mm-30 === Subject: In talk.bizarre, someone, might have been>Is it too late to put in a reservation for toroidal cats, also known as>cats oroboros?> If you check more carefully, you will discover that ordinary household> cats are toroidal. To be precise, they are fuzzy little tubes.This would explain the near instant success of my Little Orange Tabbytosses a damper on the toroidal cat >=> toroidal black hole project,since, as is well known, black holes have no hair.xanthian.I suppose now I must descend to the James Harris [=> well known polluterof sci.math <=] paradigm and start demanding my star on the walk of fameof value and don't understand the math anyway.-- === Subject: I am haunted by this confusing problem in the book Fundamentals on Complex Analysis withApplication to Ehgineering and Scienceby E. B. Saff and A. D. SniderClassify the behavior at oo for function f(z)=1/sin(z)...I got answer that when z -> oo, f(z) has an essential singularity;But the book answer says that f(z) has no isolated singularities at z ->oo...I am really confused... after many thinking I decide to resort to your help,Can anybody enlighten me to give a hand?(this is a learning process and Iwant to know why...)-Walala === Subject: >Dear all,>I am haunted by this confusing problem in the book>Fundamentals on Complex Analysis with>Application to Ehgineering and Science>by E. B. Saff and A. D. Snider>Classify the behavior at oo for function f(z)=1/sin(z)...>I got answer that when z -> oo, f(z) has an essential singularity;>But the book answer says that f(z) has no isolated singularities at z ->>oo...>I am really confused... after many thinking I decide to resort to your help,The function has singularities at n pi for all integers n. All neighbourhoodsof infinity have singularities at finite complex numbers, so there is an essential singularity at infinity (as you stated). Since every neighbourhood of infinity has a singularity apart from the singularity at infinity, then the singularity at infinity is not an isolated singularity. In fact, it is generally true that if a function f(z)is such that for every neighbourhood of z = a, f(z) has a singularityat a complex number distinct from a, then there is an essential singularityat z = a, and the singularity at z = a is not an isolated singularity.In contrast, both sin(z) and exp(z) have essential singularities at infinity,and these singularities are isolated, since they are the only singularities.>Can anybody enlighten me to give a hand?(this is a learning process and I>want to know why...)You were correct that the singularity at infinity is essential. The fact that the singularity is not isolated is extra information (as all non-isolatedsingularities are essential, but not all essential singularities are non-isolated).>-WalalaDavid McAnally-------------- === Subject: >Dear all,>>I am haunted by this confusing problem in the book>Fundamentals on Complex Analysis with>Application to Ehgineering and Science>by E. B. Saff and A. D. Snider>Classify the behavior at oo for function f(z)=1/sin(z)...>>I got answer that when z -> oo, f(z) has an essential singularity;>>But the book answer says that f(z) has no isolated singularities at z ->>oo...>>I am really confused... To say f has an isolated singularity at z means that f is analytic insome neighborhood of z, with z removed. For z = infinity this meansthat there exists R such that f is analytic in the region R < |z| < infinity. Is that true for the function 1/sin(z)?>after many thinking I decide to resort to your help,>>Can anybody enlighten me to give a hand?(this is a learning process and I>want to know why...)>-Walala>**David C. Ullrich === Subject: >Dear all,>>Please help me on this problem as I am practicing some text book>> >problems cramming for exam. The problem statement is:>> >prove that although u=ln(abs(z)) is harmonic on domain C{0}, but we>>cannot find its harmonic conjugate v such that ln(abs(z))+i*v(z) is>>analytic on domain C{0}...>[..]>> There are more interesting ways to do the problem. For example,>[...]> Or (this one is actually a version of the first solution):>> Suppose that such a v exists and let f = u + iv. You>> can use the Cauchy-Riemann equations to find what>> f'(z) is, even though you don't know v. Now recall>> that the integral of f' over a closed path is _always_>> equal to 0 (the integral of f over a closed path can>> be non-zero, depending on whether various hypotheses>> in Cauchy's Theorem are satisfied, but the Fundamental>> Theorem of Calculus shows that the integral of f' is 0>> regardless). But here it turns out that there is a closed>> path over which the integral of f' is non-zero, which is>> a contradiction.>>Yes, this way is especially appealing.It's my favorite, because it's essentially just a conciserigorous version of the original argument. YesterdayI was trying not to give too much away, but since thenI saw that the OP was posting thousands of questions,so there's no chance he's going to learn the materialregardless, so there's no point to worrying about that:What it all comes down to is that there is no Log(z)analytic in all of C{0}. One often proves that justby pointing out that Log(z) would have to be a certainthing, and look, there's no way to make that thingcontinuous. Instead one says that 1/z is not anantiderivative, since its integral over the unit circledoes not vanish - if you think about it for a secondthe non-vanishing of that integral is exactly thesame as the fact that if you continue Log aroundthe circle you don't get back to where you started,but it seems like maybe a cleaner way to put it.one year when I couldn't decide on a book Iliked for the course the fact that the integralof 1/z over the unit circle is non-zero is saidto be the source of all wisdom...)**David C. Ullrich === Subject: Have you noticed the way JSH has been addressing people by theirfull names in his posts? As in, you need to be very carefulwith the math, Nora Baron. It's funny!Mike === Subject: > Have you noticed the way JSH has been addressing people by their> full names in his posts? As in, you need to be very careful> with the math, Nora Baron. It's funny!> MikeIts full naame is James Steven Harris, if you want to go that route. === Subject: > Have you noticed the way JSH has been addressing people by their> full names in his posts? As in, you need to be very careful> with the math, Nora Baron. It's funny!> MikeThat must be why he never replies to me. === Subject: > Some of you may now see comments from mathematicians trying to hide> the logical flaw in Wiles's work where they try to claim that you> *can* indeed compare infinite sets and thereby prove a condition.> [Cantor reference snipped]There is an old math joke that goes like this:An engineer, a physicist, and a mathematician are travelling in the countryside and see a number of sheep. The engineer comments: Allof the sheep in this area must be black, after seeing a number of such sheep and no counterexamples to his claim. The physicist notes: We can only really be sure that all of the sheep we have seen are black.The mathematician says: We can really only say that all of the sheep that we have seen are black _on one side_.The error is the assumption that the sheep are single-colored.The error in picking apart a popularization of a sketch of a proof is assuming that is is a literal account of the reasoning of the proof.Cirumstantial evidence, such as a popularized, expository account of a proof cannot be criticized rather than the proof. What one shows by doing so is that there is a flaw in the _exposition_ of the proof by another person, and not a flaw in the author's proof.There are two ways to attack Wiles' proof:1. Explicitly exhibit a semistable elliptic curve which is not modular, thereby showing his result is wrong. Ergo, he makes an error in reasoning somewhere. This is the easiest form of attack, since you don't actually need to read his paper to make it. It is also a big money-maker, since then FLT is back up for grabs again.2. Refer to a page number and line number in the _original_ paper(which is available _free_ to all with an Internet connection atwww.jstor.org, search for wiles and modular) where there is a misstep.I made the following mistake the other day: I claimed that the second symmetric power of k[x,y]/(xy) was isomorphic to the second cartesian power. One way to see that this is wrong is to note that the CM-type of the second symmetric power is 2, hence it cannot be Gorenstein, but the second cartesian power is a local complete intersection, so it is Gorenstein.Another way to see this would be to see the silly mistake in the argument: the invariant subring of k[x,y,w,z]/(xy,wz) under the obvious Z2-action has as generators x+w, xw, y+z, yz. I noted that (x+w)yz=(y+z)xw=0, and claimed that those were all the relations. Anyone paying any attention can look at this argument and either 1) claim that it is incomplete, since I have not _proven_ that I found all the relations, or 2) easily exhibit another relation which I have missed.Sorry if this example is hard to follow, but you better know what Gorenstein means if you're going to go after Wiles' paper.all small deformations of products of stable varieties of general type are products of small deformations of the factors. I could say that I did this by counting deformations. Of course the deformation spaces on both sides are infinite and have the same cardinality as the complex numbers. However, using the auxilliary structure of a vector space, I can compare their dimensions. This is how I would describe the proof: by counting.Of course, there are more details which should be omitted from a brief account. Of course it is not enough for me to show that the dimensions of these vector spaces are the same to show that a particular linear map between them is an isomorphism. But it's enough if that map is injective.[rhetorical questions and polemics snipped] === Subject: === Subject: === Subject: Michael A. Van OpstallPadelford C-113opstall@math.washington.eduhttp:// www.math.washington.edu/~opstall/ === Subject: > 2. Refer to a page number and line number in the _original_ paper> (which is available _free_ to all with an Internet connection at> www.jstor.orgNo, it is free to all members of participating institutions for free.Members of the hoi polloi need to wheedle a relationship with a participating institution. === Subject: >> 2. Refer to a page number and line number in the _original_ paper> (which is available _free_ to all with an Internet connection at> www.jstor.org>> No, it is free to all members of participating institutions for free.> Members of the hoi polloi need to wheedle a relationship with a> participating institution.>Here is where you can get them for free:http://math.stanford.edu/~lekheng/flt/Jon === Subject: [... ]> Here is where you can get them for free:> http://math.stanford.edu/~lekheng/flt/> JonAs far as I can tell, the file above is over 10 Mbytes. There issmaller ~866 kbytes file at:http://modular.fas.harvard.edu/21n/papers/name= ``Wiles,Modular_Elliptic_Curves_and_Fermats_Last_Theorem.pdf The computer can read the paper aloud withAcrobat Reader 6.0 ...David Bernier === Subject: Dear mathematicians,old friend JSH; the fact is you're all 100 miles far from the truth: Iknow James Harris, and he is the most brilliant mathematician ever.To tell the truth... well... I'm not sure James will appreciate... butyou can check the following link, where you'll find:1) The social status of JSH.Now you can meet the myth ...http://www.jal.cc.il.us/math/faculty/jim.html === Subject: > Message-id: Dear mathematicians,>>old friend JSH; the fact is you're all 100 miles far from the truth: I>know James Harris, and he is the most brilliant mathematician ever.>To tell the truth... well... I'm not sure James will appreciate... but>you can check the following link, where you'll find:>>1) The social status of JSH.>>Now you can meet the myth ...>>http://www.jal.cc.il.us/math/faculty/jim.htmlIf I were him, I would consider changing my name to something more innocuous.Like Adolf Harris.--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm== ==> Dear mathematicians,> old friend JSH; the fact is you're all 100 miles far from the truth: I> know James Harris, and he is the most brilliant mathematician ever.> To tell the truth... well... I'm not sure James will appreciate... but> you can check the following link, where you'll find:> > 1) The social status of JSH.> Now you can meet the myth ...> http://www.jal.cc.il.us/math/faculty/jim.htmlwhat about this guy? http://www-ee.stanford.edu/~harris/ === Subject: >Message-id: Dear mathematicians,>>old friend JSH; the fact is you're all 100 miles far from the truth: I>know James Harris, and he is the most brilliant mathematician ever.>To tell the truth... well... I'm not sure James will appreciate... but> >you can check the following link, where you'll find:>> >1) The social status of JSH.>>Now you can meet the myth ...>> >http://www.jal.cc.il.us/math/faculty/jim.html> If I were him, I would consider changing my name to something more innocuous.> Like Adolf Harris.> --> Mensanator> 2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htmOr at least giving himself a middle intitial other than S. === Subject: > ...And are you the Dr. Virgil Stubblefield of the same faculty?GC === Subject: > Dear mathematicians,>> old friend JSH; the fact is you're all 100 miles far from the truth: I> know James Harris, and he is the most brilliant mathematician ever.> To tell the truth... well... I'm not sure James will appreciate... but> you can check the following link, where you'll find:>> 1) The social status of JSH.>> Now you can meet the myth ...>> http://www.jal.cc.il.us/math/faculty/jim.html> what about this guy? http://www-ee.stanford.edu/~harris/Who? The panda, or the one holding the panda?GC === Subject: As you know, I continue to post messages to sci.math,sci.cross-posted-negro-dumbass, alt.cross-posted.lame, messages which Ihave to revise and exuse by saying I posted in haste, because as youknow I don't take my lithium and a manic depressive nigger is sort ofa carnival attraction within the elitist circle of higher math don'tyou admit, unless your name is Nora S. Thompson in which case NoraThompson, you should read the following paragraph closer and marvel atmy ability to smoke crack.As you know, Nora, I've been contemplating the problem of thenegroes as it relates to FLT. Many in the establishment elite havebeen forced to disagree with my solution of killing all niggers,although they know this is the correct thing to do, in order toresolved the central lie taught to even you undergraduates.I must warn you that I can crosspost any replies you make to my post,and I will name you by name, and contact your place of work. I willclaim that you are a racist.So you now see how I have been underestimated by the [caucasion] worldof math. They [white mathematicians] will not admit that black peoplecan be mentally deranged also, in fact it's more likely === Subject: Taunting James Harris is an old and venerable sport around here, butfor criticism with his own behavior; please don't bring race into it.Articles like yours are uncalled-for and offensive to MANY (dare I saymost?) people, including myself.-- Wayne Brown | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: > > Taunting James Harris is an old and venerable sport around here, but> for criticism with his own behavior; please don't bring race into it.> Articles like yours are uncalled-for and offensive to MANY (dare I say> most?) people, including myself.I second that. === Subject: [crap deleted]Go away. You are not funny, and you are annoying. === Subject: === Subject: Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === Subject: == === Subject: =Arturo Magidinmagidin@math.berkeley.edu === Subject: > Yes. And each single point defines one tangent, with one slope.No it certainatly does not.A point on a curve merely describes a coordinate. It is thedifferentiable triangle AT THAT POINT that describes the tangent andslope.f(x) = y = some expression with xdy/dx = the derivative for yConsider the differentiable triangle. ds^2 = dx^2 + dy^2 ds/dx = sqrt(1 + dy^2/dx^2)ds = int sqrt(1 + f(x)^2) . dxIt is ds that describes the slope and the tangent. === Subject: > Yes. And each single point defines one tangent, with one slope.> No it certainatly does not.> A point on a curve merely describes a coordinate. It is the> differentiable triangle AT THAT POINT that describes the tangent and> slope.> f(x) = y = some expression with x> dy/dx = the derivative for y> Consider the differentiable triangle.> ds^2 = dx^2 + dy^2> ds/dx = sqrt(1 + dy^2/dx^2)> ds = int sqrt(1 + f(x)^2) . dxThis is incorrect. First, you need to intergrate both sides of the equation, and second, the right hand side was factored incorrectly. It should read s(x_f) - s(x_i) = int{sqrt[1+ (dy/dx)^2]dx}> It is ds that describes the slope and the tangent.No, ds is differential arc length. === Subject: > Yes. And each single point defines one tangent, with one slope.> No it certainatly does not.> A point on a curve merely describes a coordinate. It is the> differentiable triangle AT THAT POINT that describes the tangent and> slope.> > f(x) = y = some expression with x> dy/dx = the derivative for y> Consider the differentiable triangle.> ds^2 = dx^2 + dy^2> ds/dx = sqrt(1 + dy^2/dx^2)> ds = int sqrt(1 + f(x)^2) . dx> This is incorrect. First, you need to intergrate > both sides of the equation, and second, the right > hand side was factored incorrectly. > It should read> s(x_f) - s(x_i) = int{sqrt[1+ (dy/dx)^2]dx}> It is ds that describes the slope and the tangent.> No, ds is differential arc length.Irrespective of my typo, you are still wrong.> ds = int sqrt(1 + f(x)^2) . dxds = int sqrt(1 + f'(x)^2) . dx === Subject: >> Yes. And each single point defines one tangent, with one slope.>> > No it certainatly does not.>> A point on a curve merely describes a coordinate.No it certainly does not.A point on a curve merely describes two coordinates> It is the> differentiable triangle AT THAT POINT that describes the tangent and> slope.>> f(x) = y = some expression with x> dy/dx = the derivative for y> >> Consider the differentiable triangle.> ds^2 = dx^2 + dy^2> ds/dx = sqrt(1 + dy^2/dx^2)> ds = int sqrt(1 + f(x)^2) . dx>> This is incorrect. First, you need to intergrate> both sides of the equation, and second, the right> hand side was factored incorrectly.> >> It should read>> s(x_f) - s(x_i) = int{sqrt[1+ (dy/dx)^2]dx}> It is ds that describes the slope and the tangent.>> No, ds is differential arc length.>> Irrespective of my typo, you are still wrong.>> ds = int sqrt(1 + f(x)^2) . dx> ds = int sqrt(1 + f'(x)^2) . dxWe, regular folks say ds = sqrt(1 + f'(x)^2) dxor int{ ds }= int{ sqrt(1 + f'(x)^2) dx }or s(x_f) - s(x_i) = int{ sqrt[1+ f'(x)^2] dx }Dirk Vdm === Subject: > Starblade Darksquall a .8ecrit dans le message de> Can we find (a,b,x,z,u,v,c) in (N-{0})^7 such that:> a^2+b^2=c^2> x^2+y^2=c^2> > u^2+v^2=c^2> ab+xy=uv> ?> I.R.>> That's not arithmetic! That's a diophantine equation you've got there!> I.J.R.This may help you!All triple of positive integers (X,Y,Z) satisfying X^2+Y^2=Z^2are as follows (this is well-known) X=t(a^2-b^2), Y=2tab, Z=t(a^2+b^2) where a>b>0 and t>0 are integers such thatGCD(a,b)=1 and a-b is an odd integer.Best WishesAlireza Abdollahi === Subject: How would you know what is taught in a college math curriculum if you don'teven have a degree in mathematics? I am working on a minor in math (major inmeteorology) and I can obviously see this is flawed. Anyone with half a brainwould see this too.David Moran === Subject: > How would you know what is taught in a college math curriculum if you don't> even have a degree in mathematics? I am working on a minor in math (major in> meteorology) and I can obviously see this is flawed. Anyone with half a brain> would see this too.>> David MoranI think it is difficult to get James to focus on the details of his ownexposition. Getting his attention probably requires something larger than a 2 by4.--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: Can anyone give me a reference for the following theorem, which Iproved, but with more effort than I imagined?Suppose f(x) is a monotonically increasing differentiable functionthat goes to infty as x does. Suppose that h(x) is a function thatgoes to 0 as x goes to infty. Suppose that xf'(x)/f(x) is bounded(outside some boun interval). Then f is asymptotic to the functionwhose value at x is f(x+xh(x)).Notice that although h(x) --> 0 as x --> oo, there is no reason tosuppose that xh(x) also --> 0. For example if h(x) = 1/sqrt(x), xh(x)= sqrt(x). === Subject: > Can anyone give me a reference for the following theorem, which I> proved, but with more effort than I imagined?> Suppose f(x) is a monotonically increasing differentiable function> that goes to infty as x does. Suppose that h(x) is a function that> goes to 0 as x goes to infty. Suppose that xf'(x)/f(x) is bounded> (outside some boun interval). Then f is asymptotic to the function> whose value at x is f(x+xh(x)).> Notice that although h(x) --> 0 as x --> oo, there is no reason to> suppose that xh(x) also --> 0. For example if h(x) = 1/sqrt(x), xh(x)> = sqrt(x).Not a reference, but a proof: Mean Value Theorem.A bit more explanation. We can rephrase the problem: if x,y ->infinity and y/x -> 1, show that f(y)/f(x) -> 1. In elementarycalculus we work more commonly with differences going to 0 rather thanquotents going to 1. So change variables: say g(t) = log(f(exp(t))). Then the problem becomes: if s,t -> infinity and t - s -> 0, then showthat also g(t) - g(s) -> 0. Now we come to the MVT: (g(t)-g(s))/(t-s)= g'(c) for some point c between s and t. So to prove what we want, itis enough if g' is bounded. Translate that condition back to acondition on f.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: > > Can anyone give me a reference for the following theorem, which I> proved, but with more effort than I imagined?> > Suppose f(x) is a monotonically increasing differentiable function> that goes to infty as x does. Suppose that h(x) is a function that> goes to 0 as x goes to infty. Suppose that xf'(x)/f(x) is bounded> (outside some boun interval). Then f is asymptotic to the function> whose value at x is f(x+xh(x)).> Notice that although h(x) --> 0 as x --> oo, there is no reason to> suppose that xh(x) also --> 0. For example if h(x) = 1/sqrt(x), xh(x)> = sqrt(x).> Not a reference, but a proof: Mean Value Theorem.> Beautiful proof! I had one that was a good deal more complicated.> A bit more explanation. We can rephrase the problem: if x,y ->> infinity and y/x -> 1, show that f(y)/f(x) -> 1. In elementary> calculus we work more commonly with differences going to 0 rather than> quotents going to 1. So change variables: say g(t) = log(f(exp(t))). > Then the problem becomes: if s,t -> infinity and t - s -> 0, then show> that also g(t) - g(s) -> 0. Now we come to the MVT: (g(t)-g(s))/(t-s)> = g'(c) for some point c between s and t. So to prove what we want, it> is enough if g' is bounded. Translate that condition back to a> condition on f. === Subject: > Could anyone suggest how I might model affine transformations using> Clifford/Geometric algebra> > My background is more of a programmer than a mathematician and what I am> trying to do is to model solid objects with simple Newtonian mechanics. What> I would like to do is model all the quantities using multivectors so that> the linear and rotational properties can all be handled in one equation, I> don't want to do anything advanced like space-time or quantum theory, just> simple mechanics of solid objects. As I say I'm not a mathematician but I> would like to understand the principles behind this.> that transforms are done using the following formula:> > P2=m * P1 * m^-1> Where:> P1 = original position of point> P2 = resulting position of point after transform> m= multivector> m^-1 = inverse of m> I belive that you could look at some other of Hestenes's papers found at:http://modelingnts.la.asu.edu/html/ ComputationalGeometry.htmlandhttp://modelingnts.la.asu.edu /html/UAFCG.htmlPatrick === Subject: There are a lot of interesting documents here and I may have missedsomething but I can't find any ideas about how to model affinetransformations.Have you come across a document which covers this?Where I am stuck is, do I have to use 3,4 or 5 D vectors? Which part of themultvector holds the translation? And what goes in the other parts of themultivector?Martin === Subject: > There are a lot of interesting documents here and I may have missed> something but I can't find any ideas about how to model affine> transformations.> Have you come across a document which covers this?> Where I am stuck is, do I have to use 3,4 or 5 D vectors? Which part of the> multvector holds the translation? And what goes in the other parts of the> multivector?> MartinUnfortunately, I don't have much time to answer your technicalquestions, but I think you couldn't do any better than to start withthe paperhttp://modelingnts.la.asu.edu/pdf/CompGeom-ch1.pdfand take particular notice of the section Linearizing the Euclideangroup on page 20. As I recall, the vectors are in 4 dimensions.Patrick === Subject: You could check out Rudin's Mathematical analysis 3rd edition.>> Im taking a split undergraduate/graduate Real Analysis course which> covers Lebesgue measure and Lebesgue integral as well as radon-nikodym> theorem. What book would you recommend ??? === Subject: Let f(x) be a real C^2 function with f ''(x)>0 for all x.For h>0, let t(h) be the unique number in (0,1) such thatf(h)=f(0)+f ' (h t(h))h.Prove that limit as h-->0 of t(h) is 1/2. === Subject: >Let f(x) be a real C^2 function with f ''(x)>0 for all x.>For h>0, let t(h) be the unique number in (0,1) such that>f(h)=f(0)+f ' (h t(h))h.>Prove that limit as h-->0 of t(h) is 1/2.Note that t(h) is unchanged if you replace f(x) by f(x) - (ax+b)for constants a and b, or multiply f by a positive constant. So we can assume WLOG that f(0)=f'(0)=0 and f''(0)=2. Given epsilon > 0 there is delta > 0 such that 2-epsilon < f''(x) < 2+epsilon for |x| < delta. For convenience take h > 0 (the argument for h < 0 willbe similar). For 0 <= x < delta we have (2-epsilon) x <= f'(x) <= (2+epsilon) x and (1-epsilon/2) x^2 <= f(x) <= (1+epsilon/2) x^2. Apply these to estimate f'(h t(h)) and (f(h)-f(0))/h ... Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: > Try singular values: if V={v_1, ..., v_k} is an ordered list of> vectors in an inner product space, set up the Gram matrix G:[snip]> For example, the determinant of> [ 1, 0, -5, 0, 1]> [ 1, 2, 0, 1, 0]> [ 0, -2, -3, 1, 0]> [ 0, 4, 2, 1, -2]> [ 1, 3, 5, 1, 0]> is 1, but the first column is close to a linear combination of> the other four columns, relatively within 0.62% .It's no coincidence that the condition number of this matrix issomewhere around 1127 (which is bad - the higher the C.N., the morealmost linearly dependent the cols, and thus the more unsuitable thematrix is for numerous direct calculations.)Of course, knowing the singular values implies knowing the conditionnumber... the condition number is the ratio of the largest SV (in theabove example, 9.something) to the smallest (.00something).But while singular values can be hard to compute, decent estimates(decent enough to decide if you want to compute with that matrix) ofcondition number can be easier to come by.Of course the singular values give you oodles of information beyondjust the niceness of the matrix, but it seemed like the OP was onlyconcerned about the niceness, so that the condition number would be amore directly relevant quantity...cdj === Subject: >>> Planck's Constant>>> Previously in the thread Angular Momentum in Rotating Bodies, I>> presented an analytical framework for the interpretation of dr/dt in>> circular rotation of a point mass m at velocity v and radius r. No one>> I know of agrees with my interpretation of dr/dt. >>Lester--You should take courses in differential and vector calculus. === Subject: >> Planck's Constant>> Previously in the thread Angular Momentum in Rotating Bodies, I> presented an analytical framework for the interpretation of dr/dt in> circular rotation of a point mass m at velocity v and radius r. No one> I know of agrees with my interpretation of dr/dt. However, in the> interests of further establishing this general framework, I would like> to pursue general developement of the idea which culminates in the> analytical definition of Planck's constant.The form of dr/dt follows directly from the vector definition of r.There is no freedom for further definitions. Lester has delusionsof competency. He needs to solve his conceptual problems withvector calculus before assuming the pulpit on quantum mechanics.[Old Man] === Subject: > >> Planck's Constant>> Previously in the thread Angular Momentum in Rotating Bodies, I>> presented an analytical framework for the interpretation of dr/dt in>> circular rotation of a point mass m at velocity v and radius r. No one>> I know of agrees with my interpretation of dr/dt.>>People you don't know of are also disgusted. You are a crank, a>loud ass, and a boring ineducable lout.> Somehow I'm just not surprized.> >>http://w0rli.home.att.net/youare.swf>>[snip]> > mass and radius of rotation are inversely proportional, that is that>>Res ipsa loquiter, moron. Hey Zick - electrons are point masses and>protons are not. Why don't you tell us how a 1.67262158 x 10^(-27) kg>proton is smaller than a 9.10938188 x 10^(-31) kg electron.> Well, Al, the uncertainty is certainly smaller. There might be some> physical reason for that. Maybe that's why they use proton streams> instead of electrons for more accurate medical treatments.You are an uneducated lout who knows nothing about radiation physics,either. -- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === Subject: >> Planck's Constant>> Previously in the thread Angular Momentum in Rotating Bodies, I>> presented an analytical framework for the interpretation of dr/dt in>> circular rotation of a point mass m at velocity v and radius r. No one>> I know of agrees with my interpretation of dr/dt. However, in the>> interests of further establishing this general framework, I would like>> to pursue general developement of the idea which culminates in the>> analytical definition of Planck's constant.