September 2011 - mathfun@mailman.xmission.com

Re: [math-fun] Bealesque identities (actually another forward from Richard Guy)
by Dan Asimov 08 Sep '11

08 Sep '11

<< Twice again my messages have been bounced back, so math-fun don't know that the problem has been debunked by Stan Rabinowitz, using Geometer's Sketchpad. R. On Thu, 8 Sep 2011, Dan Asimov wrote: > Richard Guy, who is having trouble posting to math-fun, sent this message: > > -------------------------------------------------------------------------- > Apologies for butting in, but my messages to math-fun > are bounced back by a spam filter, so I don't know how > to send. What I wanted to say was: > > The following problem was given to me by Julian > Salazar, a high school student, who said he got it second-hand > from members of the US IMO team. There are several subsidiary > problems. > > The medians of a triangle divide it into six smaller > triangles. Show that the six incentres lie on an ellipse. > > 1. Who originated the problem? > > 2. Did he/she have a proof? > > 3. Experiment suggests that if the three ``cevians'' > are drawn through any point, not only the centroid, then the > result is still true, except that if the point is outside > the triangle, then the conic is a hyperbola, and if the point > is on an edge of the triangle, the conic is a pair of straight > lines. > > 4. In the original problem, is the centre of the > ellipse among Clark Kimberling's three-thousand-odd triangle > centres? > > 5. Same question as 4., but taking the defining > point to be other of the more familiar triangle centres, > besides the centroid. > > 6. If we apply Conway's extraversion to the > problem, are there any selections of six excentres which > lie on a conic? > > I must be getting old, because any attempt at > finding a proof bogs down in a morass between Euclidean and > projective geometry. Or, just possibly, it isn't true?!? > > Happy new academic year to all. R. >> Sometimes the brain has a mind of its own.

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[math-fun] about 10000 numbers, a certain document from M.I. Shamos.
by Simon Plouffe 08 Sep '11

08 Sep '11

Hello everybody, bonjour Ã tous, I received an automated notice this morning from Google, this is about a document that a certain Michael I. Shamos produced that contains about 10000 entries on real numbers. The document is available here : http://euro.ecom.cmu.edu/shamos.html http://euro.ecom.cmu.edu/people/faculty/mshamos/cat.pdf One word : wOW!, now, that's interesting!, Neil, would you agree that we should include these entries into the OEIS ? Best regards, Simon PLOUFFE

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[math-fun] Bealesque identities
by Dan Asimov 08 Sep '11

08 Sep '11

Beal's conjecture is that for integers p,q,r all > 2, then for any solution in positive integers A,B,C of A^p + B^q = C^r, at least 2 of A,B,C have a common factor. But there seem to be plenty of examples of (*) A^p + B^q = C^r where exactly one of p,q,r is equal to 2, with the other two of them still > 2, and A,B,C being pairwise relatively prime positive integers. * * * One example is 10^2 + 3^5 = 7^3. Wikipedia mentions another such, 7^3 + 13^2 = 2^9. And of course 1^p + 2^3 = 3^2, which shows that the exponent 2 can appear on the right side of (*). I wonder several things: * Do there exist methods to find all solutions to an equation like A^2 + B^5 = C^7 ??? (Is this easier if at least one solution is known?) * What is known about the set of exponents {p,q,r} = {2,s,t} such that at least one nonzero solution exists to (*) in pairwise relatively prime A,B,C > 0 ? E.g., are there infinitely many triples {p,q,r} = {2,s,t} for which a solution to (*) exists? * Is it known that for a given set of exponents {p,q,r} = {2,s,t} there can be only a finite number of triples A,B,C satisfying (*) ? * Are there examples other than 1^p + 2^3 = 3^2 where the exponent 2 appears on the right side of (*) ? --Dan Sometimes the brain has a mind of its own.

