August 2009 - mathfun@mailman.xmission.com

[math-fun] Geometric mean over a circle of complex numbers
by Dan Asimov 14 Aug '09

14 Aug '09

Puzzle: Find the geometric mean of the complex numbers lying on the circle {1 + r exp(ix) : -pi < x < pi}, as a function g(r) of r > 0. --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] Geometric proof for sum-of-powers identity
by Michael Kleber 14 Aug '09

14 Aug '09

Hi funsters, We all know that 1^3 + 2^3 + ... + n^3 = (1+2+...+n)^2, and there's a pretty geometric dissection proof (eg on p.58 of the Book of Numbers here<http://books.google.com/books?id=0--3rcO7dMYC&lpg=PP1&ots=-aWHB2jOpz&dq=boo…>, though sadly the picture is redacted). Is there a geometric dissection for its generalizations? Let S(d) be the dth-degree polynomial that is sum( i^(d-1), i=1..n ) -- um, that must have a name, aside from my off-by-1 definition, but I can't think of what. Then we have: S(2d) = sum( (-1)^i binomial(2d-2, i-1) S(i) S(2n-i), i=2...d-1 ) + (-1)^d binomial(2d-3, d-1) S(d)^2 and there's another formulation where the S(d)^2 term can be folded into the sum(..., i=2..2d-2), but you have to divide by 2 in that case. Um, and d can't be 1; these start with d=2 where it gives the S(4) = S(2)^2 mentioned above, and then you get S(6) = 4 S(2) S(4) - 3 S(3)^2, S(8) = 6 S(2) S(6) - 15 S(3) S(5) + 10 S(4)^2, and so on. Do any of you know a pretty geometric dissection for this identity? Note that, aside from S(4) = S(2)^2, the RHSs above all have mixed sign, so presumably some shuffling around of terms would be required. --Michael -- Forewarned is worth an octopus in the bush.

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Re: [math-fun] Plenty of Room at the Bottom, Part II
by Dan Asimov 14 Aug '09

14 Aug '09

Gene -- since I have a great deal of trouble wrapping my mind around quantum mechanics, I certainly appreciate any light you can shed on the subject. << Finally,a request for some feedback: It takes a considerable amount of time for me to compose, edit, and proofread these little physics essays. Are they appreciated? >> But I admit that I'm not sure what "this" refers to below: << This is a correct description of the situation. I'll add a bit more. --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] flat corners
by James Propp 14 Aug '09

14 Aug '09

I was thinking about a branched double cover of the tetrahedron. Jim

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Re: [math-fun] flat corners
by Fred lunnon 14 Aug '09

14 Aug '09

On 8/12/09, rcs(a)xmission.com <rcs(a)xmission.com> wrote: > Presumably you've noticed that all your denominators are 2K^2? > > Rich Ouch --- I hadn't --- thanks! But I had noticed that h(q) = h(p), where p = (2-2q)/(2-q) --- from which in particular the maximum h occurs where q = p, that is q = 2-sqrt2 = 0.585786438 ... though we still don't know exactly what h = 1.287188506 ... is here. And now following on from your remark, I notice that denom(h) = 2 (denom(p) denom(q))^2 in all cases [but denominators are always easier than numerators ...]. WFL Latest data: h(-sqrt2) = h(+sqrt2) = 0; h^2(-4/3) = h^2(7/5) = 59/450, h^2(-1) = h^2(4/3) = 11/18, h^2(-2/3) = h^2(5/4) = 287/288, h^2(-1/2) = h^2(6/5) = 231/200, h^2(-1/3) = h^2(8/7) = 1139/882, h^2(0) = h^2(1) = 3/2, h^2(1/4) = h^2(6/7) = 2511/1568, h^2(1/3) = h^2(4/5) = 731/450, h^2(2/5) = h^2(3/4) = 1311/800, h^2(1/2) = h^2(2/3) = 119/72. WFL

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Re: [math-fun] Plenty of Room at the Bottom, Part II
by Steve Witham 14 Aug '09

