mm-185 === Subject: penny on a corner problem. There apparently is a well-known problem that asks for the locus of the center of a penny in the corner of a box, positioned so that it touches all three perpendicular planes that meet at that corner. Can anybody refer meto a discussion of this problem?--J. Wetzel === Subject: Re: penny on a corner problem> There apparently is a well-known problem that asks for the locus of the> center of a penny in the corner of a box, positioned so that it touches all> three perpendicular planes that meet at that corner. Can anybody refer me> to a discussion of this problem?> --J. Wetzel> I haven't heard of the problem. Is the penny a disc (i.e.,2-dimensional) or a cylinder (3-dimensional)?Dale === Subject: Re: penny on a corner problemOriginator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow)>There apparently is a well-known problem that asks for the locus of the>center of a penny in the corner of a box, positioned so that it touches all>three perpendicular planes that meet at that corner. Can anybody refer me>to a discussion of this problem?I think this is discussed in Littlewood's Miscellany, but I'd have to checkto be sure.-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: penny on a corner problem>There apparently is a well-known problem that asks for the locus of the>center of a penny in the corner of a box, positioned so that it touches all>three perpendicular planes that meet at that corner. Can anybody refer me>to a discussion of this problem?> I think this is discussed in Littlewood's Miscellany, but I'd have to check> to be sure.Is Wetzel's question related to the piano movers problem? === Subject: Re: penny on a corner problem>There apparently is a well-known problem that asks for the locus of the>center of a penny in the corner of a box, positioned so that it touches all>three perpendicular planes that meet at that corner. Can anybody refer me>to a discussion of this problem?I had not heard of this before, but it shouldn't be too hard. Letthe penny (or one side of it) be a circle of radius 1 centred at p = (x_1,x_2,x_3), and the box the 'rst octant. Suppose n is a unit normal vector to the plane of the penny, and let {u,v,n} be orthonormal. Then the boundary of the penny isp + u cos(t) + v sin(t), 0 <= t < 2 pi. The minimum of the x coordinateis p_1 - sqrt(u_1^2+v_1^2) = p_1 - sqrt(1 - n_1^2), so n_1^2 = 1 - p_1^2, and similarly n_2^2 = 1 - p_2^2 and n_3^2 = 1 - p_3^2. So what we need isp_1^2 + p_2^2 + p_3^2 = 2 with 0 <= p_1, p_2, p_3 <= 1, i.e the locusis the intersection of the sphere of radius sqrt(2) centred at the originwith the cube of side 1 formed by the coordinate planes and the planes x_i = 1.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Quadratic residue formulaeI am looking for assistance on trying to prove the following statements.I have no formal number theory training, my interest is purely recreational.My knowledge and experience are therefore somewhat limited in proof andanalytical techniquesStatement 1If u=2n^2-n+2 and v=2n^2+n+2Then u^2 mod (u+v-1) =u-1 and v^2 mod (u+v-1) = u-1For all n>0 Length of fraction period in base u and v =6Length of fraction period in base u-1 and v-1 =3 Statement 2If u=2n^2-3n+3; v=2n^2-n+2; p=u+vThen u^2 and v^2 =(p-1) mod p for all u,v.Length of fraction period in base u and v =4For all nThere is also another case:Where u=n+1; v=n^2+n+1. Then u^2 and v^2 =(p-1) mod p for all u,v. Lengthof fraction period in base u and v =4For all n>0CheersRandall McDonnell === Subject: Re: constructing Hadamard Matrices> http://arxiv.org/PS_cache/math-ph/pdf/0206/0206018.pdf> which de'nes the notion of entropy for orthogonal matrices. The> entropy.> Here's an idea that I am not certain will work, but I thought I would> ask some experts: There is a law of nature which says All natural> processes act to maximize the entropy of a system.> So if we thought of an n x n orthogonal matrix as a description of> something that occurs in nature and subject to change by natural> forces, then this matrix would, as a result of these forces acting> upon it, eventually become Hadamard (assuming that there exists an n x> n Hadamard matrix. It is known that if n is not divisible by 4 and> n>2, then no Hadamard matrix exists).> The reason that I mention this is perhaps there is some algorithm> which simulates natural processes which will produce an orthogonal> matrix of maximum entropy, as de'ned in the paper that I mentioned.> And by looking at such a matrix, we could immediately determine> whether it is Hadamard or not.> If such an algorithm exists, then we could determine whether an n x n> Hadamard matrix exists for any n in real time (assuming that the> simulated natural process is fast enough).> Do any experts have any comments on this idea? I have no idea if it> could work or not, but I thought I would ask.> Thanks,> CraigCraig,The short answer is that I don't see a way to do this, although itwould be nice if one could. That's not to say that someone moreclever than I may be able to see something I don't.Please pardon me if the following discussion seems a little bitsimplistic. Mostly I'm trying to clarify in my own mind what thismaximum entropy condition means.Entropy in statistical physics is a measure of the number ofmicroscopic states. A rough way of stating the law of increase ofentropy in physical systems is that, as a system evolves in time, ittends to spend most of the time in con'gurations (of the macroscopicobservable quantities) that are associated with the greatest number ofmicroscopic states (consistent with any external constraints on thesystem). If you start the system off in a con'guration (of themacroscopic observables) that is very improbable (one that isassociated with few microscopic states), it will quickly evolve to amore probable con'guration (one that is associated with manymicroscopic states).It isn't the case that physical systems have any tendency to movetoward more probable con'gurations. It's merely that if they movefrom state to state in an unbiased way, they will more often 'ndthemselves in con'gurations with higher probability.however. In information theory, the Shannon entropy measures theminimum number of bits needed to encode a message given a set ofprobabilities on the occurrences of the letters. If there are nletters, and their probabilities of occurrence are p(1), ..., p(n),then the Shannon entropy is -Sum{i=1...n} p(i) ln p(i). (The lnshould be log base 2, but this is irrelevant in the present context.)The Shannon entropy is largest when all letters have equal probabilityof occurring. This means the number of bits needed to encode suchmessages is as large as possible, since one cannot take advantage ofsuch tricks as encoding more frequently occurring letters with shorterbit strings. Another way to see this is to note that there are moremessages in which all letters occur with equal frequency than thereare messages with any other distribution, so more information must