mm-92 === I'm teaching the beginning epsilon-delta course. Looking> ahead in the book (Ross, Elementary Analysis: the Theory> of Calculus) I see the following de?ition: 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 ?st I thought this must be a typo. But it turns out> he means it - later when he shows that this de?ition> 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 de?ition actually> standard in some circles? My impression is that theDavid,What's the other de?ition?-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 de?ition:>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 ?st I thought this must be a typo. But it turns out>> he means it - later when he shows that this de?ition>> 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 de?ition actually>> standard in some circles? My impression is that the> David,> What's the other de?ition?He told you. The other de?ition has 0 < |x-a| < delta, not just |x-a|< delta.But I think this has come up before, and the verdict was that yes, asunlikely as it may seem, this version of the de?ition is actually beingtaught in some places, notably in France.-- Dave SeamanJudge 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 de?ition: 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 ?st I thought this must be a typo. But it turns out> he means it - later when he shows that this de?ition> 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 de?ition actually> standard in some circles? My impression is that the David, What's the other de?ition? He told you. The other de?ition 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 de?ition is actually being> taught in some places, notably in France.For the de?ition of continuity of f in ait doesn't really matter of course, since 0 = |a-a| < dand |f(a)-f(a)| < ego together automatically.With limits there can be a problem since the secondexpression |f(a) - L| can obviously exceed epsilon ifwe have a function that is discontinuous *but* de?edin 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-DeltaDe?ition.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 de?ition of the limit | ( lim [ x->x0; f(x) ] = y0 )OTOH on http://mathworld.wolfram.com/ContinuousFunction.htmlthey say that continuity is sometimes de?ed in terms ofepsilon-delta and limit, and then they give an examplewhere they say (except possibly x0 itself) where theyin fact should have said (and certainly at x0 itself),since they *are* de?ing continuity here ;-)I think it is a matter of mild sloppyness. At least in thecase 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 de?ition: 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 ?st I thought this must be a typo. But it turns out> he means it - later when he shows that this de?ition> 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 de?ition actually> standard in some circles? My impression is that theDavid,What's the other de?ition?He told you. The other de?ition 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 de?ition is actually being> taught in some places, notably in France.How is the de?ition above not circular? I mean, how do you knowwhat 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 de?ition: 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 ?st I thought this must be a typo. But it turns out> he means it - later when he shows that this de?ition> 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 de?ition actually> standard in some circles? My impression is that the> David,> What's the other de?ition?He told you. The other de?ition 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 de?ition is actually being> taught in some places, notably in France.How is the de?ition 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 ?st, but the above isde?ing convergence of a function in terms ofconvergence of sequences of numbers. So sequentialconvergence must be de?ed elsewhere. - Randy === > How is the de?ition 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 de?econvergence of functions. The idea is that learning to do theseproofs, which is hard stuff (for many students), is best done ?stwith 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 de?ition:>> 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 ?st I thought this must be a typo. But it turns out>> he means it - later when he shows that this de?ition>> 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 de?ition actually>> standard in some circles? My impression is that the>> David,>> What's the other de?ition?>> He told you. The other de?ition 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 de?ition is actually being>> taught in some places, notably in France.> How is the de?ition 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 de?ition above? I'm guessing you mean the sequence de?ition,whereas the two de?itions I was discussing have nothing to do withsequences.But it's certainly possible to de?e the limit of a sequence in a waythat does not imply a particular de?ition for the limit of an arbitraryfunction at a point.-- Dave SeamanJudge 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 de?ition: 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 ?st I thought this must be a typo. But it turns out> he means it - later when he shows that this de?ition> 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 de?ition actually> standard in some circles? My impression is that the What's the other de?ition?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 de?ition:>> 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 ?st I thought this must be a typo. But it turns out>> he means it - later when he shows that this de?ition>> 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 de?ition actually>> standard in some circles? My impression is that the>What's the other de?ition?>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 botheredme was that it was |x-a| < delta instead of 0 < |x-a| < delta. (ButEdgar has pointed out I wasn't reading the notation carefully enough;the de?ition with |x-a| < delta was a de?ition of something otherthan lim_{x->a} f(x).)Ross seems to think that things will be easiest to follow if he ?sttalks about convergent sequences and then bases everythingelse on them. I'm not sure that that's right but I don't have anybig 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 ?st> 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. Equivalentde?itions are, well, equivalent, so why not just go with the ? need people versed in pedagogy to teach courses, so they'll look atthings from an educational point of view.Jon Miller === Ross seems to think that things will be easiest to follow if he ?st>> 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>de?itions are, well, equivalent, so why not just go with the ? 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 butI'm also not certain it's wrong means I'm not looking at things froman educational point of view. Huh.Yes, it certainly is true that having people who really understanda 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. (Butwouldnt |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 notationwould mean so much.-k === me was that it was |x-a| < delta instead of 0 < |x-a| < delta. (Butwouldnt |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| I am not a mathematics professor but I am curious to know, why this notation>would mean so much.De?e f(x) = 0 for all x except x = 0; let f(0) = 1. By one version of the de?ition lim_{x->0} f(x) = 0, while by the other versionthis 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 speci?allyexcludes the case x = a, while |x-a| < delta, delta > 0 does not. Itmakes a big difference when you are considering the limit of adiscontinuous function.-- Dave SeamanJudge 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 = 16a2 + 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 = 13Goal: 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 = 2First, 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 isa1 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 = 13Then 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 ?> 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 ?st 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 ism 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 = 161 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 = 171 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 = 13The 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 = -4In 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 theunknowns? m = 2+5+6 = 13 a1 = 5 - 2 = 3 a2 = 6 - 2 = 4 b1 = 0 - 5 = -5 c1 = 2 - 6 = -4I 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 ?st is speci? to your example. Changing the notation, we havey_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 ?d the best (the simplest, elegant) proof of the prime numbertheorem(a) with complex variables allowed(b) without complex variables (the so called elementary proof)? === > Where can I ?d 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 inHlawka, 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 aweakened version of the Ikehara-Wiener theorem, and then applies thelatter to give simple proofs of the (usual) prime number theorem,Hecke's prime number theorem for the Gaussian integers and the primenumber theorem for arithmetic progressions.For (b), there is a readable and detailed presentation of Selberg'sversion of the elementary proof inGioia, Anthony A. The theory of numbers. An introduction. Reprint of the 1970 original. Dover Publications, Inc., Mineola, NY, 2001. xii+207 pp. ISBN: 0-486-41449-3Unfortunately I think Gioia isn't too careful about specifying therange over which his O-estimates are valid, which leads to someconfusion near the end of the proof. But it can all be patched up byreading 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 elementaryproof. As far as I'm aware, this book represents the most thoroughtextbook-treatment of elementary prime number theory. Unfortunatelyit's been out of print for a while now.Hope this helps,Paul === > Where can I ?d the best (the