>>The form of dr/dt follows directly from the vector definition of r.>There is no freedom for further definitions. Lester has delusions>of competency. He needs to solve his conceptual problems with>vector calculus before assuming the pulpit on quantum mechanics.>[Old Man]Well, actually the form of dr/dt follows directly from the definitionof v not r. I have been roundly chastized repeatedly for consideringdr in isolation. So, I have a suggestion: get your story straight.before trying to admonish others.But thanks for the input. And the output. And the cross product. Andthe cross purpose. More or less as expected. Who'd a thunk it? === Subject: >>> [. . .]>>> This is frustrating because we're up against a deadline. So, I'd like>> to reply to just the tail end of your post now and get to the>> beginning later.>> I have a semi-free day, so...>>... if your assertion is that a stable orbit >represents changing angular momentum, where does the energy >come from/go to?>>> You know, initially I thought there was a problem in this regard. But>> now I'm not so sure. Granted that I think I'm talking about a>> constantly changing measure of angular momentum, the fact is that>> we're ever in a position to measure such changes in isolation.>> If you're talking about spin, you can't measure it in >isolation. RPM is always WRT the external Universe, neh?Well, with respect to constant motion in a straight line you cancertainly measure spin. Changes in angular momentum which are notbalanced by other changes in angular momentum should also bemeasurable I think.By the way, several years ago Jim Carr mentioned that energy is not avector quantity so it can't come or go in the context of changingangular momentum. Not sure where that leaves us. Still pondering theimplications of what I'm suggesting.>> Anyway energy is conserved. If it isn't, where're the >source/sink?> In other words, circular rotational motion can only be mechanized>> through the interaction of point masses in opposition to one another.>> Many cases to consider. Two co-orbiting objects that >aren't spinning WRT each other see each other as gravitating >masses at a constant distance.No, I'm thinking more along the lines of opposed changes in orbitalangular momenta. A planet orbiting a star for example requires each toexhibit orbital angular momenta with respect to each other. Planetaryspin wouldn't enter into orbital considerations directly.> And I think that the changes offset one another. At least this is how>> it seems to me at present and is the reason I asked to address the>> problem later.>>> The whole issue relates to the difference between constant angular>> momentum considered in purely linear contexts at constant velocity and>> cases of circular rotation. I think there has to be some analytical>> and mechanical difference between the two cases. I'm just not quite>> sure yet what that difference is and what the implications may be for>> angular mechanics in general. But I'm considering the problem.>> When you step from galactic rotation through gyroscopes >of rotation to handle. Are you trying to say they're the same?Sure. Their measures are the same in experimental terms so I don't seewhy the motion itself shouldn't be the same. We just need tounderstand why and how. Of course, we don't measure celestial angularmechanics in terms of moments of intertia and torque. But we're notreally dealing with aggregate masses either as in flywheels etc.>> Are we getting into cosmological issues here?>>>> Probably. If not right away almost certainly in the next go round. I>> don't think I'm attacking the practical implications of macro angular>> and celestial mechanics. But I do think I'm attacking the analytical>> foundation for quantum mechanical theory in pretty specific terms.>> An angular version of Mach's Principle? I don't see that >QM is going to be helpful. IIUC spin in elementary >way charge and magnetism transform; nobody (sane) thinks >electrons act like little tops.My sanity is frequently questioned. I understand that spin inangular momentum is not and that's how spin is measured. Maybe thiswill change once we get a handle on the origin of Planck's constant inanalytical terms.>> Then there're things like precession and spin-spin >quantized, but not in macro systems.We shall have to see what we shall have to see.>> Trying to think about angular motion in linear terms is >trouble. Breaking angular motion into linear bits works _if_ >you remember to refer them to an inclusive coordinate frame >_and_ to use the same coordinate frames for the bits and the >system.>>I think the best explanation I've been able to come up with is the>>circular celestial orbit with the centripetally directed gravitational>>force.>> Yes, and?>>>> Well, to me it pretty much seems to prove the case I'm making. Mainly>> because dr/dt is taken to be centripetally directed velocity which>> when combined vectorially with tangential v results in rotation of>> both vectors.>>>... I seem to recall that>>on the original Trivial Pursuit thread you remarked that the center of>>the galaxy was rotating faster than the amount of matter would>>indicate.>> Not sure that was me, but evidence seems to say that, >hence all the dark matter/energy flap.>>>> Actually, pretty much sure it was.>>> I wonder if that's true for other galaxies as well, such>>as the Andromeda.>> AFAIK yes.>>>> Good. Then I may be on the right track.>>>I've had a couple of thoughts on the subject which may>>explain the effect.>> So think out loud. You can't be any more wrong than >anyone else on the issue.>>>> I intend to. I'm just waiting for the right moment. Unfortunately, I>> have to establish one other unobvious principle before I do. I just>> wanted to make sure I'd have someone to talk it over with. This other>> principle has a well established and easily accessible phenomenon in>> experimental terms. So, I'd rather establish it before trying to apply>> the principle to galactic rotation. But it should be coming up in a>> few weeks. Nothing to do with the present subject as far as I can>> tell.>> OK, we'll see.>Let me run one other rationale by you with respect to the idea ofconstantly changing angular momentum in circular rotation.We have a cross product L = r x p. And to determine if L changes incircular rotation we analyze the time derivative of angular momentumdL/dt = dr/dt x p + r x dp/dt. At least this is how Meron analyzed theproblem and I agree with his presentation.Now L is proportional to the area between r and p. So, we have tointerpret the analysis of changes to that area in terms of dr/dt x pas reflecting time rates of change in the area according to changes inr and r x dp/dt as reflecting time rates of change in area accordingto changes in p. And the sum of both factors will reflect the overalltime rate of change to the whole area covered by L = r x p.However, if this is indeed the case, then dr/dt can only be radiallydirected. Otherwise if taken to lie along v parallel to p it cannotreflect any change in the area associated with L = r x p under anycircumstances because parallel vectors don't subtend any area.In other words, we could conceivably conclude that dr/dt is zero justas we do with dp/dt in the case of circular rotation. But we can neverconclude that dr/dt lies in the direction of v because that precludesthe possibility of any change in area related to L = r x p which iswhat the time derivative dL/dt is specifically designed to evaluate.Thus whatever magnitude we may assign to dr/dt, it has to berecognized that it is radially directed in every case in order toaccount for conceivable time rates of change in area for L = r x p.Otherwise, the analysis itself is flawed and we can conclude nothingfrom the time derivative dL/dt. === Subject: > >I'm still trying to get the post on the analytical derivation for>Planck's constant ready for early next week. So, I'm also trying to>concentrate on that.>You're just trying to drive us all crazy right? What arrogance! How >>galling!! [sputter...gasp...gulp]>>Ahem.>>You still haven't got a handle on the most basic ideas of circular >>motion (the above post is a nice summary of the extent of your >>confusion) and now you're onto an analytical derivation for Planck's >>constant?!?!>>The interesting aspect of your communications has nothing to do with >>modern physics. Alas, it is about cognitive science. Why do otherwise >>intelligent people keep on responding to you? I try to limit my own >>input but the urge is there. Sometimes it's just irresistible.>>Self control. That's what I need.>>JUST ... SAY ... NO> I know, I know. I used to experience the same thing in politics. Then> one day I learned to curb the urge. Extraordinarily frustrating but> ultimately rewarding.> I would like to say that I've sublimated politics by trolling physics> newsgroups. Just isn't the case unfortunately. Arrogance helps in the> approach but it doesn't do anything for the analysis.> As a simple bridge, let me ask in general if you consider that> everything there is to say on the subject of angular momentum and> Planck's constant has already been said. Not in detail. Just whether> you think there are things remaining to be said.> There are lots of things to be said. Start by learning a little calculus and saying: the integral of dr/dt dt is neither dr nor dr/dt. It is r.Save Planck's constant for next year.-- Joe Legris === Subject: both counts, BM does obey that property but that doesn't contradictthe theorem in question. Professor Ullrich, I like that, and I thinkit's essentially correct, but I don't think it is quite true in whatright, because when it hits an interval, say from the left, it wouldbe more likely to go next to the left than to the right. I think youcan easily fix it, either bya) instead of intervals, just look at points, orb) say Who cares if it's equally likely, any sequence still haspositive probability, the specific probability doesn't matter.and it will work out just fine. ArtflDodgr, what you have written sounds like it's correct, but whenyou say the Cameron-Martin theorem, do you mean the one that is oftencalled Girsanov's theorem? The presentation I have in my book of thattheorem is very abstract, but it seems to me that it could betranslated into saying what you use to prove the theorem. That's aslick way to do it. === Subject: > both counts, BM does obey that property but that doesn't contradict> the theorem in question. Professor Ullrich, I like that, and I think> it's essentially correct, but I don't think it is quite true in what> right, because when it hits an interval, say from the left, it would> be more likely to go next to the left than to the right. I think you> can easily fix it, either by> a) instead of intervals, just look at points, or> b) say Who cares if it's equally likely, any sequence still has> positive probability, the specific probability doesn't matter.> and it will work out just fine. > ArtflDodgr, what you have written sounds like it's correct, but when> you say the Cameron-Martin theorem, do you mean the one that is often> called Girsanov's theorem? The presentation I have in my book of that> theorem is very abstract, but it seems to me that it could be> translated into saying what you use to prove the theorem. That's a> slick way to do it.The theorem of Cameron and Martin tells us when the distribution of {B_t+f(t):0 <= t <= 1} is mutually absolutely continuous with respect to the distribution of the Brownian motion. (The (non-random ) translation function f is to be, at a minimum, continuous and f(0)=0.) Namely, absolute continuity obtains if and only if the function f is absolutely continuous and f' is square integrable on [0,1]. In the case of absolute continuity, the Radon-Nikodym derivative of the distribution of B+f with respect to that of B can be written down explicitly.Girsanov's theorem generalizes the C-M theorem by allowing f to be a random process adapted to the filtration of the Brownian motion.Here also there is a formula for the R-N derivative.-- A. === Subject: >both counts, BM does obey that property but that doesn't contradict>the theorem in question. Professor Ullrich, I like that, and I think>it's essentially correct, but I don't think it is quite true in what>right, because when it hits an interval, say from the left, it would>be more likely to go next to the left than to the right. Hmm, that's right, sorry.>I think you>can easily fix it, either by>>a) instead of intervals, just look at points, or>b) say Who cares if it's equally likely, any sequence still has>positive probability, the specific probability doesn't matter.>>and it will work out just fine. Yes.>ArtflDodgr, what you have written sounds like it's correct, but when>you say the Cameron-Martin theorem, do you mean the one that is often>called Girsanov's theorem? The presentation I have in my book of that>theorem is very abstract, but it seems to me that it could be>translated into saying what you use to prove the theorem. That's a>slick way to do it.**David C. Ullrich === Subject: Greg> both counts, BM does obey that property but that doesn't contradict> the theorem in question. Professor Ullrich, I like that, and I think> it's essentially correct, but I don't think it is quite true in what> right, because when it hits an interval, say from the left, it would> be more likely to go next to the left than to the right. I think you> can easily fix it, either by> a) instead of intervals, just look at points, or> b) say Who cares if it's equally likely, any sequence still has> positive probability, the specific probability doesn't matter.> and it will work out just fine. > ArtflDodgr, what you have written sounds like it's correct, but when> you say the Cameron-Martin theorem, do you mean the one that is often> called Girsanov's theorem? The presentation I have in my book of that> theorem is very abstract, but it seems to me that it could be> translated into saying what you use to prove the theorem. That's a> slick way to do it.> The theorem of Cameron and Martin tells us when the distribution of > {B_t+f(t):0 <= t <= 1} is mutually absolutely continuous with respect to > the distribution of the Brownian motion. (The (non-random ) translation > function f is to be, at a minimum, continuous and f(0)=0.) Namely, > absolute continuity obtains if and only if the function f is absolutely > continuous and f' is square integrable on [0,1]. In the case of > absolute continuity, the Radon-Nikodym derivative of the distribution of > B+f with respect to that of B can be written down explicitly.> Girsanov's theorem generalizes the C-M theorem by allowing f to be a > random process adapted to the filtration of the Brownian motion.> Here also there is a formula for the R-N derivative. === Subject: I am trying to find the area of the intersection of 2 sphere caps, moreprecisely:Consider a sphere of radius one. Consider two circles on the surfaceof the sphere drawn using a compass with radius set at r. What is thearea of the overlapping region of the two circles in terms of r andthe angle between the centers of the two circles ?Best,Aslan === Subject: > I am trying to find the area of the intersection of 2 sphere caps, more> precisely:>> Consider a sphere of radius one. Consider two circles on the surface> of the sphere drawn using a compass with radius set at r. What is the> area of the overlapping region of the two circles in terms of r and> the angle between the centers of the two circles ?> Best,> AslanThe area on the sphere of one circle with radius r is pi*r^2.This is valid for all r between 0 and 2. You won't be able touse a compass to draw a circle with r greater than sqrt(2)though. Unless you draw it from within the sphere ;-)Note that the angle between the center of the circle and apoint of it is given by acos(1-r^2/2).If the angle a between the centers is 0, then the overlappingregion obviously has area pi*r^2.If the angle a exceeds 2*acos(1-r^2/2) and is less than pi,then the overlapping area is 0.That's all I can come up with... I doubt if the area can beneatly expressed in terms of r and a.Dirk Vdm === Subject: > I am trying to find the area of the intersection of 2 sphere caps, more> precisely:>> Consider a sphere of radius one. Consider two circles on the surface> of the sphere drawn using a compass with radius set at r. What is the> area of the overlapping region of the two circles in terms of r and> the angle between the centers of the two circles ?> Best,> Aslan[Solution]Let denote geodesic radius of circle a = arcsin r/2.The rhombus BCD formed by centers of circles B,Dand intersection points A,C has side a anddiagonal d, which is the angle between the centers of the two circles.The area of overlapping region is twice the area of (spherical) circularsegment with central angle B and radius a, which is a differencebetween area sector ABC and isosceles triangle ABC with angles A,B,C=A:Ssegment = Ssector - Striangle.The side b of the triangle is given by Pythagorean theoremfor isosceles triangle with known height d/2:cos(b/2) cos (d/2) = cos a.Since all sides of ABC are known at this point all theangles can be calculated using cosine theorem:cos a = cos a cos b + sin a sin b cos A,cos b = cos a cos a + sin a sin a cos B,from which angles A, B, C=A can be found.The area of triangle ABC is excessive angle:Striangle = A+B+C-pi.The area of circle is 2pi(1-cos a). The area of sector isSsector =B(1-cos a). === Subject: > I am trying to find the area of the intersection of 2 sphere caps, more> precisely:>> Consider a sphere of radius one. Consider two circles on the surface> of the sphere drawn using a compass with radius set at r. What is the> area of the overlapping region of the two circles in terms of r and> the angle between the centers of the two circles ?> Best,> AslanI set this up as an integral. If you locate the center of theintersection region at the point (0,0,1), then the projection of the twocircles to the x-y plane becomes two ellipses, whose equations can befound in terms of r and the angle between the two segments that go fromthe origin to the centers of the circles. The area is the integral of theappropriate integrand (area of a surface) over this intersection ofellipses. The integral turns out to be a horrible mess that Mathematicacan't do. === Subject: >Can we find (a,b,x,z,u,v,c) in (N-{0})^7 such that:>a^2+b^2=c^2>x^2+y^2=c^2>u^2+v^2=c^2>ab+xy=uvI don't know. But I can tell you a few things to try.1. Dividing through everywhere by c, your equations are equivalent tothe question, are there three points on the (open) first quadrant ofthe unit circle, (a', b'), (x', y'), (u',v'), whose coordinates are_rational_ and for which a'b' + x'y' = u'v'. 2. Since the circle can be parameterized as the set of points(2t/(1+t^2), (1-t^2)/(1+t^2)), this simply asks whether there arethree rational numbers t1 t2 t3 in ( 0 , 1 ) for which t1(1-t1^2)/(1+t1^2)^2 + t2(1-t2^2)/(1+t2^2)^2 = t3(1-t3^2)/(1+t3^2)^2That's just _one_ equation in three unknowns.3. You can add some symmetry at any stage here: for example, if t4 = -t3,this last version asks whether there are three rational numbers t_i,two in (0,1) and one in (-1,0), with t1(1-t1^2)/(1+t1^2)^2 + t2(1-t2^2)/(1+t2^2)^2 + t4(1-t4^2)/(1+t4^2)^2 = 04. Clearing denominators, we can obtain an equivalent _polynomial_ equationin the t_i which we wish to solve in rational numbers. That equationis symmetric in the t_i so we can express it in terms ofS = t1 + t2 + t4, T = t1 t2 + t2 t4 + t4 t1, and U = t1 t2 t4.The polynomial which is to vanish I make out to be-U(U^2T-9U^2-T^3+3T^2-11T+9)+(-11U^2-5U^2T-T^3-5T^2+5T+1)S +U(5T+3)S^2+(T-1)S^3As it turns out, this is a rational (i.e. parameterizable) surface: youcan express all the solutions by taking arbitrary values of the parametersa and b in the substitution S = -(2*b^2+b^2*a^2-4+a^2)/(-2+b^2*a^2+a^2)/b T = (6+b^2*a^2+a^2-4*b^2*a-4*a)/(-2+b^2*a^2+a^2) U = -(2*b^2+b^2*a^2+4-4*b^2*a-4*a+a^2)/(-2+b^2*a^2+a^2)/bI have to caution, though, that there has been some information lost here:if you had good rational t1, t2, and t4, then you could find good rationalS,T,U, which correspond to some rational a,b. Conversely, from any rational a,b, you can get rational S,T,U, but thenthe t_i you seek will be the solutions of the cubic X^3 - S X^2 + T X - Uand there is no guarantee that these will be rational.(In algebraic-geometric terms, I've taken a surface modulo an actionof Sym(3), getting a rational surface; here, rational points map to rational points but not all rational points have rational pre-images.)Clearing denominators, it seems that all you need to do is to findtwo rational numbers a, b so that this cubic has three rational roots: 3 2 2 2 2 2 2 2 2X (b a + a - 2) b + (b a + a + 2 b - 4) X 2 2 2 2 2 2 2 2 2 + b (b a - 4 b a + a - 4 a + 6) X + (b a - 4 b a + 2 b + a - 4 a + 4)Good luck... I'm not even sure I can find choices for a and bwhich give this polynomial even ONE root in (0,1) !davePS -- someone commented that this isn't Arithmetic as claimed inSerre's book, A course in arithmetic... === Subject: I'm not familiar with (N-{0})^7... can you explain?Also, can you give more information on the background to this problem?Studying the problem domain may yield some clues on how to solve it.[snip]> Can we find (a,b,x,z,u,v,c) in (N-{0})^7 such that:> a^2+b^2=c^2> x^2+y^2=c^2> u^2+v^2=c^2> ab+xy=uv> ?> I.R.>> That's not arithmetic! That's a diophantine equation you've got there!> I.J.R. === Subject: >Clearing denominators, it seems that all you need to do is to find>two rational numbers a, b so that this cubic has three rational roots:> 3 2 2 2 2 2 2 2 2>X (b a + a - 2) b + (b a + a + 2 b - 4) X> 2 2 2 2 2 2 2 2 2> + b (b a - 4 b a + a - 4 a + 6) X + (b a - 4 b a + 2 b + a - 4 a + 4)>Good luck... I'm not even sure I can find choices for a and b>which give this polynomial even ONE root in (0,1) !With a=11/5, b=-1, for example, I get the three rational roots -1/2, 1/2 and 1. Of course that's no help for the original problem.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: 5b46a63e.0307091654.4097c298@posting.google.com... > I'm not familiar with (N-{0})^7... can you explain?Sorry, I'm french and my english is very bad.N-{0} = set of non-zero integers.I.J.RI.J.R. === Subject: ? I'm not familiar with (N-{0})^7... can you explain?? Also, can you give more information on the background to this problem?? Studying the problem domain may yield some clues on how to solve it.? ? [snip]? ? ? ? Can we find (a,b,x,z,u,v,c) in (N-{0})^7 such that:? ? ? ? a^2+b^2=c^2? ? ? ? x^2+y^2=c^2? ? ? ? u^2+v^2=c^2? ? ? ? ab+xy=uv? ? ? ? ?? ? ? ?? ? ? ?? ? ? ? I.R.? ? ?? ? ? That's not arithmetic! That's a diophantine equation you've got there!? ? ? ? ? ? I.J.R.My interpretation of the OP's problem is: Find 3 Pythagorean triangles withthe same hypotenuse c such that the area sum of the two smaller trianglesequals the area of the largest triangle.Searching up to side lengths up to 10000 I've found no solutions to thecorresponding Diophantine equation.The best relative approximation for DR = 3*abs (a*b+x*y-u*v)/(a*b+x*y+u*v)is produced bya=2175,b=1540,x=2664,y=73,u=1943,v=1824,(c=2665) which produce a relativearea mismatch of DR=2.5395*10^(-5).The best absolut approximation DA = abs (a*b+x*y-u*v) comes froma=x=63, b=y=16, u=52, v=39 (c=65) which gives DA=12More material on Pythagorean triples:http://mathworld.wolfram.com/PythagoreanTriple.html Hugo Pfoertner === Subject: proposed.--Julien Santini,CMI Technop.99le de Ch.89teau-Gombert, FranceHome page: http://www.analgebra.com === Subject: > My interpretation of the OP's problem is: Find 3 Pythagorean triangles with> the same hypotenuse c such that the area sum of the two smaller triangles> equals the area of the largest triangle.> Searching up to side lengths up to 10000 I've found no solutions to the> corresponding Diophantine equation.> The best relative approximation for DR = 3*abs (a*b+x*y-u*v)/(a*b+x*y+u*v)> is produced by> a=2175,b=1540,x=2664,y=73,u=1943,v=1824,(c=2665) which produce a relative> area mismatch of DR=2.5395*10^(-5).> The best absolut approximation DA = abs (a*b+x*y-u*v) comes from> a=x=63, b=y=16, u=52, v=39 (c=65) which gives DA=12> More material on Pythagorean triples:> http://mathworld.wolfram.com/PythagoreanTriple.html> Hugo PfoertnerYes. I agree with HP's interpretation. I searched for the triangleswithhypotenuses < 24000 but I found no solution. My best solution was withhypotenuse = 3965, the difference in areas is only = 1. L. Rodriguez === Subject: >Can we find (a,b,x,z,u,v,c) in (N-{0})^7 such that:>a^2+b^2=c^2>x^2+y^2=c^2>u^2+v^2=c^2>ab+xy=uv Yesterday I responded (inconclusively) to this thread.Today I'd like to approach this a little differently -- again inconclusively, but I've got some goodies to show off.What I observed yesterday amounts to this: we are trying to see whetherit is possible for two superspecial numbers to sum to another superspecial number. Here a number is superspecial if it is of the form t (1-t^2) / (1+t^2)^2for some rational t between 0 and 1 -- equivalently, if it ishalf the area of a rational right triangle with hypotenuse 1.Note that the mirror image of such a triangle is again such a triangle,obviously of the same area. Algebraically this becomes the observationthat t and t'=(1-t)/(1+t) lead to the same superspecial number.Using a bit of Galois theory we discover that the superspecial numbercan be written in the form (1-s)/s^2 where s = (1+t^2)/(1+t).(You don't need to know this, but s is the sum of t and t'; theirproduct is 1 - s.) This s will be rational if t is rational, andbetween 2(sqrt(2)-1) and 1 if t is between 0 and 1.So here's my new suggestion: call a number special (for s) if itis of the form (1-s)/s^2 for some rational s in the intervalnoted above. Then every superspecial number is special, but notevery special number is superspecial. (It turns out to be superspecialiff s^2 + 4 s - 4 is a rational square.) These special numbersare more common than superspecial numbers, and a little easier tomanipulate algebraically. They're also very easy to characterize inother ways; for example x is special iff 1 + 4 x is a rational squarein the appropriate interval. (Interestingly, the negative of a specialnumber need not be special -- it is, iff the number is superspecial.)Now the original problem was to solve A + B = C with threesuperspecial numbers. We can pose intermediate problems of varyinglevels of difficulty: (i) solve A + B = C with A, B, C being (merely) special(ii) solve A + B = C with A, B, C special and A superspecial(iii) solve A + B = C with A, B, C special and A and B superspecial(iv) solve A + B = C with A, B, C all being superspecial.Problem (iv) is the original one.It isn't hard to solve (i); in fact the set of all solutions may beparameterized -- we may take A to be any special number (special for s,say) and then B is special for some s' iff s' is of the forms' = 2*s*(1-r)*(1+r-s)/(s-2)^2/r for some r. (In that case C will bespecial for s'' = 2*s*(1-r)*(1+r-s)/(r*s^2-2*s^2-2*r^2-2+4*s) .)Since I can take _any_ special number for A, above, I can also find examples of (ii) by choosing A to be superspecial. For instance, 504/4225 + 11/100 = 155/676is an example, corresponding to s=65/72, s=10/11, and s=26/31respectively. Of these, 504/4225 is superspecial: (65/72)^2 + 4(65/72) - 4 is the square of 47/72. (The values of t and t' are 1/8 and 7/9.)Going one step further, suppose I choose s = (1+t^2)/(1+t) so thatA is superspecial. The condition that B also be superspecial nowbecomes the requirement that a certain polynomial in t and r bea square. (Details attached below.) As it turns out this polynomial is a quartic in r, so for fixed t I can analyze this as an elliptic curve.It turns out to have rank 2 for generic t (and torsion (Z/2)^2), meaning that there are plenty of rational points for each t (i.e. for each choice of A we have plenty of choices for r, and thus for B. For example, we can always use r = 1/2*(t^2-1)/t .) Now, we're supposedto have t in (0,1) so I just tried t = 1/2. Since we're also supposedto have s' and s'' in certain intervals, this requires r to be in a very tight interval, but there is a density theorem which guaranteeseventual success, and sure enough I looked around on the ellipticcurve and found that using r = -1591892743787576881/3999194642119499400 led to a rational solution in the right range. This gives me a solution to problem (iii) : if A=6/25 B= 131879416328454525807489724734855492459937375768041853050996110 660048190/ 267678993236039083458766684521097947172078441390134863621380201 90033150921 C= 163904381349834813220447253831030155614745499228281964499103023 906700110276/ 669197483090097708646916711302744867930196103475337159053450504 750828773025then A + B = C A and B are superspecial and C is special (but not superspecial).I haven't found an example of a solution to (iv). In an exactlyanalogous way, this requires that we find values of t and r forwhich _two_ quartics in r are squares.To tie in with yesterday's remarks, let me note again that we arelooking for nontrivial rational points on an algebraic surface X.Yesterday I noted that a quotient X/Sym(3) was a rational surface,and in particular has plenty of rational points. Today I am notingthat a different quotient X/Sym(2)^3 is rational, as is an intermediatequotient X/Sym(2)^2; a higher intermediate quotient X/Sym(2) is anelliptic surface of rank 2 and in particular has many rational points.Still unknown is whether the original surface X has any rational points.daveLong post-script follows...In case you'd like to play with these things yourself, I am attachingsome of the computations from Maple. When solving problem (iii), the problem is to choose t and r so thata certain quartic is a square, which I can write in the formt^2*(t-1)^2*(1+t^2)^2-4*t*(t-1)*(1+t^2)*(t^4-2*t^3+2*t^2 +2*t+1)*r +(2*t^8-2*t^3-8*t^2+44*t^4+2-8*t^6-2*t+2*t^5+2*t^7)*r^2 -4*(1+t)*(1+t^2)*(t^4-2*t^3+2*t^2+2*t+1)*r^3+(1+t)^2*(1+t^2)^2 *r^4 = U^2 :I can convert this to Weierstrass normal form using the substitutions{r = 1/2*(4*t^8-24*t^7+48*t^6-24*t^5+2*X*t^4-40*t^4+24*t^3-4*X*t^3+ Y*t^2+4* X*t^2+48*t^2-2*Y*t+24*t+4*X*t+4-Y+2*X)/(1+t^2)/(1+t)/X, U = -1/4*(t^2-2*t-1)^2*(16+128*t+416*t^2+640*t^3+416*t^10+16*t^12- 640*t^9 -128*t^11+240*t^4-320*t^6-512*t^5+512*t^7+240*t^8-8*Y+8*Y*t^6- 32*Y*t^5 +40*Y*t^4-40*Y*t^2-32*Y*t-8*X^2+Y^2-2*X^3+20*X^2*t^3-20*X^2*t -16*X^2*t^2 -8*X^2*t^4)/(1+t^2)/(1+t)/X^2 } :and its inverse {X = 2*(-1-2*r-9*t-20*t^2+U-9*t^3-2*t^7*r-2*t^5*r+2*t^6*r-6*r*t^3+2 *r^2*t^5 +r^2*t^6+2*r^2*t+3*r^2*t^2+4*r^2*t^3+3*r^2*t^4-6*r*t^4+26*t^4- 20*t^6 +9*t^5+9*t^7-t^8+r^2+U*t+U*t^2+U*t^3-6*r*t-10*r*t^2)/(t^2-2*t- 1)^2, Y = 4*(r+t+t^2-U+t^3+t^7*r+t^8*r+t^5*r-4*t^6*r-r*t^3+2*r^3*t^5+r^3 *t^6 +2*r^3*t+3*r^3*t^2+4*r^3*t^3+3*r^3*t^4-3*r^2*t^5+3*r^2*t^6-3*t ^7*r^2 -9*r^2*t-15*r^2*t^2-9*r^2*t^3-9*r^2*t^4+22*r*t^4+r^3-3*t^4-5*t ^6 +3*t^5+3*t^7-t^8-3*r^2+U*r*t^3+U*r*t+U*r*t^2+U*r-2*U*t-2*U*t^2 +2*U*t^3 -U*t^4-r*t-4*r*t^2)*(1+t)*(1+t^2)/(t^2-2*t-1)^3} :The resulting equation is simply Y^2 = (X+2*(t^2-2*t-1)^2)*(X+2+4*t+4*t^2-4*t^3+2*t^4)*(X+4+8*t+8*t^ 2-8*t^3+4*t^4) :It's clear that for any t this curve has three torsion elements oforder 2, those with Y=0 and X = each of -2*(t^2-2*t-1)^2, -4*t^2-2*t^4-4*t+4*t^3-2, 8*t^3-4-8*t-8*t^2-4*t^4There are also some other easily-picked-out elements which, forthe values of t I tried, tend to give two independent generators on thecurve. (The curve can have larger rank for some t.) The generators I foundare (X,Y) = [-4*t*(t^3-2*t^2+t+2), -4*(1+t^2)*(1+t)*(t-1)^2 ] [ 8*t^3-4*t^2-8*t-4 , -4*(1+t^2)*(1+t)^2*(t-1)*t] You can then generate all the additional points you'd like on the curve,using the chord-and-tangent process: if (X1,Y1) and (X2,Y2) are on thecurve, their sum is (X,Y) = [(32+256*t+832*t^2+480*t^4+1280*t^3+480*t^8-640*t^6-1024*t^5 +1024*t^7+832*t^10+32*t^12-1280*t^9-256*t^11-104*t^5*X2+20* t^8*X2+20*X1+20*X2+32*X2*t^2*X1-40*X2*t^3*X1+16*X2*t^4*X1+ 40*X2*t*X1+X1^2*X2+X2^2*X1+16*X1*X2-2*Y1*Y2+104*X1*t+208*X1 *t^2+104*X1*t^3-136*X1*t^4+20*t^8*X1+208*t^6*X1-104*t^5*X1-104 *t^7*X1-104*t^7*X2+208*X2*t^2+104*X2*t^3-136*X2*t^4+104*X2* t+208*t^6*X2)/(X1-X2)^2,-(624*Y1*X2*t^2-104*Y2*X2*t+136*Y2 *X2*t^4-104*Y2*X2*t^3-208*Y2*X2*t^2+312*Y1*X2*t-408*Y1*X2*t ^4+312*Y1*X2*t^3-20*Y2*t^8*X2+104*Y2*t^7*X2-208*Y2*t^6*X2+40* Y1*X2^2*t+3*Y1*X2^2*X1+40*Y2*X1^2*t^3-960*Y2*t^4-20*Y2*X2- 136*Y1*X1*t^4+104*Y1*X1*t^3+208*Y1*X1*t^2+104*Y1*X1*t-512* Y2*t-1664*Y2*t^2+960*Y1*t^4-2560*Y2*t^3+2560*Y1*t^3+60*Y1* X2+16*Y1*X1*X2+408*Y2*X1*t^4-312*Y2*X1*t^3-624*Y2*X1*t^2-312 *Y2*X1*t+512*Y1*t+1664*Y1*t^2+60*Y1*t^8*X2-312*Y1*t^5*X2-312* Y1*t^7*X2+624*Y1*t^6*X2-60*Y2*X1-40*Y1*X2*t^3*X1+16*Y1*X2*t ^4*X1+40*Y1*X2*t*X1-32*Y2*X2*t^2*X1+40*Y2*X2*t^3*X1-16*Y2* X2*t^4*X1-40*Y2*X2*t*X1+32*Y1*X2*t^2*X1+64*Y1-64*Y2+32*Y1* X2^2*t^2-40*Y1*X2^2*t^3+16*Y1*X2^2*t^4-3*Y2*X1^2*X2-16*Y2*X1* X2-40*Y2*X1^2*t-32*Y2*X1^2*t^2-16*Y2*X1^2*t^4+20*Y1*X1+Y1* X2^3-Y2*X1^3-16*Y2*X1^2+16*Y1*X2^2+960*Y1*t^8-1280*Y1*t^6- 2048*Y1*t^5+2048*Y1*t^7+1664*Y1*t^10+64*Y1*t^12-2560*Y1*t^9 -512*Y1*t^11+20*Y1*t^8*X1+208*Y1*t^6*X1-104*Y1*t^5*X1-104*Y1*t ^7*X1-60*Y2*t^8*X1-624*Y2*t^6*X1+312*Y2*t^5*X1+312*Y2*t^7* X1-960*Y2*t^8+1280*Y2*t^6+2048*Y2*t^5-2048*Y2*t^7-1664*Y2*t^ 10-64*Y2*t^12+2560*Y2*t^9+512*Y2*t^11+104*Y2*t^5*X2)/(X1-X2 )^3 ]:(Warning: iterating this process with the generators I gave abovewill NOT eventually produce all rational points on this surface.)As far as I can tell, small combinations of the generators won't allowus to get all three of s, s', and s'' in the right intervals untilwe take something as complicated as (gen#1)+3(gen#2)+(torsion#3),and even then only if t is between roughly 15/16 and 73/74 !That element on the elliptic curve has r = -8*(t-1)*(1+t)*(3*t^2+2*t+1)*(1+t^2)*(t^4+6*t^3-2*t-1)* (7*t^8+24*t^7+4*t^6+40*t^5+50*t^4+8*t^3-12*t^2-8*t-1)*t^2/ (t^12-136*t^11-490*t^10-664*t^9+15*t^8+624*t^7+500*t^6+144*t^5 +15*t^4 +24*t^3+22*t^2+8*t+1)/(t^8-16*t^7-28*t^6+38*t^4+16*t^3+4*t^2+ 1)The condition for C to be superspecial as well reduces to the conditionthat this expression be square: t^2*(t-1)^2*(t^2+2*t-1)^2-4*t*(t-1)*(1+t^2)^3*r+2*(t^4-t^3+2*t ^2+t+1)* (t^2+2*t-1)^2*r^2-4*(1+t)*(1+t^2)^3*r^3+(1+t)^2*(t^2+2*t-1)^2 *r^4:Using, for example, the preceding expression for r, we only need tofind a rational t (in the interval (15/16, 73/74) ) for which a certain integer polynomial of degree 84 (!) is a rational square...