beprovided to distinguish among them.A way to relate the two notions of entropy is to postulate somedynamical process that samples the space of all messages. Such asystem would tend to spend its time sampling those messages for whichthe letter distributions are nearly equal, since those messages aremuch more numerous than messages with very unequal distributions o§etters.Now Hadamard matrices can be regarded (up to normalization) as thosereal orthogonal matrices which have the special feature that all theirelements are of the same magnitude. Gadiyar, Maini, Padma, andSharatchandra are using Shannon entropy, with the squares of thematrix elements of the orthogonal matrix taking the role of theprobabilities, as a measure of the sameness of the elements. ThusHadamard matrices will maximize this entropy since all their entrieshave magnitude 1/Sqrt[n] (remembering the normalization). Any otherorthognal matrix will have elements of differing magnitudes. Theobservation of the above authors is mostly just a convenient way tode'ne Hadamard matrices in terms of the maximization of somequantity.The dif'culty in doing what you suggest is that, as members of theset of orthogonal matrices, Hadamard matrices are exceedingly*improbable* con'gurations. So using the orthogonal matrix itself asyour physical system won't work. A dynamical process that sampled thespace of orthogonal matrices in an unbiased way would never, inpractice, stumble on a Hadamard matrix. You could, of course, try todesign the dynamics of your system in such a way that it was biased infavor of Hadamard matrices, but this would amount to having an apriori notion of what Hadamard matrices ought to look like, and if youcould do this suf'ciently well, you'd already be well on your way tosolving the Hadamard conjecture.To use the Shannon entropy idea properly, the orthogonal matricesshould instead encode information about the probability distributionof some other physical sytem (which you would have to cook up). Thenby sampling this system as it evolves in time, you would measure theprobabilities. A dif'culty with this is that you would be measuringthe squares of the matrix elements, whereas the signs of the elementsare the all important thing. I don't see any solution to thisdif'culty at the moment. It's also not at all obvious what it mightmean for probability amplitudes to form an orthogonal matrix.However, a different method springs to mind for producing Hadamardmatrices using your idea. One could perform some sort of gradientascent on the space of orthogonal matrices, with the entropy as thequantity to be maximized. I suspect this method will suffer from theusual problem that there will be numerous local maxima once thematrices start to get large. But it is certainly worth doing someexperiments to 'nd out.-Will Orrick === Subject: Re: constructing Hadamard MatricesWill,Thank you for your opinion and insight. I thought about it a littlemore before I read your response and am happy that I came to the sameconclusion that you did that gradient ascent methods would be a greatidea for experiments. I am currently getting programs from NumericalRecipes in C together to try this. Lots of local extrema is the onlybarrier to this, but one can never know the extent of the barrierunless one tries it out.If it works, then it is reasonable that this method (stepping outsideof the integer space into the reals and performing nonlinearoptimization) can be used to construct other combinatorial structures.Another idea is calculating Ramsey Numbers using this method,(although the §avor is opposite that of constructing Hadamardmatrices) as a Ramsey Number can be thought of as the smallest numberin which the maximum entropy breaks down and becomes orderly,i.e., too much disorder produces order.Anyway, we'll never know unless we try.Thanks again,Craig> http://arxiv.org/PS_cache/math-ph/pdf/0206/0206018.pdf> which de'nes the notion of entropy for orthogonal matrices. The> entropy.> Here's an idea that I am not certain will work, but I thought I would> ask some experts: There is a law of nature which says All natural> processes act to maximize the entropy of a system.> So if we thought of an n x n orthogonal matrix as a description of> something that occurs in nature and subject to change by natural> forces, then this matrix would, as a result of these forces acting> upon it, eventually become Hadamard (assuming that there exists an n x> n Hadamard matrix. It is known that if n is not divisible by 4 and> n>2, then no Hadamard matrix exists).> The reason that I mention this is perhaps there is some algorithm> which simulates natural processes which will produce an orthogonal> matrix of maximum entropy, as de'ned in the paper that I mentioned.> And by looking at such a matrix, we could immediately determine> whether it is Hadamard or not.> If such an algorithm exists, then we could determine whether an n x n> Hadamard matrix exists for any n in real time (assuming that the> simulated natural process is fast enough).> Do any experts have any comments on this idea? I have no idea if it> could work or not, but I thought I would ask.> Thanks,> Craig> Craig,> The short answer is that I don't see a way to do this, although it> would be nice if one could. That's not to say that someone more> clever than I may be able to see something I don't.> Please pardon me if the following discussion seems a little bit> simplistic. Mostly I'm trying to clarify in my own mind what this> maximum entropy condition means.> Entropy in statistical physics is a measure of the number of> microscopic states. A rough way of stating the law of increase of> entropy in physical systems is that, as a system evolves in time, it> tends to spend most of the time in con'gurations (of the macroscopic> observable quantities) that are associated with the greatest number of> microscopic states (consistent with any external constraints on the> system). If you start the system off in a con'guration (of the> macroscopic observables) that is very improbable (one that is> associated with few microscopic states), it will quickly evolve to a> more probable con'guration (one that is associated with many> microscopic states).> It isn't the case that physical systems have any tendency to move> toward more probable con'gurations. It's merely that if they move> from state to state in an unbiased way, they will more often 'nd> themselves in con'gurations with higher probability.> however. In information theory, the Shannon entropy measures the> minimum number of bits needed to encode a message given a set of> probabilities on the occurrences of the letters. If there are n> letters, and their probabilities of occurrence are p(1), ..., p(n),> then the Shannon entropy is -Sum{i=1...n} p(i) ln p(i). (The ln> should be log base 2, but this is irrelevant in the present context.)