14 Aug '09

Henry Baker <hbaker1(a)pipeline.com> forwarded: >Stock Traders Find Speed Pays, in Milliseconds > >by Charles Duhigg >Friday, July 24, 2009 > >The New York Times There's a centuries-old tradition that complains that trading is tricky fast moves, and therefore unfair. The article nearly rises above that level but not quite. Hard to tell whether the people quoted were just whining, or they actually had points to make but the reporter scrambled them beyond recognition. The question boils down to whether the rules under which trading occurs are fair. If so, behavior that makes profit for the trader is also doing the market as a whole a service. If not, the system can be gamed. The question of fairness vs. gaming is not about whether any Joe off the street, or a member of just any species on the planet, no matter how slow its clock cycle, has an equal chance of making a fulfilling living as a trader. Those of us in the rest of the economy should not be trying to insure that old firms, business models, individuals, species, algorithms and chip generations are protected against upstarts or startups. I imagine there are still niches on Wall Street for smart humans. --Steve "PERnicious nonsense!" (J. Frank Parnell) "Don't let it bring you down, it's only castles burning, find someone who's turning, and you will come around." (Niel Young)

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Re: [math-fun] Plenty of Room at the Bottom, Part II
by Henry Baker 14 Aug '09

14 Aug '09

Dumb question: Given the rotation of the Earth & revolution around the Sun & so forth, is there a well-defined notion of "up" & "down" ? I guess if a photon is transmitted via a polarization-preserving optical fiber, then this notion may be well-defined, but I seem to recall that it is better if fibers _don't_ attempt to preserve polarization, since I seem to recall that ever-so-small differences in the speeds of various polarization orientations can build up over many miles of fiber length & cause bad problems -- e.g., self-interference. This is probably just an engineering problem, but since I've heard of thought experiments where Alice & Bob are separated by light-years & I didn't know how they planned to organize the coordinate axes in the first place. At 01:00 PM 8/13/2009, Thomas Colthurst wrote: >Here's a candidate scheme: >Alice is in Tokyo and Bob is in London and they receive a steady stream of >entangled particles. > >Alice waits for a very rapid price change in TYO (Toyota) to happen in Tokyo >-- a $10/share difference in 10 milliseconds, say. > >She then looks at her latest particle and if it matches the direction of the >price change, then she takes action: sells if both are up, buys if both are >down. > >Meanwhile, Bob observes all of his particles and trades on each one, buying >if his measurement is down and selling if his measurement is up. > >If we neglect transaction costs, and assume that news about Toyota breaks in >Tokyo more often than it does in London, then whenever Alice doesn't take >action, then Bob is just randomly buying and selling and whenever Alice does >take action, then Bob's trade locks in Alice's profit. > >-Thomas C > >On Thu, Aug 13, 2009 at 3:33 PM, Michael Kleber <michael.kleber(a)gmail.com>wrote: > >> My idea was specifically about Bell's Inequality violations, in which two >> measurements of an entangled pair of particles can have a higher correlation >> than is possible with any "hidden-variables" interpretation of QM (2/3 vs 5/9). >> >> --Michael >> >> On Thu, Aug 13, 2009 at 3:27 PM, Dave Dyer <ddyer(a)real-me.net> wrote: >> >> > My layman's understanding is this: >> > >> > Suppose you have a supply of entangled particles and manage to ship half of >> > each pair to New York and London. You know that if you observe one and find >> > it in "up" state, the other would instantaneously be observable in "down" >> > state. You can verify this later (after light speed delay) by comparing >> > notes. >> > >> > The problem is that you can't use this to communicate, >> > >> > (1) You can't set the state of the particle, only observe (so you can't >> > set the New York particles to "up" or "down" so London can see the effect. >> > >> > (2) You also can't tell if a particle has already been observed, so you >> > can't signal by selectively not observing. >> > >> > In the math used to describe these processes, there is no state until >> > you make an observation, so there is no loophole by which you can tease >> > more information out of the system. It remains to be seen if the >> > math, which has withstood 100 years of practical tests, is absolutely >> > correct.