(There are known to be only finitely many -- and probably none!) === Subject: > My interpretation of the OP's problem is: Find 3 Pythagorean triangles with> the same hypotenuse c such that the area sum of the two smaller triangles> equals the area of the largest triangle.> Searching up to side lengths up to 10000 I've found no solutions to the> corresponding Diophantine equation.> The best relative approximation for DR = 3*abs (a*b+x*y-u*v)/(a*b+x*y+u*v)> is produced by> > a=2175,b=1540,x=2664,y=73,u=1943,v=1824,(c=2665) which produce a relative> area mismatch of DR=2.5395*10^(-5).> The best absolut approximation DA = abs (a*b+x*y-u*v) comes from> a=x=63, b=y=16, u=52, v=39 (c=65) which gives DA=12> More material on Pythagorean triples:> http://mathworld.wolfram.com/PythagoreanTriple.html> Hugo Pfoertner> Yes. I agree with HP's interpretation. I searched for the triangles> with> hypotenuses < 24000 but I found no solution. My best solution was with> hypotenuse = 3965, the difference in areas is only = 1. L. RodriguezI can't reproduce your c=3965 result. c^2=15721225, which is expressible in13 different ways as the sum of 2 non-zero squares. The best combination ofthese 13 solutions is (715*3900)+(715*3900)=2*2788500=5577000 compared with(1525*3660)=5581500, so a*b+x*y-u*v=-4500.What you probably computed was c^2=3965 which is not a square. c^2=3965would indeed give your solution (11*62+22*59-43*46)/2=1 === Subject: Greetins,All users of newsgroup should find themselves extremely lucky, for what isto be revealed to them. Something FASCINATING on similar lines as forconstants ofnature, as we know them e.g. Pi=3.14, h, e, G, etcIf one considers that life on earth is based on Carbon (atomic number 6),then one can alsoassume that some hidden messages that determine human destiny can also beexplored by using some kind of decoding method based on 6.Let us say that each letter of English language ascend in value of +6, aarithmetical progression. for example A=6, B=12, C=18 etc and Z=156.Then if any words that have historical relevance, and have their sum addup to the 3 digit number 222, 444, 666, may help to decipher somehidden messages.For example:222= the state of the world economy.academia, egghead, GNP, hell, India, seal,444 Four=360 Time Lord tells us earthly storycensor, cross, crumble, dazzle , demerit, embroil , energy , English ,erratic , error, Godson ,Heinrich, hexagon, homicidal, Jesus , Jewish ,Joshua (6th Book),juridic, jury , landmark, London, Lucifer, messiah, nuclear, occult, opium,vixen, weapon.666= The ultimate ways of communication, including supernatural, and thefuture Human Robots to come.chartroom, communicable, computer, incoherent , indecipherable, orangutan,Sanskrit, secretarial, witchcraft, workspace.and note the integration over all the numbers, for letters A to Z is = 2025(Spooky?). (f(x)=6X is continuous function ( for x, 1,26)and also note sum of all the numbers allocated to letters A to Z = 2106(pass).222,444,666 S1,S2,Pxyz === Subject: Hey,What's the difference in linear algebra between an isomorphism and a lineartransformation? Is a linear transformation an example of an isomorphism?Could you please maybe give a basic example of both?Drew C. === Subject: > Hey,> What's the difference in linear algebra between an isomorphism and a linear> transformation? Is a linear transformation an example of an isomorphism?> Could you please maybe give a basic example of both?> An isomorphism is by definition a bijection. A linear transformation need not be a bijection. For example f:R->R given by f(x) = 0 is a linear tranformation but not an isomorphism. === Subject: An isomorphism takes you to a place that works the same as where you are.http://mathworld.wolfram.com/Isomorphism.html === Subject: > Hey,> What's the difference in linear algebra between an isomorphism and a linear> transformation? Is a linear transformation an example of an isomorphism?It's the other way around; an isomorphism is a particular kind oflinear transformation. An isomorphism is a linear transformation whichis one-to-one and onto.F(x,y,z) = (x + y, y + z, z - x) is a linear transformation from R^3 toR^3 but it is not an isomorphism because F(4,3,2) = F(5,2,3) i.e. it'snot one-to-one ( not onto either).G(x,y,z) = (x + y, y + z, z + x) _is_ an isomorphism. [G(x,y,z) = G(a,b,c) implies (x,y,z) = (a,b,c) and for any (a,b,c),G(x,y,z) = (a,b,c) has a solution for (x,y,z).]-- Paul SperryColumbia, SC (USA) === Subject: My wife and I were having this talk the night:In Star Trek of course they go from 0 to wrap 5 in the blink of an eye. Buteven if we could go the speed of light it would take awhile to reach thatspeed without becoming gue on the back wall.My question is if one had a space ship that could go the speed of light. Howlong, with a constant rate of acceleration, would it take to reach thatspeed assuming that inside our fictish space space we wanted to simulate thegavity at sea level here on earth?Also what would the equation be? === Subject: > My wife and I were having this talk the night:> In Star Trek of course they go from 0 to wrap 5 in the blink of an eye. But> even if we could go the speed of light it would take awhile to reach that> speed without becoming gue on the back wall.> My question is if one had a space ship that could go the speed of light. How> long, with a constant rate of acceleration, would it take to reach that> speed assuming that inside our fictish space space we wanted to simulate the> gavity at sea level here on earth?Infinitely long. You would never reach the speed of light.A more meaningful question would be, How long would it take to reach 90per cent of the speed of light?-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: > My wife and I were having this talk the night:>> In Star Trek of course they go from 0 to wrap 5 in the blink of an eye.But> even if we could go the speed of light it would take awhile to reachthat> speed without becoming gue on the back wall.>> My question is if one had a space ship that could go the speed of light.How> long, with a constant rate of acceleration, would it take to reach that> speed assuming that inside our fictish space space we wanted to simulatethe> gavity at sea level here on earth?>> Infinitely long. You would never reach the speed of light.>> A more meaningful question would be, How long would it take to reach 90> per cent of the speed of light?> --> Dave Seaman> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.> OK I'll go with 90% === Subject: >> My wife and I were having this talk the night:>> In Star Trek of course they go from 0 to wrap 5 in the blink of an eye.> But>> even if we could go the speed of light it would take awhile to reach> that>> speed without becoming gue on the back wall.>> My question is if one had a space ship that could go the speed of light.> How>> long, with a constant rate of acceleration, would it take to reach that>> speed assuming that inside our fictish space space we wanted to simulate> the>> gavity at sea level here on earth?>> Infinitely long. You would never reach the speed of light.>> A more meaningful question would be, How long would it take to reach 90>> per cent of the speed of light?> OK I'll go with 90%A Google search on Lorentz contraction turned up the page.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: >> My wife and I were having this talk the night:>> In Star Trek of course they go from 0 to wrap 5 in the blink of an eye.> But> even if we could go the speed of light it would take awhile to reach> that> speed without becoming gue on the back wall.>> My question is if one had a space ship that could go the speed of light.> How> long, with a constant rate of acceleration, would it take to reach that> > speed assuming that inside our fictish space space we wanted to simulate> the> gavity at sea level here on earth?>> Infinitely long. You would never reach the speed of light.>> A more meaningful question would be, How long would it take to reach 90> per cent of the speed of light?> --> Dave Seaman> Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.> OK I'll go with 90%With constant proper acceleration g the speed as seen bythe initial rest frame is v(t) = g*t / sqrt[1+(g*t/c)^2]so t = v/(g*sqrt[1-v^2/c^2])So knowing g and c you can calculate the time it takes to reach vDirk Vdm === Subject: All numbers are approximate.Speed of light = c = 300000 meters/second.Gravitational acceleration of earth = a = 9.8 meters/second/second.Time required to accelerate from 0 to c at a constant rate a (usingclassical physics; special relativity doesn't permit going this fast)= c/a = 30600 seconds = 510 minutes = 8.5 hours. === Subject: > All numbers are approximate.>> Speed of light = c = 300000 meters/second.300,000,000 meters/second>> Gravitational acceleration of earth = a = 9.8 meters/second/second.>> Time required to accelerate from 0 to c at a constant rate a (using> classical physics; special relativity doesn't permit going this fast)> = c/a = 30600 seconds = 510 minutes = 8.5 hours.Special relativity has v(t) = g*t / sqrt[1+(g*t/c)^2]so t = v/(g*sqrt[1-v^2/c^2])for v=0.9c, you get t = 63,141,945 seconds = 731 days = 2 years.Dirk Vdm === Subject: New address:http://my.tbaytel.net/forslund/index.htmlEnjoy- Bob === Subject: DavidTo answer your phone question:Q: Is Bohm's ontology equivalent to many-worlds in sense of David Deutsch's Fabric of Reality?No definitely not. Whilst you can think of all of the Bohm pilot qubit (Antony Valentini) for orthodox micro-quantum theory, you cannot do so in general as in the macro-quantum ODLRO case out of sub-quantal equilibrium avoiding heat death.Micro-quantum theory is one-way IT FROM BIT nonlocal action of wave on signal locality applying to Born statistical ensembles in which each |psi> = Sigma(i) |i>This breaks down completely in a BEC with macro-quantum ODLRO coherent superfluid signal, although it holds for the residual micro-quantum normal fluid noise, which in the virtual off-mass shell case is the zero point vacuum fluctuation.Note there are unitary frame shift to different total experimental arrangements at the micro-level though not at the macro-level.With macro-quantum ODLRO you have two-way IT FROM BIT + BIT FROM IT allowing signal nonlocality in sub-quantal non-equilibrium non-classical macro-systems. Curved space-time as c-number local Diff(4) tensor field is an example of a non-classical macro-quantum system with dark energy/matter exotic vacuum regions. === Subject: Lets say, I have the following stock return data:Stock Return === Subject: A -4B -2C -1D 3E 5Now, I want to assign weights to each stock based on the returns such thatthe stock with the highest return gets the highest weight and the stock withthe lowest return gets the smallest weight. All stocks must be assigned a+ve weight and the weights must sum to 1.If I re-scale the above range (-4 to 5) to (0 to 1) using the formula:Weight = (Return - 4)/(5-(-4)) --> where the denominator is simply the rangeI basically, get weights with values between 0 and 1. Also, the distancebetween the weights will be in line with the actual return data. But theseweghts do not sum to 1!How would I re-scale (stretch) this data so that the weights sum to 1?Jay === Subject: > Lets say, I have the following stock return data:> Stock Return> === Subject: > A -4> B -2> C -1> D 3> E 5> > Now, I want to assign weights to each stock based on the returns such that> the stock with the highest return gets the highest weight and the stock with> the lowest return gets the smallest weight. All stocks must be assigned a> +ve weight and the weights must sum to 1.> If I re-scale the above range (-4 to 5) to (0 to 1) using the formula:> Weight = (Return - 4)/(5-(-4)) --> where the denominator is simply the range> I basically, get weights with values between 0 and 1. Also, the distance> between the weights will be in line with the actual return data. But these> weghts do not sum to 1!> How would I re-scale (stretch) this data so that the weights sum to 1?> JayYou could always do: new_weight = old_weight / sum( old_weights)But some remarks. Whatever you have in mind, a search with Googleon portfolio theory / portfolio selection might help you to findnotes & lectures for free (Elton/Gruber 'Modern Portfolio Theory'is a standard). The other: try to look for some intro to statistics.One idea in that context is: on the long run you have some 'overallreturn' ( = risk free) of let us say 3% and a bell shaped curve forthe returns (take it as logarithms naturalLog(old_price/new_price),not as (old-new)/new ). Now place your stocks according to quantils(like gratings at school). This gives you some static ( ~ concurrentview) ranking and you are free to add more stocks to compare.To put it into greater context you might wish to look at variationof data (for which you only have historical except you know the future),since returns change over time. For that 'Sharpe ratio' is a keyword,one tries to compare risk versus returns. Good luck :-) === Subject: > If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?Perhaps I should not have insulted you in a huff. Any man who triesto teach himself calculus from books at a stage of life when most arecontent to watch re-runs of Barney Miller is to be praised, whateverthe outcome.However, it's probably not going to be easy.First of all, re. abuse of notation: differentials, i.e., things wewrite as dt, dx and etc. are almost fictions. Almost means weare on shaky ground, and I am hedging my bets. ;-)One definitely does _not_ take the limit as dt --> 0. dt already_is_ zero, for most purposes. If this doesn't make much sense it'sbecause, as I said, differentials themselves are a kind of fiction, tobe treated with caution -- they are both shrunk to zero, but treatedas numbers. And if we know what we are doing, we can get away withthis.The differential shorthand is for the following: In the limit asdelta t goes to zero, (delta v)/(delta t) goes to the limit dv/dt,the derivative of v with respect to t. (The deltas are written aslittle equilateral triangles, flat foot down, sometimes seen in asciiapproximation as / or /_).Now, the notation dv/dt is playing both sides of the fence: it wantsto claim in some way that it's a gestalt, an indivisible whole ofnotation. Of course, it claims, one shouldn't make anything of theapparent use of a numerator and denominator -- the whole thing issimply a number, or a function.Yeah right.dv/dt == (fully equivalent to or defined as) v'(t) or thederivative of v with respect to t suggests we consider v'(t) as theratio of little infinitesimals dv and dt, and in fact, for manypurposes, we can manipulate these infinitesimals and not writecomplete nonsense -- thermodynamics, in the usual treatment, makes awhole career of this. But we're always free to run and hide and saywe didn't really mean that! who told you you could treat dv and dtas numbers! The horror!But let's be mathematical purists. Let's say v is a function of t. Just to keep things clear, let's lable the function itself differentlyfrom its output, say f(.) . is just a placeholder for the argument(the thing replacing the placeholder, of course :-). Now we can writev = f(t), which may be read v equals f of t or v is a function oft, or we can simply call the function f, if we think we are capableof remembering that it has an argument.Now, if we *differentiate* v with respect to t, we get _another_function of t. We could call this other function whatever we liked,such as g(.) or Master Sam, but tradition suggests we label itf'(.), or simply f' . These symbols are to be read (thefunction) f prime (of t). The prime notation is just a reminderwhere we got this new function.Now, if f' is the derivative of f with respect to t, what is anantiderivative of f'?(Jeopardy music here).Er... what is ... f ???That's right! Ding, ding, ding!!! Correct.The _function_ f (of t) is an antiderivative of f' (of t).Wowee, wowee, wowee!The politest thing to be said about your opening statement is itdoens't make sense. So let me say it: It doesn't make sense.Have you perhaps picked up why we pedantically insist on a ratherthan the antiderivative? === Subject: > If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?>>Perhaps I should not have insulted you in a huff. Any man who tries>to teach himself calculus from books at a stage of life when most are>content to watch re-runs of Barney Miller is to be praised, whatever>the outcome.>>However, it's probably not going to be easy.>>First of all, re. abuse of notation: differentials, i.e., things we>write as dt, dx and etc. are almost fictions. Almost means we>are on shaky ground, and I am hedging my bets. ;-)>>One definitely does _not_ take the limit as dt --> 0. dt already>_is_ zero, for most purposes. If this doesn't make much sense it's>because, as I said, differentials themselves are a kind of fiction, to>be treated with caution -- they are both shrunk to zero, but treated>as numbers. And if we know what we are doing, we can get away with>this.>>The differential shorthand is for the following: In the limit as>delta t goes to zero, (delta v)/(delta t) goes to the limit dv/dt,>the derivative of v with respect to t. (The deltas are written as>little equilateral triangles, flat foot down, sometimes seen in ascii>approximation as / or /_).>>Now, the notation dv/dt is playing both sides of the fence: it wants>to claim in some way that it's a gestalt, an indivisible whole of>notation. Of course, it claims, one shouldn't make anything of the>apparent use of a numerator and denominator -- the whole thing is>simply a number, or a function.>>Yeah right.>>dv/dt == (fully equivalent to or defined as) v'(t) or the>derivative of v with respect to t suggests we consider v'(t) as the>ratio of little infinitesimals dv and dt, and in fact, for many>purposes, we can manipulate these infinitesimals and not write>complete nonsense -- thermodynamics, in the usual treatment, makes a>whole career of this. But we're always free to run and hide and say>we didn't really mean that! who told you you could treat dv and dt>as numbers! The horror!>>But let's be mathematical purists. Let's say v is a function of t. >Just to keep things clear, let's lable the function itself differently>from its output, say f(.) . is just a placeholder for the argument>(the thing replacing the placeholder, of course :-). Now we can write>v = f(t), which may be read v equals f of t or v is a function of>t, or we can simply call the function f, if we think we are capable>of remembering that it has an argument.>>Now, if we *differentiate* v with respect to t, we get _another_>function of t. We could call this other function whatever we liked,>such as g(.) or Master Sam, but tradition suggests we label it>f'(.), or simply f' . These symbols are to be read (the>function) f prime (of t). The prime notation is just a reminder>where we got this new function.>>Now, if f' is the derivative of f with respect to t, what is an>antiderivative of f'?>>(Jeopardy music here).>>Er... what is ... f ???>>That's right! Ding, ding, ding!!! Correct.>>The _function_ f (of t) is an antiderivative of f' (of t).>>Wowee, wowee, wowee!>>The politest thing to be said about your opening statement is it>doens't make sense. So let me say it: It doesn't make sense.>>Have you perhaps picked up why we pedantically insist on a rather>than the antiderivative?Well, if your initial apology is sincere, certainly I accept it.However, I should preface my remarks by saying that I was taughtcalculus at the U.S. Naval Academy, Annapolis MD over the course ofseveral years concluding in 1966. We seemed to have the same coursematerial given us in different guises from skinny to steam to weapons.When I used the term antiderivative I was really looking for a name togive a concept: the instantaneous or infinitesimal time integral of afunction like a. I don't recall that we were ever taught this.Probably it's just v although the subject has gotten pretty thoroughlymuddled in the course of posting.My general contention with respect to circular motion is that ifradial a is present then so is radial v of definitional necessitybecause a represents the time rate of change in v. And if radial v ispresent the actual change in radial v produced in an infinitesimaltime is itself infinitesimal. And if radial v is present the actualchange in radial r produced in an infinitesimal time is itselfinfinitesimal.However, this means that changes to radial v produced by radial a arenot zero but just infinitesimal. And similarly that changes to radialr are not zero but are just infinitesimal.This is all I'm trying to establish. I doubt you agree with me, but Icertainly don't mind discussing the matter in civil terms. Iunderstand what you outline above and am not trying to reifyinfinitesimal calculus in radical revisionist terms. The issue I'mtrying to address is the application of thoroughly conventionalcalculus to angular momentum in circular rotation. === Subject: > Well, if your initial apology is sincere, certainly I accept it.> However, I should preface my remarks by saying that I was taught> calculus at the U.S. Naval Academy, Annapolis MD over the course of> several years concluding in 1966.At the Navy Nuke School in Orlando, our math instructorshowed us a graph that was composed of line segmentsand he showed us how to calculate the area under the curveby dividing it up into rectangles and triangles.Then I said My recruiter told me I'd be learning calculus. Is this it?Our math instuctor chuckled and repliedThat's pretty funny. My recruiter told me I'd be *teaching* calculus.-- pete === Subject: [...]| First of all, re. abuse of notation: differentials, i.e., things we| write as dt, dx and etc. are almost fictions. Almost means we| are on shaky ground, and I am hedging my bets. ;-)Either this is a fine show of careful thinking on your part, or youknow something like the following, and probably both.Yes, ordinarily in a calculus course the students are advised not tothink of dy/dx as being a ratio between two things, dy and dx, butjust to think of the whole expression as an idiom, a piece of notationwhose parts aren't supposed to have meaning by themselves.Of course, the notation is based informally on the idea of thinkingof dx as being an infinitesimally small change in x, and dy as beinga corresponding infinitesimally small change in y. I suppose somestudents have found this helpful. I think Calculus Made Simple sayssomething like this. When you get to integral calculus, the integralsign was originally a tall S for sum, and the integral of f(x) dxis thought of in a similarly informal way as a sum of all the littleinfinitesimal changes in x, multiplied by f(x).I know a way of defining dx and the like so that they're nolonger fictions. In the development of differential forms, onedefines df for a function f on a manifold to be essentiallythe gradient of f. The ratio between two such gradients is thederivative in the case of a one-dimensional manifold and isn'tdefined when it's more than one-dimensional.But in the one-dimensional case, the gradient is just the derivativewith respect to whichever coordinate is being used (which transformswhen you change coordinates). So really it doesn't spare anyone fromhaving to figure out derivatives first.I've heard that calculus has been developed in infinitesimalanalysis, where the real numbers are generalized in such a wayas to include infinitesimal quantities, values r where r>0 butN*r < 1 for any positive integer N. There, I guess they'vemanaged to introduce dy and dx as being infinitesimal changesin y and x in a way that's closer to the original idea, butsome care still is needed to consider dy/dx as being the actualderivative, and not just something that's infinitesimally closeto the derivative.Keith Ramsay === Subject: >> Well, if your initial apology is sincere, certainly I accept it.>> However, I should preface my remarks by saying that I was taught>> calculus at the U.S. Naval Academy, Annapolis MD over the course of>> several years concluding in 1966.>>At the Navy Nuke School in Orlando, our math instructor>showed us a graph that was composed of line segments>and he showed us how to calculate the area under the curve>by dividing it up into rectangles and triangles.>Then I said >My recruiter told me I'd be learning calculus. Is this it?>Our math instuctor chuckled and replied>That's pretty funny. My recruiter told me I'd be *teaching* calculus.>The way the Navy was run in those days - and probably still is - ifthey said you'd be learning calculus I'm surprised you didn't wind upstudying animal husbandry. === Subject: >When I used the term antiderivative I was really looking for a name to>give a concept: the instantaneous or infinitesimal time integral of a>function like a. I don't recall that we were ever taught this.You're right, you were never taught this, and for a very good reason: thereis no such thing.-- Best Wishes,Wolf Kirchmeir, Blind River ONNot that brains are everything --you'll also need a skull to put them in. (Nancy Franklin, 1997) === Subject: >[...]>| First of all, re. abuse of notation: differentials, i.e., things we>| write as dt, dx and etc. are almost fictions. Almost means we>| are on shaky ground, and I am hedging my bets. ;-)>>Either this is a fine show of careful thinking on your part, or you>know something like the following, and probably both.>>Yes, ordinarily in a calculus course the students are advised not to>think of dy/dx as being a ratio between two things, dy and dx, but>just to think of the whole expression as an idiom, a piece of notation>whose parts aren't supposed to have meaning by themselves.