> The Shannon entropy is largest when all letters have equal probability> of occurring. This means the number of bits needed to encode such> messages is as large as possible, since one cannot take advantage of> such tricks as encoding more frequently occurring letters with shorter> bit strings. Another way to see this is to note that there are more> messages in which all letters occur with equal frequency than there> are messages with any other distribution, so more information must be> provided to distinguish among them.> A way to relate the two notions of entropy is to postulate some> dynamical process that samples the space of all messages. Such a> system would tend to spend its time sampling those messages for which> the letter distributions are nearly equal, since those messages are> much more numerous than messages with very unequal distributions of> letters.> Now Hadamard matrices can be regarded (up to normalization) as those> real orthogonal matrices which have the special feature that all their> elements are of the same magnitude. Gadiyar, Maini, Padma, and> Sharatchandra are using Shannon entropy, with the squares of the> matrix elements of the orthogonal matrix taking the role of the> probabilities, as a measure of the sameness of the elements. Thus> Hadamard matrices will maximize this entropy since all their entries> have magnitude 1/Sqrt[n] (remembering the normalization). Any other> orthognal matrix will have elements of differing magnitudes. The> observation of the above authors is mostly just a convenient way to> de'ne Hadamard matrices in terms of the maximization of some> quantity.> The dif'culty in doing what you suggest is that, as members of the> set of orthogonal matrices, Hadamard matrices are exceedingly> *improbable* con'gurations. So using the orthogonal matrix itself as> your physical system won't work. A dynamical process that sampled the> space of orthogonal matrices in an unbiased way would never, in> practice, stumble on a Hadamard matrix. You could, of course, try to> design the dynamics of your system in such a way that it was biased in> favor of Hadamard matrices, but this would amount to having an a> priori notion of what Hadamard matrices ought to look like, and if you> could do this suf'ciently well, you'd already be well on your way to> solving the Hadamard conjecture.> To use the Shannon entropy idea properly, the orthogonal matrices> should instead encode information about the probability distribution> of some other physical sytem (which you would have to cook up). Then> by sampling this system as it evolves in time, you would measure the> probabilities. A dif'culty with this is that you would be measuring> the squares of the matrix elements, whereas the signs of the elements> are the all important thing. I don't see any solution to this> dif'culty at the moment. It's also not at all obvious what it might> mean for probability amplitudes to form an orthogonal matrix.> However, a different method springs to mind for producing Hadamard> matrices using your idea. One could perform some sort of gradient> ascent on the space of orthogonal matrices, with the entropy as the> quantity to be maximized. I suspect this method will suffer from the> usual problem that there will be numerous local maxima once the> matrices start to get large. But it is certainly worth doing some> experiments to 'nd out.> -Will Orrick === Subject: why is integration harder than differentiationsomeone asked this question, and i have been thinking about it for a while. why is integration harder than differentiation?'rst of all, is it true? integration is de'ned on a larger class of functions than differentiation, so in some sense, it is easier to show the existence of the integral than the derivative.but what we really want to know is why, given a function built out of certain elementary functions, it is easy to construct the derivative in terms of those elementary functions, but hard (and sometimes impossible) to construct the antiderivative in terms of those elemntary functions.from a practical standpoint, the reason is clear: there are rules for the derivatives of the two constructions you can do to elementary functions, namely the product and the composition. if there were a rule for the integral of the composition of two functions and for the product of two functions, then from that, we could write any integrals of elementary functions in terms of elementary functions.but we cannot. so why not? what is different about integration that makes it not have these rules? on the surface, the de'nitions of differentiation and integration seem at least slightly similar: take the limit as epsilon goes to zero of some algebraic operation on your function.i wanted to mutter something about differential Galois theory, but i think that would just have been a cover for the more honest i don't know.so, what do you say?-z === Subject: Re: why is integration harder than differentiationZig wrote in> someone asked this question, and i have been thinking about it for a > while. why is integration harder than differentiation? 'rst of all, is it true? integration is de'ned on a larger class of> functions than differentiation, so in some sense, it is easier to show> the existence of the integral than the derivative.> If you are dealing with a 'nite precision machine, then integration is actually easier than differentiation. Integration is a stable process while differentiation is ill-posed.I'll leave it to you to 'nd out why (hint: think about the de'nitions). - Tim === Subject: Re: why is integration harder than differentiationPosted-And-Mailed: yes === > .... what we really want to know is why, given a function built out of > certain elementary functions, it is easy to construct the derivative > in terms of those elementary functions, but hard (and sometimes > impossible) to construct the antiderivative in terms of those elemntary > functions.> from a practical standpoint, the reason is clear: there are rules for > the derivatives of the two constructions you can do to elementary > functions, namely the product and the composition. if there were a rule > for the integral of the composition of two functions and for the product > of two functions, then from that, we could write any integrals of > elementary functions in terms of elementary functions.> but we cannot.... This is not an explanation of why it's harder to integrate, but a comment on the major historical effect of that fact. Integrals came 'rst. There are quite a few arguments in Euclid (probably due to Eudoxus) and in Archimedes, which 'nd various areas, volumes and centres of gravity by using limits of sums of little bits. Modern writers often mention that Archimedes also handled a tangent to a spiral, but this isolated case is very unlike modern differentiation. Most of what we see as calculus-style arguments in the ancient and early modern periods were integrations. But integration is hard, so each new integral was a new research problem. Then came the 17th-century development of differentiation, and the Fundamental Thgeorem of the Calculus. It was Newton and Leibniz who appreciated that differentiation admits a collection of easy algorithms, and antidifferentiating is a practical way to 'nd a lot of integrals. _That_ is the sense in which Newton and Leibniz invented the calculus. Derivatives and integrals were already there before them, but those two men showed how easy differentiation is, and what a lot of integrals it lets you 'nd. Ken Pledger. === Subject: Re: why is integration harder than differentiationZig wrote in message> someone asked this question, and i have been thinking about it for a> while. why is integration harder than differentiation?