simplest, elegant) proof of the prime> number theorem> (a) with complex variables allowedApostol's Introduction to Analytic Number Theory runs through the proof in Chapter 4. I don't know if this is the best, but itis 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 theelementary proof, which may be helpful.Bart === >Where can I ?d 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 elementaryone, 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 mewhile I was growing up (more than 25 years ago). My fater was a clerkat the railroad and did not have a college degree but he did budgetforecasting and hence worked with numbers a lot. My father broughthome the paper The Complexity of Mere Simpli?ation because he knewI had an interest in mathematics. The paper took the equation 1+1=2and extrapolated into a complex mathematical equation that includedmany identities and well known formulas in mathematics. I kept thispaper because I wanted to understand the formulas and before Igraduated from Iowa State University with a bachelors degree inmathematics, the complexity of mere simpli?ation was no wonder tome.post to this newsgroup a copy. I have long since lost this paper Iwould like to ?d 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 === MikeDoes anyone know how to generate a random vector x = (x_1, ..., x_n)> uniformly from |x| = 1? Any recommendations on what books I need toMikeWhat 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 ef?ient way). === To prove that the boundary of a manifold with boundary (what's theadjective? It can't be bounded manifold!) is well-de?ed, I need thefollowing 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 noseor the slope of the ?path?Assuming the latter, one can use simple trig. If you really meantthe former one would need a lot of additional data such as thetemperature and humidity of the air (hot air is thinner) andthe precise construction of the wing of the plane. (The speed ofsound is in there somewhere but can probably be derived from thetemperature, 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 ?d 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 onthe ground directly underneath P. (We assume here level ground.)We know the slope -- the angle GOP. (No puns of a certainelephantine nature, please. :-) ) We know the speed along thedirection 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.netIt'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 de?ition of angle might be representedArcCos(dot(A,B) / (norm(A) * norm(B)))where A and B are arbitrary vectors. This works in anydimension 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 aremeasuring an angle for only a subset of all things for whichone 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.netIt'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.netIt'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 ?ite length CF's, thus allrows 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 ?ite 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 equationsabove are rational approximations which get successivelybetter as one increases n. However, they will neverreach e -- although never is a funny term, as taking the limitessentially makes the jump, in a weird sort of twisted sense.A slightly more rigorous treatment might allow for theexistence of e and successively better constructionsby de?inge_0 = 1/0! = 1e_1 = 1/0! + 1/1! = 2e_2 = 1/0! + 1/1! + 1/2! = 2.5e_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 de?ing e(n) = e_{?)} and taking the limit as n -> +oo.It's not too dif?ult to prove that this will converge tosomething, namely e = 2.718281828459045235360287...The simplest method of proving this may be noting thatterm k -- namely, (1/k!) < (1/2^(k-1)) (for k > 1) -- and thattherefore the series converges to a number e < 3.> Dan-- #191, ewill3@earthlink.netIt'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 ?ite 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 ?ite 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?> DanFor all levels, the number e_n = 1/0! + ... + 1/n! is rational. You can usen! as denominator. This does not mean that e is rational; there are manysequences of rational numbers whose limit is irrational.Actually, one possibility to de?e the real numbers is by (inaccuratelyspeaking) saying that they are all possible limits that sequences ofrational numbers can have. So, for every real number, there are a lot ofrational 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 ?ite 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 de?e 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 thinkingabout whether a number is rational or irrational can not be deduced byany level of its iterations such as Exp(e).So basically, whatever the correct decimal length of (e) is calculatedby summing the inverse of the factorials, it will always be rationalunless 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 calculatePi?Probably not because some of the equations require the (sqrt) functionin 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 ?ite 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 de?e 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 = 1n_1 = 0.4n_2 = 0.01n_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) whichis irrational. This generalises to any other irrational.IOW the ?ite sums are approximations of the in?ite sum.-- /-- Joona Palaste (palaste@cc.helsinki.? ------------- Finland ---------- http://www.helsinki.?~palaste --------------------- rules! --------/No, Maggie, not Aztec, Olmec! Ol-mec! - Lisa Simpson === >there is a problem I ?d dif?ult. Let f:[0,1]->R be in?itely 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.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === >>there is a problem I ?d dif?ult. Let f:[0,1]->R be in?itely 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 therelevant 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 in?itely 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 ?d dif?ult. Let f:[0,1]->R be in?itely 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. MladenWhat 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 ?d dif?ult. Let f:[0,1]->R be in?itely 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 = 0If there _is_ a counterexample it seems to me thatn(x) can certainly not be piecewise constant like that -if you patch together two polynomials what you getis not in?itely 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 coef?ients in it, and therefore the function cannot be a>polynomial. === >there is a problem I ?d dif?ult. Let f:[0,1]->R be in?itely 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 ?d dif?ult. Let f:[0,1]->R be in?itely 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. MladenHere 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 de?ition 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 coef?ents of f|(a,p) and f|(p,b) so these coef?ients 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 de?ed 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 ?d dif?ult. Let f:[0,1]->R be in?itely 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. MladenHere 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 apolynomial...>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 de?ition 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 coef?ents of >f|(a,p) and f|(p,b) so these coef?ients 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 de?ed 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 somebook. I think that it can be shown that for each open set there existopen 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 ?d dif?ult. Let f:[0,1]->R be in?itely 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. MladenLet 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 ?d dif?ult. Let f:[0,1]->R be in?itely 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 peopleseem to be giving hints like this. How does the resultfollow from the Baire category theorem? === > there is a problem I ?d dif?ult. Let f:[0,1]->R be in?itely 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 bethe key.De?e a bad point as a point x such that n is unbounded in everyneighbourhood of x. The result follows easily if we can show there areno bad points. Suppose there exist bad points. The basic observationis that there is an open interval I with a bad point and an integer Msuch that every open subinterval I' of I with a bad point contains abad 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 theintersection of nested sets) with n(x) != M for all M, which isimpossible.it follows that every point in I has n(x) <= M, which contradicts theexistence of bad points in I.Michael <340b747b.0310280459.7de9f1dd@posting.google.com> === there is a problem I ?d dif?ult. Let f:[0,1]->R be in?itely 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? De?e 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 everyf^(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 openinterval 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 polynomialas 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 ?d dif?ult. Let f:[0,1]->R be in?itely 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? De?e 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 ?ling in some details, or asking forclari?ation about my answer. In case it's the latter, I'll explainmore precisely what I'm saying.We have assumed f is not a polynomial. It follows immediately thatn(x) is unbounded in every neighbourhood of certain points.(For convenience I'm taking n(x) to be the _minimal_ n such thatf^n(x) = 0. This wasn't actually said explcitly in the question, buthopefully it's clear what I mean.)I've called these points x, such that n is unbounded in everyneighbourhood of x, bad points.Then I've claimed that there is an open interval I with a bad pointand a positive integer M such that every open subinterval I' of I witha bad point contains a bad point x with n(x) = M.If this were false then that would mean that for every open interval Icontaining a bad point and every positive integer M you could ?d asubinterval I' with a bad point such that each bad point x of I'satis?s n(x) != M.OK so let I_0 be any open interval containing a bad point. (We'veassumed bad points exist.) Now set M = 1 - by the hypothesis in theabove paragraph we could ?d an open subinterval I_1 of I_0containing a bad point such that every bad point of I_1 satis?s n !