>>Of course, the notation is based informally on the idea of thinking>of dx as being an infinitesimally small change in x, and dy as being>a corresponding infinitesimally small change in y. I suppose some>students have found this helpful. I think Calculus Made Simple says>something like this. When you get to integral calculus, the integral>sign was originally a tall S for sum, and the integral of f(x) dx>is thought of in a similarly informal way as a sum of all the little>infinitesimal changes in x, multiplied by f(x).>>I know a way of defining dx and the like so that they're no>longer fictions. In the development of differential forms, one>defines df for a function f on a manifold to be essentially>the gradient of f. The ratio between two such gradients is the>derivative in the case of a one-dimensional manifold and isn't>defined when it's more than one-dimensional.>>But in the one-dimensional case, the gradient is just the derivative>with respect to whichever coordinate is being used (which transforms>when you change coordinates). So really it doesn't spare anyone from>having to figure out derivatives first.>>I've heard that calculus has been developed in infinitesimal>analysis, where the real numbers are generalized in such a way>as to include infinitesimal quantities, values r where r>0 but>N*r < 1 for any positive integer N. There, I guess they've>managed to introduce dy and dx as being infinitesimal changes>in y and x in a way that's closer to the original idea, but>some care still is needed to consider dy/dx as being the actual>derivative, and not just something that's infinitesimally close>to the derivative.>If you don't mind some inexpert speculation, I'd like to run thefollowing by you.I understand the quasi fictitious nature of elements like dr and dt.But I think the problem comes from trying to analyze such elements inisolation away from the gradient. They're real in the sense ofrepresenting infinitesimal physical increments. You just can't judgethe nature of the increment without the gradient.However, I think one can judge direction for an element like drwithout knowing the gradient apart from direction. And I think anexpression like dr/dt is an actual ratio. But I don't think that youevaluate the ratio correctly without the gradient. In other words youaren't just taking some dr and dividing it by some dt to get dr/dt.The nature of the ratio can only be determined in combinationaccording to its gradient.But that doesn't mean that the elements themselves are not physicallythere. As far as I can tell they are just infinitesimal incrementswhich behave in combination according to some specific ratio andgradient that describes their interaction in relation to one another.By the way, I appreciate your comments on the calculus ininfinitesimal analysis. I think it's a lot closer to what I have inmind. === Subject: >When I used the term antiderivative I was really looking for a name to>>give a concept: the instantaneous or infinitesimal time integral of a>>function like a. I don't recall that we were ever taught this.>>You're right, you were never taught this, and for a very good reason: there>is no such thing.>>Well, we shall have to see what we shall have to see. I'm not sure howyou can know this for a fact. But there's no point to crossing wordson the issue. Keith Ramsay offers some insight and I offer a fewcomments as well. === Subject: > >When I used the term antiderivative I was really looking for a name to>give a concept: the instantaneous or infinitesimal time integral of a>function like a. I don't recall that we were ever taught this.>>You're right, you were never taught this, and for a very good reason: there>>is no such thing.> Well, we shall have to see what we shall have to see. I'm not sure how> you can know this for a fact. But there's no point to crossing words> on the issue. Keith Ramsay offers some insight and I offer a few> comments as well.> And good luck to Keith. I'm starting my stop-watch ...... now.-- Joe Legris === Subject: >When I used the term antiderivative I was really looking for a name to>give a concept: the instantaneous or infinitesimal time integral of a>function like a. I don't recall that we were ever taught this.> You're right, you were never taught this, and for a very good reason: there> is no such thing.Well, if Mr. Zick means something like dv = a dt, then there issomething like what he is describing, and that is it. If you reallywanted to name this differential I suppose you could call it aninfinitesimial time integral, although its friends probably prefera dt or dv.But under no whatsoever circumstances should it be called anantiderivative: at least not if the person concerned wished tocommunicate with anybody else.Here -- let's use the ordinary names in a dialogue:dx: Hey dv, how's it going?dv: Oh you know, same a dt, differential day(They laugh -- infinitesimals are satisfied with infinitesimal jokes). === Subject: > I know a way of defining dx and the like so that they're no> longer fictions. In the development of differential forms, one> defines df for a function f on a manifold to be essentially> the gradient of f. The ratio between two such gradients is the> derivative in the case of a one-dimensional manifold and isn't> defined when it's more than one-dimensional.Yeah ... isn't that normally written with a capital D?That notation bothers me, since it does indeed look like adifferential (in print), but really describes a different order object(a gradient or derivative) -- unless I am missing some nuance. Itlooks as confusing as hell to me -- equations seem to match apples andoranges.Somebody should have slapped this out of the first person to use itbefore it became popular. > But in the one-dimensional case, the gradient is just the derivative> with respect to whichever coordinate is being used (which transforms> when you change coordinates). So really it doesn't spare anyone from> having to figure out derivatives first.> I've heard that calculus has been developed in infinitesimal> analysis, where the real numbers are generalized in such a way> as to include infinitesimal quantities, values r where r>0 but> N*r < 1 for any positive integer N. Cute.> There, I guess they've> managed to introduce dy and dx as being infinitesimal changes> in y and x in a way that's closer to the original idea, but> some care still is needed to consider dy/dx as being the actual> derivative, and not just something that's infinitesimally close> to the derivative.Well ... someting like that was about what I expected when I hedged:maybe in the back of my mind I was thinking of the Dirac delta;another useful object one had to make some obeisance towards themathematicians in using (of course they say this is nonsense), withthe expectation that it could or would be cleaned up eventually -- sothat physicists could go on thinking about it in the same way asalways, but assured that somebody, somewhere, had cleaned it up. :-). === Subject: > I know a way of defining dx and the like so that they're no> longer fictions. In the development of differential forms, one> defines df for a function f on a manifold to be essentially> the gradient of f. The ratio between two such gradients is the> derivative in the case of a one-dimensional manifold and isn't> defined when it's more than one-dimensional.> Yeah ... isn't that normally written with a capital D?> That notation bothers me, since it does indeed look like a> differential (in print), but really describes a different order object> (a gradient or derivative) -- unless I am missing some nuance. It> looks as confusing as hell to me -- equations seem to match apples and> oranges.> Somebody should have slapped this out of the first person to use it> before it became popular.> > But in the one-dimensional case, the gradient is just the derivative> with respect to whichever coordinate is being used (which transforms> when you change coordinates). So really it doesn't spare anyone from> having to figure out derivatives first.>> I've heard that calculus has been developed in infinitesimal> analysis, where the real numbers are generalized in such a way> as to include infinitesimal quantities, values r where r>0 but> N*r < 1 for any positive integer N.> Cute.> There, I guess they've> managed to introduce dy and dx as being infinitesimal changes> in y and x in a way that's closer to the original idea, but> some care still is needed to consider dy/dx as being the actual> derivative, and not just something that's infinitesimally close> to the derivative.> Well ... someting like that was about what I expected when I hedged:> maybe in the back of my mind I was thinking of the Dirac delta;> another useful object one had to make some obeisance towards the> mathematicians in using (of course they say this is nonsense), with> the expectation that it could or would be cleaned up eventually -- so> that physicists could go on thinking about it in the same way as> always, but assured that somebody, somewhere, had cleaned it up. :-).The matter in question is covered by the elementary (but hard to provein full generality) mean value theorem which gives meaning todifferentials is some cases but not in others. That is, differentialshave meaning in a setting of derivatives but not in integration. Inintegration, the dx is not a differential but rather a reminder thatthe measure is the identity function and that the variable ofintegration is x.Your last paragraph brings in a distribution which is completelyunrelated. Distributions are indeed interesting but different rulesapply. The definition of the derivative of a distribution is quitedifferent from the definition of the derivative of a function. This canlead to errors in manipulating distributions because signs are oftenwrong. The Dirac distribution at zero already is enough to demonstratethe type of error in minipulation I am talking about. Actually, if youread Dirac, he had a way to avoid troublesome errors in hisalso used distributions but did not trust them to give correct answers(Green's function solutions and the like).Chuck-- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@webaccess.net === Subject: > I've heard that calculus has been developed in infinitesimal> analysis, where the real numbers are generalized in such a way> as to include infinitesimal quantities, values r where r>0 but> N*r < 1 for any positive integer N. There, I guess they've> managed to introduce dy and dx as being infinitesimal changes> in y and x in a way that's closer to the original idea, but> some care still is needed to consider dy/dx as being the actual> derivative, and not just something that's infinitesimally close> to the derivative.> Keith RamsayYes, you have heard correctly. Although there are a lot of details,the basic setup is that the real line is extended to an extendedreal line. There are infinitesimals, infinitely large numbers (thereciprocals of NON-ZERO infinitesimals) and sums of these. Everyfinite number is the sum--uniquely--of an infinitesimal and anordinary real number, the latter called its ordinary part. Now, df/dxis the ordinary part of (f(x+h) - f(x))/h for an infintesimal hPROVIDED that ordinary part does not depend on h. For instance, iff(x) = x^2, then the difference quotient is ((x+h)^2 - x^2)/h, whichworks out by ordinary algebra to 2x + h. The ordinary part is 2x andthis does not depend on h, so that is the derivative.Lest anyone think that this is just a triviality, there is a lot oflogical trickery involved in showing that ANY function from the realline to itself can be extended to the extended real line. This meansall those trig functions and exponentials and logarithms and all that. So from sin(x+h) = sin x cos h + sin h cos x, together with cos h hasordinary part 1 (which has to be proved!) and (sin h)/h, the ratio oftwo infinitesimals, has ordinary part 1, you infer that the derivativeof sin x is cos x.Note that df and dx are still not well defined. You can think of dxas standing for an infinitesimal if you like and df = f(x+dx) - f(x),but dx has to be a variable ranging over ALL infinitesimals and thederivative is STILL not the quotient, but only the ordinary partthereof.In the same spirit, you can define the Riemann integral by taking aninfinite integer n (this is an element of the extended integers) and apartition of an interval into n pieces each of infinitesimal lengthand form the obvious Riemann sum and if its ordinary part does notdepend n, the partition or the point taken out of each interval thenthat is the integral. I could go through what you would do tocalculate the integral of x^2 but it would look extremely similar towhat you do in ordinary calculus and ultimately depend on the formulafor the sum of the first n squares. Until you had the fundamentaltheorem.One place this really pays dividends is the definition of continuity. f is continuous at x iff the ordinary part of f(x+h) is f(x) for everyinfinitesimal h. No epsilons or deltas, all that has been subsumedinto the logical trickery I mentioned at the beginning. === Subject: > However, I think one can judge direction for an element like dr> without knowing the gradient apart from direction. And I think an> expression like dr/dt is an actual ratio. But I don't think that you> evaluate the ratio correctly without the gradient. In other words you> aren't just taking some dr and dividing it by some dt to get dr/dt.> The nature of the ratio can only be determined in combination> according to its gradient.> But that doesn't mean that the elements themselves are not physically> there. As far as I can tell they are just infinitesimal increments> which behave in combination according to some specific ratio and> gradient that describes their interaction in relation to one another.> By the way, I appreciate your comments on the calculus in> infinitesimal analysis. I think it's a lot closer to what I have in> mind.How about if we just do finite arithmetic?If r(t) represents a vector which is changing in time,then the velocity vector at time t is, to very goodapproximation, [r(t+h)-r(t)]/h, where h is a very smallbut finite increment, say a nanosecond.Since r(t) is a vector, r(t+h)-r(t) is a vector. If youwant to know the direction of v, then it is thedirection of this difference vector. OK?Now about your infinitesimal integration I think you'rejust trying to say that you can get r(t+h) by doing this: r(t+h) = r(t) + v*hv gives you a direction of change. Multiply it by thesmall time interval h, add that vector to r(t) andyou get the position r(t+h) at time t+h.Is that what you're trying to say? If so, it's a fairlystandard naive technique for doing time evolution of systemswith computers (and frought with numerical error).Going WAY back in the thread(s), if v is directed radially,in the same direction as r(t), then that would tell youthat r(t+h) is also in the same direction. In otheroutward during that nanosecond. Clearly that doesnot correspond to what is happening in circular motion.That's why it doesn't make sense to take v as beingradial. It's not radial. That's not the direction inwhich vector r is changing. - Randy === Subject: >>>>When I used the term antiderivative I was really looking for a name to>>give a concept: the instantaneous or infinitesimal time integral of a>>function like a. I don't recall that we were ever taught this.>>You're right, you were never taught this, and for a very good reason: there>is no such thing.>>> Well, we shall have to see what we shall have to see. I'm not sure how>> you can know this for a fact. But there's no point to crossing words>> on the issue. Keith Ramsay offers some insight and I offer a few>> comments as well.>>>>And good luck to Keith. I'm starting my stop-watch ...... now.>>-- I hope it has a split second hand. === Subject: >>>When I used the term antiderivative I was really looking for a name to>> >give a concept: the instantaneous or infinitesimal time integral of a>>function like a. I don't recall that we were ever taught this.>>> You're right, you were never taught this, and for a very good reason: there>> is no such thing.>Well, if Mr. Zick means something like dv = a dt, then there is>something like what he is describing, and that is it. If you really>wanted to name this differential I suppose you could call it an>infinitesimial time integral, although its friends probably prefer>a dt or dv.You know, this is exactly what I suggested and then got sidetracked.>>But under no whatsoever circumstances should it be called an>antiderivative: at least not if the person concerned wished to>communicate with anybody else.My problem and my apology. I think what you say directly above is whatI meant. Unfortunately I started searching and plucking ideas out ofdistant memories. I appreciate the contribution.>>Here -- let's use the ordinary names in a dialogue:>>dx: Hey dv, how's it going?>>dv: Oh you know, same a dt, differential day>>(They laugh -- infinitesimals are satisfied with infinitesimal jokes).Perhaps, we should call them dj's.Once again, thanks. === Subject: >> However, I think one can judge direction for an element like dr>> without knowing the gradient apart from direction. And I think an>> expression like dr/dt is an actual ratio. But I don't think that you>> evaluate the ratio correctly without the gradient. In other words you>> aren't just taking some dr and dividing it by some dt to get dr/dt.>> The nature of the ratio can only be determined in combination>> according to its gradient.>> >> But that doesn't mean that the elements themselves are not physically>> there. As far as I can tell they are just infinitesimal increments>> which behave in combination according to some specific ratio and>> gradient that describes their interaction in relation to one another.>> >> By the way, I appreciate your comments on the calculus in>> infinitesimal analysis. I think it's a lot closer to what I have in>> mind.>>How about if we just do finite arithmetic?>>If r(t) represents a vector which is changing in time,>then the velocity vector at time t is, to very good>approximation, [r(t+h)-r(t)]/h, where h is a very small>but finite increment, say a nanosecond.>>Since r(t) is a vector, r(t+h)-r(t) is a vector. If you>want to know the direction of v, then it is the>direction of this difference vector. OK?Nay OK. :-)Please forgive the cuteness but I've just been raked over the coalsany number of times for trying to separate various components ofdifferentials.All you're really saying here is that the direction of dr/dt isdetermined by the direction of dr. This is what has been saidrepeatedly as the rationale for deciding the direction of dr/dt. Butit is incorrect not only physically but mathematically for the timederivative of L = r x p.It doesn't matter whether we're doing finite or infinitesimalarithmetic. The time derivative of L = r x p is dL/dt = dr/dt x p + rx dp/dt. In other words we are evaluating the time rate of change inangular momentum L = r x p as the time rate of change of r cross p inthe direction of r plus r cross the time rate of change in p in thedirection of p.There really isn't any choice here because this is what the derivativeof a cross product requires. It's a matter of definition. dr/dt has tolie in the direction of r in order to satisfy the definition of thetime derivative of the cross product L = r x p.If one determines that dr/dt is zero, that is one thing. But if wedon't take dr/dt in the direction of r then we aren't evaluating thederivative of the cross product L = r x p.>>Now about your infinitesimal integration I think you're>just trying to say that you can get r(t+h) by doing this:>> r(t+h) = r(t) + v*h>>v gives you a direction of change. Multiply it by the>small time interval h, add that vector to r(t) and>you get the position r(t+h) at time t+h.>>Is that what you're trying to say? If so, it's a fairly>standard naive technique for doing time evolution of systems>with computers (and frought with numerical error).It's a little hard to tell for sure. I'm talking about taking radialdr/dt and combining it vectorially with tangential v to produce asuccessive tangential v.I regard r as simply determined by the combination of radial dr/dt andtangential v. There is no change to r except as is determined by thesetwo factors. So I don't see any modification to r in itself by v ordr/dt or anything else. It is just whatever distance there is betweenm and the center of rotation as determined by radial dr/dt in relationto tangential v.>>Going WAY back in the thread(s), if v is directed radially,>in the same direction as r(t), then that would tell you>that r(t+h) is also in the same direction. In other>outward during that nanosecond. Clearly that does>not correspond to what is happening in circular motion.>That's why it doesn't make sense to take v as being>radial. It's not radial. That's not the direction in>which vector r is changing.>We've disagreed on this from the beginning. Or rather everyone hasdisagreed with me. Let me run this by you:The cross product L = r x p and is proportional to an area defined byr on one side and p on the other. Thus the time rate of change in Ldepends on the time rate of change of the area defined by r x p. Thistime rate of change depends on the rate of change in r in thedirection of r cross p plus r cross the time rate of change in p.I certainly appreciate what you are suggesting. But the time rate ofchange in L requires that we add the changes in area due to any changein r to changes in area due to any change in p. That's a matter ofdefinition. And to do that we can only consider dr/dt in the directionof r. Otherwise we are not evaluating the time derivative of the crossproduct L = r x p. === Subject: >> If a = dv/dt as dt --> 0 would not the antiderivative of a dt be dv?> Perhaps I should not have insulted you in a huff. Any man who tries> to teach himself calculus from books at a stage of life when most are> content to watch re-runs of Barney Miller is to be praised, whatever> the outcome.> However, it's probably not going to be easy.I knew it. The books LIE! It says right here, _Calculus Made Easy_. Hrrrrmph. I got stuck at chapter 9.Anyone want to offer any suggestions for other useful books? I already got _Essential Calculus with Applications_ by Silverman. I haven't tried learning anything from it yet, so I don't know if I actually need more books, but I bet I do._Calculus for Cats_ sounds fun... :)-Laurel === Subject: >>How about if we just do finite arithmetic?>>If r(t) represents a vector which is changing in time,>>then the velocity vector at time t is, to very good>>approximation, [r(t+h)-r(t)]/h, where h is a very small>>but finite increment, say a nanosecond.>>Since r(t) is a vector, r(t+h)-r(t) is a vector. If you>>want to know the direction of v, then it is the>>direction of this difference vector. OK?>>Nay OK. :-)>>Please forgive the cuteness but I've just been raked over the coals>any number of times for trying to separate various components of>differentials.That's why I got rid of the differentials. These are no longerdifferentials. The differences are finite. You could measure them withreal live lab equipment. No more abstracts or reasoning about limits.Finite, finite, finite. When I say h is a nanosecond, I'm not talkingabout differentials anymore, I'm talking about a nanosecond. You candisplay a nanosecond on an oscilloscope. OK?Finite, finite, finite. No differentials. Finite.>All you're really saying here is that the direction of dr/dt is>determined by the direction of dr.Yes. But I'm going to FINITE quantities to reason about it withoutgetting distracted by abstract differentials. Finite. Measurable. Notinfinitesimal. Finite.ASCII doesn't let us show it, but delta-r (not dr, these are finite.Measurable. Not infinitesimal. Finite.) is the vector part ofdelta-r/delta-t.> This is what has been said>repeatedly as the rationale for deciding the direction of dr/dt. But>it is incorrect not only physically but mathematically for the time>derivative of L = r x p.Proof by assertion?>It doesn't matter whether we're doing finite or infinitesimal>arithmetic. The time derivative of L = r x p is dL/dt = dr/dt x p + r>x dp/dt. In other words we are evaluating the time rate of change in>angular momentum L = r x p as the time rate of change of r cross p in>the direction of r plus r cross the time rate of change in p in the>direction of p.Yes. That is correct.>There really isn't any choice here because this is what the derivative>of a cross product requires. It's a matter of definition. dr/dt has to>lie in the direction of rThat is incorrect. Why do you make this leap?> in order to satisfy the definition of the>time derivative of the cross product L = r x p.No, the time derivative of L is, as you say r x dp/dt + dr/dt x p. Allof those vectors lie in whichever direction they lie in. Where do youget the idea that it is requiring dr/dt to lie in some direction?>If one determines that dr/dt is zero, that is one thing. But if we>don't take dr/dt in the direction of r then we aren't evaluating the>derivative of the cross product L = r x p.What the hell are you talking about?Let's look at a circular counterclockwise motion in the N,S,E,W plane.At the moment the position vector is to the East, the velocity andmomentum vectors are to the North. Here are the directions of the fourvectors in question:r: Eastdr/dt: Northp: Northdp/dt: Westr and dp/dt are 180 degrees apart. Their cross product is zero.p and dr/dt are parallel. Their cross product is zero.Thus dL/dt is a sum of two vector products which are both zero. It isequal to zero.Now where the hell do you see a requirement on the direction of dr/dtrelative to r? dr/dt is what it is. Its cross product with p iswhatever it is. dp/dt is what it is. Its cross product with r is whatit is. The vectors will work out to what they work out to, and thatwill tell you the direction and magnitude of L and dL/dt. - Randy === Subject: Just noticed some extra verbiage here which shows there's still somemajor confusion somewhere.>It doesn't matter whether we're doing finite or infinitesimal>arithmetic. The time derivative of L = r x p is dL/dt = dr/dt x p + r>x dp/dt. In other words we are evaluating the time rate of change in>angular momentum L = r x p as the time rate of change of r cross p in>the direction of rdr/dt x p is the cross product of (dr/dt) which is a vector, and p,which is another vector. Where are you getting p in the direction ofr from? A cross product is defined for any two vectors. There'snothing about p in the direction of r or cross product in thedirection of r in it.> plus r cross the time rate of change in p in the>direction of p.Again, this in the direction of p is some meaningless extra verbiageyou have added. I can take the cross product between any two vectorsno matter what their direction. There's nothing requiring any of thevectors to be in a certain direction. - Randy === Subject: >> I know a way of defining dx and the like so that they're no>> longer fictions. In the development of differential forms, one>> defines df for a function f on a manifold to be essentially>> the gradient of f. The ratio between two such gradients is the>> derivative in the case of a one-dimensional manifold and isn't>> defined when it's more than one-dimensional.>>Yeah ... isn't that normally written with a capital D?>>That notation bothers me, since it does indeed look like a>differential (in print), but really describes a different order object>(a gradient or derivative) -- unless I am missing some nuance. It>looks as confusing as hell to me -- equations seem to match apples and>oranges.>>Somebody should have slapped this out of the first person to use it>before it became popular.>> But in the one-dimensional case, the gradient is just the derivative>> with respect to whichever coordinate is being used (which transforms>> when you change coordinates). So really it doesn't spare anyone from>> having to figure out derivatives first.>>> I've heard that calculus has been developed in infinitesimal>> analysis, where the real numbers are generalized in such a way>> as to include infinitesimal quantities, values r where r>0 but>> N*r < 1 for any positive integer N. >>Cute.> There, I guess they've>> managed to introduce dy and dx as being infinitesimal changes>> in y and x in a way that's closer to the original idea, but>> some care still is needed to consider dy/dx as being the actual>> derivative, and not just something that's infinitesimally close>> to the derivative.>>Well ... someting like that was about what I expected when I hedged:>maybe in the back of my mind I was thinking of the Dirac delta;>another useful object one had to make some obeisance towards the>mathematicians in using (of course they say this is nonsense), with>the expectation that it could or would be cleaned up eventually -- so>that physicists could go on thinking about it in the same way as>always, but assured that somebody, somewhere, had cleaned it up. :-).I wonder if you could tell me what the product of two colinear vectorsis? If v and r lie in the same direction, is their product a vector,not a vector, or some other thing? === Subject: >Just noticed some extra verbiage here which shows there's still some>major confusion somewhere.>It doesn't matter whether we're doing finite or infinitesimal>>arithmetic. The time derivative of L = r x p is dL/dt = dr/dt x p + r>>x dp/dt. In other words we are evaluating the time rate of change in>>angular momentum L = r x p as the time rate of change of r cross p in>>the direction of r>>dr/dt x p is the cross product of (dr/dt) which is a vector, and p,>which is another vector. Where are you getting p in the direction of>r from? A cross product is defined for any two vectors. There's>nothing about p in the direction of r or cross product in the>direction of r in it.> plus r cross the time rate of change in p in the>>direction of p.>>Again, this in the direction of p is some meaningless extra verbiage>you have added. I can take the cross product between any two vectors>no matter what their direction. There's nothing requiring any of the>vectors to be in a certain direction.