> One major difference is that differentiation is a local operation whileintegration involves 'nite intervals -- okay, measurable sets, butintuitively it is de'ned over an extended region of the domain whiledifferentiation isn't. Norm === Subject: Re: why is integration harder than differentiation> Zig wrote in message>someone asked this question, and i have been thinking about it for a>while. why is integration harder than differentiation?> One major difference is that differentiation is a local operation while> integration involves 'nite intervals -- okay, measurable sets, but> intuitively it is de'ned over an extended region of the domain while> differentiation isn't.> Norm> differentiation is local and integration is nonlocal. hmm. i like that idea, but i am not convinced. actually, didn't i once read that one of the amazing things about deRham cohomology is that the differential forms on a manifold, which are all locally de'ned objects, can encode global information about the manifold? but i am not sure if that has any relevance here.anyway, just thinking outloud here, but the difference between integration and differentiation that makes the former hard and the latter easy (in the elementary calculus sense) is that differentiation is a sort of forward operation, and integration is an inverse operation, in a way that is not symmetric.i think the analogy with algebraic operations is good. anyone can square a small number in her head, but who can take the square root? solving the quadratic equation is hard, and solving certain quintics is impossible.but why does differentiation have to be the forward operation and integration have to be its inverse? could we make it go the other way? in other words, is there some fundamental property of integration that rules out the possibility of writing the integral of a product in terms of the integrals of the multiplicands? if we had a rule like that, and one for composition of functions, then integration would be as algorithmic as differentiation, and all elementary functions would be known.since i know that not all elementary functions have elementary integrals, then i must conclude that such a product rule cannot exist. but this is a very indirect way to see this, and sort of relies on the property i am trying to understand to explain this property. it is a bit circular. is there an obvious reason why there can be no integration rule for products? (the familiar product rule is not good enough here) === Subject: Re: why is integration harder than differentiationZig wrote in message> Zig wrote in message>someone asked this question, and i have been thinking about it for a>while. why is integration harder than differentiation?> One major difference is that differentiation is a local operationwhile> integration involves 'nite intervals -- okay, measurable sets, but> intuitively it is de'ned over an extended region of the domain while> differentiation isn't.> Norm> differentiation is local and integration is nonlocal. hmm. i like that> idea, but i am not convinced. actually, didn't i once read that one of> the amazing things about deRham cohomology is that the differential> forms on a manifold, which are all locally de'ned objects, can encode> global information about the manifold? but i am not sure if that has> any relevance here.> anyway, just thinking outloud here, but the difference between> integration and differentiation that makes the former hard and the> latter easy (in the elementary calculus sense) is that differentiation> is a sort of forward operation, and integration is an inverse> operation, in a way that is not symmetric.> i think the analogy with algebraic operations is good. anyone can> square a small number in her head, but who can take the square root?> solving the quadratic equation is hard, and solving certain quintics is> impossible.> but why does differentiation have to be the forward operation and> integration have to be its inverse? could we make it go the other way?> in other words, is there some fundamental property of integration that> rules out the possibility of writing the integral of a product in terms> of the integrals of the multiplicands? if we had a rule like that, and> one for composition of functions, then integration would be as> algorithmic as differentiation, and all elementary functions would beknown.> since i know that not all elementary functions have elementary> integrals, then i must conclude that such a product rule cannot exist.> but this is a very indirect way to see this, and sort of relies on the> property i am trying to understand to explain this property. it is a> bit circular. is there an obvious reason why there can be no> integration rule for products? (the familiar product rule is not good> enough here)>Concerning the dif'culty I guess we have to differentiate [no pun intended] between the de'niteintegral which is always with respect to a given set in the function'sdomain and the process of inde'nite integration which is just the reverseof differentiation. In an algebraic sense I don't think that inversedifferentiation is all that much more dif'cult than forwarddifferentiation. We have a set of rules to apply to elementary functionsand algebraic (and some transcendental) combinations of them todifferentiate a given function. If none of our usual rules apply, thendifferentiation is -- at least symbolically -- impossible and we're stuckwith a numerical approximation. Ditto the process of inversedifferentiation. This is generally thought to be dif'cult because mostpeople simply don't recognize the inverse rules as readily as the forwardones. There are tables to use in both cases.Concerning the inverse relationship Assuming that we de'ne the inde'nite integral in terms of the de'niteintegral, it's relatively easy to see that integration doesn't really careif a function has, for example, discrete jump discontinuities, the functionis still integrable. But that same function fails to be differentiable atthe points of discontinuity so in that case I'd have to say that the processof integration is easier than the process of differentiation. Norm === Subject: Fibonacci-type sequenceHi everybody,I was interested to see if any work had been done on the followingrecursion relation for numbers a(n),a(n) = F(n) a(n-1) + a(n-2)where F(n) is a function that depends on the labelling parameter n.The simplest example that I had looked at was F(n) = b/n, withconstant b. I can't seem to 'nd any information about its properties,so I would appreciate any input!Speci'cally, I am looking for sequences of numbers that do notoscillate too much, i.e. a(n+1) - a(n) -> 0 as n gets large. When I dothis numerically, it seems only to depend on the constant b beingpositive, independent of the starting values a(0) and a(1). Are therespecial values of a(0) and a(1) where b can be negative, and a(n) doesnot oscillate greatly?As I said, I am curious if general properties of these numbers havebeen worked out, or ideas about how to go about doing this.Thanks, Daniel === Subject: Re: Fibonacci-type sequence> Hi everybody,> I was interested to see if any work had been done on the following> recursion relation for numbers a(n),> a(n) = F(n) a(n-1) + a(n-2)> where F(n) is a function that depends on the labelling parameter n.