=1.Now set M = 2. Find an open subinterval I_2 of I_1 containing a badpoint such that every bad point of I_2 satis?s 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 arestrictly increasing/decreasing respectively, and the lengths of I_mare tending to 0. Hence the intersection of I_m is a singleton.Must this singleton x be a bad point? Yes. Otherwise n would bebounded in some neighbourhood of x, so n would be bounded in some I_mwhich contradicts the fact I_m contains a bad point. So this singletonx 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'stheorem, but I personally ?d it more intuitive to write this sort ofthing 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 positiveinteger m such that every open subinterval I' of I with a bad pointcontains a bad point x with n(x) = m.Now that means that the mth derivative of every bad point in Ivanishes. Because given a bad point x in I you can ?d bad points inI arbitrarily close to x at which the mth derivative of f vanishes.Since f^m is continuous that means that the mth derivative at everybad 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 asequence 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) = 0since f^m(x_n) = f^m(x) = 0. By induction the m+2th, m+3th derivativesetc. all vanish at the bad points in I.Now take any point x in I. We can take a maximal subinterval I' of Icontaining x on which f is a polynomial of degree n(x). Pick anendpoint x' of I' which doesn't coincide with an endpoint of I. Thenit is clear that x' is bad. Hence f^m(x') = f^m+1(x') = ... = 0 so itfollows that the degree of the polynomial in I' is < m. Hence themth,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 ?d dif?ult. Let f:[0,1]->R be in?itely 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 ?d dif?ult. Let f:[0,1]->R be in?itely 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 apolynomial in some interval some time ago, in what appears tobe the ?st reply to the OP.You say this gets things started. I don't see how: Now thatwe know f is a polynomial in some interval how does theresult follow? (Or how _might_ the result follow? What'sa possible approach to get from here to what we want toprove? I can't see any hint of a possible path from here tothere - you do?) === > there is a problem I ?d dif?ult. Let f:[0,1]->R be in?itelymany>> times differentiable such that for each x from [0,1] there existsn(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 ?st 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 thatUV has no isolated points.Then show that UV is empty (or else you'll get a contradiction).-gs- === > there is a problem I ?d dif?ult. Let f:[0,1]->R be in?itely>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 ?st 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 ?d dif?ult. Let f:[0,1]->R be in?itely>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 ?st 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)) andthat>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. weknow that there exist n0 and a,b in U with bothI:=]a,b[ inter W non-empty and a subset of Cn0. I'll show that f^(n0)=0on ]a,b[, hence ]a,b[ will be included in int(Cn0) which contradictsthe 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 easilyshown by recurrence : Let (x_k) in I^N be an injective monotonous sequencewith 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 ?d that f^(n)(x)=0 ifn>n0.Now if x is in ]a,b[ inter V, then the component V_x of x has an extremityin 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 ofdeg> there is a problem I ?d dif?ult. Let f:[0,1]->R be in?itely>>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 ?st 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 ?d 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 ?d 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 spacep(x) = 1/(sqrt(2pisigma^2))^n exp(-xx'/(2sigma^2))if we wanna do integrationon 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 easybut how about higher dimension?thanks a lot! === haha I got itit is pretty easyI just confused myselfxx' = x7^2+x6^2+...+x0^2 = r^2right?> 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$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$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 de?ing 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 JOATnot reply to spamtrap@library.lspace.orgX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-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 orneither, and that you may not know which it is. Of course, thatgreatly diminishes the utility of the term true.FWIW, I ?d both Formalism and Platonism to be unsatisfactory, but Idon't see a viable alternative to the two of them :-(-- Shmuel (Seymour J.) Metz, SysProg and JOATnot 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$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$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 TheNatural Numbers. How do you characterize statements about themoutside of some formal system? How do you determine whether a formalsystem correctly captures some of their properties?-- Shmuel (Seymour J.) Metz, SysProg and JOATnot 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 systemsto be invoked here, but it's unclear what you have in mind. Just whatis the role of formal systems in exlaining what we mean in referringto the natural numbers? For example, is the conjecture that there arein?itely 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$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft === >If you think the Peano Postulates do not follow from the de?ition I>gave, I don't think that you gave a de?ition.>The natural numbers do not form a ?ite collection, and we must be>careful about carrying over results about ?ite collections to>in?ite collections.Which is precisely why your de?ition doesn't de?e anything.>I claim that if a particular statement cannot be decided from that>de?ition, then it is the statement that is problematic, not the>de?ition.That would certainly leave you a lot of wiggle room. Certainly thereare a lot of open questions in number theory, and I certainly considerthem to be reasonable questions.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === > Despite some searching I have> not been able to ?d a de?ition of arithmetical property, Now I have. http://home.ddc.net/ygg/etext/godel/godel3.htmA relation (class) is called arithmetical, if it can be de?edsolely by means of the concepts +, . [addition and multiplication,applied to natural numbers]49 and the logical constants , ~, (x), =,where (x) and = are to relate only to natural numbers.50 The conceptof arithmetical proposition is de?ed in a corresponding way. <3f9d06e2$0$69956$edfadb0f@dread12.news.tele.dk>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oftX-Terminate: SPA(GIS) === at 12:52 PM, Michael Jrgensen said:>To me, understanding means to be able to explain to someone else.There have been cases of fallacious proofs that were accepted for along time. People were able to explain the proof to each other, butthe explanations were actually wrong. IMHO they did not understandthe purported proof.To me, you understand the proof when you understand each step alongthe way. That's not as simple as it sounds, since sometimes a step candepend on very subtle distinctions or unstated assumptions.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot 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 ? 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 in? 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 itin 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, anddecide on its correctness, than to have to gothrough 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 proofto The Machine of your choice, and have it explain its correctnessfor 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 ?hes, devices to stimulate> the imaginations of pupils.> Jose Carlos SantosBullshit.A proof begins with a truth and proceeds by logical steps to aconclusion which then must be true.It occurs to me that some of you think proofs are *created* when theyhave 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 de?ition of proof, is thatit's completely contigent upon the veracity (truth-value) andconnection 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 adabsurdum. He starts from what's true and arrives at a contradiction sosomewhere there must be a core error (and heaven forbid it should bein *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 thatyou ?d 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 dif?ult 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 dif?ult 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 producehis 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 dif?ult 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 ofevery 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 ?primeideals' but the effort doesn't seem to be worth it. I assume someposter 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 registeringoff-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. Scienti?results are veri?d when other people can reproduce themor build on them. Mathematics is advanced when other peopleaccept a proof, understand it, and build on it.Despite it's esoteric nature, Wiles proof has been understoodby a signi?ant and increasing number of people. The proofof 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 toaccept a proof with a big black cover nailed over themiddle 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 ?st version of the paperwas withdrawn when it became clear that it wasn't a validproof. You think that happened because people accepted it onface 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 validationof 