'hot potato' reactions like > those generated by http://www.oswego.edu/~baloglou/103/seventeen.html, a > purely geometrical/elementary classification of planar crystallographic > groups (wallpaper patterns) Yes, it's a nice site which I recommend to my group theory class. since we can study and classify these beasts > using group theory, since any two of them belong to the same type if and > only if their isometry groups are isomorphic, why look at other approaches? > .... Such very valuable use of symmetry groups is not at all the same > thing as the Bachmann school's approach to foundations. They use a > group-theoretic equation to express the fact that a point lies on a line, > another equation to say that two lines are perpendicular, etc. It really > is an alternative _foundation_ for geometry, Euclidean and non-Euclidean. Sounds like fun Ken Pledger. -- G.C. ==== Given that p = 2q + 1 where p and q are odd prime. Prove that the generators of the group Z*p are QNRp - { -1 }. Yes, this is an assignment question. 1) I've been able to prove that QRp cannot be generators. 2) I've also been able to prove that -1 (i.e. 2q) cannot be a generator either. 3) However, I'm unable to prove that the rest of the group members are all generators. For (1): If g is a generator, then for all b, there exist an i such that b = g^i. However, we know that g is a quadratic residue, so g=a^2. Hence, b = a^2^i = a^i^2. Hence, b must be a quadratic residue as well! So, g only generates other quadratic residue (not the whole group). For (2): This is rather trivial. (3) I'm not getting anywhere for that one. Any idea? (I would prefer a hint, not a complete solution). Robin Lavall.8ee ==== > Given that p = 2q + 1 where p and q are odd prime. Prove that the > generators of the group Z*p are QNRp - { -1 }. It's a question of counting. How many quadratic non-residues are there? How many generators does a cyclic group of a given order have? -- ==== >3) However, I'm unable to prove that the rest of the group members are >all generators. Isn't that kind of trivial, seeing as p is a prime number? John Savard http://home.ecn.ab.ca/~jsavard/index.html ==== Hallo everyone, I worked for about 3 years after college before getting into grad school. (I am currently in the first year). And I find math so much more difficult now, than I ever did; keeping up with the pace of the courses is turning out to be a nightmare. I have forgotten even some elementary results; and find myself stuck in a painstakingly slow iterative process (going back to my old Calc III notes...). My profs want me to speed up things, but I cannot seem to find an optimal path. I'd like some advice from people who have been in my position; esp. those who have made it through grad school! I know graduate school is hard for everyone, but is hard work the only solution? Is there a way of working smart instead of hard? Because at the end of the day, for first year students at least, it is the grades on the exams and the courses that really matter, irrespective of mathematical potential. Frank P.S.--I am not sure if there is any wisdom in sampling life for a few years before beginning grad school. ==== I know graduate school is hard for everyone, but is hard work the > only solution? There is no Royal Road to geometry. ==== > I know graduate school is hard for everyone, but is hard work the > only solution? There is no Royal Road to geometry. I think the OP meant that hard work *by itself* does not seem to be working for him. That I can believe. One might say the whole point of graduate education is to learn how to work smart; so the OP's desire to work smart is indeed a desideratum but it can only be achieved by (ahem) hard work. Hence the Royal Road quote. However, there are two comments I can make that might be helpful. One is, although you (the OP) probably don't yet know how to work smart, you do know enough by now to avoid certain ways of working dumb. Mind-numbing work that leads to no useful insights is necessary sometimes, but don't subject yourself to *more* of it than is necessary. You don't get any brownie points for just doing work. When you get bogged down, look to the larger picture if you possibly can; for me, that means reading more broadly so that I begin to see the *why* for all the difficult work. Thus refreshed, I can go back to it -- or perhaps, I can see what parts of the work it's safe (for now) to skip or to skim over. My other comment is a bit hard-nosed: I've no doubt you are working hard, but are you working *hard*? When the chips are down, as they seem to be, you need to throw everything and I mean everything you got at it. Take no prisoners. What I mean is, it's possible you don't yet know how hard you really can work; give it a try. (Of course this is just MHO and I should add I've never been a grad student myself, in *math* that is. I'm generalizing from my experience elsewhere, and from my own self study in math. So if my advice is misplaced, I hope that someone with more direct experience will follow up with a correction.) ==== It was really much more that I could expected! Vincenzo ==== > Theorem: If Y is a connected subspace of the connected space X and > A, B are disjoint > open sets in X <<< such that XY = A cup B, then both OP did not say and there is no need to assume Y is closed. It is enough that A and B are open in Acup B. > A cup Y and B cup Y are connected. Proof: It suffices to prove that A cup Y is connected (since the > same argument will work for B cup Y). Argue by contradiction - suppose > A cup Y is not connected. Then there is a continuous surjection > f: A cup Y --> 2. Since Y _is_ connected, f is constant on Y -- > WLOG, assume Y subset f^{-1}(0). Then f^{-1}(1) subset A and is > relatively open in A cup Y -- since A is open in X, it follows that Yes, f^(-1){A} is relatively open in Acup Y (even if Y is not closed) > f^{-1}(1) is an open subset of X. Also, there is some open O subset X > such that f^{-1}(0) = O cap (A cup Y). Extend f to F: X --> 2 by > defining F(x) = 0 if x in B. Then F is continuous (since we have > F^{-1}(1) = f^{-1}(1) and F^{-1}(0) = O cup B, both of which are open) This will also work even if Y is not closed (B is not open) because B is contained in an open set that misses all of A. > and F is a surjection (since f is). This contradicts the fact that > X is conected and establishes the result. [] > Two comments: (1) There is no need to argue by contradiction. Nowhere do you use the fact that both f(-1){0} and f(-1){1} are both non-empty until the very last line. Simply declare f is continuous and extend to F. Then argue that F is constant, hence f must be constant. Then X must be connected since f was arbitrary. (2) Much to my dismay, involving the function is just a distraction. The crux of your proof is to cover X by disjoint open sets. This is painfully obvious when you note both of which are open in order to argue that F is continuous. ==== Theorem: If Y is