>This is poorly phrased and I apologize.For dL/dt = dr/dt x p + r x dp/dt I'm referring to [(dr/dt in thedirection of r) cross p] plus [r cross (dp/dt in the direction of p)]. === Subject: >I wonder if you could tell me what the product of two colinear vectors>is? If v and r lie in the same direction, is their product a vector,>not a vector, or some other thing?A cross product is a vector by definition, last time I studied this, anyhow.Has something changed? A scalar multiplication of a vector is also a vector,BTW.The value and direction is computed, not defined. Ie, the cross product oftwo vectors of the same value will vary depending on their direction relativeto each other.HTH-- Best Wishes,Wolf Kirchmeir, Blind River ONNot that brains are everything --you'll also need a skull to put them in. (Nancy Franklin, 1997) === Subject: > >Just noticed some extra verbiage here which shows there's still some>major confusion somewhere.>It doesn't matter whether we're doing finite or infinitesimal> >>arithmetic. The time derivative of L = r x p is dL/dt = dr/dt x p + r>>x dp/dt. In other words we are evaluating the time rate of change in>>angular momentum L = r x p as the time rate of change of r cross p in>>the direction of r>>dr/dt x p is the cross product of (dr/dt) which is a vector, and p,>which is another vector. Where are you getting p in the direction of>r from? A cross product is defined for any two vectors. There's>nothing about p in the direction of r or cross product in the> >direction of r in it.> plus r cross the time rate of change in p in the>>direction of p.>>Again, this in the direction of p is some meaningless extra verbiage> >you have added. I can take the cross product between any two vectors>no matter what their direction. There's nothing requiring any of the>vectors to be in a certain direction.>> This is poorly phrased and I apologize.> > For dL/dt = dr/dt x p + r x dp/dt I'm referring to [(dr/dt in the> direction of r) cross p]And again, that's not what it means. It's dr/dt cross p, notdr/dt in the direction of r whatever that means.You take dr/dt, which is a vector, which falls in somedirection. In circular motion, that direction is not thesame direction as r.You form the cross product of this vector with the vectorp. You can form the cross product between any two vectors.There's no direction requirement. Given two vectors offixed magnitude, their cross product is greatest whenthey are at right angles, and zero if they are parallelor anti-parallel.Where are you getting this in the direction of r stuff?(dr/dt) x p means the cross product of the vector calleddr/dt and the vector called p. There's no in the directionof r there.> plus [r cross (dp/dt in the direction of p)].Same comments here.dp/dt is a vector. You can take the cross product of rwith that vector, as you can take the cross product betweenany two vectors.In the example I gave, the direction of r was East,the direction of p was North, the direction of dp/dtwas West. The cross product r x dp/dt is thecross product of an East-oriented vector, r, anda West-oriented vector, dp/dt which is most certainlynot in the direction (North) of p. - Randy === Subject: Randy, I've taken the liberty of deleting the body of this postbecause I've come up with a new line of reasoning that I hope will bepersuasive or at least address the points you raise.In circular rotation L = r x p and dL/dt = dr/dt x p + r x dp/dt.Now, we notice that r and p are not really constant. They are onlyconstant in magnitude. Thus I maintain that what we are trying toevaluate in the context of dL/dt are time rates of change in magnitudefor each of the vectors.Changes in magnitude for r are reflected in dr/dt lying along r. Andchanges in magnitude for p are reflected in dp/dt lying along p. Thisis all I can suggest as to my analytical rationale for the directionof dr/dt.The time rate of change for p is zero because its magnitude does notchange. However, the time rate of change in magnitude for r doeschange according to the infinitesimal integral of dr/dt in thedirection of r.This is the reason I also insist on the use of infinitesimals. Thetime rate of change in L is not finite. It is infinitesimal accordingto the change in r. L does not change by any finite amount over time.It only changes infintessimally as a reflection in the time rate ofchange in the magnitude of r.And finally, if you wish to evaluate dr/dt in the conventional way, Ithink you also have to extend the same analysis to dp/dt. In otherwords both rotate and the evaluation of dr/dt in the direction of pimplies the evaluation of dp/dt in the direction of r by extension.Thus no change to L would be manifested by changes to the magnitudesof r or p because dr/dt would lie in the direction of p and dp/dtwould lie in the direction of r. === Subject: [snip]> In the example I gave, the direction of r was East,> the direction of p was North, the direction of dp/dt> was West. The cross product r x dp/dt is the> cross product of an East-oriented vector, r, and> a West-oriented vector, dp/dt which is most certainly> not in the direction (North) of p.> - RandyDon't forget the Zick Uncertainty Principle: for every concept you nail down there's another one that comes unglued.-- Joe Legris === Subject: Dear friends, I am new to news group, so forgive me to ask this question. As weknow, maths question usually includes some equations, if it is simple,we can post it to news group in ordinary text, but if the equation iscomplicated, I don't know how to post it to news group. what do you usually do in this case? At first, I made it as a pdf file and posted my message as a link,but nobody give me hint, I guess perhaps I made people angry. Any hint?Best to all of youLi. === Subject: > I am new to news group, so forgive me to ask this question. As we> know, maths question usually includes some equations, if it is simple,> we can post it to news group in ordinary text, but if the equation is> complicated, I don't know how to post it to news group.>> what do you usually do in this case?Make do with ascii or if you must, simple use of TeX.> At first, I made it as a pdf file and posted my message as a link,> but nobody give me hint, I guess perhaps I made people angry.>I give links second priority, downloading pdf files last priority.> Any hint?> http://www.math-atlas.org/collection/how-to-readthat discusses writing ascii math and also the benefitsof TeX which addresses your concerns in a wider context.---- === Subject: > Dear friends,> I am new to news group, so forgive me to ask this question. As we> know, maths question usually includes some equations, if it is simple,> we can post it to news group in ordinary text, but if the equation is> complicated, I don't know how to post it to news group.> what do you usually do in this case? As you've been told, ASCII is by far preferable, with alittle bit of syntax borrowed from TeX (^ for superscript,_ for subscript, and generous use of parentheses andbrackets).> > At first, I made it as a pdf file and posted my message as a link,> but nobody give me hint, I guess perhaps I made people angry.There may have been other reasons. People have gotten responsesto linked graphics. Perhaps it was the nature of your questionrather than your formatting that caused people to miss itor not reply.What was your question about?I can tell you that the kind of question least likely toget a response, in any format, is to post a homeworkquestion with no evidence of your own thinking. - Randy === Subject: pure categories formulation of category theory, not using anunderlying foundation of sets:[1] Arturo Magidin June 19 '03 Re: A naive question about category theory http://mathforum.org/discuss/sci.math/m/516699/516747[2] George Cox June 19 '03 Re: A naive question about category theory http://mathforum.org/discuss/sci.math/m/516699/516710 In[3] F. William Lawvere home page > Chronological list of publications http://www.acsu.buffalo.edu/~wlawvere/list.htmlthere is:5. The Category of Categories as a Foundation for Mathematics, La Jolla Conference on Categorical Algebra, Springer-Verlag (1966), 1-20. Is that the proposal, or have there been others since? Or otherwriteups since if it is the same? Can someone summarize the proposal? (I don't have immediate accessto 5).-- David Libert ah170@FreeNet.Carleton.CA === Subject: > pure categories formulation of category theory, not using an> underlying foundation of sets:> [1] Arturo Magidin June 19 '03> Re: A naive question about category theory> http://mathforum.org/discuss/sci.math/m/516699/516747> [2] George Cox June 19 '03> Re: A naive question about category theory> http://mathforum.org/discuss/sci.math/m/516699/516710> In> [3] F. William Lawvere home page > Chronological list of publications> http://www.acsu.buffalo.edu/~wlawvere/list.html> there is:> 5. The Category of Categories as a Foundation for Mathematics,> La Jolla Conference on Categorical Algebra,> Springer-Verlag (1966), 1-20.> Is that the proposal, or have there been others since? Or other> writeups since if it is the same?> Can someone summarize the proposal? (I don't have immediate access> to 5).Although details vary and 1966 was before the invention of elementarytoposes, Lawvere has not changed his basic perception since. Thebasic idea is that instead of taking sets and membership as the basicatomic (undefined) terms of your foundations, you take morphism and(the partial operation of) composition. In other words a category. The axioms are the standard associativity and unitary axioms and thencome a few more to determine a set theory in order to model set theoryinside this.The specific axioms you will need to model set theory are:There is a one element set. Of course elements don't exist so yousay a set such that every set has exactly one morphism to it.Any two sets have a product (defined by the usual universal mappingcondition)Any parallel pair of morphisms (same domain and codomain) have anequalizer (also a universal mapping property).Finally, the crucial one: every set has a power set. Of course youhave to describe this in morphism theoretic terms. You begin bydefining subobject (as equivalence classes of monomorphisms, alsoneeding to be defined) and then a natural (meaning to be specified)1-1 corrspondence between morphisms A --> PB (the power object of B)and subobjects of A x B.If you stop there, you get a primitive set theory of which finite setsare a model. Usually, one adds a number of other axioms.Axiom of recursion (I won't go into detail, but essentially there is aPeano object).Axiom of choice: every epimorphism splits (has a right inverse).Axiom of two-valuedness: the one element set has exactly twosubobjects.This gives a model of set theory which is a bit weaker than the usualZFC, in that only bounded comprehension is available. I don't recallthe difference; I am just quoting what more knowledgable people havesaid. === Subject: >> Consider a graph that has a minimal cut set that is not complete. Doesthat> > imply that there exists an unchorded cycle of length > 3? (Yes it does,so> your conviction is correct.)>> Duncan>> Well, the proof is a little bit more involved than this if one wants>> LuigiWell, I wasn't presenting a proof, I was just giving you a start. I guessedyou could fill in the details :-). I think it also follows directly fromresults in Rose, D.J. (1970), where (amongst other things) he proves thatevery node in a decomposable graph is either simplicial or a member of apapers.Duncan === Subject: > Some of you may now see comments from mathematicians trying to hide> the logical flaw in Wiles's work where they try to claim that you> *can* indeed compare infinite sets and thereby prove a condition.> However, consider the following link which shows by the work of Cantor> that the set of rationals and the set of integers are in a sense the> same size.> Is JSH really Nathan the Great? === Subject: > [snip] > Then using the reasoning of Wiles, you could prove that every> rational is an integer based on the equal size of the sets.> I believe it would be more accurate to say that for each rational> there is a corresponding integer, not that the rational *is* an> integer.You are correct. Wiles's mistake is rather obvious, after all.> Remember> > when elliptic curves and modular forms are discussed that there are> INFINITE number of them:> Does it really matter how many total there are as long as there is> a method to correlate the sets? Consider the linear equation in 'x'> ax+b and the sinusoidal waves a*sin(x/b). There are an infinite number> of each, but there is also a fairly clear one-to-one correspondance> between the two sets.Yup, and if Wiles had something like that to connect elliptic curvesto modular forms then there'd be no discussion, now would there?However he compares the infinitely sized set of elliptic curves to theinfinitely sized set of modular forms, but you can't then draw aconclusion that forces a condition of modular forms upon ellipticcurves.That's why my point is that you have to have a superset. Do you seethe superset in your example?James Harris === Subject: [snip]> Yup, and if Wiles had something like that to connect elliptic curves> to modular forms then there'd be no discussion, now would there?If there were nothing to discuss, which there isn't, you'd make something up, which you have.--A good scientist must have a high tolerance for frustration, somewhat less for confusion, very littletolerance for mistakes and none at all for dishonesty.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: >> [...] Wiles's mistake is rather obvious, after all.> James, there's something terribly wrong with your medication. PLEASEconsult your psychotherapist/psychiatrist as soon as possible!F. === Subject: [...]| However, consider the following link which shows by the work of Cantor| that the set of rationals and the set of integers are in a sense the| same size.| | Then using the reasoning of Wiles, you could prove that every| rational is an integer based on the equal size of the sets.This would be true if at some point he had reasoned that two infinitesets were the same because they had the same cardinality (i.e., thesame size in the sense of Cantor). But he didn't.| Remember| when elliptic curves and modular forms are discussed that there are| INFINITE number of them:| | http://mathforum.org/isaac/problems/cantor2.html| | So why would mathematicians still argue?He didn't say that since there was a countable infinity of each,the correspondence between them was 1-1.The proof has us, in effect, approach the elliptic curve and themodular form through a sequence of simpler elliptic curves andmodular forms. (It's done with something called deformationtheory.) The place where the counting argument applies isa place where one has a finite number of choices as to how toadjust each of them to get a new pair with the elliptic curvemore like the original one you're given in certain respects.So it's not all elliptic curves and all modular forms being countedall at once; it's a selected subset of them which turns out tobe helpful in proving the final result.This is still a completely crude outline of how it goes, but ifthere's interest, especially from someone who's motivated bysomething other than a desire to make Wiles look bad, we couldhave an interesting time digging up more details.Keith Ramsay === Subject: > [...]> | However, consider the following link which shows by the work of Cantor> | that the set of rationals and the set of integers are in a sense the> | same size.> | > | Then using the reasoning of Wiles, you could prove that every> | rational is an integer based on the equal size of the sets.> This would be true if at some point he had reasoned that two infinite> sets were the same because they had the same cardinality (i.e., the> same size in the sense of Cantor). But he didn't.> | Remember> | when elliptic curves and modular forms are discussed that there are> | INFINITE number of them:> | > | http://mathforum.org/isaac/problems/cantor2.html> | > | So why would mathematicians still argue?> He didn't say that since there was a countable infinity of each,> the correspondence between them was 1-1.> The proof has us, in effect, approach the elliptic curve and the> modular form through a sequence of simpler elliptic curves and> modular forms. (It's done with something called deformation> theory.) The place where the counting argument applies is> a place where one has a finite number of choices as to how to> adjust each of them to get a new pair with the elliptic curve> more like the original one you're given in certain respects.> So it's not all elliptic curves and all modular forms being counted> all at once; it's a selected subset of them which turns out to> be helpful in proving the final result.That is, lifting. > This is still a completely crude outline of how it goes, but if> there's interest, especially from someone who's motivated by> something other than a desire to make Wiles look bad, we could> have an interesting time digging up more details.> > Keith RamsayThose on sci.math who kept up with FLT discussion heard a lot aboutlifting from a guy who claimed to have proven FLT with a Hensel lift.Mathematicians, he said, told him that lifting didn't work. Hedisagreed.I agree that in his case it didn't work.Suffice it to say though that you can see where the discussion movesbeyond the range of people who don't know a lot of advancedmathematics.But that's a dodge. The logical problem remains, which is that Wilesattempts to prove a condition with an association.The assertion you hear now is that there's a lifting where results ona finite subset apply across infinity.Now then that leaves you having to trust that mathematicians did theirjobs, but now ask about computer checking and notice the reaction.Suddenly math proofs--logical objects--aren't amenable to computerchecking as it seems the *judgement* of mathematicians is still moreimportant or supposedly hard to program.It simply doesn't smell right. The most rigid solution would be tohave computer programs--expert systems--do what editors do and reviewmathematical works.Expert systems work quite well in so many other areas, yet supposedlythey fail for anything *important* in mathematics.Gee, I wonder why.James Harris === Subject: >Some of you may now see comments from mathematicians trying to hide>the logical flaw in Wiles's work where they try to claim that you>*can* indeed compare infinite sets and thereby prove a condition.>>However, consider the following link which shows by the work of Cantor>that the set of rationals and the set of integers are in a sense the>same size.>>Then using the reasoning of Wiles, you could prove that every>rational is an integer based on the equal size of the sets. Tee-hee. Exactly where does Wiles use such reasoning, claimingthat every is a because the size of the set of thisesis the same as the size of the set of thats?(We need to see where _Wiles_ _used_ _that_ reasoning, notsome informal description that says showed each one of thesecorresponds to one of those by a counting argument - theactual proof is not included in that informal description.)Giggle.>Remember>when elliptic curves and modular forms are discussed that there are>INFINITE number of them:>>http://mathforum.org/isaac/problems/cantor2.html> >So why would mathematicians still argue?>>Well, ask yourself, do you find yourself still believing them?>>Or if you're finally hesitant, are you *still* hesitant despite being>an experienced logician?>>Why?>>Come on, admit it, as long as mathematicians keep *telling* you>there's more to it, you want to give them the benefit of the doubt,>against logic in which you may have thought you had complete belief.>>But I've shown you how you have a weakness inside your own head, in>terms of your own rationality, as human beings are *wired* to believe>based on social pressures. You're possibly a victim of that very>versatile but sometimes intriguingly limited human brain of yours.Indeed. We should all be thankful to you for pointing out the weaknessin our heads. Funny about that - when one notices that _everyone_seems to share the same weakheadedness one might wonder whether...naah, never mind.>James Harris**David C. Ullrich === Subject: > Some of you may now see comments from mathematicians trying to hide> the logical flaw in Wiles's work where they try to claim that you> *can* indeed compare infinite sets and thereby prove a condition.> However, consider the following link which shows by the work of Cantor> that the set of rationals and the set of integers are in a sense the> same size.> Then using the reasoning of Wiles, you could prove that every> rational is an integer based on the equal size of the sets. Remember> when elliptic curves and modular forms are discussed that there are> INFINITE number of them:> http://mathforum.org/isaac/problems/cantor2.html> So why would mathematicians still argue?> James,The weight of your arguments is crushing. We give up. It's really just a big conspiracy, organized by the freemasons. We've never thought anyone would see through it, but then we weren't prepared for your towering intellect.--Andras === Subject: [...]| > So it's not all elliptic curves and all modular forms being counted| > all at once; it's a selected subset of them which turns out to| > be helpful in proving the final result.| | That is, lifting.Very good. It has a family resemblance to the process one would useto solve a congruence mod increasing powers of a prime. To solvex^2 = -1 mod 5 requires selecting among 5 possibilities, of whichtwo work. Then solving x^2 = -1 mod 5^2, starting from one of thesolutions to x^2 = -1 mod 5 requires choosing from among 5 furtherpossibilities, and so on.[...]| Now then that leaves you having to trust that mathematicians did their| jobs, but now ask about computer checking and notice the reaction.|| Suddenly math proofs--logical objects--aren't amenable to computer| checking as it seems the *judgement* of mathematicians is still more| important or supposedly hard to program.It's not suddenly anything.If you doubt whether it's many times more tedious to write a proof so asto be checkable by computer than it is to write it up to usual standardsof informal proof, I suggest you try it sometime.Keith Ramsay === Subject: >[...]>>Now then that leaves you having to trust that mathematicians did their>jobs, but now ask about computer checking and notice the reaction.>>Suddenly math proofs--logical objects--aren't amenable to computer>checking as it seems the *judgement* of mathematicians is still more>important or supposedly hard to program.>>It simply doesn't smell right. The most rigid solution would be to>have computer programs--expert systems--do what editors do and review>mathematical works.>>Expert systems work quite well in so many other areas, yet supposedly>they fail for anything *important* in mathematics.>>Gee, I wonder why.Your proof of FLT has not yet been verified by computer. Why not?>James Harris**David C. Ullrich === Subject: > Suppose you have a plane passing through three 3d points, a b & c,> which form a triangle in that plane. So then you have the plane> normal n = (b-a) X (c-a) / ||(b-a) X (c-a)||, and the plane in 4d> homogenous coords is [n_x, n_y, n_z, -n dot a], where n_x is the x> coordinate of the normal, and so on. Now suppose a b & c are> transformed by matrices A B & C, and the new points are a' = A*a, b' => B*b and c' = C*c. (And A B & C could be limited to orthogonal> matrices, if necessary). These three points may define a new plane.Allowing different matrices to be used for each point means that a'and friends are essentially completely random points with no relationto each other or to a, b and c. You could still formally compute what the new plane coefficients would be, in terms of the old ones andthe three matrices, of course, but just redoing the calculations fromscratch using a', b' and c' is bound to be faster.-- Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de)Even if all the snow were burnt, ashes would remain. === Subject: > Suppose function f(x)=x^2 is given and we have to find out half range> fourier series, we will carry out in Maple like this> > a0 := 2/(2*2)*int(x^2, x = 0..2);> an := 2/2*int(x^2*cos(n*Pi*x/2), x = 0..2);> S1 := a0 + sum(an*cos(n*Pi*x/2), n = 1..10);> plot(S1, x = -2..2, scaling = constrained);> bn := 2/2*int(x^2*sin(n*Pi*x/2), x = 0..2);> S2 := sum(bn*sin(n*Pi*x/2), n = 1..10);> plot(S2, x = -2..2, scaling = constrained);> So my question is what inference can be drawn if plot sum of the two> half range series like this> plot((S1+S2), x = -2..2);> or it is just meaningless.> S1 is the Fourier series of the even extension of x^2. S2 is the Fourier series > of the odd extension of x^2. Add the two together - it is the Fourier series of> f(x) = 2x^2 if 0 f(x) = 0 if -2 R such that(1) p(f) is positive or zero for all f in [0,B](2) int_0^B p(f) df = P (P and B are some fixed positive real numbers)(3) p(f) is continuous on [0,B]LetQ: S-> R such thatQ(p(f)) = int_0^B log( p(f) + u(f) ) dfwhere u(f) is a fixed continuous function which is positive for all f in [0,B]It is easy to show that Q is bounded on S, so there's a least upper bound. However, S is not (in general) compact. I can therefore not prove the following conjecture:Conjecture: that there exists p(f) in S such thatQ(p(f)) = sup_S Q.Note: just an existence proof is needed! === Subject: >>Let >>S = the set of all functions p(f): [0,B] -> R such that>>(1) p(f) is positive or zero for all f in [0,B]>>(2) int_0^B p(f) df = P (P and B are some fixed positive real numbers)>>(3) p(f) is continuous on [0,B]>>Let>>Q: S-> R such that>>Q(p(f)) = int_0^B log( p(f) + u(f) ) df>>where u(f) is a fixed continuous function which is positive for all f in >[0,B]>>It is easy to show that Q is bounded on S, so there's a least upper bound. >However, S is not (in general) compact. I can therefore not prove the >following conjecture:>>Conjecture: that there exists p(f) in S such that>>Q(p(f)) = sup_S Q.Taking B = 1 for simplicity, and letting U = int u:If P >= U then there is a unique u which makesu + p constant, and this is the u that gives thesupremum. Because u + p = U + P, and Jensen'sinequality shows that for any other u' we haveint log(u' + p) <= log(int (u' + p)) = log(U + P) = int log(u + p).So it could be that when P < U the extremal u is one whichmakes u + p as close as possible to constant, in somesense (and then depending on what sense that is itshould be clear whether there is such a u...)>Note: just an existence proof is needed!>**David C. Ullrich === Subject: >Let >S = the set of all functions p(f): [0,B] -> R such that>(1) p(f) is positive or zero for all f in [0,B]>(2) int_0^B p(f) df = P (P and B are some fixed positive real numbers)>(3) p(f) is continuous on [0,B]>Let>Q: S-> R such that>Q(p(f)) = int_0^B log( p(f) + u(f) ) df>where u(f) is a fixed continuous function which is positive for all f in >[0,B]>It is easy to show that Q is bounded on S, so there's a least upper bound. >However, S is not (in general) compact. I can therefore not prove the >following conjecture:>Conjecture: that there exists p(f) in S such that>>Q(p(f)) = sup_S Q.>>Note: just an existence proof is needed!It seems to me fairly clear that the maximum is obtained withp(f) = { c - u(f) if u(f) < c { 0 otherwisewhere the constant c is chosen so that int_0^B p(f) df = P.The simplest existence proof may be just to show that thisis the maximum. A main ingredient is the concavity of ln.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israelUniversity of British ColumbiaVancouver, BC, Canada V6T 1Z2 === Subject: >Let > >S = the set of all functions p(f): [0,B] -> R such that>(1) p(f) is positive or zero for all f in [0,B]>(2) int_0^B p(f) df = P (P and B are some fixed positive real numbers)>(3) p(f) is continuous on [0,B]>Let>Q: S-> R such that> >Q(p(f)) = int_0^B log( p(f) + u(f) ) df>where u(f) is a fixed continuous function which is positive for all f in >[0,B]>It is easy to show that Q is bounded on S, so there's a least upper bound. >However, S is not (in general) compact. I can therefore not prove the >following conjecture:>Conjecture: that there exists p(f) in S such that>>Q(p(f)) = sup_S Q.>>Note: just an existence proof is needed!> It seems to me fairly clear that the maximum is obtained with> p(f) = { c - u(f) if u(f) < c> { 0 otherwise> where the constant c is chosen so that int_0^B p(f) df = P.> The simplest existence proof may be just to show that this> is the maximum. A main ingredient is the concavity of ln.> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2Call your function p_0(x) (I can't bring myself to use f for theindependent variable!). For any other fixed admissible function p(x),definep_t(x) = p_0(x) + t*dp(x) where dp(x) = p(x) - p_0(x) and 0 <= t <= 1.Note that int_0^B dp(x) dx = 0. Compute the derivatives of G(t) =Q(p_t) with respect to t:G' = int_0^B dp(x) / (p_t(x) + u(x)) dxNote that p_0(x) + u(x) >= c for all x in [0, B]. This implies thatG'(0) <= int_0^B dp(x) / c = 0Furthermore,G''(t) = int_0^B {-(dp(x))^2 / (p_t(x) + u(x))^2} dx < 0 for all t in[0, 1].This shows that G(1) > G(0), or in other words, Q(p_0) > Q(p), so p_0is the unique maximum of Q.John Mitchell === Subject: > ...>Conjecture: that there exists p(f) in S such that>>Q(p(f)) = sup_S Q.>> >Note: just an existence proof is needed!> It seems to me fairly clear that the maximum is obtained with> p(f) = { c - u(f) if u(f) < c> { 0 otherwise> > ...> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2> ...> This shows that G(1) > G(0), or in other words, Q(p_0) > Q(p), so p_0 ^^^^^^^^^^^ This should have been G(0) > G(1).John Mitchell === Subject: This argument looks correct, but won't it work for any candidate p_0(x) which is bounded above by some constant c? What is it that makes p_0(x) so special?> Call your function p_0(x) (I can't bring myself to use f for the> independent variable!). For any other fixed admissible function p(x),> define> p_t(x) = p_0(x) + t*dp(x) where dp(x) = p(x) - p_0(x) and 0 <= t <= 1.> Note that int_0^B dp(x) dx = 0. Compute the derivatives of G(t) => Q(p_t) with respect to t:> G' = int_0^B dp(x) / (p_t(x) + u(x)) dx> Note that p_0(x) + u(x) >= c for all x in [0, B]. This implies that> G'(0) <= int_0^B dp(x) / c = 0> Furthermore,> G''(t) = int_0^B {-(dp(x))^2 / (p_t(x) + u(x))^2} dx < 0 for all t in> [0, 1].