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 onefrom people when you suggested somebody else should invest themonths of effort to build the machine implementation becauseyou couldn't be bothered. You don't see why that should getan 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 currentmath 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 youto 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, super?ially, 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 thecriticism of a proof, perhaps having to do with your innate biastowards the competence of high-school teachers? If you didn't knowthat the OP is a high-school teacher would you have been so criticalof 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 thanunderstanding 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'snot clear that he didn't understand the proof.If the basis of the proof is color however you like and then ? theproblems, then there needs to be a recognition that there might be acircularity in the repair procedure. There had been a prior thread inwhich someone used the notion of distance (which is not a priori relevanthere) to circumvent the problem. How, exactly, the introduction ofdistance saves the day takes a bit of re?n. Did this poster seewhy the introduction of lengths helps? My memory of the thread atthe time was that the answer was not really; now on second thoughtI'll just say I can't tell.But yes, I do think it's fair to take into account a person'sbackground when assessing whether or not they understand a proof. Idon't know this teacher but I do teach others who are about to becomehigh school math teachers, and if they turned in this proof I would bemuch more critical than I would if you (B.D.) or another person oftrusted mathematical strength were giving this proof. Is that reallysurprising? To give another example, consider this gem: is there an equilateraltriangle whose vertices lie on the integer lattice? The answer is noand my proof is: Pick's Theorem. Cool, huh? A proof which is certainly short enough to remember! Now, does a person who hears thatunderstand the proof? Frankly this sort of thing happens to me allthe 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 convincedthat 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 Icompletely agreed that he didn't understand the proof. Perhaps itwould 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, thelatter 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), fromhttp://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 authordidn't understand the proof, esp. without bias towards H.S. teachers.Surely a novice might make the same objections regarding a proofby an expert, only because the novice regarded some missing stepas 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 matureaudience of sci.math (versus a less mature forum, say k12.ed.math).In summary, I don't think it is fair to criticize the correctnessof the proof, since it would be deemed correct by a suf?ientlymature audience. Instead, one can only argue that the author mightbe 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 thatirrationality 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 toohas 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 arelattice points. So connecting the midpoints dissects T into 4 smallersuch 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 evenAn 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 generalizationhttp://google.com/groups?selm=6loo52%24foo%241% 40nnrp1.dejanews.comhttp://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 [*]. QEDI 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 allbetter 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 proofBUT including one more level of detail for every step.Leslie Lamport once said that he had found he needed to write downthe proof so that it was completely obvious how to get from each stepto 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 dif?ult for logic students. After all, supposeyou're studying a purely formal proof of, say, one of the De Morganlaws. 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 takeeverything at its most literal meaning -- not that I have any examplesat 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 ?ally 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 dif?ult 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 onlyapply in such cases that there is at least one more level of detail.Examples of things like this would bethe proof of FLT given by Mr. Harris orthe proof of the orbifold theorem given by Dr. Thurston.disrespect by my statement, both those individuals have more energythan I ever will, those were ONLY two examples that came to mind.X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-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 equivalentto 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 thediagonal argument is inapplicable.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot 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 (alsoequivalent 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 de?ed explicitly?Have a tolerable existence. EliIncredibly naive question coming up, since the answer is obviously no,but for the life of me I can't ?ure out what's wrong with it... whycan't we de?e the ordering a < b in the seemingly obvious way? Sincereal numbers have decimal expansions, choose for each real number theterminating expansion if possible and compare digits until you ?d adifference and order that way? === > Incredibly naive question coming up, since the answer is obviously no,> but for the life of me I can't ?ure out what's wrong with it... why> can't we de?e 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 ?d 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 thestandard ordering, for the same reason. There are subsets (such as theset of positive rationals) that have no least element. However, it isknown that Q is countable. Choose a bijection f: N -> Q, and de?e anew 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 thereforede?es 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 anordering on the naturals. But it's conceivable that there might be someuncountable ordinal alpha and a bijection f: alpha -> R. That's wherethe question lies. In the absence of the axiom of choice, it's possiblethat no such ordinal exists.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Out of curiosity, how does one de?e an ?ordering' without formulating a> bijection with a countable set like the naturals? Syntactically, I mean..> I'm clueless, which is why I askOthers have given you good partial answers, so I'll just comment on thesyntactically 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 realsis 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 notconcern 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 productof A with itself, i.e. the set of all tuples (x_1,...,x_n) withx_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.?Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === > Out of curiosity, how does one de?e an ?ordering' without formulating a> bijection with a countable set like the naturals? Syntactically, I mean..> I'm clueless, which is why I ask-- > QuaternionThe usual ordering of the reals does not require a bijection with any countable set. For example, using upper Dedekind cuts to de?e 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 ternaryrepresentations, are uncountable by a number of proofs, and one of them isCDA. However, with WOP, a roster notation of the reals is possible, so takethat roster and list it. Now, we have a (perhaps uncountable) listing ofthe reals, but Cantor says we cannot have that. For example, with the setof ordinals, less than w_2, CDA would be meaningless. Using CDA on awell-ordering of the reals would likely produce a number that, while in thelist, is uncountably far away.I have also heard that you need to able to give a ?ite number to>correspond to any number I name, but this can be solved by putting the>de?able numbers ?st.>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 willassume you're just babbling nonsense when they realize they haveno idea what you're talking about.>>I have also heard that you need to able to give a ?ite number to>>correspond to any number I name, but this can be solved by putting the>>de?able numbers ?st.>>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 arewell-ordered,>then they can be listed, so what about Cantor's DiaganolizationArgument? 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 themis>CDA. However, with WOP, a roster notation of the reals is possible, sotake>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 inthe>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 thinkthat my question was answered about 10 posts ago when someone said that youcannot create real number with an uncountable number of digits.I have also heard that you need to able to give a ?ite number to>correspond to any number I name, but this can be solved by putting the>de?able numbers ?st.>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 ?ite number to>>correspond to any number I name, but this can be solved by putting the>>de?able numbers ?st.>>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 uncountablelisting of reals, or why you think that Cantor's diagonalization argumentapplies 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 somespeci? ordinal, the cardinality of the reals. So let C be the cardinalityof the reals, and for every ordinal alpha less than C, let r_alpha be thereal corresponding to ordinal alpha.How do you think that you are going to diagonalize to come up witha real r that is unequal to r_alpha, for every ordinal alpha?You can certainly diagonalize out of the ?st omega-many reals: Youcan de?e 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_nBut that only proves that r is unequal to r_alpha when alpha is ?ite.It doesn't eliminate the possibility that r = r_alpha for some in?iteordinal alpha.You can't diagonalize using more than the ?st omega many reals inthe listing r_alpha, because real numbers only have omega many decimalplaces.