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Re: [math-fun] [seqfan] about 10000 numbers, a certain document from M.I. Shamos.
by Simon Plouffe 08 Sep '11

08 Sep '11

Hello, yes, I saw another error too with the golden ratio, on page xiv or page 16 : 1 61803397498948482... as the golden ratio, just a typo. nevertheless, the document is quite interesting, 715 pages, this represent a fait amount of work.! bonne journée, have a good day. Simon Plouffe Le 08/sept./2011 15:47, Maximilian Hasler a écrit : > Hello Simon, > > thanks for that interesting link. > However, there is a typo in the first non-integral value: > > 1 .00000027557319224027... > = sum( k=0...oo, 1/(10k)! ) > > and not, as written, sum( k=1...oo , ... ) > > Maximilian > > > > > On Thu, Sep 8, 2011 at 4:48 AM, Simon Plouffe<simon.plouffe(a)gmail.com> wrote: >> >> Hello everybody, bonjour à tous, >> >> I received an automated notice this morning >> from Google, this is about a document >> that a certain Michael I. Shamos produced >> that contains about 10000 entries on real >> numbers. >> >> The document is available here : >> http://euro.ecom.cmu.edu/shamos.html >> >> http://euro.ecom.cmu.edu/people/faculty/mshamos/cat.pdf >> >> One word : wOW!, >> >> now, that's interesting!, >> >> Neil, would you agree that we should include these >> entries into the OEIS ? >> >> Best regards, >> Simon PLOUFFE >> >> >> _______________________________________________ >> >> Seqfan Mailing list - http://list.seqfan.eu/ >>

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[math-fun] Ruminations on negative genus
by David Wilson 06 Sep '11

06 Sep '11

We know that a sphere has topological genus 0, and if we punch a hole in the sphere, we get a donut with genus 1, and we increase the genus by 1 for each additional hole we punch. So the "punch a hole" operation increases genus by 1, ergo the inverse "fill a hole" operation, like filling the hole in a donut to get a sphere, decreases genus by 1. Likewise, if we start with the donut, and slice through one side, it becomes topologically equivalent to the sphere again, so the "cut apart" operation decreases genus by 1 (given that our cut has a simple closed curve as a profile, that is, slices through a single solid piece, not through any holes or bubbles in the object). Likewise, the inverse operation of "join together" increases genus by 1. If we stretch a sphere of genus 0 into a rope and join the ends into a donut, we increase the genus by 1. In fact, "punch a hole" and "join together" perform the topological operation "add a handle", while "fill a hole" and "cut apart" perform "remove a handle" ... supposing a handle exists. What happens if we cut a sphere into two pieces? A single sphere has genus 0, the "cut apart" operation should therefore produce two spheres together having genus -1. If we cut one of those spheres, we end up with 3 spheres having genus -2. By induction, n spheres together have genus 1-n. The consistent conclusion is that 0 spheres, that is, empty space, has genus 1. Again, start with a sphere, this time flatten it into a sheet and blow it up like a balloon, it still has genus 0. Now use the "fill a hole" operation to close the opening of the balloon, producing a hollow sphere. Since "fill a hole" reduces genus by 1, the hollow sphere, or "sphere with a bubble", has genus -1. You can shrink the bubble, and repeat the operation, blowing another bubble in the sphere, closing the hole, and getting a sphere with two bubbles and genus -2. By induction, a sphere with n bubbles has genus -n. Now consider that n spheres have genus 1-n, while a sphere with n bubbles, that is, n spherical holes carved out of it, has genus -n. This suggests that complementing a figure with respect to an enclosing sphere decreases its genus by 1 (leastwise, we have an example for every negative n). We will call this the "hollowing out" operation, and its inverse would be the "make a cast" operation. So let's start with nothing, which we agree has genus 1. We hollow this nothing out of a sphere, which decrements the genus and produces a sphere of genus 0 (consistent so far). We hollow this sphere out of another sphere, producing a hollow sphere of genus -1, in agreement with our earlier assessment. We hollow this hollow sphere out of a yet bigger sphere, and we get a hollow sphere with a sphere inside, of genus -2. If we take a hollow sphere of genus -1, and pull a little knob out from its inside surface, and cut the knob off ("cut apart"), we likewise obtain a hollow sphere with a sphere inside of genus -2, so everything seems to be consistent. So what is the genus of the Borromean rings? Well, these are three intertwined donuts. If we cut each of the 3 donuts, the genus is reduced by 3, the rings fall apart, each donut becomes a sphere, and we get 3 spheres with genus -2. This means that the Borromean rings had genus 1. In fact, this same argument goes through for any configuration of 3 links, no matter how knotted. One the 3 links are cut, the genus is reduced by 3, the links become 3 spheres of genus -2, so the original links together had genus 1. Indeed, this same argument goes through for any number of links, no matter how knotted. Suppose there are n links. Once the n links are cut, the genus is reduced by n, the knot falls apart, and the n links become n spheres, togther with genus 1-n. This means that the original set of links had genus 1, that is, every knot or linkage has genus 1, including the unknot (donut) and the empty knot (empty space). You can also start with two donuts, each of genus 1. If you join them together, you increase their genus by 1, and end up with a two-hole donut, of genus 2. This means that the two donuts originally must have had genus 1. In fact, you can join n donuts together with n-1 joins, increasing their genus by n-1, to get an n-hole donut of genus n, meaning that all n donuts together had genus 1. This just verifies that any number of links together have genus 1. We can also surmise that the "juxtapose" operation of considering an object of genus n together with an object of genus m as a single object seems to yield an object of genus n+m-1. For example, a donut has genus 1, so two donuts together have genus 1+1-1 = 1. Juxtaposing this 2-donut object with a third donut again gives genus 1+1-1 = 1, and again we confirm our theory about knots and linkages all having genus 1. Likewise, a sphere juxtaposed with a hollow sphere has genus 0 + -1 - 1 = -2, as we earlier determined for a sphere inside a hollow sphere. Juxtaposing any object of genus n with empty space should result in the original object with the same genus n, confirmed by juxtaposed genus n+1-1 = n. So far, I haven't found any inconsistencies in the concept of negative genus. Supposing it is consistent, I must assume it is well known to the topology crowd. At any rate, I have heard it said that a topologist is someone who can't tell a donut from a cup of coffee. My condition is worse, I can't tell a dozen donuts from a cup of coffee.