> This shows that G(1) > G(0), or in other words, Q(p_0) > Q(p), so p_0> is the unique maximum of Q.> > John Mitchell === Subject: > Suppose a function f: R^n to R^n is differentiable at a point x, with> an invertible derivative.> Is there necessarily an open neighbourhood, U, of x, such that f is> injective on U, and such that the inverse of f is differentiable on> f(U)?> The answer is probably no, but I think counterexamples are hard to> come by.> Paul Epsteinf(x) = x + x^2*sin(1/x) on R^1You can check that f'(0) = 1, but f'(1/2PI*n) = 0 for all positiveintegers n, so the inverse function is not differentiable on anyneighborhood of 0.John Mitchell === Subject: If a group is cyclic, then it is generated by an element, say a. G = a^n, n= 1,2,3,4....Well, is a * a *a*....*a any different than a*a*a*...*a ?Lurch> I am doing some self study of Group Theory just because I missed out whenI> had the chance>> to study it properly and feel so ignorant about it. I am currently messing> with quotient groups.>> I have been trying to prove some things and would love some feedback.>> Let Z(G) be the center of group G, show that> a) Z(G) is a normal subgroup of G.> b) If G/Z(G) is cyclic, then G is Abelian.>> I finished part a), but am struggling with part b). This is what I have> tried:>> Let x and y be elements of G.> G/Z(G) is cyclic, let aZ(G) with a an element of G be its generator. (In> the following, I will use>> Z rather than the typographically more complicated Z(G).>> x, y elm's of G => xZ, yZ elm's of G/Z.>> Since G/Z is cyclic, xZ = (aZ)^j = (a^j)Z and yZ = (aZ)^k = (a^k)Z forsome> int's j and k.>> (xy)Z = (xZ)(yZ) (since Z is normal subgroup of G)> = (a^j)Z(a^k)Z> = (a^(j+k))Z> = (a^k)(a^j)Z> = (a^k)Z(a^j)Z> = (xZ)(yZ)> = (yx)Z>> This seems to be making some progress toward G being Abelian, but can I> conclude here that>> xy = yx giving commutativity. I realize that (xy)Z and (yx)Z are cosets> that have the same>> elements, but I don't think this gives me what I need. Do I have to bring> in some other fact at>> this point, or is the whole approach incorrect.> Ken> === Subject: Can anyone point me to a book or website which contains some information onthe density of Sophie-Germain primes?GREG === Subject: While its true that all derivations are technically proofs, most people seemto distinguish proofs that can be accomplished by applying a series ofmechanical operations in succession, from something that requires, say,explanation in English. And yes I realize this is all very vague. At whatline does a proof go from being a derivation to being a 'proof'-proof? Itdoesn't seem like there is one (that can defined formally at least). I hopesomeone knows what I'm talking about.l8r, Mike N. Christoff === Subject: > > Well David Ullrich is a math professor, and apparently has a big ego. > > Now if you're patiently listening to him and showing what he sees as> respect, then why should he attack you?> > I have disagreed with him in previous posts. He defended his view but> I did not see this as an attack. I do not expect everyone to think> exactly the same as I do.> Nor do I. That's not my point.> But if you challenge him, he's likely to get vicious.> And if you argue with him, he'll reply, reply, reply until you tire of> the exchange.> If he tired me, I would stop - no problem. > Well consider my experience. When I stopped replying to him, he kept> replying to me, stalking me in threads, and even went so far as to go> to newsgroups like alt.writing and others to do the stalking.> Let's imagine that I started posting in reply to you anywhere you went> on Usenet, don't you think that'd be a bit odd?> And remember I'd *stop* replying to him, but he has kept up his> stalking to this day.As I said already, I do not intend to discuss David anymore. > search at groups.google.com where David Ullrich is the author, where> > he uses the word idiot.> For another aspect of David Ullrich, check where he's the author and> > he uses the word rape.> I am a computer programmer myself. I do not come to this group for> computing advice and tips. Either I come to enjoy maths that is> nothing to do with computers or if it does concern computers then it> the maths element and not the computing element which brings me here.> The newsgroup isn't bad for getting information. But it shouldn't> surprise you that there are certain posters who have a rep for talking> to just about everyone as if they actually know the subjects well,> when they have a hidden agenda.> Don't take my word for it, do the search I recommended on posts> authored by David Ullrich where he uses the word idiot and see his> true colors.> Just remember there's always something in it for the person replying> to you.> > Here I clearly have an axe to grind about Ullrich, so I'm warning you> about him in a way that sends a message to a lot of others.> > Yet there are people who daily use computers to handle such problems,> and I'm not talking about the four color problem. You may *assume*> certain things about the state of the art, but how do you know?> I am quite familiar with the state of the art of computers. I visit> the labs of a major manufacturer quite frequently and attend numerous> conferences and exhibitions. It is part of my job to keep track of> the subject.> That's not the subject.No it is not the subject of the post but nor is anything you write. You appeared to be questioning my knowledge of computers. That is whyI made those comments. > My guess is that you're annoyed at my implication that you've been> taken in by David Ullrich. Rather than consider the evidence, you're> reacting as if you personally feel attacked.No, I do not feel that I have been taken in by him. If I am annoyedat anyone, it is you for all this off topic stuff you are posting tomy thread. > However, doesn't it even bother you slightly that you just fibbed> about your own subject line for a thread *you* created?> Now consider how much you know now versus what you might know if you> dug in and checked with credible and truly knowledgeable sources.> As I just said, checking knowledgeable sources is part of my job. > Then why did you create the thread asking questions?The knowledge that I am claiming is of computers and not mathematics. It is the mathematics that I was asking about. > Was it as a ploy so that you can start sharing your knowledge with the> group after first behaving as if you were ignorant?No. I have some knowledge of computers and a rather rusty knowledgeof mathematics. I come here for fun and to try and clean some of therust of my maths knowledge. > I think you've become defensive.This comment of yours above You may *assume* certain things about thestate of the art, but how do you know? seemed rather offensive. Especially with the asterisks. > That point is key: Now consider how much you now know versus what you> might know if you dug in and checked with credible and truly> knowledgeable sources.> > You answer is basically that you already have such sources, which> makes me wonder why you created the thread.I have easy access to reliable information on computers but not formathematics. If it became important, I could drive 30 or so miles tomy old university and look in the library. > That's why it's also important that you don't get waylaid or> distracted by posters like David Ullrich, as you will wade through a> lot of useless information, when it's better to find good sources.> I will make my own judgements on who I will let distract me. So far I> have been quite happy to be distracted by David's posts.> Of coure you will. And besides, I've revealed my agenda which is not> about just talking to you.> However, the fact is that you've already displayed that you don't seem> to have learned much on the subject of this thread, but look at how> many times Ullrich has already posted.> Have I? I thought that I had learned something. > Well it didn't sound like it to me, which is subjective, but your> reply sounded like you still didn't know of routine programs for doing> integrations, when from what I've picked up in posts, much of the math> world uses them routinely, and they don't necessarily rely on> difference techniques.See below. > You see that's part of the problem, the newsgroup readers don't> necessarily feel a responsibility to save you from people like David> Ullrich, a math professor, who has repeatedly demonstrated a lack of> knowledge of even basic computer tools for math.> > Just remember that just because someone replies and *sounds* credible,> it doesn't mean they actually know of what they speak.> I may have been out of university quite a while but I did spend quite> > a while studying maths. I do not think that I would be too easily> duped.> Spoken like you're new to Usenet and the Internet.> Why do you think that? I am quite a long term user of Usenet and the> Internet.> You don't sound like it to me.I may be more modest and polite than the average Usenet user. I amalso British rather than American. Maybe you confuse these thingswith naivety.> Welcome to the new Wild Wild West.> I am a long term resident. > Good for you.> James Harris> This will be my last post on the subject of David Ullrich. This is a> newsgroup for discussing maths and not David.> Yet somehow he does manage to come up as the subject quite often. > I'd be interested in someone with basic knowledge on the subject> giving an overview of the actual state of the art in the area of> computer analysis with regard to differential equation, and difference> equations, including some of the known math experts in the field.Others have also posted to this thread. > James HarrisThis is now my last post on the subject of me. I will no longer replyto you unless your post is relevant to the thread and interesting. Ido not wish to discuss me, David or anyone else any further. This isa maths newsgroup.J === Subject: >>I thought that a good pseudo random number generator could generate two>uncorrelated sequences of random numbers with two different seeds. Let>the random number generator be rand() and it takes a seed to initialize>it. So I thought rand(-1) and rand(-2) should produce two sequences of>uncorrelated random numbers. But my professor told me that it was not>guaranteed and the two sequences were possibly correlated or even highly>correlated. He said only two sequences of random number generated with>the same seed could be uncorrelated. What do you think about this? I>doubt what he said. If you doubt what he said why didn't you ask him to explain?Anyway, almost every random number generator in the universegenerates a sequence of numbers by saying x[0] = the seedand x[n+1] = f(x[n]) for some function f. Now think aboutwhat happens if you use two seeds, and the second seedhappens to be f(the first seed)... (are the two sequences1, 5, 8, 5, 6, 0, 6, 8, 0, ...5, 8, 5, 6, 0, 6, 8, 0, 3, ...uncorrelated?)> If a random number is good enough, such as ran1 and>ran2 in Numerical Recipes, it should produce two uncorrelated sequences of>random numbers with two different seeds.>>David**David C. Ullrich === Subject: Given integer n>=2 , what is the maximal number of pairs of points withEuclidean distance 1 in a set of n points in the plane ? === Subject: > Given integer n>=2 , what is the maximal number of pairs of points with> Euclidean distance 1 in a set of n points in the plane ?>I think the best you can do is pack the points in as vertices of atesselation of the plane by equilateral triangles. If that isthe case, then the maximum number of pairs would befloor(3n - sqrt(12n-3)).For n of the form 3k^2 + 3k + 1, the points can be arranged in ahexagonal shape, and in this case 3n - sqrt(12n-3) is an integer. === Subject: > Given integer n>=2 , what is the maximal number of pairs of points with> Euclidean distance 1 in a set of n points in the plane ?> >> I think the best you can do is pack the points in as vertices of a> tesselation of the plane by equilateral triangles. If that is> the case, then the maximum number of pairs would be> floor(3n - sqrt(12n-3)).> For n of the form 3k^2 + 3k + 1, the points can be arranged in a> hexagonal shape, and in this case 3n - sqrt(12n-3) is an integer.This is a famous problem of Erd.9as. The best known lower bound is Erd.9as', n^{1+c/loglog n} for some c, and comes from packing points in a square lattice with spacing 1/q where the factorization of q involves many primes congruent to 1 mod 4. The best upper bound is quite far from this, O(n^{4/3}). See http://cs.smith.edu/~orourke/TOPP/P39.html for a more detailed discussion.If you required distance 1 to be the smallest distance in the point set you would get answers more similar to the one you posted.-- David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine, School of Information & Computer Science === Subject: > Given integer n>=2 , what is the maximal number of pairs of points with> Euclidean distance 1 in a set of n points in the plane ?>> I think the best you can do is pack the points in as vertices of a> tesselation of the plane by equilateral triangles. If that is> the case, then the maximum number of pairs would be> floor(3n - sqrt(12n-3)).> For n of the form 3k^2 + 3k + 1, the points can be arranged in a> hexagonal shape, and in this case 3n - sqrt(12n-3) is an integer.When you said packing, does this discriminate between two differenttypes of equivilent packing, which involve shifting the planes uponwhich the vertices were placed differently with respect to eachother?Or is it the same result every time?Also, what if this is only an asymptotic value and not a real value?(...Starblade Riven Darksquall...) === Subject: > Given integer n>=2 , what is the maximal number of pairs of points with> Euclidean distance 1 in a set of n points in the plane ?>> I think the best you can do is pack the points in as vertices of a> tesselation of the plane by equilateral triangles. If that is> the case, then the maximum number of pairs would be> floor(3n - sqrt(12n-3)).> For n of the form 3k^2 + 3k + 1, the points can be arranged in a> hexagonal shape, and in this case 3n - sqrt(12n-3) is an integer.Do you assume that the points are distinct ?Tim === Subject: See David Eppstein's reply in this thread. It appears that I amsomewhere out in left field, if I'm in the ball park at all.> Given integer n>=2 , what is the maximal number of pairs of pointswith> > Euclidean distance 1 in a set of n points in the plane ?> I think the best you can do is pack the points in as vertices of a> tesselation of the plane by equilateral triangles. If that is> the case, then the maximum number of pairs would be> floor(3n - sqrt(12n-3)).>> For n of the form 3k^2 + 3k + 1, the points can be arranged in a> hexagonal shape, and in this case 3n - sqrt(12n-3) is an integer.>> Do you assume that the points are distinct ?>> Tim === Subject: >> Given integer n>=2 , what is the maximal number of pairs of points with>> > Euclidean distance 1 in a set of n points in the plane ?>> >> I think the best you can do is pack the points in as vertices of a>> tesselation of the plane by equilateral triangles. If that is>> the case, then the maximum number of pairs would be>> floor(3n - sqrt(12n-3)).>>Do you assume that the points are distinct ?Yes, if you don't assume the points are distinct then the answer is verywell-known. Turan's theorem tells us that the number of pairs is at mostthe number of edges of a complete tripartite graph on n vertices which isabout n^2/3. If the points are not required to be distinct, this boundis trivially achievable: split the n points equally at the corners of anequilateral triangle, and we get a complete tripartite graph. -- Erick === Subject: >> >> My gripe is the OP hasn't specifed how g is on one hand a function and on >> the other a distribution. We now enter into the realm of: Guess what the OP >> is asking. (Why do we play this game? Isn't it enough that we answer the >> damned questions? LOL.)>>>Apologies. I didn't want to give too many unnecessary details about my>problem, but looks like I gave away too little!>>Here is my problem. The actual integral is >>int_{R^n} h f^ g^>>Here, ^ means (distributional) Fourier transform. I know that both f>and g are continuous functions that can be viewed as distributions.>They satisfy>>h**1/2 f^ in L^2(R^n)>>and >>h**1/2 g^ in L^2(R^n)>>so the integral certainly exists. What I'd really like to do is 'unhat>the g' so that I can rewrite the integral as>>int_{R^n} (h f^)check g.>>Here, check denoted the inverse Fourier transform (IFT). I am>assuming in this that (h f^) is L^2 so that I can take it's IFT in>fact in this work (h f^)check is assumed to have compact support.>However, the unhating is not legal (at first site at least) because g>is not L^2! I'm confident there is some distributional argument that>will make this move legal.>>My first thoughts were to approximate (h f^) by a sequence of test>functions phi_j then hope to write:>>int phi_j g^ = [g^,phi_j] = [g,phi_j^] = int phi_j^ g >>and then 'go to the limit'. Here, [ , ] is the action of a>distribution on a test function. I don't know if the first equality is>legal - this was intended to be the substance of my first post. Whether the first equality is legal is something we can't saywithout more information, it seems to me. (I mean, at thetop you say that ^ is the _distribution_ FT. If that's what the^ in g^ means then int phi_j g^ does not _officially_ makeany sense, just as statements like h**1/2 g^ in L^2(R^n)official make no sense...say something about how _if_ this and that distributionis actually given by integration against this or that function...now it turns out that you're not sure whether it is or not.We can't say whether it is or not - whether or not g isgiven by integration against some function is part ofthe _definition_ of g.)>The>second equality here is legal by definition of the FT. The 3rd>equality is legal here as in this work I also know that g in L^1_loc.Aargh. Here's a quote from your first post: Keep in mind that I don't know that g is in L^1_loc...>I hope things have been made clearer. Am I on the right tracks? >>Ta,>>T**David C. Ullrich === Subject: >>Sorry for this confusion but you should notice that the g in my first>>post is not>>the g of the second post. >>> For heaven's sake, how was one supposed to notice this?>>> I give up.>>>I concur, I've made quite a mess of things in my explanation. So, I>think I'll>back out of this thread gracefully now. I assure you that my intention>wasn't to irritate you David. Don't worry about the irrittation - I guess I was in fact irritated,but just a little. But do note the part you snipped about why Idon't see how various things you're saying make any sense:If you're assuming one thing then the question answersitself while if you're not assuming that then I honestly don'tsee what something else _means_... it's very possible I'mjust being stupid there, but.>It's back to Rudin's Functional Analysis>for me to get to grips with the subtleties of distributions which I>clearly lack.>>Ta,>>T**David C. Ullrich === Subject: >> Theorem. Let>> f(x) = a_n x^n + a_n-1 x^(n-1) + ... + a_0>> be a polynomial with integer coefficients. If there exists a prime>> number p such that>> a_n-1 = a_n-2 = ... = a_0 = 0 (mod p)>> but a_n not congruent 0 (mod p) and a_0 not congruent 0 (mod p^2),>> then f(x) is irreducible over the field of rational numbers.>> Ok, not so hard to prove.> Corollary. If p is prime, then the polynomial>> phi(x) = x^(p-1) + ... + x + 1>> is irreducible over the field of rational numbers.>> Egads, how does the corollary follow from the theorem?Hard to say :-)>> -- Fundamental Theorem of Algebra>> Any polynomial of positive degree with complex coefficients>> has a complex root.>> What method is used to show the fundamental theorem?Lots are possible (Gauss found 4, essentially distinct) Simplest useLiouville's theorem (any bounded analytical function is constant)> The related material below seems inadequate.>> DeMoivre's Observation.>> For any positive integer n, the equation z^n = 1>> has n distinct roots in the set of complex numbers.> cos 2k.pi/n + i sin 2k.pi/n, for 1 <= k <= n.>> Kronecker's Theorem.>> Let F be a field and f(x) a nonconstant polynomial in F[x].>> Then there exists an extension field E of F>> and an element u in E such that f(u) = 0.>> Show any such extension field of the complex field C is C?> Ouch, doesn't seem the way to go.Well, it is possible (if you add the idea that any odd degree polynomial ofR[X] has a real root : Euler-Lagrange, with a twist) but certainly notobvious (as the fist was well-known since deMoivre :-), and the second wasadmitted as obvious in Euler and Gauss times)>> ---- === Subject: Equivalent random variables implies thatP[X neq Y] = 0However I don't understant completely this definition. Do you have a different one?Does it implies that... P[x = y] = 0 for some distributions? Diego Andr.8es === Subject: > Equivalent random variables implies that> P[X neq Y] = 0> However I don't understant completely this definition. Do you have a> different one?> Does it implies that... P[x = y] = 0 for some distributions? > Suppose X is uniformly distributed in [1,2] and Y is uniformlydistributed in [3,4]. Then of course P[X=Y] = 0 because X=Yis never true.Or let X be uniformly distributed in [0,1] and let Y=-X.Then the event [X=Y] is the event [X=0] which has probability zero, andin this case P[X=Y]=0 again. Even though that event is not empty, itstill has probability zero. === Subject: Greetings,Could anyone give me an example of a stochastic process (or random process)that is ergodic but not stationary?Sincerely, rjc.--Ryan CassidyElectrical Engineering Graduate StudentStanford University === Subject: [snip]> It seems to me that clearing you out of the discipline is necessary.Harrigance - the guy on the balcony strikes again: http://users.pandora.be/vdmoortel/dirk/Stuff/Arrogance1.jpg http://users.pandora.be/vdmoortel/dirk/Stuff/Arrogance2.jpg Dirk Vdm === Subject: If we choose randomly and uniformly two points p,q in the n-dimensionalunit cube in R^n what is the expected Euclidean distance d(p,q) ? === Subject: >If we choose randomly and uniformly two points p,q in the n-dimensional>unit cube in R^n what is the expected Euclidean distance d(p,q) ?I can give you a method of computing it numerically withone integration and a reasonable amount of computing.The moment generating function of the square of the distanceis m(t) = (6(exp(t) - 1 - t - t^2/2)/t^3)^n, and the Laplacetransform is just m(-t). Now compute int -m'(-t)/sqrt(t) dt/sqrt(pi),the integral going from 0 to infinity, and this is the answer.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Deptartment of Statistics, Purdue University === Subject: >> If we choose randomly and uniformly two points p,q in the n-dimensional> unit cube in R^n what is the expected Euclidean distance d(p,q) ?If you only need an approximate value, note that a Taylor expansionaround the mean gives: E sqrt(Z) =~ sqrt(mu) - 1/(8*mu^(3/2)) * sigma^2for a random variable Z with mean mu and variance sigma^2.The sum Z of squared distances in each component has mean n/6 andvariance 7n/180 giving an approximate expected distance of: sqrt(n/6) - (6/n)^(3/2) * 7*n/1440Simulation shows this approximation appears to be good to within about1% for all n.-- Kevin === Subject: I'm looking for a function g such that for all positive integer a, b,c we have thatIntegral(0 -> 1) { g(ax)*g(bx)*g(cx) } dx = 1 if a = b = c = 0 otherwiseTo clarify, the integrand is the product of the three quantitiesg(ax), g(bx) and g(cx) (where, as usual, ax denotes the product a*x)and the limits of the integral are 0 and 1. === Subject: > Integral(0 -> 1) { g(ax)*g(bx)*g(cx) } dx = 1 if a = b = c> = 0 otherwiseAre you familiar with the Dirac delta function?This is an improper function (strictly speaking a funcional) that satisfies:Integral(-infinity -> infinity) { f(x) delta(x) } dx = f(0)Informally, delta(x) is zero everywhere except for an infinite spike at zero.Let w be an irrational number greater than zero and less than one.w = 1 / sqrt(2) will do as an example.Let h(x) = Sum(k = 1 -> infinity) { delta(x-k) }Informally, h has a spike at every positive integer,and is zero elsewhere.Now g(x) = cuberoot( h(x - w) )For any positive integers a and b, g(ax) g(bx) = 0.For any positive integer k,g(kx)^3 has k spikes between zero and one,but each is only 1/ktimes as wide as the original delta.Now, if you don't like delta functions, you can do the same thingwith very narrow rectangular spikes, provided a, b and c are not too large. === Subject: > You know I am with you. I see more questions than answers but oh what> questions they are. :)> I'm now at over load point. I'll need time to recharge.> others.> I missed some of your points and made errors due to my hurry. > Yet, since I have risked talking to others I have gone from thinking> 3x+1 was the only system to x+y and x-y, GCD's, rings, dynamical> systems and more. Not bad even if made me look foolish.The fools are those who don't learn from their mistakes.> Ernst>>Hey would you check [5x+7,x/2] That is what is called Divergent right?>> >Ok, but it will be tomorrow night before I can get the web page updated.>> Forget everything I've previously said.> I need to take a closer look at my program, it may not be doing what I think> it's doing.> > I didn't check _every_ one of the 2765 attractors I found in 3a+b, but every> one I checked seemed legit. But then I noticed some funny things.> For 3a+5, three of the attaractors found were, with parity vector:> 19 10101000> 31 10100010> 49 10001010> but these three are actually part of the same loop> 19 62 31 98 49 152 76 38 19> If we actually extend this loop to more than one cycle> 19 62 31 98 49 152 76 38 19 62 31 98 49 152 76 38 19 62 31 98 49 152 76 38 > 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 > 1 0 1 0 0> we get a different parity vector depending on which odd number we start on. And> in the course of testing every permution of parity vector, the program> encountered this loop from each of the three possible starting numbers. And> since it really is part of a loop, that must be why I got an integer crossover.I've changed my program. After finding an integer crossover point, Icalculate the rest of the odd numbers in the sequence and record theminimum one. I get a record that looks like:system: [3x+5]crossover: 31attractor: 19parity vector: [1, 3, 1](The parity vector list just shows the count of 0s in the vector. Itis implied that there is a single 1 to the left of each block of 0s.)The 19 62 31 98 49 152 76 38 19 sequence will also produce two otherinteger crossover points:system: [3x+5]crossover: 19attractor: 19parity vector: [1, 1, 3]system: [3x+5]crossover: 49attractor: 19parity vector: [3, 1, 1]This gives me the best of both worlds. I've three differentcrossovers/parity vectors if I'm looking for patterns there, but asingle attractor if that's what I'm interested in.> Even stranger things are happening in 5a+b. I'm finding attractors that lead> into a loop without actually being part of it> In 5a+5 I have 5 and 31 as attractors, but the sequence is> 31 160 80 40 20 10 5 30 15 80 40 20 10 5 30 15 80 40 20 10 5> I'm not sure what to make of that.What happened here was that I inadvertently recorded 3a+b data as5a+b. Naturally, when I put the 3a+b attractors into the 5a+b Collatzprocess, strange things resulted. Strange because 5 just happens to bean attractor in BOTH 3a+5 AND 5a+5, so the above sequence almostworks. Good lesson here on the importance of verifying your computerprograms. It was an operator error, not the program's. I told it tocompute 3a+b and that's what it did. It wasn't the program's faultthat I directed the output to the 5a+b file.> I won't be putting up anymore charts and will remove the ones that are there> now. === Subject: I have found a structure to the attractors. There is more to do butthe basic structure I see.Ernst === Subject: >> I want to find a 3d point (x, y, z) in the 3d Cartesian space using> the following data that I have:> - Cylinder parameters: two 3d points that describes cylinder axis,> let call these p0 and p1;> - Radius of the cylinder, let call R;>(x1,y1,z1)-p0 = m(p1-p0) eqn of line, x1,y1,z1= any point on linesolve for m for each component and let m=m=m for eqn of line.((x,y,z)-(x1,y1,z1))*(p1-p0) =0 dot product, eqn of cylinder.R = |(x,y,z)-(x1,y1,z1)|alternatively, R|((x,y,z)-(x1,y1,z1))x(p1-p0)|=|(x,y,z)-(x1,y1,z1)||(p1-p0) | cross product,eqn of cylinder. let (x1,y1,z1)=m(p1-p0) + p0 in the bottom equation and solve for m.then use this in the dot product equation for the final form. === Subject: > Yes, and an r-generator finite abelian group is a product of> r cyclics. This drops out immediately from the theory of the Smith> normal form.Which Smith?-- Timothy Murphy tel: +353-86-233 6090 === Subject: >> Yes, and an r-generator finite abelian group is a product of>> r cyclics. This drops out immediately from the theory of the Smith>> normal form.> Which Smith?Is there more than one Smith?It's Henry John Stephen Smith of Balliol College.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === Subject: > Yes, and an r-generator finite abelian group is a product of> r cyclics. This drops out immediately from the theory of the Smith> normal form.>> >> Which Smith?>>Is there more than one Smith?Apparently yes. D.A. Smith, `A basis algorithm for finitely generated abelian groups',Math. Algorithms, 1 (1966), 13-26.But that's the wrong Smith!H.J.S. Smith, `On systems of linear indeterminate equations and congruences'Philos. Trans. Royal Soc. London, CLI (1861), 293-326.is the correct Smith. (I didn't know he was of Balliol though!)Derek Holt.>It's Henry John Stephen Smith of Balliol College.>>-- >Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html> The League of Gentlemen === Subject: >> Yes, and an r-generator finite abelian group is a product of>> r cyclics. This drops out immediately from the theory of the Smith>> normal form.> Which Smith?