a connected subspace of the connected space X and > A, B are disjoint > open sets in X <<< such that XY = A cup B, then both OP did not say and there is no need to assume Y is closed. > It is enough that A and B are open in Acup B. > True -- I misremembered the result when I was thinking about this ... and (seemingly) never bothered to recover :-) > A cup Y and B cup Y are connected. Proof: It suffices to prove that A cup Y is connected (since the > same argument will work for B cup Y). Argue by contradiction - suppose > > A cup Y is not connected. Then there is a continuous surjection > f: A cup Y --> 2. Since Y _is_ connected, f is constant on Y -- > WLOG, assume Y subset f^{-1}(0). Then f^{-1}(1) subset A and is > relatively open in A cup Y -- since A is open in X, it follows that Yes, f^(-1){A} is relatively open in Acup Y (even if Y is not closed) f^{-1}(1) is an open subset of X. Also, there is some open O subset X > such that f^{-1}(0) = O cap (A cup Y). Extend f to F: X --> 2 by > defining F(x) = 0 if x in B. Then F is continuous (since we have > F^{-1}(1) = f^{-1}(1) and F^{-1}(0) = O cup B, both of which are open) This will also work even if Y is not closed (B is not open) > because B is contained in an open set that misses all of A. > and F is a surjection (since f is). This contradicts the fact that > X is conected and establishes the result. [] Two comments: (1) There is no need to argue by contradiction. > Nowhere do you use the fact that both f(-1){0} and f(-1){1} > are both non-empty until the very last line. > Simply declare f is continuous and extend to F. > Then argue that F is constant, hence f must be constant. > Then X must be connected since f was arbitrary. > OK -- I'll buy that ... (2) Much to my dismay, involving the function is just a distraction. > The crux of your proof is to cover X by disjoint open sets. > This is painfully obvious when you note both of which are open > in order to argue that F is continuous. but ... I don't buy that -- I guess that's why Baskin-Robbins has to have 31 flavors -- I really *do* understand what's going on better when the proof is presented in this particular way. YMMV, of course ... ==== >What are some good books? Depends on the flavor and level you want. _An Introduction to the Theory of Numbers_, 5th Edition, by Ivan Niven, Herbert Zuckerman, and Hugh Montgomery, starts with notions of divisibility, primes, and the binomial theorem, and continues through congruences, quadratic forms and quadratic reciprocity, arithmetic functions and the Mobius inversion formula, some diophantine equations, Farey fractions and irrational numbers, simple continued fractions, a bit about multiplicative number theory (Dirichlet series and something on primes in arithmetic progressions), partitions, densities, and even a bit about algebraic number fields (particularly quadratic number fields). Pretty good first introduction, and covers a lot of stuff, I think. If you want a look at Analytic Number Theory, there's Tom Apostol's _Introduction to Analytic Number Theory_, Springer Verlag Undergraduate Texts in Mathematics. Also starts with divisibility, gcds, primes, and the Euclidean algorithm, but then quickly moves through Arithmetical functions, Dirichlet multiplication, averages of arithmetical functions, and some theorems on the distribution of primes, before moving to congruences, finite abelian groups and their characters, and Dirichlet's Theorem on primes in arithmetic progressions. Then he deals with Gauss sums, quadratic reciprocity, primitive roots, Dirichlet series and Euler products, the zeta function, and an analytic proof of the Prime Number Theorem, before finishing up with partitions. An introduction to Algebraic Number Theory, though a bit more advanced than Apostol's intro to analytic number theory, is Kenneth Ireland and Michael Rosen's _A Classical Introduction to Modern Number Theory_, 2nd Edition (Springer Verlag Graduate Texts in Mathematics); I've seen it used for an undergraduate course once (at Berkeley). Starts with Unique factorizaion in Z, k[x], talks about PIDs, then the ring of Gaussian and of Eisenstein integers; then applications of unique factorization, congruences, the units of the integers modulo n, quadratic reciprocity, then quadratic Gauss sums, finite fields, general Gauss sums and Jacobi sums, cubic and biquadratic reciprocity, equations over finite fields, then the zeta functions, and then dives right into Algebraic Number Theory: unique factorization, ramification and degree, quadratic and cyclotomic fields, the Eisenstein Reciprocity Law, Bernoulli numbers, Dirichlet L-functions, diophantine equationes, and even three final chapters on Elliptic curves, the Mordell-Weil Theorem, and New Progress in Arithmetic Geometry (at least, new as of 1981). Finally, if you know enough abstract algebra (groups, rings, fields, basic Galois Theory), then I find that an excellent introduction to algebraic number theory is Daniel Marcus's _Number Fields_ (Springer Verlag Universitext). It is chock full of exercises, usually around 50 per chapter, which involve doing actual implementations of sundry theorems. It starts by discussing a special case of Fermat's Theorem (basically, Lame's idea and how one could try to make it work, leading to what is essentially a proof of Fermat's Theorem for the regular primes in Case I, though not quite). Then defines number fields and rings of integers, decomposition in number fields, Galois Theory applied to the decomposition, the ideal class group and the unit group, distribution of ideals, Dedekind's Zeta function and the class number formula, and the distribution of primes. Finally, a somewhat quick overview of what Class Field Theory is all about. The heart of the book is the exercises, and they give you a very solid grounding on algebraic number theory at a level around that of the end of a first graduate course in Algebraic Number Theory. But the prerequisites are steeper than for the other books. ====================================================================== 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 ==== The first half of Ireland-Rosen was written for Brown's undergraduate math major number theory course, and is often used for that purpose. It presupposes a basic undergrad algebra course (groups, rings, a little about fields, but not Galois theory). The latter half was added later, and is suitable for a beginning graduate course. Another very enjoyable, short, elementary introduction to number that as a paperback, it's relatively inexpensive. Well, looks like $30 on Amazon, so maybe not all that cheap. Anyway, when I took number theory (from Rosen, using Ireland-Rosen), he also had us buy Davenport's book to get another look at the subject. JHS ------------------------ than Apostol's intro to analytic number theory, is Kenneth Ireland and > Michael Rosen's _A Classical Introduction to Modern Number Theory_, > 2nd Edition (Springer Verlag