--Daryl McCulloughIthaca, 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> speci? 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 thateach member corresponds to ordinal less than w, a set of cardinalityalpha can be enumerated so that each member corresponds to an ordinalless 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 ?st omega-many reals: You> can de?e a real r by its decimal expansion as follows:Any well-ordering of an uncountable set necessarily has a member thatmaps 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 ?ite.> It doesn't eliminate the possibility that r = r_alpha for some in?ite> ordinal alpha.> You can't diagonalize using more than the ?st omega many reals in> the listing r_alpha, because real numbers only have omega many decimal> places.-- Dave SeamanJudge 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>> speci? 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, thenthe set of reals can be enumerated so that each real corresponds tosome ordinal alpha less than C.--Daryl McCulloughIthaca, 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> speci? 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 SeamanJudge 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 havea listing labeled by the natural numbers, because the labelsfor the digits are also the natural numbers, so we get a diagonalwhere the label of the number equals the label of the digit. But if wehave an uncountable labelling for the real numbers, we cannot de?e adiagonal 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 ?ite number to>correspond to any number I name, but this can be solved by putting the>de?able numbers ?st.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 de?e an ?ordering' without formulating a> bijection with a countable set like the naturals? Syntactically, I mean..> I'm clueless, which is why I askPerhaps 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 SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-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 doyou assume that you can't de?e a topology that yields theconvergence that you have de?ed?Isn't this analogous to the situation in a metric space, where thereis no topology, yet the metric lets you de?e a unique topologycompatible with it?-- Shmuel (Seymour J.) Metz, SysProg and JOATnot 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......ahahansonHanson, how long has it been since you had your meds adjusted? Perhapsyou 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 + 432which has all the terms and all nonzero integer factors.--Lynn === >>What is the ?st 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 ?d 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 asB. Bonds can hit a homerun from simply just tossing the ball in theair from homeplate and then seeing whether a homerun can be hit. Thosewho have played baseball know this test in that they do ?ldingpractice in many instances where a batter with bat in one hand andwith 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 thepitcher 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 sluggerof baseball can hit a homerun from a 0 speed ball from homeplate.I heard from the announcers that the ?ld park in Florida for theMarlins is vastly deeper than the park in New York. But I am wonderingof a Barry Bonds test of a 0 speed ball and whether Bonds can hit thatball a homerun?? If he can hit a 0 speed ball from Yankee stadium as ahomerun then my opinion is that baseball stadiums are grosslyunderdistanced and that they need to be altered such that no hitter inbaseball is able to hit a homerun from a 0 speed ball.In fact, my opinion is that the slowest speed pitch possible, what isit??? Is it 10 mph that a pitcher in the major leagues can get acrossthe plate?? My opinion is that all pitches of this slowest speedshould not be able to be hit as a homerun in any major league baseballpark and that these parks should be re landscaped such that theirdistances make it impossible for any slugger to hit a 10 mph pitch asa 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. Iwould love to hear the sound of a pitch hit from a aluminum bat. Butif these bats can cause death and injury to players then keep the banon aluminum bats.If a Barry Bonds is unable to hit a 0 speed ball as a homerun thenthat indicates to me that the Optimal Strategy Pitching in Baseball isnot in the direction of fast pitching but in the direction of slowpitching.I need a test of the average slow speed of a baseball that gets itacross the plate as a strike. I am simply guessing of 10 mph but needan accurate number.And the reason that no pitchers in modern baseball do a slow pitch isbecause of social antagonism that they demand and expect all pitchersto deliver fast pitches. But if the rules in Baseball allow slowpitches then that will be the future trend.And the distance of all baseball parks out?ld should be aphysics-science determination. Obviously a Barry Bonds is able to hithomeruns from 90 mph pitches in virtually every baseball park. But ishe able to hit homeruns in every park from a 30 mph pitch?? So thedistance of every major league baseball park should be determined bythe speed of pitch and those parks that are inadequate should berelandscaped and designed better. So how can you justify to anybaseball fan that a Barry Bonds can hit a homerun in their stadiumfrom a simple hand tossed 0 pitch speed?? You cannot justify that andthat the ?ld distance needs to be ?ed. Almost as if a Grandmothercan go out to that ?ld and hit a homerun.So that in the future, when a Barry Bonds comes to the plate andscared of him hitting a homerun, then a stream of slow pitches isrendered instead of a forced walk to the base. In fact, if a pitcherin the future perfects a slow pitch with a spin on the ball thatalmost every batter will have an impossible time of hitting a homerun.But whether the scores in baseball will become larger because so manysingles or doubles are hit remains a question for slow pitching.The trouble with baseball at the moment is that science and physicshave never rally given the sport any inputs and that the rules andagenda of baseball as a sport has been governed more by the tillationeffect on fans. Most every fan of baseball is unaware at the moment ofthese writings that the pitcher is the dominate feature of baseballwhere most fans immediately think of the batting slugger. And what ifa machine, a pitching machine that is anchored on the pitching mounddesigned to deliver a fast ball at a speci?d speed in every pitch.The fans would balk and baseball lose much of its appeal. And in thatcircumstance we can truly say that the slugger in baseball is of allimportanceand not the pitcher.So I have raised many important questions above. (1) what is theslowest speed todeliver in major league pitches (2) can the present day sluggers hit ahomerun in the stadiums from a 0 speed toss up (3) are most stadiumsinadequate in distance of homerun (4) link and connect speed of pitchto out?ld distance (5)research the advantages of slow speed pitchingin baseballArchimedes Plutoniumwhole entire Universe is just one big atom where dotsof 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 + dMaple 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 systemto a ?st-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' youthen get an equation for v' and y' in terms of y and v. And thendv/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 secondtype, class B.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Lately I found some simpli?ation coming out from model.To give better overview about my problem I start from begining by describingproblem 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 ?d out average of Z molecule, in function of time. (In truththis is not my original problem, which is larger, but only sub problem. Inmy original problem ratio k1 is not constant but vary depending on time(there are more reaction that produce X), I can calculate this function, butassume that this is constant. At the end I am looking for variance notaverage)How I get equations:P_{x}(t) = probability that in time t there is x molecules type X. [similarP_{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 moleculesY 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) + k1Ey'(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 + dy' = a*y + b*x*y + dYesterday I observe one more thing in my model, that we can calculateaverage 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 + dWhich 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 + dButa = -v2b = -v1d = v2*k3and f(t) >=0 as it is E_{x+z}so (b*f(t)+a)^2 + 4v2*v2*k3 > 0and 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 thatI 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 + dI 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 + dI 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.De?e solve.Since the equations are autonomous (there is no t) you can solvethe problems graphically -- not in terms of t but at least showinghow x and y will vary together. Simply make little marks inthe 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 combinationsof (x,y) which can possibly be attained as t varies. (Of course, you'llonly be interested in the one curve passing through your initialpoint (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) + dbut of course numerical analysts can suggest much more robust approximations.What you probably can't do, except for very fortunate combinations ofa,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 + dI 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 + ddoes not seem that simple to me.is that a textbook problem?> thanks,> Dargen === > The ?st eqn can be written as:> x' = y' + cxwell, yeah, and the second can be written as y' - (a + bx)y = dbut 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 ?st-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 ?st order theories you might appreciate Wanda Szmielew, A German school centred on Friedrich Bachmann produced a version of> geometry based ?mly 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 ?d second-hand copies.I looked into the three English volumes some years ago and I recall thatchapter. 