> The simplest example that I had looked at was F(n) = b/n, with> constant b. I can't seem to 'nd any information about its properties,> so I would appreciate any input!> Speci'cally, I am looking for sequences of numbers that do not> oscillate too much, i.e. a(n+1) - a(n) -> 0 as n gets large. When I do> this numerically, it seems only to depend on the constant b being> positive, independent of the starting values a(0) and a(1). Are there> special values of a(0) and a(1) where b can be negative, and a(n) does> not oscillate greatly?> As I said, I am curious if general properties of these numbers have> been worked out, or ideas about how to go about doing this.Thanks, Daniel==According to Poincare-Perron Theorem, there exists two particular solutions x_n(1) and x_n(2) such that x_{n+1}(k) lim_{n-->infty} ----------- =(-1)^k for k in {1,2} . x_n(k)See [1] Gelfond A.O., ,,Calculus of Finite Differences , (Russian), Nauka ,Moskow,1957 [2] Milne-Thompson L.M., ,,The Calculus of Finite Differences,Macmillan,London,1951 === Subject: Re: Fibonacci-type sequenceOriginator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow)>a(n) = F(n) a(n-1) + a(n-2)>where F(n) is a function that depends on the labelling parameter n.>The simplest example that I had looked at was F(n) = b/n, with>constant b. I can't seem to 'nd any information about its properties,I don't know the answers to any of your speci'c questions, but yourrecurrence has the same form as the three-term recurrence for a continuedfraction, so perhaps that literature will be helpful. Since your F(n)isn't necessarily a positive integer, the number-theoretic literatureprobably won't be as useful as the analysis literature (Pade approximationand so forth).-- Tim Chow tchow-at-alum-dot-mit-dot-eduThe range of our projectiles---even ... the artillery---however great, willnever exceed four of those miles of which as many thousand separate us fromthe center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Fibonacci-type sequence>I was interested to see if any work had been done on the following>recursion relation for numbers a(n),>a(n) = F(n) a(n-1) + a(n-2)>where F(n) is a function that depends on the labelling parameter n.>The simplest example that I had looked at was F(n) = b/n, with>constant b. I can't seem to 'nd any information about its properties,>so I would appreciate any input!Well, you can write the general solution of your example as a(n) = 2 p_n(b) a(0)/n! + q_n(b) a(1)/n! where p_n(b) = b p_{n-1}(b) + n (n-1) p_{n-2}(b) and similarly for q,p_0(b) = 1/2, p_1(b) = 0, q_0(b) = 0, q_1(b) = 1. For n >= 2,p_n and q_n are monic polynomials of degree n-2 and n-1 respectively,with nonnegative integer coef'cients;p_n is an even function if n is even and an odd function if n is odd,q_n is an odd function if n is even and an even function if n is odd.>Speci'cally, I am looking for sequences of numbers that do not>oscillate too much, i.e. a(n+1) - a(n) -> 0 as n gets large. When I do>this numerically, it seems only to depend on the constant b being>positive, independent of the starting values a(0) and a(1). Are there>special values of a(0) and a(1) where b can be negative, and a(n) does>not oscillate greatly?The generating function f(x) = sum_{n=0}^in'nity a(n) x^n satis'esthe differential equation (1-x^2) f'(x) - (b + 2 x) f(x) = a(1) - b a(0)with initial condition f(0) = a(0). And then the generating function of a(n) - a(n-1) is (1 - x) f(x).The solution of the differential equation, according to Maple, is / /x (- 1/2 b - 1) (1/2 b - 1) | | (x - 1) (x + 1) | | (-a(1) + b a(0)) | | |/0 (- 1/2 b) (1/2 b) a(0) | (_z1 + 1) (_z1 - 1) d_z1 + --------------------| / 1 | exp|-- I (b + 2) Pi|| 2 //This has singularities (in general, branch points) at x = 1 and x = -1.Now basically a singularity like (x-1)^p, (x+1)^p, (x-1)^p ln(x-1) or (x+1)^p ln(x+1) with p < -1 will result in terms whose absolutevalue goes to in'nity as n -> in'nity; p > -1 will make terms go to 0; (x-1)^(-1) has terms that are constant, (x+1)^(-1) has terms that oscillate but are bounded. At _z1 = -1, the integrand is const (_z1+1)^(-b/2) + O((_z1+1)^(1-b/2))When integrated, this leads to (1-x) f(x) of the form const (x+1)^(b/2 - 1) (1 + O(x+1)) + O(1). This if b < 0, there is a singularity at x=-1 which will result inunbounded oscillation, at least unless the constant turns out to be 0.For example, if b = -2 I getf(x) = ((3 - 4 ln(2)) a(0) + (1 - 2 ln(2)) a(1)) (x+1)^(-2) + O(1)as x -> -1, so this will make a(n) - a(n-1) unbounded unless (3 - 4 ln(2)) a(0) + (1 - 2 ln(2)) a(1) = 0. At _z1 = 1, the integrand is const (_z1-1)^(b/2) + O((_z1-1)^(b/2+1))and (1-x) f(x) should be const (1-x) + const (1-x)^(-b/2) + ...This would make a(n) - a(n-1) unbounded if b > 2.However, again there may be values of a(0) and a(1) making the constant 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: Paper published by Geometry and TopologyThe following paper has been published:Geometry and Topology, Volume 8 (2004) Paper no. 11, pages 475--509URL:http://www.maths.warwick.ac.uk/gt/GTVol8/paper11. abs.htmlTitle:Permutations, isotropy and smooth cyclic group actions on de'nite 4-manifoldsAuthor(s):Ian Hambleton, Mihail TanaseAbstract:We use the equivariant Yang-Mills moduli space to investigate therelation between the singular set, isotropy representations at 'xedpoints, and permutation modules realized by the induced action onhomology for smooth group actions on certain 4-manifolds.AMS Classi'cation Numbers. Primary: 58D19, 57S17Secondary: 70S15Keywords:Gauge theory, 4-manifolds, group actions, Yang-Mills, moduli spaceReceived: 29 July 2003Revised: 17 January 2004Accepted: 9 February 2004Published: 16 February 2004Proposed: Ronald FintushelSeconded: Ronald Stern, Robion KirbyAuthor(s) address(es):Department of Mathematics and Statistics McMaster University, Hamilton, ON L8S 4K1, CanadaEmail: ian@math.mcmaster.ca, tanasem@math.mcmaster.ca === Subject: Matrix OptimizationHello,I'm looking for pointer to papers or books to help me with a optimizationproblem where I want to 'nd the optimum matrix. The basic form of my equationise^2 = x^H (T - BS)^H (T - BS) xwhere e is error which I'm trying to minimize, ^H is the complex conjugatetranspose, x is an nx1 vector, T and S are projection matrices with equal ranksbut the rank is less than n, and B is the nxn matrix that I'm trying to 'nd.Further, T and S are diagonal matrices where only set of the diagonal elementare equal to 1 and the remaining elements are equal to 0. The set of diagonalelements equal 1 for T and S are not equal but few elements may be in commonand hence, we cannot 'nd a B such that T=BS. The sets have the same number ofelements and thus, the rank of T and S are equal.Does anyone have any suggestions how to solve this problem?Thanks,Scott === Subject: topology and analysis/differential geometryCan anyone of you provide me with information about the link betweentopology and analysis / differential geometry (maybe give me somelinks or reference to good books)? === Subject: Minimization of L0 norm via dynamic programming linux)Cancel-Lock: sha1:OmwbS4nDJ3zzEP9q2T/L0RAXiJw=Hello,I need to compute the vector f that minimizes a functional J(f), whereone of the terms is the L0 norm of f (or possibly another Lp normwhere p < 1). In this case J(f) is non convex and the minimisationalgorithm is a dynamic programming procedure.I would appreciate any link or reference on the topic.Thanks,-- J.8er.99me Kalifa === Subject: Re: derived tensor product Cancel-Lock: sha1:10RGsxVN7ZqYTArmGuSrdlBsFNw=> El Tue, 10 Feb 2004 21:16:42 +0100, Axel Vogt vas> dir: Let R be a commutative ring. Does there exist a nonzero object X in> the (unbounded) derived category for R so that the derived tensor> product of X with itself is zero? (I think that this is impossible if> R is noetherian, but what if it isn't?) For any n>=2, does there> exist an object X so that the n-fold derived tensor product of X with> itself is zero, but the (n-1)-fold derived tensor product is not?> Here's a related non-example: notice that as Z-modules, Q/Z tensor Q/Z> is zero. In the derived category of Z, though, I think that the> derived tensor product of Q/Z with itself is isomorphic to Q/Z in> degree 1 (or degree -1, depending on how you grade things). This is> correct, right? (I'm not very good with the derived tensor product,> but it's legitimate to compute it using §at resolutions, isn't it?) --> J. H. Palmieri> Dept of Mathematics, Box 354350 mailto:palmieri@math.washington.edu> University of Washington http://www.math.washington.edu/~palmieri/> Seattle, WA 98195-4350>If nobody else does i try a sketch based on my dusty Hartshorne [H]>Residues and Duality, Chap II 4 and 5.>For derived tensors (or local hyperTor) i think you need some left>boundedness (as you need §at resolutions). > No: you don't need boundedness.> This hypothesis is not necessary in order to derive functors in the> category of complexes.Right, so this makes a spectral sequence argument problematic: thatis, if you were trying to use a spectral sequence to show that if X isnonzero in the derived category, then so is X tensor X, there wouldprobably be serious convergence issues.-- J. H. PalmieriDept of Mathematics, Box 354350 mailto:palmieri@math.washington.eduUniversity of Washington http://www.math.washington.edu/~palmieri/Seattle, WA 98195-4350 === Subject: Re: 8th grade> Hi there,> I'm going to be teaching a lesson on slope for my methods class next week.> The teacher tells me the kids are average, and doing ok in their textbook.> My question is when is it appropriate to introduce Greek letters?> Since I'm going to be talking about calculating slope, of course the words> change in x and change in y are going to come up a lot.> Do you think they will understand if I tell them that DELTA x means change> in x?The notation f(x + h) - f(x) k --------------- = - (x + h) - x his _much_ better than f(x + delta x) - f(x) delta y ---------------------- = -------- x + delta x - x delta xBut you should brie§y explain the use of delta in case they see itelsewhere.-- G.C.Note ANTI, SPAM and invalid to be removed if you're e-mailing me.-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: 8th gradeThe discussion of slope of a line in an 8th grade class seems to me tohave veered off from the main point. First, introducing a Greek letter (delta) is a useless distraction. Greek letters are never needed in elementary math - they just distractcertain students, or cause a bit of confusion, or reinforce the notionsome have that math is hard, abstract, and pointless. We should bespending our time on teaching about interesting math, not seeminglyexotic notation. Ignore Greek letters entirely in K-14.Second, the main point about ïslope' is that it is an INVARIANTquantity, a constant, that is computable using ANY pair of points on aline L, given a Cartesian coordinate system X - Y in which L is notparallel to the Y axis. Such a fact would in physics be called aConservation Law. Everthing about the connection between lines andlinear equations falls out of this one fact, as do the equations of uniform linear motion (parametric representation of a line).We should always make a big fuss whenever we come upon invariants inmath -- it's an important and useful idea. It's the simplest kind ofpattern to notice. For example, circumference/diameter for allcircles is an invariant called pi, (area of inscribed circle)/(area ofsquare) is invariant pi/4, (volume of cone)/(volume of prism) isinvariant 1/3 for a cone inscribed in a prism [same base of any shape,same height], a sequence is arithmetic iff its 'rst differences areconstant [invariant is amount of increase between successive terms], asequence is geometric iff the ratios of its successive terms isconstant [growth factor], a sequence is shifted geometric iff theratios of its successive 'rst differnces is invariant, a sequence isquadratic iff its second differences are constant, a sequence is cubiciff its third differences are constant, 'rst differences of how far aball falls (or rolls down a slope, see Galileo) during succesive timeperiods (seconds, etc) is 2*(distance travelled in 'rst period), etc.But if the kids don't know about similarity, how will I convince themthat slope is indeed invariant? Well I might make a fewtransparencies for an overhead projector, each one having a Cartesiangrid and a single line L drawn on it, and on the line I would havemarked maybe 6 points A,B,C,D,E,F and beside each one the pair of itscoordinates -- which I should also be able to see from the grid andtic marks on the axes. There are then C(6,2)=15 pairs of points thatcan be used to calculate (change in Y)/(change in X) when moving fromone point to the other, and these calculations could be carried out bystudents in small groups assigned to use a couple of pairs. Everybodygets the same number (we hope) for 1 line, so that is its invariantslope, and then we try it for some other lines, each yielding its owninvariant slope calculation.(Note that I only teach college students.)