Graduate Texts in Mathematics); I've seen > it used for an undergraduate course once (at Berkeley). ==== > What are some good books? Hardy and Wright, The Theory of Numbers. Niven, Zuckerman, and Montgomery, An Introduction to the Theory of Numbers. LeVeque, Fundamentals of Number Theory. Uspensky and Heaslett, Elementary Number Theory. Roberts, Elementary Number Theory. Ireland and Rosen, A Classical Introduction to Modern Number Theory. The Andrews book and the Rosen book others have recommended are also good. -- ==== What are some good books? Hardy and Wright, The Theory of Numbers. > Niven, Zuckerman, and Montgomery, An Introduction to the Theory > of Numbers. > LeVeque, Fundamentals of Number Theory. > Uspensky and Heaslett, Elementary Number Theory. > Roberts, Elementary Number Theory. > Ireland and Rosen, A Classical Introduction to Modern Number Theory. The Andrews book and the Rosen book others have recommended are also > good. What about Fundamental Number Theory with Applications by Richard A. Mullin? ==== > What about Fundamental Number Theory with Applications by Richard A. > Mullin? Again, I'm biased here, since Mollin is a real good friend of mine, and I was one of the poor souls that got roped into proofreading the text and the solutions manual, but: The good points about the texts are that there are tons of exercises and copious historical notes. It is modern, including up to date material about factoring and cryptology. It depends what the OP wants as to whether it's an appropriate text for his use. Mollin's book starts from scratch in that it spends quite a bit of time developing things from the axioms, working through arithmetic and so on, before getting on to number theory proper. (Hence the name Fundamental, instead of Elementary.) So in the hands of an instructor, who can guide the learning as appropriate for the particular students' background, this is fine and admirable. But for self-study, one could become bogged-down in stuff that one already knows, stuck on hard exercises that aren't really relevant, when one really wants to get on with it. Along the same lines, the last section of every chapter deals with an optional thread quadratics, which is Mollin's research specialty. A self-study student should probably skip that stuff, but might not know whether he should or not. It depends on what the OP's unstated purposes were. I wouldn't hand it to a student for self-study without knowing the student and then marking the sections that I think he should skip. Otherwise, it's a nice book for classroom use, with lots of good, challenging exercises, and well-researched historical notes. Bart ==== It depends on what the OP's unstated purposes were. I wouldn't hand > it to a student for self-study without knowing the student and then > marking the sections that I think he should skip. Otherwise, it's > a nice book for classroom use, with lots of good, challenging exercises, > and well-researched historical notes. > My intension is self-study. Therefore, I want a book that has it all, and that has a clear and easy to follow structure. Really good math textbooks are like good novels. You almost can't put them down, or you can't wait to apply what you've just read, to excercises. /David ==== (M y_i + q_i y_i)^T x = 0, i=1,...,n, (1) where x is an unknown vector in R^3 with constraint ||x||=1, M is a 3-by-3 matrix, y_i, i=1,...,n are known vectors in R^3, and q_i > 0, i=1,...,n are known parameters. Practically, there is noise (perturbation) in q_i. I need to slove Eq. (1) for x. To this end, I construct the following optimization problem min ||Ax||, s.t. ||x||=1, (2) where A is an n-by-3 matrix, each row of A is of (M y_i + q_i y_i)^T. This problem can be solved by SVD, x is equal to the singular vector associated with the smallest singular value of A. The solution of (1) or (2) for non-zero q_i makes sense to me. However, I find that, due to noise in q_i, the solution of (2) frequently approximates the one under the case of q_i=0 for all i, i.e. the one satisfying (M y_i)^T x = 0, i=1,...,n, but this is a trivial solution to me. How to solve (1) or (2) robustly to the noise in q_i, for the non-trivial x? Is it useful by increasing the amount of equations? Is there any additional regularization methods? Zhaozhong Wang ==== I understand that I can use Pascal's Triangle to complete a binomial expansion, but it's very long-winded. What is the quickest way of working out the coefficient of x^5 in the expansion of (x^2 - (3/x))^7? Please bear in mind that I have a scientific calculator which I am prepared to use (Casio fx-991MS) and I have a test soon in which speed is definitely an issue (i.e. my working counts for nothing). ==== > I understand that I can use Pascal's Triangle to complete a binomial > expansion, but it's very long-winded. What is the quickest way of working > out the coefficient of x^5 in the expansion of (x^2 - (3/x))^7? Please bear in mind that I have a scientific calculator which I am > prepared > to use (Casio fx-991MS) and I have a test soon in which speed is > definitely > an issue (i.e. my working counts for nothing). According to the Binomial Theorem, the factor for the nth term in a power 7 expansion is 7! / n! * (7-n)! You also need to factor in the -3, which will be raised to the (n - 1)th power. In this case, the nth term will give the coefficient of x^(14 - 2*(n - 1) - (n - 1)), or x^(14 - 3 * (n - 1)). You want the coefficient of x^5, so 14 - 3(n - 1) = 5, or n = 4. So, your coefficient is: 7!/3!4! * (-3)^3 = 35 * -27 = -945 -- ==== >I understand that I can use Pascal's Triangle to complete a binomial >expansion, but it's very long-winded. What is the quickest way of working >out the coefficient of x^5 in the expansion of (x^2 - (3/x))^7? The coefficient of a^i*b^{n-i} in (a+b)^n, i=0,....,n, is given by the binomial coefficient (n choose i). For (a-b)^n it is (-1)^{n-i}*(n choose i) The general formula for (n choose i) is n!/ i!(n-i)! where for integer m, m! is the factorial of m, m! = 1*2*3*...*m The coefficient of x^5 in the expansion of (x^2-(3/x))^7 corresponds to what term? If a=x^2, b=(3/x), then the term with a^i b^{n-i} has x^{2i-(n-i)} = x^{3i-n}. and n=7 So you want 3i-7 = 5; 3i=12, i=4. So you are dealing with the term a^4*b^3 = (x^2)^4 *(3/x)^3, so the answer is that the coefficient is (-1)^3*(7 choose 4) = -( 7*6*5)/(3*2) = -35. Unless I screwed up somewhere... ====================================================================== 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 ==== >I understand that I can use Pascal's Triangle to complete a binomial >expansion, but it's very long-winded. What is the quickest way of working >out the coefficient of x^5 in the expansion of (x^2 - (3/x))^7? The coefficient of a^i*b^{n-i} in (a+b)^n, i=0,....,n, is given by the > binomial coefficient (n choose i). For (a-b)^n it is (-1)^{n-i}*(n choose i) The general formula for (n choose i) is n!/ i!