July 2003 - mathfun@mailman.xmission.com

[math-fun] A027424
by Allan C. Wechsler 11 Jul '03

11 Jul '03

Is it not the case that a number appears in the N x N multiplication table precisely when none of its prime factors exceed N, and the number itself does not exceed N^2? Not that I'm claiming this helps to calculate A027424(N). -A

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Re: [math-fun] "Fractional" composition?
by asimovd＠aol.com 11 Jul '03

11 Jul '03

Mike Stay asks: << Let g(n,x) be the nth iterate of f(x), i.e. g(0,x) = x g(1,x) = f(x) g(2,x) = f(f(x)) etc. If f is the log function, is there a nice function that behaves like g when n is integral, but allows for non-integer values (something like the gamma function wrt factorial)? >> Mike, this is a question that has occupied some of us on math fun (alone and jointly) for many years! Usually the question has been posed about exp(x) or exp(z). There's a lot to say about it. One thing is that there are infinitely many ways to define such an iteration, using the fact that R / (x ~ exp(x)) defines a real analytic structure on the circle. There are infinitely ways to express its equivalence g: R/(x~exp(x)) --> R/Z by a real analytic diffeomorphism g. Lift this equivalence to a diffeomorphism from h:R --> R, and that can be used to define a "flow" of exp(x) via exp_t(x) = h^-1 ( h(x) + t ). Integral t will give the expected integer iteration. This is kind of an embarrassment of riches, since there doesn't seem to be a clear way to pick out the "right" g. (This was discovered by Bill Thurston the same weekend that I noticed it independently.) If one wants to find a complex analytic embedding of exp in a flow near certain points in C, then there are infinitely many, but at least *discrete* choices to make, which will give such a thing. Alas, it doesn't seem to extend in any natural way to the reals. (This is from work I've done, though quite likely others have found the same thing; I'm not sure.) For each real c in (1, exp(1/e)) (and probably a complex neighborhood thereof) the function f(x) = c^x (for x positive real and some complex neighborhood of R+, there are only *two* methods that appear to be the rightest ways to embed f(x) in a complex analytic flow (a flow of analytic functions which is also analytic in the flow (time) parameter), but no apparent way to choose between these two. (This is from work I've done, and closely related if not overlapping things have been discovered by Rich Schroeppel. Dean Hickerson also did related research, which included demonstrating that the two methods mentioned are distinct, after I'd convinced myself they yielded identical answers.) There is little doubt that much of the above has been found by other people at another time, but I'm not aware of such research. Maybe I'd better let it go at that for the time being. --Dan A.

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Ignore: Re: [math-fun] Can the alternating sum of reciprocal primes be 1/pi ?
by asimovd＠aol.com 11 Jul '03

11 Jul '03

Scratch that! I lost a few digits somewhere. dan

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[math-fun] Can the alternating sum of reciprocal primes be 1/pi ?
by asimovd＠aol.com 11 Jul '03

11 Jul '03

Does 1/pi = 1/2 - 1/3 + 1/5 - 1/7 + 1/11 - 1/13 + ... ??? Just wondering. --Dan

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[math-fun] "Fractional" composition?
by Mike Stay 10 Jul '03

10 Jul '03

Let g(n,x) be the nth iterate of f(x), i.e. g(0,x) = x g(1,x) = f(x) g(2,x) = f(f(x)) etc. If f is the log function, is there a nice function that behaves like g when n is integral, but allows for non-integer values (something like the gamma function wrt factorial)? -- Mike Stay staym(a)clear.net.nz http://www.xaim.com/staym

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[math-fun] Re: [PrimeNumbers] "Contact" primes
by Edwin Clark 10 Jul '03

10 Jul '03

If one converts the prime: 1412007000395936567688325529242177821814480981887773276739818172174237144350800185420854858836262254435789289718814792571362186754193456509548811926587255022099203909145779862540980335509923043069239333581971669175500801 to binary, arranges the bits into a 27 x 27 matrix, and then replaces each 1 by * and each 0 by a blank one obtains the following pattern: * * * * * * * * * * * * * * * * * * * * * * * * * --Edwin

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Re: [math-fun] misc + (Riemann surfaces & string theory)
by asimovd＠aol.com 08 Jul '03