>>It's Henry John Stephen Smith of Balliol College.It's a rather disgusting way of expressing the Structure Theoremfor Finite Abelian Groups, isn't it?Who showed one could express such a groupas the product of cyclic groups of prime-power order?Maybe it was Jones ...-- Timothy Murphy tel: +353-86-233 6090 === Subject: === Subject: > What is the formula for determining the surface are a of a truncated> cone?I suppose you're talking about a frustum. If so, see. David === Subject: > What is the formula for determining the surface are a of a truncated> cone?In general, the volume is always the same when the vertex moves ontoany point in a plane parallel to base and distance H from base area Aof frustum. Equals to AH/3. BTW, who/when enunciated it first ? === Subject: 1) Given from one to five points on x-axis on the orthogonal cartesianco-ordinations (x_1, x-2, x_3, x_4, x_5). At least one of those pointsexists on the x-axis.(Shifting (or moving) the x-axis (parallel to the the original x-axis) inthe direction of y-axis (+ or -) you can recognize more points on theparallel line to the original x-axis.)2) It is - because of the dense real line - evident that you can drawinfinite linear equations through a point x_i on the x-axis, which is one ofthose points (x_1, x-2, x_3, x_4, x_5). Or, is it?4) Proof or disproof that you cannot find the point x_i on the x-axis, whichis one of those points (x_1, x-2, x_3, x_4, x_5), though you have infinitemany linear equations to use, namely y=kx+a. (ref.: Dedekinds Cut).In other words: Proof or disproof that although a point x_i is on thex-axis, it's not algebraic point on the x-axis. What kind of point is apoint x_i, as it is clearly on real line?Tapio === Subject: I think the function for the Gabrial Horn problem is f(x)=1/x. I was easilyable to show that the surface area from 1 to 00 of f(x)=1/x is infinite andthat the volume of f(x)=1/x from 1 to 00 is finite. This was verymechanical. Now I understand that strange things happen at 00 but this oneis real strange. I asked some top mathematicians to tell me what's going onhere and they all walked away from this one. So I am asking you folks to tryto explain what's going on without too much hand waving. I teach thisproblem in the calculus sequence and I am not happy at all with what I tellmy students. === Subject: > I think the function for the Gabrial Horn problem is f(x)=1/x. I was easily> able to show that the surface area from 1 to 00 of f(x)=1/x is infinite and> that the volume of f(x)=1/x from 1 to 00 is finite. This was very> mechanical. Now I understand that strange things happen at 00 but this one> is real strange. I asked some top mathematicians to tell me what's going on> here and they all walked away from this one. So I am asking you folks to try> to explain what's going on without too much hand waving. I teach this> problem in the calculus sequence and I am not happy at all with what I tell> my students.I doubt top mathematicians walked away because they couldn't explain it. Think of a cylinder of height h and base radius r. Ignoring the end pieces, the surface area is (2Pi)r*h and its volume is Pi*r^2*h. The ratio of surface area to volume is thus 2/r, which for small r is very large. So even with bounded simple objects like cylinders, the surface area and volume can get out of whack. If you now allow unbounded objects, like the graph of y = 1/x revolved about the x-axis, you can naturally expect strange things to happen. Note that the same phenomenon is occuring with Gabriel's Horn as with the cylinders: The radius of the cross-section -> 0 as x -> oo.If you revolved y = 1/sqrt(x) about the x-axis, then both the surface area and volume would be infinite. The reason is that the radius of the cross section doesn't go to 0 fast enough.If you think the above is strange, consider the function f(x) = x*|sin(1/x)|, x in [0,1] (set f(0) = 0). Now revolve this graph about the x-axis. Then you end up with a bounded surface of infinite surface area. Your students will love it. === Subject: > I think the function for the Gabrial Horn problem is f(x)=1/x. I was> easily able to show that the surface area from 1 to 00 of f(x)=1/x is> infinite and that the volume of f(x)=1/x from 1 to 00 is finite. This> was very mechanical. Now I understand that strange things happen at 00> but this one is real strange. I asked some top mathematicians to tell> me what's going on here and they all walked away from this one. So I> am asking you folks to try to explain what's going on without too much> hand waving. I teach this problem in the calculus sequence and I am not> happy at all with what I tell my students.One source of discomfort with the result might be distrusting theresult -- is this just some weirdness resulting from calculus itself?The same sort result can be found just by considering series --1/1 + 1/2 + ... 1/n diverges, but 1/1 + 1/4 + ... 1/n^2 converges.You can make these series correspond (more or less) to an under-estimateof the Horn's surface area (still comes out infinite) and an over-estimateof the volume (still comes out finite) -- think of the surface area andvolume of a series of unit-length cylinders approximating the Horn.The other source of discomfort is likely to be the way oureveryday interpretations and analogies of volume and surfacearea jar with their ideal mathematical counterparts. It seemswe could fill the Horn with a finite amount of ink, but requirean infinite amount of ink to paint the surface.Real-world ink, of course, is made up of discrete atoms which cannotbe arbitrarily stretched or compressed -- just as well, givencounter-intuitive objects like the Horn :-).Consider the volume of the horn -- it's finite, but of infiniteextent. This is easily imagined via an analogy with a finite piecemathematical chewing gum - put the gum on a wall and have along-lived friend delicately pinch the gum and start running off toinfinity. The infinitely stretchy gum keeps its finite volume andjust gets very, very thin as you move away from the wall. Finitevolume, arbitrarily long length - easy. But what happens to thesurface area? It becomes arbitrarily large too -- somehow thisseems easier to imagine as a limiting process (as the friendgets further and further away) rather than a mysterious resultwhen the friend reaches infinity.It may be simpler to consider a version of the paradox in fewerdimensions by considering the curve y=1/x^2 from 1 to 00 -- the areaunder the curve is finite, but the length of the curve is infinite.We need a finite amount of ink to colour the area, but an infinite amountto draw the boundaries. If you think of this as a limiting process,at each step where we add the section from k <= x < k+1 we're addinga negligible amount of area but a non-negligible amount of length.In real-world colouring, border lines have thickness -- they're areas.In mathematics, a unit line length takes 0 area of ink to colourbut 1 length of ink to colour. The extreme version of this exampleis the area under the curve y=0; at each step we add 0 area and 1 length.All this is one reason why I prefer discrete mathematics :-). === Subject: > I think the function for the Gabrial Horn problem is f(x)=1/x. I was easily> able to show that the surface area from 1 to 00 of f(x)=1/x is infinite and> that the volume of f(x)=1/x from 1 to 00 is finite. This was very> mechanical. Now I understand that strange things happen at 00 but this one> is real strange. I asked some top mathematicians to tell me what's going on> here and they all walked away from this one. So I am asking you folks to try> to explain what's going on without too much hand waving. I teach this> problem in the calculus sequence and I am not happy at all with what I tell> my students.What do you tell them?If the question is How can one be finite and the other infinite? thesimplestexplanation is that volume is not surface area. Other examples comefrom fractal geometry, where a plane figure with finite area maybounded by a curve ofinfinite length.On the other hand, if they complain that it contradicts their ideasaboutthe physical world, reply that these are abstractions. Clearly, it isimpossible to physically build Gabriel's Horn.Another favorite of mine along this line:The Banach-Tarski paradox, or how to decompose a sphere, userotations/translations to re-assemble the pieces to form two spheresidentical to the first. Mathematically sound, physically impossible. === Subject: > I think the function for the Gabrial Horn problem is f(x)=1/x. I was easily> able to show that the surface area from 1 to 00 of f(x)=1/x is infinite and> that the volume of f(x)=1/x from 1 to 00 is finite. This was very> mechanical. Now I understand that strange things happen at 00 but this one> is real strange. I asked some top mathematicians to tell me what's going on> here and they all walked away from this one. So I am asking you folks to try> to explain what's going on without too much hand waving. I teach this> problem in the calculus sequence and I am not happy at all with what I tell> my students.I don't think it seems strange, stated that way. The cute way tosay it, which *does* make it seem a little paradoxical, is You can fill it with paint, but you can't paint it.So let's look a little closer at the intuitions to which thisexample does violence. Ordinarily, we would expect that fillingsomething with paint *would* paint it, because the paint wouldstick to it. We also expect that to paint something of infinitearea would require an infinite amount of paint, because paintcan be spread only so thin -- pace Dave Barry, if you dividepaint finely enough, it's no longer made up of paint.But now the paradox goes away. If we're talking about ordinaryreal-world paint made up of molecules, then you can't fill thehorn with paint -- when it gets thinner than a single molecule,the paint will stop there. On the other hand, if we have a supplyof lovely theoretical paint that can be spread arbitrarily thin,then you *can* paint the horn, by spreading the paint thinnerand thinner as you go down. With either kind of paint, there'sno conflict. === Subject: > I think the function for the Gabrial Horn problem is f(x)=1/x. I was easily> able to show that the surface area from 1 to 00 of f(x)=1/x is infinite and> that the volume of f(x)=1/x from 1 to 00 is finite. This was very> mechanical. Now I understand that strange things happen at 00 but this one> is real strange. I asked some top mathematicians to tell me what's going on> here and they all walked away from this one. So I am asking you folks to try> to explain what's going on without too much hand waving. I teach this> problem in the calculus sequence and I am not happy at all with what I tell> my students.I'm not sure what to tell you. You've derived the results (which aretrue). That's the least hand-waving way to go about explaining anunintuitive result. What I mean is, there's really nothing toexplain.What I think you might mean is that you wish to reconcile yourintuition with the result you've gotten. Well, given that the Riemannintegral is the limit of approximations of rectangles under the curve,and that as x -> oo, f' -> 0, we can come up with an approximatingrectangle of arbitrary length. But that doesn't matter since f = 0where this approximation is acurate. So the area under the curve inthat interval is zero.But I fear that even this is not enough. Please call again if thiswas insufficient.Alex SollaJuniorReed College === Subject: > I think the function for the Gabrial Horn problem is f(x)=1/x. I was easily> able to show that the surface area from 1 to 00 of f(x)=1/x is infinite and> that the volume of f(x)=1/x from 1 to 00 is finite. This was very> mechanical. Now I understand that strange things happen at 00 but this one> is real strange. I asked some top mathematicians to tell me what's going on> here and they all walked away from this one. So I am asking you folks to try> to explain what's going on without too much hand waving. I teach this> problem in the calculus sequence and I am not happy at all with what I tell> my students.Two bakers working in a wedding cake factory received a surprise orderfrom a *very* eligible bachelor - he was suddenly not so eligible,having become engaged to be married! He had selected their company tobake the wedding cake, because of the abnormally large number of mathsPhD's among the cooking staff.The bachelor was a little eccentric - he said I want a wedding cake in the shape of Gabriel's Horn (he laterexplained that the name of the love of his life was Gabriella).The two bakers went to work. One was responsible for the icing, theother for the filling of the cake. They knew they would need a lot ofingredients for such a large cake, so the ordered truckloads of icingsugar, eggs, flour... and so on.Soon the trucks began to arrive, and the bakers went to work. Afteronly a week, they had no more icing sugar! (They had used about 40% ofthe flour and eggs). Looks like we need to call in another trucksaid the cake man. The icing man shrugged, and reached for the phone.The next day, another truck of icing sugar arrived, and the two bakerspitched back to work. However, by the end of the day, the cake wasonly 1.648 times higher, and they had used up all the icing sugaragain! There were still plenty of eggs and flour from the first day.The next day, the same thing happened. and again. By the end of thefirst week, the icing man was very frustrated - every day they hadcalled for more icing sugar. And although the eggs and flour werebeginning to run out, the cake maker seemed to use less and less eachday, so there was always enough.By the end of the second week, the cake was 665 times as tall as onthe first day. However, even though the bakery had called in 14truckloads of icing sugar, they were still using flour and eggs fromthe first (refrigerated, of course) truck!The icing man, exhausted, looked and saw that they had actually used99.96% of the original truckload. Surely tomorrow... he consoledhimself Surely tomorrow it will be my friend who runs out of eggs!But tomorrow came and went. Although the truck had been only 0.04%full, and although they increased the height of the cake by another64.8% until the icing ran out, the cake man had to make so little caketo fill the tower, that he only had to scrape a half-thimbleful offlour, and a few drops of egg white - before rolling the dough into avery long noodle - to fill the cake.The end of the month came. The bachelor was getting impatient. The eggand flour trucks looked empty, but with the help of a magnifyingglass, the cake maker was able to find enough grains of flour tocomplete his share of the work. The icing man had fallen into aroutine by now, and the truck drivers all called him by name - afterall, he saw them every day, and every day the cake grew taller (mostlyicing, of course).One day, the icing man's frustrated wife told her cousin about herpoor overworked husband. The cousing told her work colleague, who toldher uncle, who told his nephew's brother-in-law who happened to be agood friend of the bachelor's fiancee's second cousin's half-sister.The bachelor heard the news.The next day he dropped by to check on the progress of the cake (heicing man sood nervously. The bachelor eyed the cake, stretching upinto the sky.He stared, and eyed, and eyed and stared. Then he said.... How much flour and eggs will you need to reach infinity?The cake maker replied that they had calculated the amount in advance,and ordered just enough.And how much icing?The icing man looked at the floor - ashamed he had not done theintegral before starting the project. The cake man replied...A lot, sir.How much, exactly?Well, we would have to use limits to answer that sir. It's not aneasy...How much?Infinitely much, sir.The bachelor looked at the exhausted icing man. Did you tell yourcolleague this?The cake man knew he could not lie and get away with it.Er, no sir.The bachelor gazed at the men - the cake - the men... Finally, hespoke.Good. The cake is tall enough now. Now I want you to expand thebase.Pardon me, sir? the cake man asked.Expand the base. Still in the shape of Gabriel's horn, but builddownwards. Every day, order a truck of eggs and flour, and of icing ifyou need it. But build downwards to expand the base of my cake.Downwards, sir? said the cake-man, small drops of sweat slowlyforming on his brow.Yes, downwards. You see the radius now is 1 foot, yes? And the slopeat the base is 45 degrees, yes? Well, build downwards, so the edge ofthe cake is still on the curve y=1/x. I know you are mathematicalbakers and understand what I mean. I'll talk to your supervisor so hemakes sure you do what I want. My wedding is still two months away.Keep building downwards until that day.And the bachelor left, pausing only to give a friendly wink to thetired icing man - for whom the first rays of uncomprehending hope werebeginning to dawn upon his weary eyes....Downwards?? said the cake man, all strength gone from hisvoice.... === Subject: > I think the function for the Gabrial Horn problem is f(x)=1/x. I was easily> able to show that the surface area from 1 to 00 of f(x)=1/x is infinite and> that the volume of f(x)=1/x from 1 to 00 is finite. This was very> mechanical. Now I understand that strange things happen at 00 but this one> is real strange. I asked some top mathematicians to tell me what's going on> here and they all walked away from this one. So I am asking you folks to try> to explain what's going on without too much hand waving. I teach this> problem in the calculus sequence and I am not happy at all with what I tell> my students.Error in my story:After only a week, they had no more icing sugarshould readAfter a day... === Subject: >I think the function for the Gabrial Horn problem is f(x)=1/x. I was easily>able to show that the surface area from 1 to 00 of f(x)=1/x is infinite and>that the volume of f(x)=1/x from 1 to 00 is finite. This was very>mechanical. Now I understand that strange things happen at 00 but this one>is real strange. I asked some top mathematicians to tell me what's going on>here and they all walked away from this one.Right.> So I am asking you folks to try>to explain what's going on without too much hand waving. I teach this>problem in the calculus sequence and I am not happy at all with what I tell>my students.Well, the first question is why is there anything that needsexplaining? The area is infinite and the volume is finite, so what?The answer is that it seems like a surface of infinite area shouldrequire infinitely much paint to cover it. But why does it seem thatway? Because when you think of painting a wall what you havein mind is covering it with a layer of paint of uniform thickness.If you have a wall with infinite area you can easily cover it withfinitely much paint, if the thickness of the layer of paint isnot required to be constant - the layer of paint gets thinner andthinner, so it takes only finitely much of it to cover that infinitearea. And that's what happens here - when you imagine filliingthe inside of the surface with finitely much paint the thicknessof the layer of paint tends to zero as x -> infinity.**David C. Ullrich === Subject: > What I think you might mean is that you wish to reconcile your> intuition with the result you've gotten. Well, given that the Riemann> integral is the limit of approximations of rectangles under the curve,> and that as x -> oo, f' -> 0, we can come up with an approximating> rectangle of arbitrary length. But that doesn't matter since f = 0> where this approximation is acurate. So the area under the curve in> that interval is zero.That makes no sense at all to me. === Subject: > Has anyone earned a bacalaureate degree in something while young, and> more than two decades later returned to earn a higher degree in> something else (and achieved it)?> If you reduce that to one decade then a famous example is Bernard Dwork:http://www.ams.org/notices/199903/mem-dwork.pdf-- John Leohttp://www.halfaya.org/leo/ === Subject: Found this morning in Yahoo science. Science - ReutersGeniuses, Criminals Do Best Work in Their 30s Wed Jul 9, 5:54 PM ETLONDON (Reuters) - Geniuses and criminals may not seem to havemuch in common but they both do their best work in their 30s -- andmainly to impress the opposite sex. When Satoshi Kanazawa, of the Universityof Canterbury in New Zealand, studiedbiographies of prominent, mostly malescientists he discovered that they madetheir key discovery before their mid 30s,around the same age that criminal behaviorpeaks. He believes the male competitive urge toattract females is a driving force for thescientific and criminal achievements,according to New Scientist magazine. They do whatever they do in order to get laid, said Kanazawa. He added that the competitive drive decreases with age and as men'spriority shifts from competing for women to taking care of theiroffspring. Kanazawa also found that marriage dampens the drive in botharenas, the magazine added. === Subject: >Now we are getting somewhere in resolving this. It's not really that>math is a young man's game. More like proving a highly original>mathematical result is usually a young man's game (where man can be>he or she, of course), subject to any number of various exceptions>mentioned by other posters.Of course there is the possibility that the causality goes the otherway: staying absorbed in mathematics tends to keep you young. THere'ssomething refreshingly child-like about mathematicians and theirabsorption with their discipline -- it seems like as long as theykeep at the mathematics, they never quite grow old and weary.Musicians too: Son: Dad, I've decided to be a musician when I grow up. Father: Now, son, you can't have it both ways.dave === Subject: In gp-parilet q1=Qfb(3,-1,15) and q2=Qfb(8,4,36)gp give q1*q2=Qfb(12,-4,24)q1 is 3x^2-xy+15y^2q2 is 8x^2+4xy+36y^2How by hand find 12x^2-4xy+24y^2 === Subject: > In gp-pari> let q1=Qfb(3,-1,15) and q2=Qfb(8,4,36)> gp give q1*q2=Qfb(12,-4,24)> q1 is 3x^2-xy+15y^2> q2 is 8x^2+4xy+36y^2> How by hand find 12x^2-4xy+24y^2 The same result (q1*q2) may be obtained using qc=qfbcompraw(q1,q2) qfbred(qc) With qfbcompraw(x,y): 'Gaussian composition without reduction of thebinary quadratic forms x and y' and qfbred(x) the 'reduction of thebinary quadratic form x' (that is the unique reduced form equivalent tox). Of course you could examine directly the source code (functioncomp_gen(GEN z,GEN x,GEN y) with x=q1=[3,-1,15] and y=q2=[8,4,36] of thefile src/basemath/arith2.c) to find out what qfbcompraw() is doing. A better way should be to look at Fernando Rodriguez-Villegasexplications bforms.txt (and gp code!) in 'Binary quadratic forms' here: http://www.ma.utexas.edu/users/villegas/cnt/ cnt-no-frames.html (his function compf() corresponding to pari's qfbcompraw() ) But concerning pari-gp Henry Cohen's book A course in ComputationalAlgebraic Number Theory is probably the best reference (Chapter 5 titleis Algorithms for Quadratic Fields and 5.4.2 is about Reduction andComposition of Quadratic Forms). I found there that : The set of classes of quadratic forms is in natural bijection with theclass group so that we may give a group structure to classes ofquadratic forms (Gauss 1798 : composition of quadatic forms). Note thatsince we work with classes of forms we need a reduction procedure(qfbred) but let's come to composition : Let (a1,b1,c1) and (a2,b2,c2) be two quadratic forms with the samediscriminant D= bk^2-4*ak*ck, and consider the corresponding ideals Ik =ak Z +(-bk+sqrt(D))/2 Z (k=1,2) then... set s=(b1+b2)/2, d=gcd(a1,a2,s)and let u,v,w be integers such that u*a1+v*a2+w*s=d then I1.I2= d(A Z+(-B+sqrt(D))/2 Z) where A= d0*a1*a2/d^2, B=b2+2*a2/d*(v*(s-b2)-w*c2) and d0=1 if at least one of the forms (a1,b1,c1) or (a2,b2,c2) isprimitive (gcd(ai,bi,ci)=1 if i=1 or 2) and in general d0=gcd(a1,a2,s,c1,c2,n) whew n=(b1-b2)/2 (The Ideal I3=I1.I2 is generated as a Z-module by the four products ofthe generators of I1 and I2, i.e. by g1=a1*a2,g2=(-a1*b2+a1*sqrt(D))/2, g3=(-a2*b1+a2*sqrt(D))/2 andg4=((b1*b2+D)/2-s*sqrt(D))/2 ) To resume : (a3,b3,c3)= (d0*a1*a2/d^2, b2+2*a2/d*(v*(s-b2)-w*c2),(b3^2-D)/(4*a3)) modulo the action of Gamma_oo (not defined here..) Hoping it helped a little, Raymond === Subject: 3F0DF384.57F7D9F9@free.fr...>> In gp-pari> let q1=Qfb(3,-1,15) and q2=Qfb(8,4,36)> gp give q1*q2=Qfb(12,-4,24)> q1 is 3x^2-xy+15y^2> q2 is 8x^2+4xy+36y^2> How by hand find 12x^2-4xy+24y^2>> The same result (q1*q2) may be obtained using> qc=qfbcompraw(q1,q2)> qfbred(qc)>> With qfbcompraw(x,y): 'Gaussian composition without reduction of the> binary quadratic forms x and y' and qfbred(x) the 'reduction of the> binary quadratic form x' (that is the unique reduced form equivalent to> x).>The composition of binary quadratic form who have not same discriminant isit possible?Again thanks. === Subject: > The composition of binary quadratic form who have not same discriminant is> it possible?> Again thanks. I'm (very) far from being an expert in these topics but I think no andthat the composition is only defined for a fixed discriminant D. Anyway the ideal generators (g1=a1*a2, g2=(-a1*b2+a1*sqrt(D))/2..)don't seem to apply if D is replaced by Dk so that... pari was probablyto generous to return you any answer for your product (political choiceI link otherwise...). Raymond === Subject: I'd be much obliged if someone pointed out some of the most well-knownapplications of Hahn-Banach theorems (extension or separation) inmathematics. It could be some not-so-difficult theorems in proofs of whichHBT is used. I'd be happy hearing of even a few examples.olej === Subject: > I would like to know if it is possible to solve an elliptic integral> of the second kind with maple and how it is possible.> The integral I have to solve is like the following:> EllipticE(((h0^2+2-2*sin(t1))/h0^2)^(1/2),(h0^2/(h0^2+4))^(1/2 ))> Claudia**What do you mean by solve? We solve equations; integralsare evaluated. If you are asking whether an ellipticintegral can be expressed as a finite combination ofelementary functions, the answer is no. _________________________________________________________ Eric J. Wingler (wingler@math.ysu.edu)Dept. of Mathematics and StatisticsYoungstown State UniversityOne University PlazaYoungstown, OH 44555-0001330-941-1817 === Subject: Dear All,I saw this wonderful documentary showcasing the inventions ofHeron (or Hero) of Alexandria. He apparently made a numberof inventions for temples and also theatre. He may very wellbe history's first special effects expert. These web-sitesshow pictures of his many inventions:steam 'sphere' invention:http://www.smith.edu/hsc/museum/ancient_inventions /steamengine2.htmlholy water dispenser:http://www.smith.edu/hsc/museum/ancient_inventions /hsc18b.htmself-moving stand:http://www.smith.edu/hsc/museum/ancient_inventions/ hsc20b.htmApparently, Heron also gave us his formula know as Heron's formula.Can anyone out there tell me if Heron's formula applies to anyof his inventions and/or his applications to theatre?If so, I would love to see an example.Tony ScottUniv. of BielefeldGermany === Subject: > Can anyone out there tell me if Heron's formula applies to any> of his inventions and/or his applications to theatre?>The area of a triangle with sides a, b and c is:sqrt(p*(p-a)*(p-b)*(p-c)) where p = (a+b+c)/2.Here is your Heron's formula....--Julien Santini,CMI Technop.99le de Ch.89teau-Gombert, FranceHome page: http://www.analgebra.com === Subject: Utilisateur1> Can anyone out there tell me if Heron's formula applies to any> > of his inventions and/or his applications to theatre?>> The area of a triangle with sides a, b and c is:> sqrt(p*(p-a)*(p-b)*(p-c)) where p = (a+b+c)/2.> Here is your Heron's formula....I read somewhere that Archimedes (287-212 BC) was in possession of thisformula, long before Heron (c. 60 AD). I've seen it used in calculationsabout sphere-packing and lattices in 3-space, but that's all.Larry === Subject: Is it possible to strectch this logic to Spherical Trigonometry ?[ sides of the spherical triangle a,b,c now are a-> a/R, b->a/R, c->c/R where R is radius of sphere ] Is the area of a sphereical triangle with sides a, b and c equal to sqrt(p*(p-a)*(p-b)*(p-c)) where p = (a+b+c)/2 ? === Subject: > Is it possible to strectch this logic to Spherical Trigonometry ?> [ sides of the spherical triangle a,b,c now are a-> a/R, b->a/R, c->c/R> [ where> R is radius of sphere ]> Is the area of a sphereical triangle with sides a, b and c equal to> sqrt(p*(p-a)*(p-b)*(p-c)) where p = (a+b+c)/2 ?No, as any fule kno, the area is A + B + C - pi(A, B and C angles).-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === Subject: Sorry if this problem seems very simple, I'm a student and I've used all other possible resources I know of, and none have helped.reCDVst>> >In a city, electricity charges are calculated as follows:Units used Charges per unit (cents)first 20 25next 20 35the rest 45If the electricity charge is x cents and the number of units used is n, write down a formula connecting x and n.>> > === Subject: I dont know why, but i cant seem to get my head around the idea thatyou can create two identical shapes, using the same parts yet get ahole from nowhere ..its probably something really stupid, but i dont understand how eachshape, being made from the same shapes, dont have equal surfacearea's.surely not ? the implications are crazy .. (i think :) ).to see what im going on about, checkout;http://www.simeonmagic.com/triangle/triangle1.htm Quack Quack Quack === Subject: > I dont know why, but i cant seem to get my head around the idea that> you can create two identical shapes, using the same parts yet get a> hole from nowhere ..> its probably something really stupid, but i dont understand how each> shape, being made from the same shapes, dont have equal surface> area's.> surely not ? the implications are crazy .. (i think :) ).> to see what im going on about, checkout;> http://www.simeonmagic.com/triangle/triangle1.htm> Quack Quack QuackWhat makes you think that the shapes are the same? === Subject: > I dont know why, but i cant seem to get my head around the idea that> you can create two identical shapes, using the same parts yet get a> hole from nowhere ..