(n-i)! where for integer m, m! is the factorial of m, m! = 1*2*3*...*m The coefficient of x^5 in the expansion of (x^2-(3/x))^7 corresponds > to what term? If a=x^2, b=(3/x), then the term with a^i b^{n-i} has x^{2i-(n-i)} = > x^{3i-n}. and n=7 So you want 3i-7 = 5; 3i=12, i=4. So you are dealing with the term a^4*b^3 = (x^2)^4 *(3/x)^3, so the > answer is that the coefficient is (-1)^3*(7 choose 4) = -( 7*6*5)/(3*2) = -35. Unless I screwed up somewhere... ... well... you did forget the 3^3 that should be in there somewhere. -- Clive Tooth http://www.clivetooth.dk ==== > So you are dealing with the term a^4*b^3 = (x^2)^4 *(3/x)^3, so the > answer is that the coefficient is (-1)^3*(7 choose 4) = -( 7*6*5)/(3*2) = -35. Unless I screwed up somewhere... I think you may have. I have a sample test in which I have five possible answers to this question: a) -945 b) -210 c) 0 d) 243 e) 1415 It must be one of the above, unless I've made a mistake somewhere. ==== >> So you are dealing with the term a^4*b^3 = (x^2)^4 *(3/x)^3, so the >> answer is that the coefficient is >(-1)^3*(7 choose 4) = -( 7*6*5)/(3*2) = -35. >Unless I screwed up somewhere... I think you may have. I have a sample test in which I have five possible >answers to this question: a) -945 >b) -210 >c) 0 >d) 243 >e) 1415 It must be one of the above, unless I've made a mistake somewhere. Yeah, I made a slight mistake. What I got was that the term that has an x^5 is -35*(x^2)^4*(3/x)^3 = -35 * x^8 * 3^3/x^3 = -35*3^3*x^5 (remember, I figured out the coefficient of a^i*b^{n-i} in (a-b)^n, and then set a=x^2, b=(3/x), n=7, found that we were talking about i=4, and then calculated that coefficient). Which means that the coefficient of ->x^5<- is -35 * (3)^3 = (-35)*27 = -945. ====================================================================== 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 ==== For the gaussian pdf integration suppose x=(x7,x6,...,x0) in R^8 space p(x) = 1/(sqrt(2pisigma^2))^n exp(-xx'/(2sigma^2)) F(x) = int_{x in a n dim sphere(n>4)} p(x) so change x axis to be shpere axis p(r) = 1/(sqrt(2pisigma^2))^n exp(-r^2/(2sigma^2)) then what's the jacobian matrix is |J|=1? ==== Given that x^2 + y^2 = 6xy, find an equivalent expression to 2 log(x + y). How does the first expression relate to the logarithmic expression? ==== >Given that x^2 + y^2 = 6xy, find an equivalent expression to 2 log(x + y). How does the first expression relate to the logarithmic expression? 2*log(x+y) = log[ (x+y)^2 ] = log[ x^2+2xy+y^2 ] = log[ (x^2+y^2) + 2xy] Can you take it from here? ====================================================================== 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 ==== How can you tell if a set of simultaneous equations has solutions or not? e.g. 12x - 4y = 8, -9x + 3y = 6 or x^3 - y^2 = 0, x = y As far as I can tell, you can work out a value for x and a value for y every time...? So what defines whether there are solutions or not? ==== > How can you tell if a set of simultaneous equations has solutions or not? e.g. 12x - 4y = 8, -9x + 3y = 6 or x^3 - y^2 = 0, x = y As far as I can tell, you can work out a value for x and a value for y > every > time...? So what defines whether there are solutions or not? I may be missing something in the question, but if the equations are linear, then, generally, a pair of simultaneous equations a1x + b1y = c1, a2x + b2y = c2 have a solution if the determinant |a1 b1| |a2 b2| evaluates to non-zero. For larger numbers of variables the determinant gets more complicated (bigger). In your example 12*3 - (-4)*(-9) = 0, therefore no solution. -- ==== >How can you tell if a set of simultaneous equations has solutions or not? >e.g. 12x - 4y = 8, -9x + 3y = 6 For linear systems such as this, use linear algebra. >or >x^3 - y^2 = 0, x = y This is a nonlinear system. In general it may be hard to decide whether solutions exist, especially if you're interested only in real solutions. Groebner basis techniques may be useful. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >How can you tell if a set of simultaneous equations has solutions or not? e.g. 12x - 4y = 8, -9x + 3y = 6 or x^3 - y^2 = 0, x = y As far as I can tell, you can work out a value for x and a value for y every >time...? So what defines whether there are solutions or not? There are solutions if you can find a set of values for the variables so that each of the equations is true. e.g. for your second set of equations, x = 0, y = 0 is the only solution. the first set doesn't have any solutions: 12x -4y =8 => 3x-y=2 => y = 3x-2 substitute this into the second equation: -9x + 3y=6 => -9x + 3(3x-2) = 6 => -9x +9x-6 = 6 => -6 = 6. Now -6 does not equal 6 no matter which values you choose for x and y, so the first set of equations has no solution. If you make the first set a bit different: 12x-4y=8 and -9x+3y=-6 then you have an infinite number of solutions. just pick a value for x, and let y = 3x-2. e.g. x = 1, y = 1. In general, a set of simultaneous equations can have no solutions, exactly one solution, or an infinite number of solutions. Gareth <2brtpvsln95dj6c39r8tv25jtosupr9vfg@4ax.com> ==== > In general, a set of simultaneous equations can have no solutions, > exactly one solution, or an infinite number of solutions. This is not exactly true, unless you're talking about a set of LINEAR equations. For example, this set of equations has exactly two solutions: x^2-y = 4 -x^2-y = -4 (The solutions are x=-2, y=0, and x=2, y=0.) -- Derrick Coetzee ==== thanks for the boilerplate, but you still haven't answered the question about shouldness, that is to say, some sort of proof of this lack of algebraic integers (of course, some numbers are not algebraic integers, like those with non-integer coefficients, I guess). your mass Usenet audience is awaiting signs of a brain in your correpondence; can you read us? on the wayside, there's nothing wrong with the definition of atom, although the orginal Greek meaning really applies to molecule, since it had only to do with the sensible qualities of them. > If we are interested in the set of numbers which are > roots of monic polynomials with integer coefficients, > what does it mean that the set should include numbers > that are not such roots? > You've also never answered this one: It is possible to > form a product ab = c with a and c in the set Z but b > not in the set Z. Does that mean the set Z is incomplete > and there is an error in the definition? (Ex: a=3, c=5, > b = 5/3). What I've found is a problem with their set of algebraic integers, as > unfortunately, despite what many mathematicians think, it's too small. > Excerpt from http://mathforprofit.blogspot.com --les ducs d'Enron! ==== > >>good demonstration, Randy Poe. >Algebraic integers are defined to be roots of monic polynomials with >>integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where >>monic refers to the leading coefficient. >>If we are interested in the set of numbers which are >roots of monic