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Re: [math-fun] Plenty of Room at the Bottom, Part II
by Dave Dyer 13 Aug '09

13 Aug '09

My layman's understanding is this: Suppose you have a supply of entangled particles and manage to ship half of each pair to New York and London. You know that if you observe one and find it in "up" state, the other would instantaneously be observable in "down" state. You can verify this later (after light speed delay) by comparing notes. The problem is that you can't use this to communicate, (1) You can't set the state of the particle, only observe (so you can't set the New York particles to "up" or "down" so London can see the effect. (2) You also can't tell if a particle has already been observed, so you can't signal by selectively not observing. In the math used to describe these processes, there is no state until you make an observation, so there is no loophole by which you can tease more information out of the system. It remains to be seen if the math, which has withstood 100 years of practical tests, is absolutely correct.

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[math-fun] Old bernpuzzle solved
by rwg＠sdf.lonestar.org 13 Aug '09

13 Aug '09

---------------- Date: Sat, 24 Jan 1998 03:11-0500 Reply-To: rwg(a)NEWTON.macsyma.com Subject: Bernoulli recurrences To: math-fun(a)optima.CS.Arizona.EDU Message-Id: <"19980124081150.3.rwg@SWEATHOUSE"@SWEATHOUSE.macsyma.com> Bernoulli numbers and polynomials are often defined by a "full history recurrence" whose calculation involves all the predecessors down to B_0. At the other extreme is Ramanujan's *first* order recurrence, but it involves the infinite product over primes for a ratio of zetas. A "one sixth history recurrence" for the polynomials is [slightly simplified from original email] sum(binomial(n+2,n-6*k)*bernpoly(y,n-6*k)/(2*k+1),k,0,floor(n/6)) = (y^(n+2)+(y-1)^(n+2))/2-'realpart((y+(-1)^(2/3))^(n+2)) n floor(-) 6 ==== binom(n + 2, n - 6 k) B (y) \ n - 6 k > --------------------------------- / 2 k + 1 ==== k = 0 n + 2 n + 2 y + (y - 1) 2/3 n + 2 = --------------------- - Re((y + (- 1) ) ) 2 (for real y. Otherwise expand the Re first.) For y=-1/2, this give a neat recurrence for the Bernoulli numbers: j floor(-) 6 6 k - j + 1 ==== binomial(j + 2, 6 k + 2) B (2 - 1) \ j - 6 k > ---------------------------------------------------- = / 2 k + 1 ==== k = 0 j/2 j/2 + 1 %pi j ((- 1) + 3 ) cos(-----) 2 -------------------------------- . j + 2 2 Can we do better than every 6th? ------------------------ The mystery was: Why aren't there more of these? Where are the rest? Answer: This was a fluke! What's usually on the rhs is a linear combination of Bernpolys(y + various roots of unity), but in this case, the Bernpolys (nearly) cancel in contiguous pairs due to the integer separations between 1^(k/6). The underlying identity is really 6*x^2*'sum(binomial(n,6*k+2)*x^(6*k)*bernpoly(y,n-6*k-2),k,0,floor(n/6)) = -(bernpoly(y+(-1)^(2/3)*x,n)+bernpoly(y-(-1)^(2/3)*x,n))/(-1)^(1/3)-(-1)^(1/3)*(bernpoly(y+(-1)^(1/3)*x,n)+bernpoly(y-(-1)^(1/3)*x,n))+bernpoly(y+x,n)+bernpoly(y-x,n) n floor(-) 6 ==== 2 \ 6 k 6 x > binomial(n, 6 k + 2) x B (y) / n - 6 k - 2 ==== k = 0 2/3 2/3 B (y + (- 1) x) + B (y - (- 1) x) n n = - --------------------------------------- + B (y + x) + B (y - x) 1/3 n n (- 1) 1/3 1/3 1/3 - (- 1) (B (y + (- 1) x) + B (y - (- 1) x)). n n The 2k+1 in the denominator was a red herring from a gratuitous integration. Similar identities with any desired sparseness can be had from multisecting the standard identity x^n*sum(binom(n,k)*bernpoly(y,k)/x^k,k,0,n) = bernpoly(y+x,n) n ==== \ n - k > binomial(n, k) x B (y) = B (y + x) / k n ==== k = 0 and linearly combining eadem(w*x) for w = various roots of unity. --rwg Actually, the Continuum Hypothesis is false. Between C and 2^alpha_0 lies the number of ways to spell "Chebyshev".