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 ?d a copy of Fundamentals.... in the ?st place.I know that Coventry Public Library (in the UK) has it. I shallobediently return to it. Ken Pledger.-- G.C. === A German school centred on Friedrich Bachmann produced a version of>geometry based ?mly 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 classi?ation 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 samething as the Bachmann school's approach to foundations. They use agroup-theoretic equation to express the fact that a point lies on a line,another equation to say that two lines are perpendicular, etc. It reallyis an alternative _foundation_ for geometry, Euclidean and non-Euclidean. Ken Pledger. === > A German school centred on Friedrich Bachmann produced a version of>geometry based ?mly 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 classi?ation 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 thegenerators 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 agenerator either.3) However, I'm unable to prove that the rest of the group members areall 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, gonly 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 ahint, 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 Savardhttp://home.ecn.ab.ca/~jsavard/index.html === Hallo everyone,I worked for about 3 years after college before getting into gradschool. (I am currently in the ?st year). And I ?d math so muchmore dif?ult now, than I ever did; keeping up with the pace of thecourses is turning out to be a nightmare. I have forgotten even someelementary results; and ?d myself stuck in a painstakingly slowiterative process (going back to my old Calc III notes...). My profswant me to speed up things, but I cannot seem to ?d 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 theonly solution? Is there a way of working smart instead of hard?Because at the end of the day, for ?st year students at least, it isthe grades on the exams and the courses that really matter,irrespective of mathematical potential.FrankP.S.--I am not sure if there is any wisdom in sampling life for afew 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 seemto be working for him. That I can believe.One might say the whole point of graduate education is to learnhow to work smart; so the OP's desire to work smart is indeeda 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 worksmart, you do know enough by now to avoid certain ways of workingdumb. Mind-numbing work that leads to no useful insights isnecessary sometimes, but don't subject yourself to *more* of itthan is necessary. You don't get any brownie points for justdoing work. When you get bogged down, look to the larger pictureif you possibly can; for me, that means reading more broadly sothat I begin to see the *why* for all the dif?ult work. Thusrefreshed, I can go back to it -- or perhaps, I can see what partsof 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 areworking hard, but are you working *hard*? When the chips aredown, as they seem to be, you need to throw everything and Imean everything you got at it. Take no prisoners. What I meanis, it's possible you don't yet know how hard you really canwork; give it a try.(Of course this is just MHO and I should add I've never been agrad student myself, in *math* that is. I'm generalizing from myexperience elsewhere, and from my own self study in math. So ifmy advice is misplaced, I hope that someone with more directexperience 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 bothOP 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 suf?es 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 thatYes, 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> de?ing 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 bothOP 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 suf?es 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 thatYes, 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> de?ing 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 ? -- 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 ?and level you want._An Introduction to the Theory of Numbers_, 5th Edition, by IvanNiven, Herbert Zuckerman, and Hugh Montgomery, starts with notions ofdivisibility, primes, and the binomial theorem, and continues throughcongruences, quadratic forms and quadratic reciprocity, arithmeticfunctions and the Mobius inversion formula, some diophantineequations, Farey fractions and irrational numbers, simple continuedfractions, a bit about multiplicative number theory (Dirichlet seriesand something on primes in arithmetic progressions), partitions,densities, and even a bit about algebraic number ?lds (particularlyquadratic number ?lds). Pretty good ?st introduction, and covers alot of stuff, I think.If you want a look at Analytic Number Theory, there's Tom Apostol's_Introduction to Analytic Number Theory_, Springer VerlagUndergraduate Texts in Mathematics. Also starts with divisibility,gcds, primes, and the Euclidean algorithm, but then quickly movesthrough Arithmetical functions, Dirichlet multiplication, averages ofarithmetical functions, and some theorems on the distribution ofprimes, before moving to congruences, ?ite abelian groups and theircharacters, and Dirichlet's Theorem on primes in arithmeticprogressions. Then he deals with Gauss sums, quadratic reciprocity,primitive roots, Dirichlet series and Euler products, the zetafunction, and an analytic proof of the Prime Number Theorem, before?ishing up with partitions.An introduction to Algebraic Number Theory, though a bit more advancedthan Apostol's intro to analytic number theory, is Kenneth Ireland andMichael Rosen's _A Classical Introduction to Modern Number Theory_,2nd Edition (Springer Verlag Graduate Texts in Mathematics); I've seenit used for an undergraduate course once (at Berkeley). Starts with Unique factorizaion in Z, k[x], talks about PIDs, then thering of Gaussian and of Eisenstein integers; then applications ofunique factorization, congruences, the units of the integers modulo n,quadratic reciprocity, then quadratic Gauss sums, ?ite ?lds,general Gauss sums and Jacobi sums, cubic and biquadratic reciprocity,equations over ?ite ?lds, then the zeta functions, and then divesright into Algebraic Number Theory: unique factorization, rami?ationand degree, quadratic and cyclotomic ?lds, the EisensteinReciprocity Law, Bernoulli numbers, Dirichlet L-functions, diophantineequationes, and even three ?al chapters on Elliptic curves, theMordell-Weil Theorem, and New Progress in Arithmetic Geometry (atleast, new as of 1981).Finally, if you know enough abstract algebra (groups, rings, ?lds,basic Galois Theory), then I ?d that an excellent introduction toalgebraic number theory is Daniel Marcus's _Number Fields_ (SpringerVerlag Universitext). It is chock full of exercises, usually around 50per chapter, which involve doing actual implementations of sundrytheorems. It starts by discussing a special case of Fermat's Theorem(basically, Lame's idea and how one could try to make it work, leadingto what is essentially a proof of Fermat's Theorem for the regularprimes in Case I, though not quite). Then de?es number ?lds andrings of integers, decomposition in number ?lds, Galois Theoryapplied to the decomposition, the ideal class group and the unitgroup, distribution of ideals, Dedekind's Zeta function and the classnumber formula, and the distribution of primes. Finally, a somewhatquick overview of what Class Field Theory is all about. The heart ofthe book is the exercises, and they give you a very solid grounding onalgebraic number theory at a level around that of the end of a ?stgraduate course in Algebraic Number Theory. But the prerequisites aresteeper 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 Magidinmagidin@math.berkeley.edu === The ?st half of Ireland-Rosen was written for Brown's undergraduatemath major number theory course, and is often used for that purpose.It presupposes a basic undergrad algebra course (groups, rings, alittle about ?lds, but not Galois theory). The latter half was addedlater, and is suitable for a beginning graduate course.Another very enjoyable, short, elementary introduction to numberthat as a paperback, it's relatively inexpensive. Well, looks like $30on Amazon, so maybe not all that cheap. Anyway, when I took numbertheory (from Rosen, using Ireland-Rosen), he also had us buyDavenport'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 proofreadingthe text and the solutions manual, but:The good points about the texts are that there are tons of exercisesand copious historical notes. It is modern, including up to datematerial 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 spendsquite a bit of time developing things from the axioms, working through arithmetic and so on, before getting on to number theoryproper. (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, thisis ?e and admirable.But for self-study, one could become bogged-down in stuff thatone already knows, stuck on hard exercises that aren't reallyrelevant, when one really wants to get on with it.Along the same lines, the last section of every chapter deals withan optional thread quadratics, which is Mollin's research specialty.A self-study student should probably skip that stuff, but might notknow whether he should or not.It depends on what the OP's unstated purposes were. I wouldn't handit to a student for self-study without knowing the student and thenmarking the sections that I think he should skip. Otherwise, it'sa 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 a3-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 followingoptimization 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 vectorassociated with the smallest singular value of A.The solution of (1) or (2) for non-zero q_i makes sense to me.However, I ?d 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 thenon-trivial x? Is it useful by increasing the amount of equations? Isthere any additional regularization methods?Zhaozhong Wang === I understand that I can use Pascal's Triangle to complete a binomialexpansion, but it's very long-winded. What is the quickest way of workingout the coef?ient of x^5 in the expansion of (x^2 - (3/x))^7?Please bear in mind that I have a scienti? calculator which I am preparedto use (Casio fx-991MS) and I have a test soon in which speed is de?itelyan 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 coef?ient of x^5 in the expansion of (x^2 - (3/x))^7? Please bear in mind that I have a scienti? calculator which I am > prepared> to use (Casio fx-991MS) and I have a test soon in which speed is > de?itely> 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 is7! / 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 coef?ient of x^(14 - 2*(n - 1) - (n - 1)), orx^(14 - 3 * (n - 1)).You want the coef?ient of x^5, so 14 - 3(n - 1) = 5, or n = 4.So, your coef?ient 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 coef?ient of x^5 in the expansion of (x^2 - (3/x))^7?The coef?ient of a^i*b^{n-i} in (a+b)^n, i=0,....,n, is given by thebinomial coef?ient (n choose i). For (a-b)^n it is (-1)^{n-i}*(n choose i)The general formula for (n choose i) isn!