polynomials with integer coefficients, >what does it mean that the set should include numbers >that are not such roots? >>You've also never answered this one: It is possible to >form a product ab = c with a and c in the set Z but b >not in the set Z. Does that mean the set Z is incomplete >and there is an error in the definition? (Ex: a=3, c=5, >b = 5/3). > What I've found is a problem with their set of algebraic integers, as > unfortunately, despite what many mathematicians think, it's too small. Too small for *what*? That's like saying there's a problem with a floppy disk because it's too small. It's not that the floppy is too small, but you are trying use it for something it wasn't intended to do. That's it. The definition they use is too small to do what they think > it does, which is include all these interesting numbers with special > properties. What, pray tell, do we think the definition does? What special properties? But because they think it's big enough, mathematicians have an error > in their discipline based on their false assumption, as they've come > up with more arguments based on that assumption, which then aren't > actually proven. Do you realize that at this point you've provided no true content? You've spent a great number of words to fail to say anything. It's like when the Greeks with their word atom thought they had the > smallest thing, and later our civilization used it, and broke atoms > apart, though part of the original Greek definition is that they are > indivisible, as people can define things, and later refine their > definitions. So we should be calling quarks, or whatever we're down to, atoms to return to the Greek notion? Excerpt from http://mathforprofit.blogspot.com apparently post no feedback. Have you gotten any positive feedback? -- Will Twentyman ==== >good demonstration, Randy Poe. >>Algebraic integers are defined to be roots of monic polynomials with >integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where >monic refers to the leading coefficient. >If we are interested in the set of numbers which are >>roots of monic polynomials with integer coefficients, >>what does it mean that the set should include numbers >>that are not such roots? >>You've also never answered this one: It is possible to >>form a product ab = c with a and c in the set Z but b >>not in the set Z. Does that mean the set Z is incomplete >>and there is an error in the definition? (Ex: a=3, c=5, >>b = 5/3). > What I've found is a problem with their set of algebraic integers, as > unfortunately, despite what many mathematicians think, it's too small. Too small for *what*? That's like saying there's a problem with a > floppy disk because it's too small. It's not that the floppy is too > small, but you are trying use it for something it wasn't intended to do. Well it looks like posters are switching to more talk about the posts than about the math, which isn't a surprise. My problem with that is that it's MATH, and it's not some ego contest, or about debates or anything silly of that nature. That's it. The definition they use is too small to do what they think > it does, which is include all these interesting numbers with special > properties. What, pray tell, do we think the definition does? What special properties? I give an excerpt. The full post has the explanation. But because they think it's big enough, mathematicians have an error > in their discipline based on their false assumption, as they've come > up with more arguments based on that assumption, which then aren't > actually proven. Do you realize that at this point you've provided no true content? > You've spent a great number of words to fail to say anything. And what have you done Will Twentyman? It's like when the Greeks with their word atom thought they had the > smallest thing, and later our civilization used it, and broke atoms > apart, though part of the original Greek definition is that they are > indivisible, as people can define things, and later refine their > definitions. So we should be calling quarks, or whatever we're down to, atoms to > return to the Greek notion? I note that atom has a definition that has been refined. Excerpt from http://mathforprofit.blogspot.com Check it out!!! > apparently post no feedback. Have you gotten any positive feedback? And once again you see the *social* nature of math society. What I can do is trace out my argument step-by-step, showing that it begins with a truth and proceeds by logical steps to a conclusion that then must be true. So, feedback in one sense is irrelevant. While, of course, in terms of proper recognition of my work, it's VERY important. But that's my business, and not that of Will Twentyman. James Harris ==== Excerpt from http://mathforprofit.blogspot.com apparently post no feedback. Have you gotten any positive feedback? Someone sent him a nickel so he could buy himself a clue. He didn't. ==== > Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. If the definition is so problematic, what should the definition be? The problem has to do with *assuming* that the ring doesn't have certain interesting properties, so it's not like the definition needs to be changed. It's just that its limitations must be noted. Here the problem is that the emphasis on monic polynomials clips some numbers. Not a big deal, if you know that's happening, but mathematicians didn't realize it for over a hundred years, so now it's a big deal. My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. 2. The important tool I use is a polynomial: P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) where m varies in the ring of algebraic integers. > Demonstrate your claim using P(m)=m^3-m+6. [remainder cut] What? Are you wondering if that polynomial can be factored into non-polynomial factors? The answer is, yes, it can be. What exactly do you think I'm claiming that you just toss out some polynomial? I'm curious what your reasoning is. Remember, all I do basically is factor a polynomial into non-polynomial factors with a neat technique, look at constant terms of those factors, divide the polynomial by some integer, and consider the constant terms again to draw a rather direct and obvious conclusion. James Harris ==== > Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. > 1. First the problematic definition: > Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. If the definition is so problematic, what should the definition be? The problem has to do with *assuming* that the ring doesn't have > certain interesting properties, so it's not like the definition needs > to be changed. It's just that its limitations must be noted. What are those limitations? Here the problem is that the emphasis on monic polynomials clips some > numbers. Not a big deal, if you know that's happening, but mathematicians > didn't realize it for over a hundred years, so now it's a big deal. > > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. > 2. The important tool I use