-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: 8th gradeLadnor Geissinger wrote in message> The discussion of slope of a line in an 8th grade class seems to me to> have veered off from the main point.> First, introducing a Greek letter (delta) is a useless distraction.> Greek letters are never needed in elementary math - they just distract> certain students, or cause a bit of confusion, or reinforce the notion> some have that math is hard, abstract, and pointless. We should be> spending our time on teaching about interesting math, not seemingly> exotic notation. Ignore Greek letters entirely in K-14.How can you call something (use of Greek alphabet) that is mainstay inmathematics exotic notation?Are you implying pi should not be introduced until at least yr. 15??? Or doyou instead replace the Greek letter with something else, and call itsomething else?> Second, the main point about ïslope' is that it is an INVARIANT> quantity, a constant, that is computable using ANY pair of points on a> line L, given a Cartesian coordinate system X - Y in which L is not> parallel to the Y axis.Stop shouting (I thought this was a moderated group...I shouldn't have to bethe one to say this)Slope is not necessarily constant. For a determined line, perhaps, but whymust everything in k-12 math be a determined line in the context of slope.> Such a fact would in physics be called a> Conservation Law. Everthing about the connection between lines and> linear equations falls out of this one fact, as do the equations of> uniform linear motion (parametric representation of a line).> We should always make a big fuss whenever we come upon invariants in> math -- it's an important and useful idea. It's the simplest kind of> pattern to notice. For example, circumference/diameter for all> circles is an invariant called pi,You shouldn't be allowed, according to your own statements, to use any Greekhere. Seems its OK for you to do it, though, just not anyone else.Or did you forget pi is a Greek letter?Pi, is what you call ïinvariant' because it's a _constant_. Why not justcall it a constant, since that's what it is. In this context, its not anydifferent from any other constants (0, 1, etc.)_Slope_ is not necessarily constant. Even in the context of lines, itdoesn't have to be. Consider a general line of the form y=mx+b where m canbe any real. There. The slope is _not_ constant. (area of inscribed circle)/(area of> square) is invariant pi/4, (volume of cone)/(volume of prism) is> invariant 1/3 for a cone inscribed in a prism [same base of any shape,> same height], a sequence is arithmetic iff its 'rst differences are> constant [invariant is amount of increase between successive terms], a> sequence is geometric iff the ratios of its successive terms is> constant [growth factor], a sequence is shifted geometric iff the> ratios of its successive 'rst differnces is invariant, a sequence is> quadratic iff its second differences are constant, a sequence is cubic> iff its third differences are constant, 'rst differences of how far a> ball falls (or rolls down a slope, see Galileo) during succesive time> periods (seconds, etc) is 2*(distance travelled in 'rst period), etc.> But if the kids don't know about similarity, how will I convince them> that slope is indeed invariant?Because maybe its just _not_ so invariant as you think it is. You seem tobe using invariant differently than most. Normally, in mathematics whenwe say something is an invariant, we mean that it doesn't really changeafter going through some sort of transformation. You seem to be using it asa synonym for constant. They are really not the same concept. Invariantscan be variable; constants are just, well, constants. Well I might make a few> transparencies for an overhead projector, each one having a Cartesian> grid and a single line L drawn on it, and on the line I would have> marked maybe 6 points A,B,C,D,E,F and beside each one the pair of its> coordinates -- which I should also be able to see from the grid and> tic marks on the axes.Again, you have a determined line. Why do you think any discussion of aline at all must involve such a determined line? Can't we (and in fact,_don't_ we) speak of general cases, where the slope has not been speci'ed? There are then C(6,2)=15 pairs of points that> can be used to calculate (change in Y)/(change in X) when moving from> one point to the other, and these calculations could be carried out by> students in small groups assigned to use a couple of pairs. Everybody> gets the same number (we hope) for 1 line, so that is its invariant> slope, and then we try it for some other lines, each yielding its own> invariant slope calculation.> (Note that I only teach college students.)Are they juniors yet (so that you can show them Greek letters?) Justcurious. You said Greek letters should not be introduced at all in k-14.Let me get this straight. You teach _college_ students how to 'nd theslope of a given line by giving them a pair of points on the line? so when_do_ you get around showing them any Greek? Grad school?-- Darrell-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: 8th grade> The discussion of slope of a line in an 8th grade class seems to me to> have veered off from the main point. > First, introducing a Greek letter (delta) is a useless distraction. > Greek letters are never needed in elementary math - they just distract> certain students, or cause a bit of confusion, or reinforce the notion> some have that math is hard, abstract, and pointless. We should be> spending our time on teaching about interesting math, not seemingly> exotic notation. Ignore Greek letters entirely in K-14.Greek letters aren't scary. If you imply that using greek letters isgrown up (_We_ call the change in x delta x) it might even makethem want to learn it. Then USE deltas all over the place. My nephews memorized the names of lots of Yu-Gi-Oh! monsters at the agesof six and seven, and Pokemon before that, and so did all theirfriends. If they can comprehend the concept of change in x, I don'tthink that the name delta x is going to be a problem.> Second, the main point about ïslope' is that it is an INVARIANT> quantity, a constant, that is computable using ANY pair of points on a> line L, given a Cartesian coordinate system X - Y in which L is not> parallel to the Y axis. Such a fact would in physics be called a> Conservation Law. Everthing about the connection between lines and> linear equations falls out of this one fact, as do the equations of > uniform linear motion (parametric representation of a line).