April 2015 - mathfun@mailman.xmission.com

[math-fun] Observation on Collatz
by David Wilson 26 Apr '15

26 Apr '15

I was looking a A182137 in the OEIS. For n >= 0, define the Collatz iterate as c(n) = n/2 if n even; (3n+1)/2 if n odd. Note that this is the "normalized" iterate that rolls together the 3x+1 and subsequent x/2 steps for odd numbers. We say n shrinks with b steps if the trajectory of n under c reaches a value <= n within b steps. It is well known that if m == n (mod 2^b), then the first b iterations of c on m and n, respectively, make the same choices of n/2 or (3n+1)/2 on the first b steps. This means that if m == n (mod 2^b), m shrinks within b steps iff n shrinks within b steps. This implies that on any 2^b consecutive nonnegative integers, there is a fixed number a(b) that shrink within b steps. If my calculations are correct, a = A182137 (my computations agree up to a(22)), though the definition of that sequence seems somewhat dense to me. At any rate, if A182137 and my calculations are indeed correct, I have made interesting observation. We see that in general A182137(n+1) >= 2*A182137(n). But for a mysterious set of indices S = (2,5,8,10,13,16,18,21,24,27,29,32,35,36,..), we have equality A182137(n) == 2*A182137(n/2). I would think this means that there is no n for which the trajectory of n reaches a values <= n in exactly b steps for b in S. But this would seem to imply that there are no multiples of 3 in S, since the trajectory of 1 reaches a value <= 1 on steps that are a multiple of 3. Perhaps my calculations are wrong, or there is some subtle problem with my logic.

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[math-fun] What's your favourite maths puzzle?
by Alex Bellos 24 Apr '15

24 Apr '15

Hi all, Due to the success of the Cheryl’s birthday logic problem last week in the Guardian (my two stories received well over 5 million views, making it the biggest story of the week), the paper has asked me to curate a maths puzzle slot on its website. (The Guardian is the second biggest newspaper site in the world) I thought I would reach out to you to ask for what are your favourite math puzzles. Any suggestions? I think a good math puzzle for a newspaper is one that is easy to state and tricky to solve, or at least that requires lateral thinking, but not too tricky and definitely not technical as the readership will largely be non-mathematicians. Maybe some of you have your own puzzles, or twists on classic puzzles, you might want to share. My hope is that the Guardian’s puzzle page becomes a reference in puzzle-land for the dissemination and discussion of mathematical conundrums, posers, brainteasers and the like... Please share on the list, or email me direct… Thankyou! Alex

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[math-fun] What's your favorite maths puzzle?
by Warren D Smith 24 Apr '15

24 Apr '15

As a general suggestion about puzzles... there are two kinds. (a) There are those with a slick canned pre-made answer. Those can be fun, and serve an educational purpose. But they do not result in new mathematics and new discoveries. (b) There are open-ended problems, which nevertheless somebody who is well short of being Albert Einstein, can nevertheless make progress on, thanks to, e.g. computer searches they program. For these, the readers of the puzzle column might well be able to make new discoveries and find new records, and outdo professional mathematicians. I would recommend a mix of both kinds of puzzle. Martin Gardner understood this. http://rangevoting.org/PuzzlePage.html is a collection of democracy-related math puzzles mainly by me. Puzzle #66 is of type (b). Puzzle #0 is of type (a). Puzzle #2 is intermediate type.

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[math-fun] "new" kind of fourier series
by Warren D Smith 24 Apr '15

24 Apr '15

> However, it actually looks like any real-valued series in Warren's form can > be > converted to an equivalent real-valued usual Fourier series, if you include > both positive and negative frequencies (technically, positive and negative > for > every coordinate except the first - more on that later) with the same > number > of coefficients as the new kind of Fourier series. --aha. Neil Bickford has redeemed himself! :) So, originally, I thought the "new" kind of Fourier series had about 2^D or 2^(D-1) or whatever (depends if real or complex, and whether I count right...) times more coefficients than the "old" kind -- and that is definitely true if the old kind happened to consist of exactly one nonzero term, as you can show by expressing exp as a sum of two trigs and expanding out a product. However, Bickford points out that kinda also works in reverse. Namely, if the "new" series happens to consist of exactly one nonzero term, then its "old"-form equivalent, will consist of about 2^D terms! ("About" means up to a constant factor, since I'm not carefully counting.) And this you can also show, albeit the demonstration is somewhat messier. His idea is basically to write a trig as a sum of two exps, then expand out product. Oh. So then, you realize the "new" and "old" Fourier series actually are equivalent. However, the equivalence can cause either the new or the old thing to have up to about 2^D times more nonzero terms than the other, in the right circumstances. So, given that 2^D can be pretty large, that could under the right circumstances be a serious savings. Also, in practice one would truncate the series at some point, using only a finite set of terms, and that finite number might actually be comparable or even much smaller than 2^D... for example if D=40... in which case the two kinds of series could exhibit considerably different practical behaviors. So, that was interesting, and I thank Neil Bickford for figuring that out. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] parametrizing the Koch snowflake curve
by Bill Gosper 24 Apr '15

24 Apr '15

Fourier epicycles thru nth: (d16) sum(((product((1 - (( - 1)^j * tan(((%pi * (2 * k + (1/3)))/(2^j)))/(sqrt(3)))),j,1,inf)) * (%e)^(%i * (6 * k + 1) * t)/((2 * k + (1/3))^2)),k, - n,n) See http://www.tweedledum.com/rwg/cog.htm Also http://gosper.org/fst.dvi Pix: http://gosper.org/fst1.png ,http://gosper.org/fst2.png , http://gosper.org/fst3.png ,http://gosper.org/fst4.png . (For parametric, take Re and Im.) Rich's base 4 recursion loops for rational arg. Find the fixed point of the pending stack ops to simulate infinite recursion, and get an exact val on the curve. Equate to this Fourier series for a *non-obvious* identity. --rwg > If you view the starting triangle as running from 0 to 3, > and write the fraction base 4, you can use the > fraction digits to select which part of the (sub)edge > you are on. The orientation of the edge can be calculated > from the 1s and 2s of the high order portion of the fraction, > and that should give a somewhat kludgy sum for the (x,y) > coordinates of the image point. > RWG likely has some nice FFT clock sum expression. > > Rich > > > -----Original Message----- > From: math-fun-bounces(a)mailman.xmission.com [mailto: math-fun-bounces(a)mailman.xmission.com] On > Behalf Of N. J. A. Sloane > Sent: Thursday, February 25, 2010 12:43 PM > To: math-fun(a)mailman.xmission.com > Cc: njas(a)research.att.com > Subject: Re: [math-fun] parametrizing the Koch snowflake curve > > Jim Propp asked: > >> Does anyone (Bill Gosper, maybe?) know an explicit parametrization > of the Koch snowflake curve? > > The Cantor set consists of the points 0 <= x <= 1 which have a base 3 "decimal" expansion that > does not contain a 1. Is there a similar characterization of the Koch snowflake? > > Neil Sloane > >

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Re: [math-fun] California water crisis
by Henry Baker 24 Apr '15

24 Apr '15

I seem to recall someone suggesting doing this using the ocean's own pressure: stick a 900' pipe with an RO membrane on the end vertically into the ocean & suck all the seawater out of it. You now have to pump a much smaller amount of fresh water up from the bottom of the pipe & keep the pipe empty (i.e., full of just air at atmospheric pressure). Perhaps you don't have to go down as far as 900' for this version to work; I haven't done the math. At 05:49 PM 4/23/2015, rcs(a)xmission.com wrote: >During the 197x Energy Crisis, there was an article in Science pointing >out the osmotic pressure difference between fresh water & sea water was >equivalent to a 900 foot dam. The authors had a proposal to extract >some of that as usable energy, building some dam-like gadget at the >mouth of a river. Points in favor: sizable new energy source, tolerable >cost, known physical chemistry. Against: serious eco-damage, possible >membrane issues.

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Re: [math-fun] water crisis
by Guy Haworth 24 Apr '15

24 Apr '15

And coincidentally, apols if this has already been mentioned ... In the Volvo Ocean Race, yacht Dongfeng nearly had to pit-stop because it's powered water-maker sprang a leak, and the crew were having to hand-pump for over 5 hrs/day to make basic water supplies. They've fixed it with a combination of carbon strapping and glue - but it does suggest that there should be a few more water-maker spare parts on board as part of the 'one design' provision. http://www.volvooceanrace.com/en/video/38706_Water-maker-fix.html refers ... Guy

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[math-fun] "new" kind of fourier series
by Warren D Smith 24 Apr '15

24 Apr '15

> From: Neil Bickford <techie314(a)gmail.com> > Unfortunately (if I've interpreted the above correctly), it looks like > expanding the D-bit version over individual bits of T just results in a > product of a whole bunch of phase-shifted sine waves, scaled by > some amount*. > > This can't represent arbitrary functions (or even as many as > any individual Fourier transform can), because it has zeroes > all over the place.** > > *Since C_{A,0...}*Sin[i*x1*a1]*(rest of the sum...) + > C_{A,1...}*Cos[i*x1*a1]*(rest of the sum...) > = > Sqrt[c1^2+c2^2]*Sin[i*x1*a1 + ArcTan[c2/c1]]*(rest of the sum...). > > ** Unless the series expresses a constant, in which case it's, um, > well, not extraordinarily interesting. > > --Neil Bickford --WDS responds: The fact that sin(x) and cos(x) have many zeros is no obstacle to SUM_k a[k] * sin(x*k) + b[k] * cos(x*k) representing any 2*pi-periodic nice-enough function. This sum is, in fact, a standard 1-dimensional fourier series, which DOES represent any 2*pi-periodic nice-enough function. Therefore I fail to understand your "This can't represent arbitrary functions (or even as many as any individual Fourier transform can), because it has zeroes all over the place... unless a constant." If you expand exp(i*x) = cos(x)+i*sin(x) and exp(i*x+i*y) = [cos(x)+i*sin(x)]*[cos(y)+i*sin(y)] = cos(x)*cos(y)+i*sin(x)*cos(y)+...-sin(x)*sin(y) etc, then the usual fourier series in the form USUAL FOURIER = SUM_A C_A * exp(i*A.X) summed over integer vectors A -- which will represent any nice-enough complex-valued function of D variables that is periodic in a D-dimensional box with sidelengths 2*pi -- is converted via the expansion process to a form like my second kind of fourier series. In fact, it is the same as my second kind of fourier series, EXCEPT that the coefficients of the new series are determined (via the expansion process) from a much small number of coeffs of the old series, so the "extra information" in all those extra coefficients is illusory. But if we now permit the extra coeffs to be altered independently so that that extra information really is there, then we get a wider class of series -- which was my point. And it still seems to me to be a good point, I'm not seeing anything wrong about it. This "new" kind of fourier series seems likely to me to be more useful than the old plane-wave kind for many applications. Which is more useful? Probably depends on the application.

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Re: [math-fun] California water crisis
by Henry Baker 24 Apr '15

24 Apr '15

Allan: The wicking "space cup" for the astronauts might be slightly relevant: https://www.wired.com/2015/01/coffee-cup-designed-let-astronauts-sip-espres… A Coffee Cup Designed to Let Astronauts Sip Espresso in Space Liz Stinson Design 01.27.15 6:30 am "The cupÂs shape is oddÂ-a little like a plastic baby bootÂ-and was determined by mathematical models. Every curve and geometric shape is designed to encourage the controlled movement of liquid. YouÂll notice a pointed corner in the center of the cup; this strange bit of design is what makes it possible to drink liquids in low gravity. The corner essentially acts like a wick, using surface tension to guide liquid toward your mouth. As soon as an astronaut touches her mouth to the lip of the cup, a capillary connection is formed and the liquid travels up the vessel and forms sippable balls of coffee." At 12:39 PM 4/22/2015, Allan Wechsler wrote: >I have a blue-sky invention, that I have been cherishing in the back of my >mind for a decade or so, that is relevant to this discussion. > >It's a simple nanotech application -- actually within reach with present >methods. > >I call it "black tape" because I imagine that is what it would look like. >The tape would be coated with a substance, to be designed that would >capture ambient water molecules and transport them preferentially in one >direction (probably it would be marked with arrows every few centimeters to >show the transport direction). When the tape was wetted anywhere, the >downstream end would soon start to drip water. The transported water would >be reasonably pure, because the surface coating would not be designed to >transport, say, salt. One would dip one end in salt water, and the other >end would soon start producing fresh water.

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Re: [math-fun] California water crisis
by Henry Baker 24 Apr '15

24 Apr '15

Thanks, Gene for the terrific analysis. Let's say that RO requires 5 kWh/m^3. 'A typical "150 watt" solar module is about a square meter in size. Such a module may be expected to produce 0.75 kWh every day, on average, after taking into account the weather and the latitude, for an insolation of 5 sun hours/day.' http://en.wikipedia.org/wiki/Photovoltaic_system 1 cubic meter = 264.172052 US gallons Each person requires ~ 50 gallons = ~1/5 m^3 of water per day, or about 1kWh per day for energy for desalinization. So each person requires about 2 m^2 = ~21 sqft (~ size of a large 3'x7' bed) of solar cells just to power his/her desalinization needs. Clearly, there is more than enough roof area on most (California) homes to provide enough power to desalinize the water for its occupants. So even if we're off by a factor of 2-4x, this is still reasonable. So perhaps the answer is piping seawater to people's homes and doing distributed desalinization? At 06:40 PM 4/22/2015, Eugene Salamin via math-fun wrote: >According to [ http://en.wikipedia.org/wiki/Desalination ], reverse osmosis requires 3-5.5 kWh/m^3, just for the reverse osmoses, and excluding the energy needed to operate the facility.

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