08 Jul '03

1. Rich writes: << One of my favorite word puzzles is finding "consecutive" words, where the last letter is incremented to make a new word. CONVEX CONVEY ... >> Just wondering in what sense you mean "consecutive" (since e.g. CONVEXITY would appear between these words alphabetically). ------------------------------------------------------------- 2. The NY Times had a nice short article by Dennis Overbye in Sunday's Week in Review section (June 29) that is a paean to the joys of math. (It's nice to have at least one math fan writing for the public.) This is freely available through next Saturday on the web; sorry I can't dig up the URL right now (but just go to www.nytimes.com and it should be easy to find, esp. with advanced search). Overbye mentions that string theory has made the subject of Riemann surfaces a hot topic. Can anyone please explain this remark to me? (Please bear in mind that I have only a vague metaphoric idea of what string theory is about.) (First guess: String theory as I understand it posits little simple closed curves in physics space (11 dimensional?) a what underlie elementary particles. Maybe now someone has suggested replacing the s.c.c.'s with surfaces?) --Dan

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Re: [math-fun] circular discs in corners
by Henry Baker 07 Jul '03

07 Jul '03

I think that the right way to think about this is to keep the straw stationary, and wiggle a right-angled corner around on the end of it. The corner itself traces out the surface of a hemisphere of radius r sqrt2 around the center of the end of the straw. But I disagree that you think that these are two different problems. The plane determined by the end of the straw intersects the corner in a triangle, and I contend that the end of the straw is the incircle of that triangle. Question: Does the base of the hemisphere (described above) traced out by the corner form the circumcircle of the triangle? That doesn't seem right, since the circumcircle isn't always concentric with the incircle. So what _is_ the geometric significance of a circle sqrt(2) bigger than the incircle? At 04:14 AM 7/7/03 -0400, Dan Hoey wrote: >About my remark that > >> > any way you lean a disc of radius r on the floor against two >> > walls, the center will be sqrt(2)r from the corner. > >Henry Baker <hbaker1(a)pipeline.com> says: > >> You are absolutely right! I checked this using a symbolic algebra >> package on the plane to Boston today. From one of my previous >> forays, I already had a good method for finding the incenter of a >> circle in 3D, so I found the incenter of the triangle (x,0,0), >> (0,y,0), (0,0,z). I then found the distance^2 to the origin. I then >> divided by the radius^2 of the incircle, and voila'! The ratio is >> exactly _2_ ! Taking the square root, we get a ratio of sqrt(2) for >> the ratio of the distance to the radius, as you point out. > >Well, that is even more startling, because it's a solution to a >different problem! Your calculation is of a disc supported by three >perpendicular lines; mine is of a disc supported by three >perpendicular planes. Does this mean what I suspect it means--that if >you support the disc on three perpendicular planes, the lines from the >points of tangency to the origin must be mutually perpendicular? I'm >becoming motivated to make sure I get this solved, just to verify that >conjecture. > >> A man in the street could understand this -- nail one end of a >> string into a corner of a room. Take a large circular table with a >> small hole in the center, and put the string through the center. >> Now push the table into the corner, and pull the string taut. You >> can move the table around -- up to and including flat against one >> wall, and you won't break the string, and the string won't become >> slack either! > >I thought of a gadget something like that, only with a rigid strut >holding the disc center to the corner. Then you don't have to watch >it to know it doesn't go loose. Of course, neither of these tethers >lets you turn the disc over, as you can in theory. (If the above >plane-support=line=support conjecture is true, then you could also >mount a tripod of perpendicular struts from the corner so that they >meet the disc at the points of tangency.) > >I wonder if this linkage has any useful qualities as a somewhat >universal joint? Say one that provides heavy-duty support (from the >disc resting on the planes) and light-duty tension (from the strut). >The strut can be connected with ball-and-socket joints on both ends, >since it isn't subjected to great force. I don't know if there's a >better way to support the disc than with sliding edges. If the planes >are backed off, a carriage could be inserted that slides on the plane >and accepts the disc in a groove. Optionally, a hole through the >carriage could maintain alignment with the tripod leg, if that >conjecture works out. This would be interesting just to watch. > >Meanwhile, I'm still trying to solve the original problem (which is >the key to all the other stuff.) I looked at the two-dimensional >problem to see if I could get anywhere. I think I've seen rectangles >inscribed in rectangles before, but I'm not sure where, so I worked on >it from scratch. Anyway, for an a x b rectangle is inscribed in a >c x d rectangle, I get > > 0 = (a^2 - b^2)^2 - (ad - be)^2 - (ae - bd)^2 > >which means solving a quartic (unless I've overlooked some >simplication.) > >(I find it interesting that this equation can also be written > > 0 = (a^2 - b^2)^2 - (a^2 + b^2) (d^2 + e^2) + 4 abde > >which is only quadratic in the reparameterization of the rectangles >from (width,length) to (hypotenuse squared,area). And then the two >kinds of parameters are nicely segregated from each other, though the >a^2b^2 term is of ambiguous kind.) > >I still haven't worked out a way of using this to solve the >three-dimensional problem. > >Dan >Hoey(a)AIC.NRL.Navy.Mil