is a polynomial: > P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) > where m varies in the ring of algebraic integers. > > Demonstrate your claim using P(m)=m^3-m+6. [remainder cut] What? Are you wondering if that polynomial can be factored into > non-polynomial factors? The answer is, yes, it can be. > I still want to see your claim using P(m)=m^3-m+6. You claim it can be factored into non-polynomial factors and I want to SEE those factors. > What exactly do you think I'm claiming that you just toss out some > polynomial? I'm curious what your reasoning is. > I am trying to be fair. I looked at your full argument, but I find difficult to follow. I am hoping that a demonstration with P(m)=m^3-m+6 will clear up my confusion. > Remember, all I do basically is factor a polynomial into > non-polynomial factors with a neat technique, look at constant terms > of those factors, divide the polynomial by some integer, and consider > the constant terms again to draw a rather direct and obvious > conclusion. > James Harris ==== >> Luckily for me the mathematical argument that proves that I've been >> correct all along can be further simplified by the use of *numbers* >> instead of variables, as while algebra as an idea is well-founded, so >> letter symbols should do, when mathematicians are lying about the >> mathematics, you need to use what you can, and pray. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with >> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where >> monic refers to the leading coefficient. If the definition is so problematic, what should the definition be? The problem has to do with *assuming* that the ring doesn't have >certain interesting properties, so it's not like the definition needs >to be changed. Huh??? You've been going on for months about how there's an error with this definition. Now in spite of that it doesn't need to be changed? You wonder why people have trouble following your arguments. It's because over and over you don't say exactly what you mean. >It's just that its limitations must be noted. Here the problem is that the emphasis on monic polynomials clips some >numbers. Not a big deal, if you know that's happening, but mathematicians >didn't realize it for over a hundred years, so now it's a big deal. Uh, right. We define algebraic integers as roots of monic polynomials, and nobody has realized for over a hundred years that that means that some numbers are not algebraic integers. Guffaw. My assertion is that the over hundred year old definition excludes >> numbers that have to be included to keep from having contradiction >> i.e. mathematical inconsistency. 2. The important tool I use is a polynomial: P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) where m varies in the ring of algebraic integers. >> Demonstrate your claim using P(m)=m^3-m+6. [remainder cut] What? Are you wondering if that polynomial can be factored into >non-polynomial factors? The answer is, yes, it can be. What exactly do you think I'm claiming that you just toss out some >polynomial? I'm curious what your reasoning is. Remember, all I do basically is factor a polynomial into >non-polynomial factors with a neat technique, look at constant terms >of those factors, divide the polynomial by some integer, and consider >the constant terms again to draw a rather direct and obvious >conclusion. >James Harris ************************ ==== Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. If the definition is so problematic, what should the definition be? The problem has to do with *assuming* that the ring doesn't have >certain interesting properties, so it's not like the definition needs >to be changed. Huh??? You've been going on for months about how there's an > error with this definition. Now in spite of that it doesn't need to > be changed? The definition doesn't need to be changed. I've already given the fix, which is the object ring. Algebraic integers are included in the object ring. Now then, here's the math argument emphasizing *CONSTANT TERMS* which is key in showing there is a problem. Notice how I'll be strongly emphasizing constant terms all the way down. P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 which has a constant term that is 1078. Well P(x) can also be written out as P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 so I can factor to get P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7) where the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Notice it *appears* that the constant terms for the three factors are all 7, which can't be right, as the constant term of P(x) is 1078, so setting x=0, reveals P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic defining the a's with x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2 for 0, so that leaves a_3 with a value of 3 when x=0. So let a_3 = b_3 + 3, where I keep indices matched. Then I have P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 5(3) + 7) P(x) = (5 a_1 + 7)(5 a_2 + 7)(5 b_3 + 22) and now my constant terms work out correctly. But P(x) has 49 as a factor as every term in P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078 has 49 as a factor, so I can divide by 49, and dividing 1078 by 49 gives me 22, as the new constant term. Well that means that P(x)/49 = (5 a_1/7 + 1)(5 a_2/7 + 1)(5 b_3 + 22) is the only way that the constant terms keep matching. James Harris ==== >> Luckily for me the mathematical argument that proves that I've been >> correct all along can be further simplified by the use of *numbers* >> instead of variables, as while algebra as an idea is well-founded, so >> letter symbols should do, when mathematicians are lying about the >> mathematics, you need to use what you can, and pray. 1. First the problematic definition: Algebraic integers are defined to be roots of monic polynomials with >> integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where >> monic refers to the leading coefficient. If the definition is so problematic, what should the definition be? The problem has to do with *assuming* that the ring doesn't have >certain interesting properties, so it's not like the definition needs >to be changed. Oh, that clears it all up. Certain interesting properties. I see. So, all you have to do is tell mathematicians to look over their work and if they claim that algebraic integers don't have certain properties that are interesting then their work might be wrong. Are these any particular certain interesting properties or just certain interesting properties in general? Have mathematicians been assuming that the ring doesn't have certain uninteresting properties as well or is this blind spot limited to interesting properties only? Alan -- Defendit numerus ==== > Luckily for me the mathematical argument that proves that I've been > correct all along can be further simplified by the use of *numbers* > instead of variables, as while algebra as an idea is well-founded, so > letter symbols should do, when