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[math-fun] tennis anyone?
by David P. Moulton 06 Sep '11

06 Sep '11

Here's a nice solution from my friend Bryan Clair at Saint Louis University. It doesn't use integrals, and I think the calculations are easier to understand than in Gareth's solution. ---------- Forwarded message ---------- Let's say W wins, beating R in the first round. W beats X in the championship. X beats Y in the semis Y beats Z in the 2nd round. Z beats U in round 1. We want to know if R is better than Z. So: There are 16 players. Pick W. Now there are 15. There are 6 players in W's bracket besides R. They don't matter. That leaves 9 players, and we want to know if R is better than the 3rd/4th finisher from the other 8. Well, 1/9 of the time, R is the best of them. Otherwise (8/9 of the time), X and the three players on X's side of the bracket don't matter. Then we have 5 players left, and we want to know if R is better than the 2nd place finisher from the remaining 4. 1/5 of the time, R is the best of them. Otherwise (4/5 of the time), Y and the player Y beats in round 1 don't matter. Then we're left with R, Z, and U. Then from these 3 players, is a random player better than the winner between the other two? Yes, 1/3 of the time. So: 1/9 + (8/9)*(1/5 + (4/5)*(1/3)) = 71/135 Pretty clear you could do some kind of induction and handle more than 16 players, or treat some (maybe all?) other comparisons. Bryan

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[math-fun] tennis anyone?
by Erich Friedman 05 Sep '11

05 Sep '11

suppose 16 tennis players enter an unseeded single-elimination tournament, and that the better player wins every match. after 4 stages and 15 matches, the best player (who we call W) wins. on average, which of the two following players is better? a) the player who loses to W in the first round, or b) the player who won the first match, but then was beaten by the player who was beaten by the player who lost to W in the final round? the answer, obtained by a 16-fold integral, surprised me. erich

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[math-fun] quintic trivia
by Bill Gosper 05 Sep '11

05 Sep '11

"Exhaustive" search (81 cases) sez there are no solvable irreducible quintics with coeffs restricted to {-1,0,1}. After x^5-2 (modulo root negation and reciprocation), the smallest, in some sense, seems to be -1 - 2 x - x^2 + x^3 + x^4 + x^5, whose five solutions are -1/5 +w*u_00+w^2*u_01+w^3*u_10+w^4*u_11, where w = e^(2 pi i n/5), n an integer mod 5, and u_jk := j j k (-1) 1371 - 145 (-1) (sqrt 5) + 65 (-1) sqrt(470 (1 - ------)) sqrt 5 R(-----------------------------------------------------------) 12500 where R is the *real* fifth root of a real: R(#):=#/(#^4)^(1/5). If you use the default complex 5th root, the negative ones will need a correction factor of e^(4 i pi/5). Beware: This is the kind of 5th root they mean in the Wikipedia article<http://en.wikipedia.org/wiki/Quintic_function> and in the examples in the (glitchy) Intelligencer article<http://www.math.carleton.ca/%7Ewilliams/papers/pdf/244.pdf> . (InputForm in case Gmail murdered the above: (1371 - 145*(-1)^j*Sqrt[5] + 65*(-1)^k*Sqrt[470]*Sqrt[1 - (-1)^j/Sqrt[5]])/12500 ). The u_jk^5 are the (real) roots of the quartic 1 + 325793750 u + 4331113281250 u^2 - 41839599609375 u^3 + 95367431640625 u^4 (Recall that the quintic only had 2s and 1s.) So a convenient way to exhibit the solution to an irreducible quintic ax^5+bx^4+... is the constant -b/(5a) + one of these quartics. Recall that quartic solutions in general have quadratic surds inside cubic surds inside quartic surds, so for the quintic we appear doomed to suffer quadratic surds inside cubic surds inside ... inside quintic surds. But no! For some magic Galwological reason, these quartics seem always to solve down to (real!) quadratics in quadratics, at worst. Sometimes, as with -3 + 3 x + 3 x^2 + 3 x^3 + 3 x^4 + x^5, the quartic even factors: (59049 - 12543750 u + 9765625 u^2) (-7776 + 1050000 u + 9765625 u^2). But usually it doesn't even decompose, so your quartic solver needs to be smart. --rwg Is there a name for these nice quartics? They're not usually biquadratic. How about "nice"?-) PS, for some reason, Preview saturates my Mac CPU while trying to scroll the later portions of the Intelligencer article<http://www.math.carleton.ca/%7Ewilliams/papers/pdf/244.pdf>.

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Re: [math-fun] To David Re: Ruminations on negative genus
by David Wilson 02 Sep '11

02 Sep '11

I'm not a topologist, which I suppose does not excuse my incorrect nomenclature. What I am talking about is solid objects, so if "ball" is the right word, then use it. Sorry for using the word "sphere" which would certainly be misleading, as would using "torus" instead of "donut" had I done that. When I talk about "cutting apart", I mean stretching a portion of the object into a thin strand and cutting that strand. It is understood that the cross section of the cut is simple closed curve, not a cut through a 3D hole (bubble) that might be inside the object. If this strand happens to be part of a topological handle, cutting the strand removes the handle, decreasing the object's genus by 1. "Join together" would be the opposite operation, pinching the object in two places and pulling out two strands which you then join at the ends, creating a handle as it were, and increasing the object's genus by 1. "punching a hole" is stretching part of the object into a sheet and punching a hole in the sheet, so it's adding a 2D hole in your nomenclature, increasing the genus by 1. "Filling a hole" would be the inverse operation, pinching off a 2D hole, decreasing the genus by 1. My observation is that "cutting apart" is a local operation. If you see only the small section of strand you are cutting, you don't know if you are cutting a handle, leaving the object connected, or cutting the object into two disconnected pieces. In the former case, a topologist would say the cut reduces the genus by 1. I was exploring the consequences of having a "cut apart" operation always reduce the genus by 1. Performing a "cut apart" operation in the obvious way on a donut turns the donut into a ball, a topologist would say this is removing a handle from the donut, which decreases its genus by 1 from 1 to 0. Now start with a ball, of genus 0. If you stretch it into a strand and cut the strand, the result is two balls. If you ignore the fact that you did not cut a handle, and simply agree that the "cut apart" operation always reduces the genus by 1, you would conclude that the two balls, considered as a single topological object, have genus -1. If you had instead cut the strand at n-1 places, you would end up with n balls, and your n-1 cuts would have decremented the genus of the original ball n-1 times, so that logically, the n balls considered as a single topological object, should have genus 1-n. Similarly, if you start with a sphere, flatten it into a sheet, and blow a bubble in it, so that it looks like a balloon, it remains genus 0. If you tie off the balloon, which is locally equivalent to pinching off a 2D-hole, you end up with a hollow ball, a ball with a "3D hole" or bubble in it. If pinching off the hole had eliminated a handle, a topologist would have agreed it reduced the genus by 1. If we ignore the topologist and agree that pinching off a hole always reduces genus by 1, we conclude that the genus of a hollow ball should logically be -1. If we blow another bubble in the hollow ball and pinch it off, just as we did before, we end up with a ball with two bubbles (ball-shaped cavities). Our second pinching off operation should again have reduced the genus by 1, so a ball with two ball-shaped cavities should logically have genus -2, and by induction, a ball with n ball-shaped cavities should have genus -n. Now above, we concluded that n balls, considered as a single object, have genus 1-n. Also, a ball with n ball-shaped cavities have genus -n. One would be tempted to generalize that if we start with a ball, and remove from its interior an object N of genus n, resulting in a ball with an internal N-shaped cavity, that this latter object would have genus n-1. For example, a donut has genus 1, so a ball with an internal donut-shaped cavity would have genus 0. We confirm that if we punch a hole from the exterior of this ball to its internal donut shaped cavity, its genus should be increased by 1, and indeed the resulting object is topologically equivalent to a donut of genus 1. If we accept that for any positive n, n balls considered as a single object have genus 1-n, the obvious extrapolation is that 0 balls, that is, the empty object, has genus 1. Likewise, by playing around with these operations, we can convince ourselves that if we have have two objects X and Y of genus x and y, respectively, that the operation of juxtaposition (or would aggregation be better?), that is, considering X and Y together as a single topological object, results in an object of genus x+y-1. For example, a ball has genus 0, an object consisting of 2 balls has genus 0+0-1 = -1, aggregating another ball gives genus -1+1-1 = -2 for an object consisting of 3 balls, and in general, genus 1-n for an aggregation of n balls, consistent with our earlier conclusion. An easy induction will show that any nonnegative number of donuts aggregated together have genus 1. Since any knot or link is no more than an aggregation of donuts (links homeomorphic to the circle), it follows that any knot or linkage has genus 1. On 9/2/2011 8:33 PM, Dan Asimov wrote: > David, > > Part of the time you seem to be talking about solid objects, > and part of the time, about surfaces. > > I can't follow your descriptions. It is standard in topology > that sphere means the surface, and ball means the filled surface. > > Also, you refer to punching out, or filling, holes. > > There are 1D and 2D holes. A 2D hole (like a closed curve, > or a hole punched from a piece of paper by an office hole > punch) can be filled by shrinking its boundary to a point. > > A 3D hole is like the hollow inside a sphere (or inside any > closed surface in 3-space), and this can be filled, well, > in the obvious way by filling it with solid stuff. > > Maybe you can help me out by clarifying what you meant. > > Thanks, > > Dan > > Sometimes the brain has a mind of its own. >

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Re: [math-fun] Connectedness
by Dan Asimov 02 Sep '11

02 Sep '11

I thought I knew what you meant by the original question, but now I'm confused. What exactly is a cut/slice? Anything like a subset? --Dan David M. wrote: << Forgive me, I should have added such that the original set of dimension d is a dimension d cut/slice of the new set of 1 more (integer) dimension. >> Sometimes the brain has a mind of its own.

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