/ i!(n-i)!where for integer m, m! is the factorial of m, m! = 1*2*3*...*mThe coef?ient of x^5 in the expansion of (x^2-(3/x))^7 correspondsto 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=7So 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 theanswer is that the coef?ient 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 Magidinmagidin@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 coef?ient of x^5 in the expansion of (x^2 - (3/x))^7? The coef?ient of a^i*b^{n-i} in (a+b)^n, i=0,....,n, is given by the> binomial coef?ient (n choose i). For (a-b)^n it is (-1)^{n-i}*(n choosei) 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 coef?ient 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 coef?ient 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 Toothhttp://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 coef?ient 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 ?e possibleanswers to this question:a) -945b) -210c) 0d) 243e) 1415It 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 coef?ient 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 ?e possible>answers to this question:a) -945>b) -210>c) 0>d) 243>e) 1415It 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 ?ured out the coef?ient 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 abouti=4, and then calculated that coef?ient).Which means that the coef?ient 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 Magidinmagidin@math.berkeley.edu === For the gaussian pdf integrationsuppose 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 axisp(r) = 1/(sqrt(2pisigma^2))^n exp(-r^2/(2sigma^2))then what's the jacobian matrixis |J|=1? === Given that x^2 + y^2 = 6xy, ?d an equivalent expression to 2 log(x + y).How does the ?st expression relate to the logarithmic expression? === >Given that x^2 + y^2 = 6xy, ?d an equivalent expression to 2 log(x + y).How does the ?st 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 Magidinmagidin@math.berkeley.edu === How can you tell if a set of simultaneous equations has solutions or not?e.g. 12x - 4y = 8, -9x + 3y = 6orx^3 - y^2 = 0, x = yAs far as I can tell, you can work out a value for x and a value for y everytime...? So what de?es 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 de?es 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 = 6For linear systems such as this, use linear algebra. >or>x^3 - y^2 = 0, x = yThis is a nonlinear system. In general it may be hard to decide whethersolutions exist, especially if you're interested only in real solutions.Groebner basis techniques may be useful.Robert Israel israel@math.ubc.caDepartment 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 = 6orx^3 - y^2 = 0, x = yAs far as I can tell, you can work out a value for x and a value for y every>time...? So what de?es whether there are solutions or not?There are solutions if you can ?d a set of values for the variablesso that each of the equations is true.e.g. for your second set of equations, x = 0, y = 0 is the onlysolution. the ?st set doesn't have any solutions:12x -4y =8 => 3x-y=2=> y = 3x-2substitute 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 ?st set of equations has no solution. If you make the ?st set a bit different:12x-4y=8 and -9x+3y=-6then you have an in?ite number of solutions. just pick a value forx, 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 in?ite number of solutions.Gareth <2brtpvsln95dj6c39r8tv25jtosupr9vfg@4ax.com> === > In general, a set of simultaneous equations can have no solutions,> exactly one solution, or an in?ite number of solutions.This is not exactly true, unless you're talking about a set of LINEARequations. 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, butyou still haven't answered the question about shouldness,that is to say, some sort of proof of this lackof algebraic integers (of course,some numbers are not algebraic integers,like those with non-integer coef?ients, I guess). your mass Usenet audience is awaiting signs of a brainin your correpondence; can you read us?on the wayside,there's nothing wrong with the de?ition of atom, althoughthe orginal Greek meaning really applies to molecule, sinceit 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 coef?ients,> 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 de?ition? (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://mathforpro?.blogspot.com--les ducs d'Enron! === > >>good demonstration, Randy Poe.>Algebraic integers are de?ed to be roots of monic polynomials with>>integer coef?ient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where>>monic refers to the leading coef?ient.>>If we are interested in the set of numbers which are>roots of monic polynomials with integer coef?ients,>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 de?ition? (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 ?disk because it's too small. It's not that the ?is too small, but you are trying use it for something it wasn't intended to do.That's it. The de?ition 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 de?ition 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 de?ition is that they are> indivisible, as people can de?e things, and later re?e their> de?itions.So we should be calling quarks, or whatever we're down to, atoms to return to the Greek notion?Excerpt from http://mathforpro?.blogspot.comapparently post no feedback. Have you gotten any positive feedback?-- Will Twentyman === >good demonstration, Randy Poe. >>Algebraic integers are de?ed to be roots of monic polynomials with>integer coef?ient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where>monic refers to the leading coef?ient. >If we are interested in the set of numbers which are>>roots of monic polynomials with integer coef?ients,>>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 de?ition? (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 > ?disk because it's too small. It's not that the ?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 poststhan 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 de?ition 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 de?ition 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 de?ition is that they are> indivisible, as people can de?e things, and later re?e their> de?itions.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 de?ition that has been re?ed.Excerpt from http://mathforpro?.blogspot.comCheck 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 itbegins with a truth and proceeds by logical steps to a conclusion thatthen must be true.So, feedback in one sense is irrelevant. While, of course, in termsof 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://mathforpro?.blogspot.comapparently 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 simpli?d 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 de?ition:Algebraic integers are de?ed to be roots of monic polynomials with> integer coef?ient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where> monic refers to the leading coef?ient.If the de?ition is so problematic, what should the de?ition be?The problem has to do with *assuming* that the ring doesn't havecertain interesting properties, so it's not like the de?ition needsto be changed.It's just that its limitations must be noted.Here the problem is that the emphasis on monic polynomials clips somenumbers.Not a big deal, if you know that's happening, but mathematiciansdidn't realize it for over a hundred years, so now it's a big deal. My assertion is that the over hundred year old de?ition 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 intonon-polynomial factors? The answer is, yes, it can be.What exactly do you think I'm claiming that you just toss out somepolynomial?I'm curious what your reasoning is.Remember, all I do basically is factor a polynomial intonon-polynomial factors with a neat technique, look at constant termsof those factors, divide the polynomial by some integer, and considerthe constant terms again to draw a rather direct and obviousconclusion.James Harris === > Luckily for me the mathematical argument that proves that I've been> correct all along can be further simpli?d 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 de?ition:> Algebraic integers are de?ed to be roots of monic polynomials with> integer coef?ient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where> monic refers to the leading coef?ient.If the de?ition is so problematic, what should the de?ition be?The problem has to do with *assuming* that the ring doesn't have> certain interesting properties, so it's not like the de?ition 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 de?ition 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 canbe factored into non-polynomial factors and I want to SEE thosefactors.