> http://www.simeonmagic.com/triangle/triangle1.htmWhat are the areas of the(i) red triangles(ii) green trangles(iii) the complete first diagram(iv) the second complete diagram?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === Subject: > I dont know why, but i cant seem to get my head around the idea that> you can create two identical shapes, using the same parts yet get a> hole from nowhere ..> its probably something really stupid, but i dont understand how each> shape, being made from the same shapes, dont have equal surface> area's.> surely not ? the implications are crazy .. (i think :) ).> to see what im going on about, checkout;> http://www.simeonmagic.com/triangle/triangle1.htm> Quack Quack Quackstraight line. === Subject: >>I dont know why, but i cant seem to get my head around the idea that>you can create two identical shapes, using the same parts yet get a>hole from nowhere ..>>its probably something really stupid, but i dont understand how each>shape, being made from the same shapes, dont have equal surface>area's.>>surely not ? the implications are crazy .. (i think :) ).>>to see what im going on about, checkout;>>http://www.simeonmagic.com/triangle/triangle1. htm>>Quack Quack QuackSomeone should mention the Banach-Tarski paradox here,just to add to the confusion. (I'm sure not about to, though...)**David C. Ullrich === Subject: > I dont know why, but i cant seem to get my head around the idea that> you can create two identical shapes, using the same parts yet get a> hole from nowhere ..All such puzzles are based on an illusion. Mostcommonly, something that appears to be a triangleis not a triangle. That's the case here. Notice thatthe big (red) triangle has a slope of 3:8 and the little(blue-green) triangle has a slope of 2:5.> its probably something really stupid, but i dont understand how each> shape, being made from the same shapes, dont have equal surface> area's.They do. The amount of colored area in both cases isthe same. The illusion is that you think both outlines(the big triangles the pieces create) are the same,and they aren't.There are many such illusions. This is a particularlyclever one.> surely not ? the implications are crazy .. (i think :) ).> to see what im going on about, checkout;> http://www.simeonmagic.com/triangle/triangle1.htmHere's the trick. As I said, the little triangle hasa slope of 2:5 = 0.4. The big red triangle has a slopeof 3:8 = 0.375. It's slightly lower in slope. So theupper shape has a hypotenuse that bends slightlyinward, while the lower shape has a hypotenuse thatbends slightly outward. That outward bend accountsfor the extra area.If you print this puzzle out and put your eye downclose to the paper and look along the hypotenuseat a very low angle, you'll see the bend I'm talkingabout. - Randy === Subject: > straight line.I think you'll find that it's actually 3/5 which isn't equal to 5/8 :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === Subject: this is a pretty old problem; unfortunately,I couldn't gues tha answer when I first saw it, andI can't recall the nifty answer, either. > http://www.simeonmagic.com/triangle/triangle1.htm--A church-school McCrusade (Blair's ideals?):Harry-the-Mad-Potter want's US to kill Iraqis?...For a 1000-year anglo-american hegemony?HEY, JIMMY; LET'S US and SU FIGHT -then-PM of England & Zbiggy http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX= === Subject: >>I dont know why, but i cant seem to get my head around the idea that>you can create two identical shapes, using the same parts yet get a>hole from nowhere ..>>its probably something really stupid, but i dont understand how each>shape, being made from the same shapes, dont have equal surface>area's.>>surely not ? the implications are crazy .. (i think :) ).>>to see what im going on about, checkout;>> >http://www.simeonmagic.com/triangle/triangle1.htm>> >Quack Quack Quack> Someone should mention the Banach-Tarski paradox here,> just to add to the confusion. (I'm sure not about to, though...)I think we are all glad that you didn't ;o)RonL === Subject: > Someone should mention the Banach-Tarski paradox here,> just to add to the confusion. (I'm sure not about to, though...)If that's the one I think it is, why not?It would be a cheap way to whip up a second sun, using onlyexisting materials conveniently at hand, to set aside forwhen this one burns out.xanthian, long range worry-wart.http://www.y5b.com/-- === Subject: (strange with the timezone, i started this thread with my question,but my post is half-way down the list, its not listed first.. howweird! - im not a regular google.groups user, still unsure how it allworks)Nice explanation!but ... im still having trouble believing that to be the case, as thefirst thing i did is go and create my own, on my own grid - in anaccurate CAD drawing program.Step 1:First i created a triangle 13,5. (not cut up as the puzzle describes)- so i *know* the hypotenuse is an exact line and it is infact aperfect triangle.Step 2:I then drew cut lines on this shape as in the puzzle and colored itin. (how fun :) ).Now i know this is a perfect triangle, and the shapes are measured asthey should be, as the puzzle describes them.Step 3:i then took a copy of this, re-arranged the shapes to create triangle#2. and yes, the hole is there.Step 4:I then took both of these, and doing a pixel-by-pixel comparison, theyare infact an exact fit. the total shapes are exactly the same, andthe hyptonuse is *not* bent, as i started with a perfect triangle tobegin with. (so even my final measurements or comparisons in thatrespect are irrelevant) - i know it, because i drew it.so i still cant believe its an illusion of a 'bent' hypotenuse..theremust be another explanation.(the hypotenuse never changes, it started as an uncut straight line,and finishes as an uncut straight line)draw one yourself, in the order i did - it removes the possibilitythat it is not a true triangle. ( i think :) ):)Quack Quack Quack> I dont know why, but i cant seem to get my head around the idea that> you can create two identical shapes, using the same parts yet get a> > hole from nowhere ..> All such puzzles are based on an illusion. Most> commonly, something that appears to be a triangle> is not a triangle. That's the case here. Notice that> the big (red) triangle has a slope of 3:8 and the little> (blue-green) triangle has a slope of 2:5.> its probably something really stupid, but i dont understand how each> shape, being made from the same shapes, dont have equal surface> area's.> They do. The amount of colored area in both cases is> the same. The illusion is that you think both outlines> (the big triangles the pieces create) are the same,> and they aren't.> There are many such illusions. This is a particularly> clever one.> surely not ? the implications are crazy .. (i think :) ).> to see what im going on about, checkout;> http://www.simeonmagic.com/triangle/triangle1.htm> Here's the trick. As I said, the little triangle has> a slope of 2:5 = 0.4. The big red triangle has a slope> of 3:8 = 0.375. It's slightly lower in slope. So the> upper shape has a hypotenuse that bends slightly> inward, while the lower shape has a hypotenuse that> bends slightly outward. That outward bend accounts> for the extra area.> > If you print this puzzle out and put your eye down> close to the paper and look along the hypotenuse> at a very low angle, you'll see the bend I'm talking> about.> > - Randy === Subject: >(strange with the timezone, i started this thread with my question,>but my post is half-way down the list, its not listed first.. how>weird! - im not a regular google.groups user, still unsure how it all>works)>Nice explanation!>>but ... im still having trouble believing that to be the case, as the>first thing i did is go and create my own, on my own grid - in an>accurate CAD drawing program.>>Step 1:>First i created a triangle 13,5. (not cut up as the puzzle describes)>- so i *know* the hypotenuse is an exact line and it is infact a>perfect triangle.Then the two small triangles are not right triangles and willnot fit together exactly. If they are triangles with exact rightangles, then one with sides of 2 and 5 will not line up with one withsides 3 and 8.You can NOT divide a triangular shape up into shapes of exactly theshapes and sizes you think you have here without fudging. >>Step 2:>I then drew cut lines on this shape as in the puzzle and colored it>in. (how fun :) ).>>Now i know this is a perfect triangle, and the shapes are measured as>they should be, as the puzzle describes them.>>Step 3:>i then took a copy of this, re-arranged the shapes to create triangle>#2. and yes, the hole is there.>>Step 4:>I then took both of these, and doing a pixel-by-pixel comparison, they>are infact an exact fit. the total shapes are exactly the same, and>the hyptonuse is *not* bent, as i started with a perfect triangle to>begin with. (so even my final measurements or comparisons in that>respect are irrelevant) - i know it, because i drew it.>>so i still cant believe its an illusion of a 'bent' hypotenuse..there>must be another explanation.You've created a different gap. There's a different puzzle where youstart with an 8x8 square (area 64) and reassemble it into a 5 x 13rectangle (area 65) with no apparent gap. The gap is a slight onealong the main diagonal.>(the hypotenuse never changes, it started as an uncut straight line,>and finishes as an uncut straight line)Something is not exactly where you think it is. I guarantee it, by thePythagorean theorem. Probably as I said your two triangular pieces arenot right triangles and/or your two rectangular pieces do not haveexactly square corners.coordinates of the lines that created the large shape and the smalltriangles, and I'll show you where you had to fudge. - Randy === Subject: > (strange with the timezone, i started this thread with my> question, but my post is half-way down the list, its not listed> first.. how weird! - im not a regular google.groups user, still> unsure how it all works)Your posts are disseminated immediately, but show up in Google's own list only later. Sometimes a thread decomposes into two or more chunks that are still grouped by their topic but may appear mangled.> Step 1:> First i created a triangle 13,5. (not cut up as the puzzle> describes) - so i *know* the hypotenuse is an exact line and it> is infact a perfect triangle.> Step 2:> I then drew cut lines on this shape as in the puzzle and colored> it in. (how fun :) ).> Now i know this is a perfect triangle, and the shapes are> measured as they should be, as the puzzle describes them.They cannot be. Try what I did after you got me wondering:1. Draw the individual pieces-- light green 200x100 + 300x200 orange 200x200 + 300x100 dark green 500x200 rectangular red 800x300 rectangular;2. put them together like on the webpage;3. draw a crop line (say, in blue) from bottom left to top right;4. realize that the hypotenuse does *not* pass through the vertex where red and dark green meet. If it does in the construction you described above, something is wrong with your parts.Martin === Subject: > (strange with the timezone, i started this thread with my question,> but my post is half-way down the list, its not listed first.. how> weird! - im not a regular google.groups user, still unsure how it all> works)> Nice explanation!> but ... im still having trouble believing that to be the case, as the> first thing i did is go and create my own, on my own grid - in an> accurate CAD drawing program.> Step 1:> First i created a triangle 13,5. (not cut up as the puzzle describes)> - so i *know* the hypotenuse is an exact line and it is infact a> perfect triangle.In the upper figure, you have a red triangle with corners (0,0),(8,0) and (8,3).You drew a straight line of slope 5/13 starting at (0,0).This will not go through the point (8,3). It goes throughthe points (7.8,3) and (8,3.0769). So in fact you can notmake a cut as shown on this hypotenuse at the point (8,3)since (8,3) is in the interior of your perfect triangle.What are the coordinates of the point where you made yourcut on the hypotenuse?I guarantee you there is a slight misalignment somewhere,a long narrow gap or a line bent which you think isstraight. But as you've changed the puzzle, you've changedwhere the illusion is. - Randy === Subject: > (strange with the timezone, i started this thread with my question,> but my post is half-way down the list, its not listed first.. how> weird! - im not a regular google.groups user, still unsure how it all> works)> Nice explanation!> > but ... im still having trouble believing that to be the case, as the> first thing i did is go and create my own, on my own grid - in an> accurate CAD drawing program.> Step 1:> First i created a triangle 13,5. (not cut up as the puzzle describes)> - so i *know* the hypotenuse is an exact line and it is infact a> perfect triangle.> Step 2:> I then drew cut lines on this shape as in the puzzle and colored it> in. (how fun :) ).> Now i know this is a perfect triangle, and the shapes are measured as> they should be, as the puzzle describes them.Nope, they're not, as I pointed out in previous messages. Theunits don't add up.I haven't seen your response yet (long Google delay) so Imade some assumptions. Here's how it works out with myassumptions:1. Draw a perfect right triangle with legs 13 and 5.2. Draw a vertical line at x=8. This will intersect the hypotenuseat y = 3.0769 (not 3).3. Draw a horizontal line. This creates a small trianglewhich has legs of 5 (as described) and 1.9231 (not 2).4. Outline the lower L-shaped piece just as described:the upper edge goes right 2 units, up 1 unit, right 3 units.5. The upper L-shaped piece has segment lengths as follows,counterclockwise from lower left: right 2 units, up 1 unit,right 3 units, up 1.0769 units, left 5 units, down 2.0769units.These will not fit together as neatly as shown in thesecond diagram. I'm guessing that when you reassembledthem in your CAD package, you had some overlap. A pieceof something is lying partly on top of one of theother pieces. The way I described the cut, you'retrying to fit an L-shaped piece with an edge of 2.0769units next to a triangle with an edge of 2.0 units.It will stick up. Where the two edges of the L's meet,you have one edge of length 1.9231 and one of length2.0. They won't match up either. - Randy === Subject: > > (strange with the timezone, i started this thread with my> question, but my post is half-way down the list, its not listed> first.. how weird! - im not a regular google.groups user, still> unsure how it all works)> Your posts are disseminated immediately, but show up in Google's own > list only later. Sometimes a thread decomposes into two or more > chunks that are still grouped by their topic but may appear mangled.Google appears to sort by title. This leads to two problems: 1. If you use a title that was used earlier, even yearsago, the new message becomes part of the old thread. Thishappens a lot with generic titles like Help or Mathor Prime numbers. 2. Sometimes people edit the subject lines. Google breaksthe thread and starts a new one.Also occasionally google shows a thread multiple timeswith the same title but slightly different messages inthe list. I think their database is breaking and I'vewritten them about other weird symptoms, but I don't knowif they believe me. - Randy === Subject: For a set D where a member of D is of the form (x,y,z)and represents a point in 3D cartesian space would it be reasonableto assume the following? That if D has 3 or less members then all members lie within a single plane. That if D has 4 members then co-planarity can be confirmed by calculating an appropriate determinant( A 'fast' method for calculating determinants would be appreciated.) and testing for a non-zero value.(If non zero the points do not all lie on the same plane. The reason for asking is that I was because the supply of a set of non co-planar points confuses some computer graphics engines when generating 'polygons'.As the application I was designing allows users to select random pointsto form a polygon some kind of check was needed.An additonal question should perhaps be, would limiting user selectionto 4 sided shapes be a good idea in such an application? Or is there an extension to the checks that can be made to allows more points to beselected?FMNT80 === Subject: >For a set D where a member of D is of the form (x,y,z)>and represents a point in 3D cartesian space would it be reasonable>to assume the following?> That if D has 3 or less members then all members lie within> a single plane.Certainly.> That if D has 4 members then co-planarity can be confirmed by> calculating an appropriate determinant( A 'fast' method for > calculating determinants would be appreciated.) and testing > for a non-zero value.(If non zero the points do not all lie on the same> plane.Yes. With the appropriate caveat that if this is done in floating-pointthe result is not likely to be exactly 0 even when it really should be,so you have to decide how small a number is to be considered as 0.> The reason for asking is that I was because the supply of a set of non> co-planar points confuses some computer graphics engines when generating> 'polygons'.>As the application I was designing allows users to select random points>to form a polygon some kind of check was needed.>An additonal question should perhaps be, would limiting user selection>to 4 sided shapes be a good idea in such an application? Or is there >an extension to the checks that can be made to allows more points to be>selected?In general, for n points you want to check that the 3 x (n-1) matrix A whose columns are the vectors from one point to the others has rank <= 2. This is true if and only if det(A A^T) = 0.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: >For a set D where a member of D is of the form (x,y,z)>and represents a point in 3D cartesian space would it be reasonable>to assume the following?> That if D has 3 or less members then all members lie within> a single plane.> Certainly.Certainly 3 points in general position give a single plane,and 3 points are always co-planar, but if they are co-linear or coincident, they don't define a unique plane.> That if D has 4 members then co-planarity can be confirmed by> calculating an appropriate determinant( A 'fast' method for> calculating determinants would be appreciated.) and testing> for a non-zero value.(If non zero the points do not all lie on the same> plane.> Yes. With the appropriate caveat that if this is done in floating-point> the result is not likely to be exactly 0 even when it really should be,> so you have to decide how small a number is to be considered as 0....> ... is there>an extension to the checks that can be made to allows more points to be>selected?> In general, for n points you want to check that the 3 x (n-1) matrix A> whose columns are the vectors from one point to the others has rank <= 2.> This is true if and only if det(A A^T) = 0.One could use three points to compute a,b,c,d for equation of planeax+by+cz+d=0 and then test whether other points satisfy that equation.I think this uses very nearly the same number of adds and multiplies,so probably won't be noticably faster or slower, but it has the advantage of allowing one to designate three master points that others are tested against, and others can be accepted or rejected individually rather than en masse.-jiw === Subject: I got this:x^2 = -251 (mod 300)so i realize it's nothing else butx^2 = 49 (mod 300)and it seems to be more than a coencidense that i getsquare on both sides of the =. Nice, i think to myself.Then, i don't get any further. I suspect i has somethingto do with the fact that 49 and 300 are relatively primebut it'd be nice if somebody poke me in right direction.-- V.8anligenKonrad--------------------------------------- ------------phone #1: (+46/0) 708 - 70 73 92phone #2: (+46/0) 704 - 79 96 95url: http://konrads.webbsida.com--------------------------------- --Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be lazy--------------------------------------------------- === Subject: =>I got this:>x^2 = -251 (mod 300)>so i realize it's nothing else but>x^2 = 49 (mod 300)>and it seems to be more than a coencidense that i get>square on both sides of the =. Nice, i think to myself.>Then, i don't get any further. I suspect i has something>to do with the fact that 49 and 300 are relatively prime>but it'd be nice if somebody poke me in right direction.Write the equation as (x-7)(x+7) = 0 (mod 300)So x-7 and x+7 are two integers whose product is divisible by 300. Consider the prime factorization of 300...Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: >x^2 = 49 (mod 300)> Write the equation as> (x-7)(x+7) = 0 (mod 300)>> So x-7 and x+7 are two integers whose product is divisible by 300.> Consider the prime factorization of 300...I find solutions like x = 157, 257 157^2 = 24649; 257^2 = 66049Anything simplier than the ensuing?--if n|300 and (x-7)(x+7) = 0 (mod 360)x-7 = n + 300k for some kx+7 = 300/n + 300j for some jn + 300k + 7 = x = 300/n + 300j - 7n^2 + 14n + 300n(k-j) - 300 = 0(n+7)^2 = 300 + 300n(j-k) + 49x = n + 7 = sqr(49 + 300i) for some i i x 0 7 6 43 38 107 68 143 82 157 124 193 220 257 286 293FOR n = 0 TO 300 s = SQR(49 + 300 * n) IF INT(s) = s THEN PRINT n, sNEXT---- === Subject: Sorry, still nothing...I get this:(x-7)(x+7) = 0 mod (2 * 2 * 3 * 5 * 5) === Subject: > Sorry, still nothing...> I get this:> (x-7)(x+7) = 0 mod (2 * 2 * 3 * 5 * 5)> x = 293 or x = 307> but furhter on i can't see. Also, the hint about factorization> of 300 in primes isn't used here. What did i miss?Given that a product of two factors (x-7)(x+7)is divisible by (2 * 2 * 3* 5 * 5), you can conclude that each of those primes must divide one ofthe two factors.For example, one possible solution is for (x-7) to be divisible by (2 *3) and for (x+7) to be divisible by (2 * 5 * 5).Can you think of an x that makes that happen? Can you think of any otherways to make (x-7)(x+7) divisible by (2 * 2* 3 * 5 * 5)?-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: > Sorry, still nothing...> I get this:> (x-7)(x+7) = 0 mod (2 * 2 * 3 * 5 * 5)> x = 293 or x = 307> but furhter on i can't see. Also, the hint about factorization> of 300 in primes isn't used here. What did i miss?> KonradThe ring Z/(300) is not entire because 300 is not prime. This meansthat there can be a and b not equal to zero but so that a*b=0 mod(300)For example, consider a=2*2*3 and b=5*5 (see the factorization above).You can calculate all a and b for which this can happen from thefactorizatiion of 300 and then look for which combinations a-b=14holds. === Subject: > Sorry, still nothing...> I get this:> (x-7)(x+7) = 0 mod (2 * 2 * 3 * 5 * 5)> x = 293 or x = 307> but furhter on i can't see. Also, the hint about factorization> of 300 in primes isn't used here. What did i miss?> Konrad 300 = 2^2 3 5^2. x-7 and x+7 must both be even (otherwise, one is divisible by 2^2 and the other is odd - but (x+7)-(x-7) = 14 ... ). Also, they can't both be divisible by 5 (as _that_ would force 14 to be a multiple of 5). So there are just 4 ways to distribute the prime factors of 300 between x-7 and x+7: x-7 x+7 ------- ------- 2,3,5^2 2 2,3 2,5^2 2,5^2 2,3 2 2,3,5^2 First case: x = 150k + 7 Then x^2 = 75*300k^2 + 7*300k + 49 which is _always_ = 49 (mod 300). Second case: x = 50k - 7, so x^2 = 2500k^2 - 700k + 49. You need 100k^2 - 100k = 0 (mod 300) or k(k-1) = 0 (mod 3). If k = 3j, then x = 150j - 7 and x^2 = 49 (mod 300). But, _also_, if k = 3j+1, x = 150j + 43 and x^2 = 75*300j^2 + 43*300j + 1849 so x^2 = 49 (mod 300). Third case: pretty much like the second. You get either x = 150j + 7 or x = 150j + 43. Fourth case: x = 150k - 7 and this always works. So (to sum up) x^2 = 49 (mod 300) iff either x = 150k +/- 7 or x = 150j +/- 43 for k, j integers ... === Subject: >> Does anyone know of a notation or a scheme for describing >>rotational paths (locus of a point on a sphere)?>> In three-dimensions a single Pi/2 rotation is nicely captured by a >>3x3 orthogonal matrix, and that one can compose a sequence of them by >>multiplying the appropriate matrices in the same order.>> The question is, if one were to compose a sequence of these >>rotations, is there a notational scheme that's been devised to annotate >>each type of rotation (with a symbol) and to then describe such composed >>paths as a word, made up of the symbols?>> Would it be possible to associate one word with a unique path, or >>would there sometimes be multiple distinct paths satisfying a particular >>word?>> Anyone know of a good reference for this type of problem?>>MK> Euler angles and quaternions are ways of describing these types of> transformations.> --> John E. Prussingon two texts, and of them I second seems most promising so far.The readings I've found are:1. 'Quaternionic and Clifford Calculus for Physicists and Engineers', K. Gurlebeck & W. Sprossig, Wiley 19972. 'Rotations, Quaternions, and Double Groups', S. Altmann, Oxford 1986MK === Subject: >>> Does anyone know of a notation or a scheme for describing >rotational paths (locus of a point on a sphere)?> In three-dimensions a single Pi/2 rotation is nicely captured by a >3x3 orthogonal matrix, and that one can compose a sequence of them by >multiplying the appropriate matrices in the same order.> The question is, if one were to compose a sequence of these >rotations, is there a notational scheme that's been devised to annotate >each type of rotation (with a symbol) and to then describe such composed >paths as a word, made up of the symbols?> Would it be possible to associate one word with a unique path, or >would there sometimes be multiple distinct paths satisfying a particular >word?> Anyone know of a good reference for this type of problem?>>>MK>>>>> Euler angles and quaternions are ways of describing these types of>> transformations.>> -->> John E. Prussing>on two texts, and of them I second seems most promising so far.>The readings I've found are:>1. 'Quaternionic and Clifford Calculus for Physicists and Engineers', K. >Gurlebeck & W. Sprossig, Wiley 1997>2. 'Rotations, Quaternions, and Double Groups', S. Altmann, Oxford 1986>MKI forgot to include the general category of direction cosine matrices.These are 3x3 matrices whose elements are dot products of orthogonalunit vectors in the rotated frame and unrotated frame. Sequences ofrotations are represented by products of these matrices. --John E. PrussingUniversity of Illinois at Urbana-ChampaignDepartment of Aerospace Engineeringhttp://www.uiuc.edu/~prussingI guess there's no one smart enought to evaluate the Smart Model... I got no response.S. Enterprize Co. (Membership)http://www.s-enterprize.com/S. Enterprize (Science Journal)http://smart1234.s-enterprize.com/ === Subject: > I guess there's no one smart enought to evaluate the Smart Model...>> I got no response.Do you mean no submission to the web host or no feedback on the current sites.I have 7 publishable theories at www.adamskingdom.com under [Intellectual Feats]and several more requiring editing before I do another upload. If you providea hyperlink to an actual model I'll evaluate it, other than that I give you 8.3 out of 10for the com name s-enterprize and its development.Herc> S. Enterprize Co. (Membership)> http://www.s-enterprize.com/> S. Enterprize (Science Journal)> http://smart1234.s-enterprize.com/>> === Subject: > > I guess there's no one smart enought to evaluate the Smart Model...>> I got no response.>> Do you mean no submission to the web host or no feedback on the current sites.> I have 7 publishable theories at www.adamskingdom.com under [Intellectual Feats]> and several more requiring editing before I do another upload. If you provide> a hyperlink to an actual model I'll evaluate it, other than that I give you 8.3 out of 10> for the com name s-enterprize and its development.>> Herc>afterthought, ask Mitch from sci.logic to prepare some work, he is a brilliant logicianwithout the resources for a website and showed some interest in making one.Herc === Subject: >I guess there's no one smart enought to evaluate the Smart Model...Maybe we're all too smart to pay attention to anyone who thinks there's a z in Enterprise?-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/You find yourself amusing, Blackadder.I try not to fly in the face of public opinion. === Subject: > I guess there's no one smart enough to evaluate the Smart Model...No sorry, as we're all too smart, we're over qualified.> I got no response.Except from the elite few who are dumb enuf to respond.However to finish our devaluation we need toknow if the Smart Model is tall and blond. === Subject: Perhaps, on the other hand, everyone is TOO smart to do so....Chris Lomont> I guess there's no one smart enought to evaluate the Smart Model...>> I got no response.> S. Enterprize Co. (Membership)> http://www.s-enterprize.com/> S. Enterprize (Science Journal)> http://smart1234.s-enterprize.com/> === Subject: = Find the greatest positive integer p and smallest integer q , such that (1+ 1/pi)^{pi + p/3} < pi < (1+ 1/pi)^{pi + q/2} .It is true that (p,q) in { (3,4) , (3,5), (4,5) } ? === Subject: >> A. Yu. Olshanskii showed in 1982 (Groups of bounded period with> subgroups of prime order, Algebra Logic 21(1983), 369--418 [= English> translation]) that if p is a sufficiently large prime, there is an> infinite simple p-group all of whose proper subgroups are of order p.> Groups, is published by Kluwer; unfortunately, last time I checked> the price was $436. If you're interested in this stuff, it would be> less expensive to try to find a university library that has some of> his papers.> the original question, not so much detail. But this may not be possible.I just would add that Olshanskii's web page ishttp://www.math.vanderbilt.edu/~olsh/. Unfortunately, his old papers arenot available online, but the recent papers are, giving an adequateimpression about ideas and techniques involved in these constructions. === Subject: This is just a specific example of a recently posted result of mine,but I find it just barely interesting enough to post here.(True?)For 0 < y < 1, integral{0 to y} floor(y/x) (x/(1-x^2)^2) dx=(1/4) (1 - pi y cot(pi y))Leroy Quet === Subject: