February 2014 - mathfun@mailman.xmission.com

Re: [math-fun] oscillating series
by Warren D Smith 02 Feb '14

02 Feb '14

Veit Elser: h(z)=e^(-2^(2z))-e^(-2^(2z+1)) Make a plot of h. Convince yourself it vanishes exponentially for z -> -\infty and super-exponentially for z -> +\infty. Now look back at the formula for f(y) as a semi-infinite sum of h(m-y) and convince yourself that f(y) approaches a periodic function of y for large y. In fact, the periodic limit is exactly the periodic function you get by extending the sum to be doubly-infinite: p(y)=\sum_{m=-\infty}^\infty h(m-y) This is where you use the Poisson summation formula, to express p(y) in terms of its Fourier amplitudes: p(y)=\sum_{k=-\infty}^\infty e^(i 2pi k y) a_k a_k=\int_{-\infty}^{+\infty} e^(i 2pi k y) h(y) So the Fourier amplitudes of the periodic limit (in the logarithmically stretched variable y) is just the Fourier transform of h(y) evaluated at integers. Apart from a_0=1/2, these turn out to be very small: a_1=0.0000789616 + 0.00137234 i --WDS: brilliant. Now, change variables u=4^z and z=log_4(u) and du/u=(ln4)dz then h(z) =e^(-u)-e^(-2u) and then a_k=\int_{-\infty}^{+\infty} e^(i 2pi k y) h(y) dy = (ln4) \int_{-\infty}^{+\infty} u^(i 2pi k/ln4) [exp(-u) - exp(-2u)] h(y) du/u which now can be expressed in terms of gamma functions, this is just Euler's integral definition of the gamma function with a few minor tweaks. So Elkies was right that all the a_k can be expressed in closed form using gamma function of imaginary arguments.

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Re: [math-fun] physics thought experiment
by Adam P. Goucher 02 Feb '14

02 Feb '14

> Yet, I'm surprised to learn that 2D Brownian motion (say, on a disk, and which reflects when reaching the boundary) will cover a positive area with probability one. Can you point me to a reference, or did I misinterpret what you said? I would have guessed that with probability one, such a 2D Brownian motion will have 2D measure = 0, even after t = oo. At t = infinity, the path forms a dense subset of R^2 (i.e. impregnates every open ball). This is not the same as being a space-filling curve (it could have measure 0, for instance), but does pass within every epsilon > 0 of every point. This is equivalent to a two-dimensional random walk on Z^2 almost surely reaching every point, which is in turn equivalent to a two-dimensional random walk on Z^2 being recurrent (almost surely returning to the origin infinitely often). Sincerely, Adam P. Goucher > --Dan > > > On 2014-02-01, at 1:22 PM, Adam P. Goucher wrote: > > > Trace out a random walk (Brownian motion, but bounded within the confines > > of the galaxy), accelerating towards c so that the proper time elapsed is > > bounded. In particular, accelerate such that when the galaxy reaches an > > age of infinity, you have only aged by one hour. > > > > Now, probabilistically you should collide infinitely often with absolutely > > everything, including your buddy, within one hour of proper time. > > > > This also works in a two-dimensional unbounded universe, but not within > > a three-dimensional unbounded universe. > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] various
by Adam P. Goucher 02 Feb '14

02 Feb '14

> You wrote: > > Re Dyer, it seems to me you are arguing about infinity*zero. > I don't think I should accept that... I'm arguing that if v(t) is your speed as a function of t (where t is the time in the reference frame of the galaxy), then the integral from 0 to infinity of: (1 - v^2/c^2)^1/2 dt can be made to converge. This will work if you set v(t) = (1 - exp(-t))c, for example. http://en.wikipedia.org/wiki/Proper_time Sincerely, Adam P. Goucher

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Re: [math-fun] physics thought experiment
by Dave Dyer 02 Feb '14

02 Feb '14

I think we have to impose other constraints, for example you have to complete the search before the heat death of the universe, and you can't collide with any stellar masses or black holes. How much acceleration is "reasonable" so for example you could make one round trip to C in one year of subjective time, so you could expect to make 100 such jaunts (with Obamacare).