> We should always make a big fuss whenever we come upon invariants in> math -- it's an important and useful idea. It's the simplest kind of> pattern to notice. For example, circumference/diameter for all> circles is an invariant called pi, (area of inscribed circle)/(area of> square) is invariant pi/4, (volume of cone)/(volume of prism) is> invariant 1/3 for a cone inscribed in a prism [same base of any shape,> same height], a sequence is arithmetic iff its 'rst differences are> constant [invariant is amount of increase between successive terms], a> sequence is geometric iff the ratios of its successive terms is> constant [growth factor], a sequence is shifted geometric iff the> ratios of its successive 'rst differnces is invariant, a sequence is> quadratic iff its second differences are constant, a sequence is cubic> iff its third differences are constant, 'rst differences of how far a> ball falls (or rolls down a slope, see Galileo) during succesive time> periods (seconds, etc) is 2*(distance travelled in 'rst period), etc.> But if the kids don't know about similarity, how will I convince them> that slope is indeed invariant? Well I might make a few> transparencies for an overhead projector, each one having a Cartesian> grid and a single line L drawn on it, and on the line I would have> marked maybe 6 points A,B,C,D,E,F and beside each one the pair of its> coordinates -- which I should also be able to see from the grid and> tic marks on the axes. There are then C(6,2)=15 pairs of points that> can be used to calculate (change in Y)/(change in X) when moving from> one point to the other, and these calculations could be carried out by> students in small groups assigned to use a couple of pairs. Everybody> gets the same number (we hope) for 1 line, so that is its invariant> slope, and then we try it for some other lines, each yielding its own> invariant slope calculation.> (Note that I only teach college students.)Math should be fun. Calculating the slope of a single line 'fteentimes sounds dreadful. If you really must show that the slope of aline doesn't change, why not just cut out a right triangle and slideit along the line? Not a terribly exciting demonstration but it getsit over with more quickly so you can discuss more interesting topics.--Jeff-- Ho, ho, ho, hee, hee, heeand a couple of ha, ha, has;That's how we pass the day away,in the merry old land of Oz.-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Solving the stated problem not the problem you think up > Re: Solving the stated problem not the problem you think up > My comment:> I wasn't educated in the US, and American skepticism of testing has always > amused me.> Here's an example of why. Schools in Colorado are subject to both federal testing mandated by No Child Left Behind, and state CSAP testing.One school in the news recently recieved a perfect score on the federal tests. It may be closed soon due to inadaquete scores on the CSAP.(Actually, this is more an example of what happens when someone other than teachers designs the tests; the CSAP expects students to know things they won't be taught for another two years. Thus, skeptism that it serves any useful purpose)-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Solving the stated problem not the problem you think up > Re: Solving the stated problem not the problem you think up > My comment:> I wasn't educated in the US, and American skepticism of testing has always > amused me.> IMHO, no testing instruments can possibly be perfect. You can always 'nd > fault with a test. But if I fail a test, should I blame that on that §aw> and ignore my own share of the blame? hmmm. (same with other areas of > life...do we blame outcomes of athletic competitions or job interviews on the > scoring systems...academics is the one being singled out for nitpicking in > that regard)If there are biases in the test questions that re§ect something otherthan knowledge of the subject then the test isn't testing for what itpurports to be testing - it is testing that other something. Inathletic competitions, the testing is of very simple abilities. Ifthey added your shoe size into your high jump results then jumpingalone wouldn't necessarily get you the gold medal - the test would bebiased towards people with long feet. Sometimes people with shortfeet would do well in this new high jump, but they would have to workharder for reasons that had nothing to do with their jumping ability.The current academic tests can be like this new form of high jumpscoring and many people feel it would make sense to make the testsmore like the original form of high jump scoring.> Besides, where I grew up, testing serves a very important function:> It helps enforce a uniform curriculum so the rich and the poor students are > more or less learning the same thing (more than can be said > for the American schools!)Each state has an organization which sets the curriculum for eachgrade level and subject area, that should keep all schools in eachstate teaching all children the same things. How well the subjectsare taught (and how well the students learn them) is dependent on alot of other factors that may directly (fewer or inferior resources)or indirectly (psychological biases) re§ect economic level.A recent study, done in Arizona, showed that high stakes tests do notcorrelate with - in fact may be counterproductive to - future successin academia and the real world. Humans are not like widgets and thewhole premise that you can quality control their education by testingthem more appears to be faulty.> It gives poor and uneducated parents a hope of > getting their kids a chance to go up the socioeconomic ladder.The point is not to give people just a hope of this, but to try tomake it so that there's NO correlation between where you start off andwhere you end up. That is supposed to be goal of the US educationalsystem and that is a valid perspective from which to critiquestandardized tests.> To those > parents, the testing becomes the symbol of that hope. And my parents were > those who believed in it. That's how my brother and I managed to go to > college. We were taught not to blame anyone --not tests, not teachers or > schools-- but ourselves for our academic results. If a student can maintain > that attitude, no one can stop this child from excelling.> If I were taught in an American school, given my socioeconomic status back > then, I'd be in an inner-city school, with limited access to an advanced > curriculum. I wouldn't have learned as much.Testing is not a magic bullet for improving the schools. There aremany issues involved and it is a thorny political problem. But, thatsaid, there are many inner-city kids who do excel and who do learn alot even here in the US.> Completely off the topic: given the American emphasis on equality and equity, I > 'nd the discrepancy between the education a rich American kid and a poor > American kid receives to be absolutely shocking.There are many ideals which attract more talk than action in the US.--Jeff-- Ho, ho, ho, hee, hee, heeand a couple of ha, ha, has;That's how we pass the day away,in the merry old land of Oz.-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: learning disabilitiesI am trying to 'nd information on the Everyday Math curriculum and ifthere is a parallel curriculum for students with LearningDisabilities, as the students I have are being left behind by theregular Everyday Math curriculum. Thank you!-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html