mathematicians are lying about the > mathematics, you need to use what you can, and pray. >1. First the problematic definition: >Algebraic integers are defined to be roots of monic polynomials with > integer coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where > monic refers to the leading coefficient. If the definition is so problematic, what should the definition be? The problem has to do with *assuming* that the ring doesn't have > certain interesting properties, so it's not like the definition needs > to be changed. It's just that its limitations must be noted. Here the problem is that the emphasis on monic polynomials clips some > numbers. Not a big deal, if you know that's happening, but mathematicians > didn't realize it for over a hundred years, so now it's a big deal. > My assertion is that the over hundred year old definition excludes > numbers that have to be included to keep from having contradiction > i.e. mathematical inconsistency. >2. The important tool I use is a polynomial: >P(m) = 49((2401 m^3 - 147 m^2 + 3m) 5^3 - 3(-1 + 49 m )5 + 7) >where m varies in the ring of algebraic integers. > > Demonstrate your claim using P(m)=m^3-m+6. [remainder cut] What? Are you wondering if that polynomial can be factored into > non-polynomial factors? The answer is, yes, it can be. What exactly do you think I'm claiming that you just toss out some > polynomial? I'm curious what your reasoning is. Remember, all I do basically is factor a polynomial into > non-polynomial factors with a neat technique, look at constant terms > of those factors, divide the polynomial by some integer, and consider > the constant terms again to draw a rather direct and obvious > conclusion. > James Harris He's just wanting to see your method applied, there's no need to get uptight about it. You should be honored that he's taken an interest. Instead you're ungrateful, as usual. David Moran ==== > What I've found is a problem with their set of algebraic integers, as > unfortunately, despite what many mathematicians think, it's too small. That's it. The definition they use is too small to do what they think > it does, which is include all these interesting numbers with special > properties. You seem to think that algebraic integers should be a field. They're not. No mathematician ever claimed they are. Algebraic integers form a ring, which is not terribly difficult to prove, and algebraic numbers are a field, which is not terribly difficult to prove either. > But because they think it's big enough, mathematicians have an error > in their discipline based on their false assumption, as they've come > up with more arguments based on that assumption, which then aren't > actually proven. Nonsense. ==== Can someone suggest a good book on Homological algebra?? ==== Does anybody know of any examples of co-groups? and know if they are widely studied?? Any info on them would be appreciated. ==== >Does anybody know of any examples of co-groups? and know if they are widely >studied?? _Cogroups and Co-rings in Categories of Associative Rings_, by George M. Bergman and Adam O. Hausknecht. Mathematical Surveys and Monographs, Volume 45, American Mathematical Society (1996). MR 97k:16001. A search in MathSciNet for Title: cogroup yielded 23 hits, but I suspect that it may be that there are several notions involved. In particular, I am doubtful, given the description in the Math Review, that the notion of cogroup in J.E. Eaton's 1940 paper Theory of Cogroups, Duke J. Math. 6, (1940). 101--107; is the same as the notion of cogroup in Bergman and Hausknecht... For the latter, if V is a variety of algebras (in the sense of General Algebra), then a co-group object in V is an object of V that represents a covariant function from Group to V. ====================================================================== 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 ==== I was wondering if any one knows a clever way to solve the following integral: Integrate[ 1/(x^2+1+(x+z)^2) * 1/(y^2+1+(y+z)^2) * Sqrt[2 + z^2] / (2 + z^2 - lz^2), {z,-Infinity,Infinity}] I can solve it with a lot of algebra and factoring the fraction, but if some one knows a more elgant way of doing this, I'd be greatful. ==== Could you please check (and help) the following when trying to find a solution (using recurrence relations): A) A_n = A_n-1 + 3 where A_0 = 1 ANS: A_n = 3 + A_n-1 = 3+(3 + A_n-2) = (3+3) + A_n-2 = (3*3)+A_n-2 = (3*3)+(3 + A_n-3) = (3*3) + A_n-3 . . . = (n*3)+A_(n-n) = (n*3)+A_0 = (n*3)+1 = 3n + 1 I think Im corect on this prev one but the next one Im not sure what to do because the initial form seems strange to me. B) A_n = -A_(n-1) where A_0 = 5 Do I do something like ... <3F9C5A9D.ECC0A6F0@yahoo.com> <3F9C6EDF.CD8B5D@hate.spam.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 05:03 PM, Uncle Al said: >Metric models are geometric and have geodesics. Affine models are >electromagnetic in form and have no geodesics. I hope that you mean that the geodesics are not relevant to the Physics, rather then that they don't exist. An affine connection is all that you need to define geodesics; you don't need a metric. >Einstein made no geometric mistakes, neither did Euclid. Euclid did; his axioms and postulates were not adequate to prove all of the statements he listed as theorems. >Euclid is incomplete by trivial demonstration. Yes, but not in the sense you mean. >You cannot deep sea navigate using Euclid, nor >can you survey large tracts of land. Sure you can; Euclid covered Solid Geometry. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== That's a bit confusing, one can set the Lorentz >Force to generally vanish ie. >f_u = q*F_uv *U^v =0 >and in effect decribe EM geodesics, then put >q*F_uv in the metric forming a non-symmetrical >metric. The problem with that is that both sides of that equation are tensors. If they vanish in one frame, then they must vanish in all frames. Geodesics are based on affine connections, which are not tensors, so they have non-homogeneous transformation properties. Moreover, geodesic acceleration should be independent of mass, which applies to forces such as centrifugal, coriolis, and of course, gravitation. Acceleration due to Lorentz force will be inversely dependent on mass, making it a no go. >>All metric theory >>predictions are included in affine theories. Geodesics are a >>convenience. This may be semantics, but geodesics are the >*solution* of the motion equation, and need >to be provided in a manner demonstrable >in any CS. Maybe a brief example from >Newtons 1st law could be provided? It's actually a bit more than semantics. Not all motion is geodesic motion. See remarks above. ==== >That's a bit confusing, one can set the Lorentz >>Force to generally vanish ie. >>f_u = q*F_uv *U^v =0 >>and in effect decribe EM geodesics, then put >>q*F_uv in the metric forming a non-symmetrical >>metric. The problem with that is that both sides of that equation are tensors. >If they