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[math-fun] oscillating series
by Cris Moore 01 Feb '14

01 Feb '14

Define f(x) = x-x^2+x^4-x^8+... As x->1, f(x) becomes Grandi's series 1-1+1-1+..., and it oscillates more and more rapidly around 1/2. These oscillations are periodic in log(1-x): one way to see that is to notice that f(x) = x-f(x^2) or roughly f(x) = 1-f(x^2). So squaring x, which (when x is close to 1) increments log(1-x) by log 2, flips f(x) around 1/2. Hardy knew about these oscillations (see http://en.wikipedia.org/wiki/Summation_of_Grandi%27s_series) But my question is: how do we calculate their amplitude? Numerically, f(x) ranges from 1/2+delta to 1/2-delta, where delta = 0.00274 or so. Does anyone know how to obtain this number? - Cris

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Re: [math-fun] physics thought experiment
by Adam P. Goucher 01 Feb '14

01 Feb '14

Trace out a random walk (Brownian motion, but bounded within the confines of the galaxy), accelerating towards c so that the proper time elapsed is bounded. In particular, accelerate such that when the galaxy reaches an age of infinity, you have only aged by one hour. Now, probabilistically you should collide infinitely often with absolutely everything, including your buddy, within one hour of proper time. This also works in a two-dimensional unbounded universe, but not within a three-dimensional unbounded universe. Sincerely, Adam P. Goucher > ----- Original Message ----- > From: Dave Dyer > Sent: 02/01/14 06:35 PM > To: math-fun > Subject: [math-fun] physics thought experiment > > You and your buddy have each woken up at some random, unknown point > in the milky way galaxy. > > You have a very bright, omni-directional beacon that can easily be seen > from any point inside it's light cone. > > You can travel in any direction, at any speed up to C (but not exceeding > it of course). > > Your goal is to locate and rendezvous with your buddy for one last drink, > sometime before one of you dies of old age. > > What's your strategy. > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] physics thought experiment (Dave Dyer)
by Warren D Smith 01 Feb '14

01 Feb '14

> You and your buddy have each woken up at some random, unknown point > in the milky way galaxy. > > You have a very bright, omni-directional beacon that can easily be seen > from any point inside it's light cone. > > You can travel in any direction, at any speed up to C (but not exceeding > it of course). > > Your goal is to locate and rendezvous with your buddy for one last drink, > sometime before one of you dies of old age. > > What's your strategy. --other answers were wrong (or Dyer posed it wrong) since problem statement says only you, not you buddy, can move, and only you, not buddy, have beacon. Also, the beacon seems not useful unless you get within a few light years of buddy, otherwise he'll never see it before dies of old age. Also, if you are more than 100 lightyears from buddy then even if you travel at C he'll die before you reach him. Also, Goucher's brownian motion solution is technically wrong in the sense brownian motion has infinite path length so even if you travel at C and do not mind infinite acceleration, you'll still never get anywhere. So anyhow... as posed, there is no solution. If however Dyer really meant BOTH you & buddy can move then: cover the galaxy with a wireframe grid at grid spacing 1 meter (say) then traverse all wires in the grid at speed C while buddy does same on his grid. Continue until you collide. (Assume you both wider than 1 meter.) No beacon needed, but easy to avoid collision by using...

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[math-fun] physics thought experiment
by Dave Dyer 01 Feb '14

01 Feb '14

You and your buddy have each woken up at some random, unknown point in the milky way galaxy. You have a very bright, omni-directional beacon that can easily be seen from any point inside it's light cone. You can travel in any direction, at any speed up to C (but not exceeding it of course). Your goal is to locate and rendezvous with your buddy for one last drink, sometime before one of you dies of old age. What's your strategy.

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Re: [math-fun] Maximum number of triples of orthogonal directions
by Allan Wechsler 01 Feb '14

01 Feb '14

f(8) is at least 4, because ABC+CDE+EFG+GHA can be realized. I don't think it would be too much work to show that 5 is not possible. 1,1,2,2,3,4, unsurprisingly, has hundreds of hits on OEIS, so we need more terms. I have a suspicion that all the graphs of the class you describe (no four vertices with a "lune", AB+AC+BC+BD+CD) are "realizable" as orthogonality graphs in the projective plane. If they are different, then your function f probably won't diverge from the corresponding Ramsey Theory question (largest number of triangles in a luneless graph) for quite a while. We should think about trying to prove the realizability conjecture. I have only the very vaguest inklings about how to try that. On Thu, Jan 30, 2014 at 8:35 PM, Warren D Smith <warren.wds(a)gmail.com>wrote: > I posted this to math-fun also, but it has not been "approved" yet by RCS: > > I can prove f(6)=2 and f(7)=3. > > An upper bound on F(n) which shows this is: > the maximum number of triangles in an n-vertex graph G, where > only G which do not contain <|> > as a 4-vertex subgraph, are allowed. > > This function of n is also interesting by itself, and I do not know what > it > is. >

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