January 2008 - mathfun@mailman.xmission.com

[math-fun] Edward Tufte one-day class
by James Propp 28 Jan '08

28 Jan '08

Does anyone have an opinion of Edward Tufte's one-day class on presenting data and information? http://www.edwardtufte.com/tufte/courses The issue for me is whether it's worth the time of attending (as opposed to the issue of whether it's worth the money). Jim Propp

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Re: [math-fun] Reflections on Orientation
by Eugene Salamin 27 Jan '08

27 Jan '08

----- Original Message ---- From: Fred lunnon <fred.lunnon(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Sunday, January 27, 2008 5:25:36 AM Subject: Re: [math-fun] Reflections on Orientation ... The Lie group corresponding to Euclidean isometries has 4 connected components rather than 2. ... ----------------------------- How can there be 4 connected component? I thought the general Euclidean isometry is (R,t) where R is a rotation, t is a translation, and the action is (R,t)x = Rx + t. Two connected components correspond to R being proper or improper. What are the other 2 components? Are you confusing this with the case of Lorentz transformations, where space reflections and time reflections are independent? Or, is it me that is confused? Gene ____________________________________________________________________________________ Never miss a thing. Make Yahoo your home page. http://www.yahoo.com/r/hs

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Re: [math-fun] Reflections on Orientation
by Dan Asimov 27 Jan '08

27 Jan '08

<< You might have hoped you'd heard the last of this topic. However, a discussion with David Gale has inspired the following vicious conundrum. Consider a transparent sheet of overhead projector film, onto which has been glued the legend "P A B L / N H Ø O" in plastic symbols, coloured green on the front and red on the back [from a numerate classicist toddler's alphabet: there would have been 10 symbols, but the swastika's been eaten]. Flatten it out on the table [the sheet, not the toddler], with the lettering legible, and coloured green. Let's call this positively oriented. Now turn it over, as a page from a book, and slide it back again. It occupies the same space as before; and the legend is illegible and coloured red. The orientation of the sheet is now negative. Then begin again, but instead lay it against a mirror. The image again occupies the same space (more or less) as the original. The legend in the reflected image is legible and coloured red [or if you foozled it, illegible and coloured green --- whichever]. Question: is the reflected orientation positive, or is it negative? If anyone comes across a previous reference to something related, I should like to hear about it. [Martin Gardner's "The Ambidextrous Universe" might perhaps be a good place to start looking.] >> I've been rading this thread with the hopes of jumping at some point, since I feel orientation is a concept I have a good handle on. But I can't, since at no point do I feel that any clearly defined mathematics question has been asked. I will say this: * There is no such thing as a "positive" or "negative" orientation. Orientations of the same manifold can be compared with each other and declared "same" or different". But they cannot be labeled "positive" or "negative" in any consistent way. * An orientation on a Euclidean space R^n does not determine an orientation on a lower-dimensional subspace thereof. * There is, however, a way for an orientation on R^n to determine an orientation on an affine R^(n-1) that does not pass through the origin of R^n, and vice versa: such a subspace can determine an orientation on the R^n in which it lies. Since I don't really understand the question(s) under discussion, I don't know if these facts will be of any help. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Correction Re: Orientation
by Dan Asimov 27 Jan '08

27 Jan '08

I wrote (for a smooth manifold X): << As a topological space, [the frame bundle] V(X) is either connected or not. When it is not connected, it has exactly two connected components. Definition: If V(X) has only one component, it's called nonorientable. Definition: When V(X) has two connected components, X is called *orientable*. Definition: If X is orientable, the choice of one of the components of V(X) is called an *orientation* of X. >> Correction: The first definition should read: << Definition: If V(X) has only one component, X is called *nonorientable*. >> --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Orientation
by Dan Asimov 27 Jan '08

27 Jan '08

Fred wrote: << . . . to define the concept of orientation at all! >> For a smooth manifold there are many equivalent definitions of orientation. For Euclidean space the definitions are especially simple (though more complex than one might expect for such a basic idea"). Consider the set B of all ordered bases (v_1,...,v_n) of R^n as a real vector space. B can be considered as a topological space (and metric space), as a subset of the space (R^n)^n = R^(n^2). As such, it has 2 connected components. Definition: An *orientation* of R^n is the choice of one of these components. ---------------------- Given any fixed ordered basis f = (f_1,...,f_n) of R^n, then each element b of B can be expressed uniquely in terms of f, yielding an nxn matrix M_f(b) of real numbers. Then the two components of B can be distinguished by the sign of the determinant of M_f(b). ---------------------- In the more general case of an arbitrary smooth n-manifold X: Each point x of X has a tangent space T_x that naturally has the vector space structure of R^n. So for each x in X there is a space of ordered bases B_x of T_x. Taking the disjoint union of B_x for all x in X, one gets the set of *all* ordered bases at *all* points of X. This is a topological space in a natural way (e.g., as a result of any reasonable metric structure on it). This is called the "Stiefel variety" of X and denoted V(X). (It's also called the "frame bundle" of X, and can be considered as a fibre bundle over X with fibre = the Lie group GL(n).) As a topological space, V(X) is either connected or not. When it is not connected, it has exactly two connected components. Definition: If V(X) has only one component, it's called nonorientable. Definition: When V(X) has two connected components, X is called *orientable*. Definition: If X is orientable, the choice of one of the components of V(X) is called an *orientation* of X. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Reflections on Orientation
by Fred lunnon 26 Jan '08