> 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 ?ddif?ult to follow. I am hoping that a demonstration withP(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 simpli?d 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 de?ition:Algebraic integers are de?ed to be roots of monic polynomials with>> integer coef?ient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where>> monic refers to the leading coef?ient.If the de?ition is so problematic, what should the de?ition be?The problem has to do with *assuming* that the ring doesn't have>certain interesting properties, so it's not like the de?ition needs>to be changed.Huh??? You've been going on for months about how there's anerror with this de?ition. Now in spite of that it doesn't need tobe changed?You wonder why people have trouble following yourarguments. It's because over and over you don'tsay 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 de?e algebraic integers as roots of monic polynomials,and nobody has realized for over a hundred years that that meansthat some numbers are not algebraic integers. Guffaw. My assertion is that the over hundred year old de?ition 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 simpli?d 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 de?ition:Algebraic integers are de?ed to be roots of monic polynomials with> integer coef?ient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where> monic refers to the leading coef?ient.If the de?ition is so problematic, what should the de?ition be?The problem has to do with *assuming* that the ring doesn't have>certain interesting properties, so it's not like the de?ition needs>to be changed.Huh??? You've been going on for months about how there's an> error with this de?ition. Now in spite of that it doesn't need to> be changed?The de?ition doesn't need to be changed.I've already given the ?, which is the object ring.Algebraic integers are included in the object ring.Now then, here's the math argument emphasizing *CONSTANT TERMS* whichis key in showing there is a problem.Notice how I'll be strongly emphasizing constant terms all the waydown.P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078which has a constant term that is 1078.Well P(x) can also be written out asP(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3so I can factor to getP(x) = (5 a_1 + 7)(5 a_2 + 7)(5 a_3 + 7)where the a's are roots ofa^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 areall 7, which can't be right, as the constant term of P(x) is 1078, sosetting x=0, revealsP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic de?ing the a's with x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1 and a_2for 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 haveP(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 + 1078has 49 as a factor, so I can divide by 49, and dividing 1078 by 49gives me 22, as the new constant term.Well that means thatP(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 simpli?d 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 de?ition:Algebraic integers are de?ed to be roots of monic polynomials with>> integer coef?ient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where>> monic refers to the leading coef?ient.If the de?ition is so problematic, what should the de?ition be?The problem has to do with *assuming* that the ring doesn't have>certain interesting properties, so it's not like the de?ition 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 workand if they claim that algebraic integers don't have certain propertiesthat are interesting then their work might be wrong.Are these any particular certain interesting properties or just certaininteresting properties in general?Have mathematicians been assuming that the ring doesn't have certainuninteresting properties as well or is this blind spot limited tointeresting properties only?Alan-- Defendit numerus === > Luckily for me the mathematical argument that proves that I've been> correct all along can be further simpli?d 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 de?ition:>Algebraic integers are de?ed to be roots of monic polynomials with> integer coef?ient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, where> monic refers to the leading coef?ient. If the de?ition is so problematic, what should the de?ition be? The problem has to do with *assuming* that the ring doesn't have> certain interesting properties, so it's not like the de?ition 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 de?ition 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 HarrisHe's just wanting to see your method applied, there's no need to get uptightabout it. You should be honored that he's taken an interest. Instead you'reungrateful, 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 de?ition 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 ?ld. They're not. No mathematician ever claimed they are. Algebraic integers form a ring, which is not terribly dif?ult to prove, and algebraic numbers are a ?ld, which is not terribly dif?ult 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 widelystudied??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 GeorgeM. Bergman and Adam O. Hausknecht. Mathematical Surveys andMonographs, Volume 45, American Mathematical Society (1996). MR97k:16001.A search in MathSciNet for Title: cogroup yielded 23 hits, but Isuspect that it may be that there are several notions involved. Inparticular, I am doubtful, given the description in the Math Review,that the notion of cogroup in J.E. Eaton's 1940 paper Theory ofCogroups, Duke J. Math. 6, (1940). 101--107; is the same as thenotion of cogroup in Bergman and Hausknecht...For the latter, if V is a variety of algebras (in the sense of GeneralAlgebra), then a co-group object in V is an object of V thatrepresents 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 Magidinmagidin@math.berkeley.edu===I was wondering if any one knows a clever way to solve the followingintegral: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,-In?ity,In?ity}]I can solve it with a lot of algebra and factoring the fraction, butif 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 ?d asolution (using recurrence relations):A)A_n = A_n-1 + 3 where A_0 = 1ANS: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 + 1I think Im corect on this prev one but the next one Im not sure what to dobecause the initial form seems strange to me.B) A_n = -A_(n-1) where A_0 = 5Do I do something like ... <3F9C5A9D.ECC0A6F0@yahoo.com> <3F9C6EDF.CD8B5D@hate.spam.net>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft === at 05:03 PM, Uncle Al said:>Metric models are geometric and have geodesics. Af?e models are>electromagnetic in form and have no geodesics. I hope that you mean that the geodesics are not relevant to thePhysics, rather then that they don't exist. An af?e connection isall that you need to de?e 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 allof 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 JOATnot 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 af?e connections, which are not tensors, sothey have non-homogeneous transformation properties. Moreover,geodesic acceleration should be independent of mass, which applies toforces 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 af?e 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 geodesicmotion. 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 af?e connections, which are not tensors, so>they have non-homogeneous transformation properties. Ah, but the geodesic equation is the absolute derivativeof the 4 velocity U^u = dx^u/ds, ieDU^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 truefor relatively large masses, that's why accelerationis a better term than force in GR, otherwise you'llneed to include the term k*t_uv >0 we ?uring 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 doesnot, so the stated objection evaporates.>All metric theory>predictions are included in af?e 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 viewof QT, meaning all motion must be geodesical,otherwise absolute acceleration exists.Ken S. Tucker === >>Metric models are geometric and have geodesics. Af?e 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_kwith 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 ?d 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. Af?e 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_kwith 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 vectorsgot a little skewed). The ?st term would be gravitation and thesecond the Lorentz, but how you'd ever get a cross product out of asymmetric connection is a bit of a stretch. === >Metric models are geometric and have geodesics. Af?e 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 ?ure this is the solution from G_uv =0 (with theenergy density being zilch, by the speci?ation of apoint) 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 addedwithin your connections, and may very well be ok, is a signi?ant departure from conventionalmathematics. I think QT excludes continuous acceleration off_u = q*F_uv *U^v =0may 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 ?st 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 proportionalityexists between the Magnetic ?ld and the angular momentum of the mass the charge creating the currentand B-?ld is attached (at rest) to. Wouldn't this suggest that the metric describingangular momentum and it's direct analog be equal?Ken S. Tucker === said:>Knowing all the variables is a rare luxury. IN this case we're addressing variables that are known. What needs tobe determined is which, if any, have had a signi?ant impact, andwhat 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 datasuggest a genetic in?ut not the effects of that in?therdata suggest hat the in?sn't nearly as large as you believe.Given our current understanding of genetics, the DNA evidence has tobe 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,>re?ing it down and considering all the variables I could ?d, I>wouldn`t be publishing it in a newsgroup.Well, at least not an unmoderated one.-- Shmuel (Seymour J.) Metz, SysProg and JOAT ===