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[math-fun] Old bernpuzzle solved
by rwg＠sdf.lonestar.org 13 Aug '09

13 Aug '09

---------------- Date: Sat, 24 Jan 1998 03:11-0500 Reply-To: rwg(a)NEWTON.macsyma.com Subject: Bernoulli recurrences To: math-fun(a)optima.CS.Arizona.EDU Message-Id: <"19980124081150.3.rwg@SWEATHOUSE"@SWEATHOUSE.macsyma.com> Bernoulli numbers and polynomials are often defined by a "full history recurrence" whose calculation involves all the predecessors down to B_0. At the other extreme is Ramanujan's *first* order recurrence, but it involves the infinite product over primes for a ratio of zetas. A "one sixth history recurrence" for the polynomials is [slightly simplified from original email] sum(binomial(n+2,n-6*k)*bernpoly(y,n-6*k)/(2*k+1),k,0,floor(n/6)) = (y^(n+2)+(y-1)^(n+2))/2-'realpart((y+(-1)^(2/3))^(n+2)) n floor(-) 6 ==== binom(n + 2, n - 6 k) B (y) \ n - 6 k > --------------------------------- / 2 k + 1 ==== k = 0 n + 2 n + 2 y + (y - 1) 2/3 n + 2 = --------------------- - Re((y + (- 1) ) ) 2 (for real y. Otherwise expand the Re first.) For y=-1/2, this give a neat recurrence for the Bernoulli numbers: j floor(-) 6 6 k - j + 1 ==== binomial(j + 2, 6 k + 2) B (2 - 1) \ j - 6 k > ---------------------------------------------------- = / 2 k + 1 ==== k = 0 j/2 j/2 + 1 %pi j ((- 1) + 3 ) cos(-----) 2 -------------------------------- . j + 2 2 Can we do better than every 6th? ------------------------ The mystery was: Why aren't there more of these? Where are the rest? Answer: This was a fluke! What's usually on the rhs is a linear combination of Bernpolys(y + various roots of unity), but in this case, the Bernpolys (nearly) cancel in contiguous pairs due to the integer separations between 1^(k/6). The underlying identity is really 6*x^2*'sum(binomial(n,6*k+2)*x^(6*k)*bernpoly(y,n-6*k-2),k,0,floor(n/6)) = -(bernpoly(y+(-1)^(2/3)*x,n)+bernpoly(y-(-1)^(2/3)*x,n))/(-1)^(1/3)-(-1)^(1/3)*(bernpoly(y+(-1)^(1/3)*x,n)+bernpoly(y-(-1)^(1/3)*x,n))+bernpoly(y+x,n)+bernpoly(y-x,n) n floor(-) 6 ==== 2 \ 6 k 6 x > binomial(n, 6 k + 2) x B (y) / n - 6 k - 2 ==== k = 0 2/3 2/3 B (y + (- 1) x) + B (y - (- 1) x) n n = - --------------------------------------- + B (y + x) + B (y - x) 1/3 n n (- 1) 1/3 1/3 1/3 - (- 1) (B (y + (- 1) x) + B (y - (- 1) x)). n n The 2k+1 in the denominator was a red herring from a gratuitous integration. Similar identities with any desired sparseness can be had from multisecting the standard identity x^n*sum(binom(n,k)*bernpoly(y,k)/x^k,k,0,n) = bernpoly(y+x,n) n ==== \ n - k > binomial(n, k) x B (y) = B (y + x) / k n ==== k = 0 and linearly combining eadem(w*x) for w = various roots of unity. --rwg Actually, the Continuum Hypothesis is false. Between C and 2^aleph_0 lies the number of ways to spell "Chebyshev".

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