vanish in one frame, then they must vanish in all frames. >Geodesics are based on affine connections, which are not tensors, so >they have non-homogeneous transformation properties. Ah, but the geodesic equation is the absolute derivative of the 4 velocity U^u = dx^u/ds, ie DU^u/ds = U^u;v * U^v =0. >Moreover, >geodesic acceleration should be independent of mass, which applies to >forces such as centrifugal, coriolis, and of course, gravitation. True for acceleration of point masses, (but not true for relatively large masses, that's why acceleration is a better term than force in GR, otherwise you'll need to include the term k*t_uv >0 we figuring G_uv). >Acceleration due to Lorentz force will be inversely dependent on mass, >making it a no go. motion must respect Quantum Theory, Lorentz force does not, so the stated objection evaporates. >All metric theory >predictions are included in affine theories. Geodesics are a >convenience. >>This may be semantics, but geodesics are the >>*solution* of the motion equation, and need >>to be provided in a manner demonstrable >>in any CS. Maybe a brief example from >>Newtons 1st law could be provided? It's actually a bit more than semantics. Not all motion is geodesic >motion. See remarks above. Classically true, but Lorentz force vanishes in view of QT, meaning all motion must be geodesical, otherwise absolute acceleration exists. Ken S. Tucker ==== >>Metric models are geometric and have geodesics. Affine models are >>electromagnetic in form and have no geodesics. That's a bit confusing, one can set the Lorentz >Force to generally vanish ie. >f_u = q*F_uv *U^v =0 >and in effect decribe EM geodesics, then put >q*F_uv in the metric forming a non-symmetrical >metric. I'm always tempted to make the connections non-scalar. If the geodesic equation is a_i = -{i,jk} v_j v_k with a=d^2(x)/dt^2, v=dx/dt, {i,jk} an ASCII-compatible symbol for the connection, you can multiply through by an m and call -m*{i,jk} the force of gravity. So I always want to put some column vectors in there, like a_i = -(1) {i,jk1} v_j v_k - (0) {i,jk2} v_j v_k (0) (1) and then multiply through by (m,q). But maybe that's just dumb. -- Let us learn to dream, gentlemen, then perhaps we shall find the truth... But let us beware of publishing our dreams before they have been put to the proof by the waking understanding. -- Friedrich August Kekul.8e ==== >>Metric models are geometric and have geodesics. Affine models are >electromagnetic in form and have no geodesics. >>That's a bit confusing, one can set the Lorentz >>Force to generally vanish ie. >>f_u = q*F_uv *U^v =0 >>and in effect decribe EM geodesics, then put >>q*F_uv in the metric forming a non-symmetrical >>metric. I'm always tempted to make the connections non-scalar. If the geodesic >equation is a_i = -{i,jk} v_j v_k with a=d^2(x)/dt^2, v=dx/dt, {i,jk} an ASCII-compatible symbol for the >connection, you can multiply through by an m and call -m*{i,jk} the force >of gravity. So I always want to put some column vectors in there, like a_i = -(1) {i,jk1} v_j v_k - (0) {i,jk2} v_j v_k > (0) (1) and then multiply through by (m,q). But maybe that's just dumb. Yeah, I see what you're trying to do (even though your column vectors got a little skewed). The first term would be gravitation and the second the Lorentz, but how you'd ever get a cross product out of a symmetric connection is a bit of a stretch. ==== >Metric models are geometric and have geodesics. Affine models are >>electromagnetic in form and have no geodesics. >>That's a bit confusing, one can set the Lorentz >Force to generally vanish ie. >f_u = q*F_uv *U^v =0 >and in effect decribe EM geodesics, then put >q*F_uv in the metric forming a non-symmetrical >metric. >>I'm always tempted to make the connections non-scalar. If the geodesic >>equation is >> a_i = -{i,jk} v_j v_k >>with a=d^2(x)/dt^2, v=dx/dt, {i,jk} an ASCII-compatible symbol for the >>connection, you can multiply through by an m and call -m*{i,jk} the force >>of gravity. I have no objection provided m is small relative to the gravitating body, such as Mercury is to the Sun, but check out Dover's P of R, A.E.'s GR Eq.(45), and read geodesic is the motion of the point. I figure this is the solution from G_uv =0 (with the energy density being zilch, by the specification of a point) as opposed to the G_uv =k*T_uv, in the case like solutions. Opinions Please. >>So I always want to put some column vectors in there, like >> a_i = -(1) {i,jk1} v_j v_k - (0) {i,jk2} v_j v_k >> (0) (1) >>and then multiply through by (m,q). Very interesting how the indexes 1,2 are added within your connections, and may very well be ok, is a significant departure from conventional mathematics. I think QT excludes continuous acceleration of f_u = q*F_uv *U^v =0 may be reasonable. ((Knowing Mr. Hansen has R.C. Tolmans, Relativity, Thermodynamics and Cosmology, he may want to glance Eq. (103.1) and (103.2).)) >>But maybe that's just dumb. Well it's more complex than I usually do. >Yeah, I see what you're trying to do (even though your column vectors >got a little skewed). The first term would be gravitation and the >second the Lorentz, but how you'd ever get a cross product out of a >symmetric connection is a bit of a stretch. The cross-product is primarily a magnetic effect, from relative charge motion considered from a rest frame. If I'm not mistaken, a direct proportionality exists between the Magnetic field and the angular momentum of the mass the charge creating the current and B-field is attached (at rest) to. Wouldn't this suggest that the metric describing angular momentum and it's direct analog be equal? Ken S. Tucker <3F82F182.992FEC19@hate.spam.net> <3F832520.71D6778@hate.spam.net> <3f89f614$40$fuzhry+tra$mr2ice@news.patriot.net> <1482GpA5WNj$EwIC@aglow.demon.co.uk> <3f8fd502$6$fuzhry+tra$mr2ice@news.patriot.net> <3f9865ad$12$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 01:53 AM, Jan said: >Knowing all the variables is a rare luxury. IN this case we're addressing variables that are known. What needs to be determined is which, if any, have had a significant impact, and what the nature of that impact was. >Apart from the question of whether the IQ thing is true or not, this >fails to take into account the craniofacial evidence I have >mentioned before, It also fails to take into account the phases of the moon. Those data suggest a genetic influx, but not the effects of that influx. Other data suggest hat the influx wasn't nearly as large as you believe. Given our current understanding of genetics, the DNA evidence has to be considered more compelling than gross anatomical features. >and very possibly direct genetic evidence. The genetic data point the other way. >If I spent some years on a piece of research, >refining it down and considering all the variables I could find, I >wouldn`t be publishing it in a newsgroup. Well, at least not an unmoderated one. -- Shmuel (Seymour J.) Metz, SysProg and JOAT ====