26 Jan '08

You might have hoped you'd heard the last of this topic. However, a discussion with David Gale has inspired the following vicious conundrum. Consider a transparent sheet of overhead projector film, onto which has been glued the legend "P A B L / N H Ø O" in plastic symbols, coloured green on the front and red on the back [from a numerate classicist toddler's alphabet: there would have been 10 symbols, but the swastika's been eaten]. Flatten it out on the table [the sheet, not the toddler], with the lettering legible, and coloured green. Let's call this positively oriented. Now turn it over, as a page from a book, and slide it back again. It occupies the same space as before; and the legend is illegible and coloured red. The orientation of the sheet is now negative. Then begin again, but instead lay it against a mirror. The image again occupies the same space (more or less) as the original. The legend in the reflected image is legible and coloured red [or if you foozled it, illegible and coloured green --- whichever]. Question: is the reflected orientation positive, or is it negative? If anyone comes across a previous reference to something related, I should like to hear about it. [Martin Gardner's "The Ambidextrous Universe" might perhaps be a good place to start looking.] Fred Lunnon

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[math-fun] Expected time to get any of several hits
by Bernie Cosell 26 Jan '08

26 Jan '08

I've been fiddling with the recursion for the expected number of probes [without replacement] to find any of M specific values among N objects. When M=1 the recursion is simple enough: E(N) = 1/N + (N-1)/N * (1+E(N-1))) and it reduces fairly easily to E(N) = (N+1)/2. But for general M it seems a lot harder. The recursion I have is: Em(N) = m/N + (N-m)/N * (1 + Em(N-1)) And I can't see how to simplify it. Am I missing something obvious, or some trick, or is this just a hard one? Thanks!! /bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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Re: [math-fun] Expected time to get any of several hits
by Dan Asimov 26 Jan '08

26 Jan '08

Oops, it has been kindly pointed out to me that Bernie's questions is about sampling *without* replacement, which I somehow managed to miss. Sorry about that. In this case I don't think there's any neat closed formula for the expected number of trials before first choosing one of the M items out of N. It's just E(M;N) = E(# trials to the first success) = 1*[1]*(M/N) + 2*[(1-(M/N))]*(M/(N-1)) + 3*[(1-(M/N)*(1-M/(N-1))]*(M/(N-2)]+. . . + (N-M+1)*[(1-(M/N))*(1-M/(N-1))*...*(1-M/(N-M+1))]*(M/(N-(N-M)) where the last fraction is of course = 1. ----------------------------------------- There might be an interesting asymptotic formula for E(k*M;k*N) as k -> oo. (My guess, however, is that this just approaches the expectation for sampling *with* replacement, i.e., N/M.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Expected time to get any of several hits
by Dan Asimov 25 Jan '08

25 Jan '08

Bernie wrote: << I've been fiddling with the recursion for the expected number of probes [without replacement] to find any of M specific values among N objects. When M=1 the recursion is simple enough: E(N) = 1/N + (N-1)/N * (1+E(N-1))) and it reduces fairly easily to E(N) = (N+1)/2. But for general M it seems a lot harder. The recursion I have is: Em(N) = m/N + (N-m)/N * (1 + Em(N-1)) And I can't see how to simplify it. Am I missing something obvious, or some trick, or is this just a hard one? Thanks!! >> << Just to clarify a bit, my "intuition" for the answer seems straightforward: if there are M items, you'd expect them to be about equally distributed, and so you'd expect to have to go, on average, around 1/(M+1)-th of the way to hit the first on, and on average M/(M+1)-th of the way to get all of them. [and so if you were looking for five items, you'd expect to go have to check about one-sixth of them before you ran into the first one and have to go through five-sixths to find them all]. I was just trying to see if I could derive that explicitly. >> I'm guessing that the intended question is to find the expected number of binomial trials to the first success. If so, the only relevant number here is the probability of success in each trial, p. So here, p = M/N. By definition, this expectation is E(p) = Sum{k=1..oo} k * (1-p)^(k-1) * p, i.e., = p * Sum{k=1,oo} k * q^(k-1) (where q = 1-p). = p * (d/dq)(Sum{1,oo) q^k) = p * (d/dq)(q/1-q) = p * 1/(1-q)^2 = 1/p. So here the expected number of trials to the first success is N/M. (Since 1/p is such a simple answer, I'd expect there's a much cleverer way to obtain it.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Expected time to get any of several hits
by Dan Asimov 25 Jan '08

25 Jan '08

Bernie wrote: << I've been fiddling with the recursion for the expected number of probes [without replacement] to find any of M specific values among N objects. When M=1 the recursion is simple enough: E(N) = 1/N + (N-1)/N * (1+E(N-1))) and it reduces fairly easily to E(N) = (N+1)/2. But for general M it seems a lot harder. The recursion I have is: Em(N) = m/N + (N-m)/N * (1 + Em(N-1)) And I can't see how to simplify it. Am I missing something obvious, or some trick, or is this just a hard one? Thanks!! >> << Just to clarify a bit, my "intuition" for the answer seems straightforward: if there are M items, you'd expect them to be about equally distributed, and so you'd expect to have to go, on average, around 1/(M+1)-th of the way to hit the first on, and on average M/(M+1)-th of the way to get all of them. [and so if you were looking for five items, you'd expect to go have to check about one-sixth of them before you ran into the first one and have to go through five-sixths to find them all]. I was just trying to see if I could derive that explicitly. >> I'm guessing that the intended question is to find the expected number of binomial trials to the first success. If so, the only relevant number here is the probability of success in each trial, p. So here, p = M/N. By definition, this expectation is E(p) = Sum{k=1..oo} k * (1-p)^(k-1) * p, i.e., = p * Sum{k=1,oo} k * q^(k-1) (where q = 1-p). = p * (d/dq)(Sum{1,oo) q^k) = p * (d/dq)(q/1-q) = p * 1/(1-q)^2 = 1/p. So here the expected number of trials to the first success is N/M. (Since 1/p is such a simple answer, I'd expect there's a much cleverer way to obtain it.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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