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Re: [math-fun] triple star
by Henry Baker 06 Jul '03

06 Jul '03

Actually "entropy" doesn't have much meaning in a reversible system as a whole. It only starts taking on meaning when you look at _part_ of the system, such that you have only partial information. I actually have an enormous nit to pick with the way thermodynamics is taught. Physicists focus on "energy" and "entropy", whereas "information" is really much more important. And, _no_, "information" is much more than "negative entropy", because physicists never really get around to being precise about entropy in the first place. E.g., "temperature" is defined as being some sort of integrating function in some textbooks! For example, when physicists talk about a "closed" thermodynamic system, they talk about it being insulated from "heat" (whatever that is!) and insulated from the transfer of "energy" (at least _that_ seems to be reasonably well defined!). However, one can conceive of such systems as being insulated in the normal physical sense, but _not_ insulated from the transfer of information in/out. Due to this imprecision, most of the papers I've read about Maxwell's Daemon are nonsensical. Maxwell's Daemon works perfectly well so long as the energy cost of gathering the information is less than the energy gained. As a situation becomes more complex, the energy variations (hence the amount to be gained) becomes smaller, while the cost to gain the information becomes larger, and it eventually becomes not worth the effort. All "life" is a version of Maxwell's Daemon in action. Modern quantum physics makes it essentially impossible to have a "closed" subset of a quantum universe. What quantum physics is really telling us is that information about the evolution of the system seems to pervade all of the universe. An interesting question is whether even black holes can hide the information within -- I conjecture not even temporarily. (See Ahranov-Boehm (sp?) effect.) So it may be that "thermodynamics" needs even more of a reworking for quantum physics than it has gotten so far. At 06:32 PM 7/5/03 EDT, asimovd(a)aol.com wrote: >"Entropy increases" is not a law of physics, but rather a consequence of initial conditions. In our universe it can be traced back to our local stars' having "advanced" rather than "retarded" radiation, or more accurately to their radiation being emanated toward us in the same direction that we perceive ourselves to grow older. (So if there are stars with "retarded" radiation, i.e., being emanated in the time direction we call negative, their local life forms would see those stars just as we see ours, timewise.) > >--Dan

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Re: [math-fun] circular discs in corners
by Henry Baker 06 Jul '03

06 Jul '03

You are absolutely right! I checked this using a symbolic algebra package on the plane to Boston today. From one of my previous forays, I already had a good method for finding the incenter of a circle in 3D, so I found the incenter of the triangle (x,0,0), (0,y,0), (0,0,z). I then found the distance^2 to the origin. I then divided by the radius^2 of the incircle, and voila'! The ratio is exactly _2_ ! Taking the square root, we get a ratio of sqrt(2) for the ratio of the distance to the radius, as you point out. I've also been racking my brain trying to come up with a way to "see" this without any algebra, but nothing has yet come to mind. This is a very cool fact. A man in the street could understand this -- nail one end of a string into a corner of a room. Take a large circular table with a small hole in the center, and put the string through the center. Now push the table into the corner, and pull the string taut. You can move the table around -- up to and including flat against one wall, and you won't break the string, and the string won't become slack either! I wonder if this has any meaning for a circularly polarized wave hitting a corner reflector... Now that many tests allow the use of calculators, I wonder if they're ready to start allowing access to Macsyma/Mathematica/Maple/... on tests? At 06:22 PM 7/3/03 -0400, Dan Hoey wrote: >Meanwhile, I'm intrigued by the straw problem. So far, the best I've >done is to express the length as sqrt(tt+uu+vv) where (t,u,v) >satisfy equations > > (x-t)^2 (tt+uu+vv) = 4 r^2 (uu+vv) > (y-u)^2 (tt+uu+vv) = 4 r^2 (tt+vv) > (z-v)^2 (tt+uu+vv) = 4 r^2 (tt+uu) > >in the box dimensions (x,y,z) and the straw radius r. Adding them >up, I'm startled (possibly not for the first time) that any way you >lean a disc of radius r on the floor against two walls, the center >will be sqrt(2)r from the corner. Is there an easy geometrical >proof? > >But I'm pretty much flummoxed about solving the whole system. Maybe I >should have gone to high school in New York. > >Dan >Hoey(a)AIC.NRL.Navy.Mil

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