April 2011 - mathfun@mailman.xmission.com

[math-fun] Testing 1-2-3
by Fred lunnon 30 Apr '11

30 Apr '11

Where is everybody? Don't tell me they've all been watching Kate & Wills ... WFL

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Re: [math-fun] Testing 1-2-3
by Bill Gosper 30 Apr '11

30 Apr '11

On 4/29/2011 6:10 PM, Fred lunnon wrote: > Where is everybody? > > Don't tell me they've all been watching Kate& Wills ... > > WFL > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com <https://wm.sdf.org/src/compose.php?send_to=math-fun%40mailman.xmission.com> > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > I have 0 interest in Kate & Willie. Here is a speculation on something nonmathematical. It often happens that we can't recall a word or someone's name but "know that we know" it. Consider the memory space taken up in two cases: 1. We recall the name with no problem. 2. We can't recall it, but it still takes up the same amount of memory, plus the space needed to realize we can't recall it. Brain economy would, I think, prefer the easier situation. So why does it ever happen that we know that we know it but can't recall it? Admittedly a very naive view, but this and other evidence convinces me that storing and retrieving are very different operations. Steve Gray Anomic aphasia. http://www.anomiapress.com/ http://www.pagetutor.com/jokebreak/210.html --rwg

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Re: [math-fun] How can I accurately calculate (p/q)^n mod 1,
by Steve Witham 27 Apr '11

27 Apr '11

>From: Dan Asimov <dasimov(a)earthlink.net> > >Maybe I should be more specific. > >Mathematica can do a loop of length 10000, >say calculating > > Floor[100*Mod[(3/2)^k,1]] [correction from later post] > >for 10001 <= k < 20000, > >but when I try to do something like that for >a loop of size 100,000 or 1,000,000, it seems >to take forever (or hang, I'm not sure which). Oh, you mean a million large, not 2^4096 large. Could it be that Mathematica's rational arithmetic is trying to reduce the fraction at every step in the exponentiation? Try maybe Floor[100*Mod[3^k,2^k]/2^k] ? For that matter, try 3^k >From: Bill Gosper <billgosper(a)gmail.com> > >Oops, luckily, Julian was eavesdropping! > >---------- Forwarded message ---------- >From: Julian Ziegler Hunts <julianj.zh(a)gmail.com> > >Maybe you want Floor[10^16*Mod[1.5^k,1]], which diverges from >the exact computation at k=21, or, with 10000 (decimal) digits of precision, >at k=56758. You mean diverges from the precision Dan asked for, right? To get the *exact* answer you need about three times as many digits (and when q = 2 you would need to calculate in binary). If you want d digits of precision at the end, log10(p/q)*n + d digits is enough, right? (Seems to be what Julian is calculating.) Eg. for k = 1,000,000 you need 176,092 + d digits. Btw, we're talking about precision *mod 1* right? So .995 might come out .99 or .00 and that would be okay? >From: Dan Asimov <dasimov(a)earthlink.net> > >Yes, Julian was right, in fact >what I was calculating was > > Floor[100*Mod[(3/2)^k,1]] > >for long sequences of consecutive k. That must be a lot faster to calculate in an incremental way, even in Mathematica, right? I don't have Mathematica but I wondered what might take it so long, so I timed something similar in Python. Python has big ints but only IEEE floats built in, so I took (p^n mod q^n) / q^n (This doesn't have the wraparound problem, it just takes more time and space.) def p_over_q_to_the_n_mod_1(p, q, n): pn = p ** n qn = q ** n m = pn % qn # Can't convert a number > 2**1023 to IEEE float, so... nbits = int(log(qn, 2)) if nbits > 1000: scale = 2 ** (nbits - 1000) m = m / scale qn = qn / scale return float(m) / qn $ ./p_over_q.py 3 2 1000000 pn = p^n... 0.974992990494 sec. qn = q^n... 0.0310521125793 sec. m = pn mod qn... 22.5215368271 sec. scale down... 0.12483215332 sec. r = float(m) / qn... 8.70227813721e-05 sec. Total: 23.6525011063 sec. 0.327352766151 The odd thing in Python's case is that it spent most of its time taking a number mod a power of two! Adding a special case for that... if q == 2: m = pn & (qn - 1) # mod using binary mask else: m = pn % qn $ ./p_over_q.py 3 2 1000000 pn = p^n... 0.976704120636 sec. qn = q^n... 0.0312979221344 sec. m = pn mod qn... 0.00104093551636 sec. scale down... 0.117168188095 sec. r = float(m) / qn... 8.20159912109e-05 sec. Total: 1.12629318237 sec. 0.327352766151 If that helps... --Steve

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Re: [math-fun] DO sweat it
by Bill Gosper 26 Apr '11

26 Apr '11

VE>The radiative heat flow to the droplets goes like their area on the glass, but the heat required to raise their temperature by one degree goes like their volume. The small drops therefore heat up faster than the large drops and their surface tension diminishes faster, causing them to become unstable (against gravity) first. Veit Sorry, no, the larger ones still fall first, as before. I was merely asking why they only fell when warm, and you said the magic words that *surface tension* declines with temperature. So the smaller ones need to get hotter. Perfect. Thanks! --rwg On Apr 25, 2011, at 7:18 PM, Bill Gosper wrote: > Sprinkle some water on the horizontal inside surface of the opened glass > door of > a room-temperature toaster oven. Close the door, and wait half a minute for > the > larger drops to run down. Press TOAST. What happens to the remaining > drops? > > > I would have expected them to evaporate. Instead, they run down in > decreasing > order of size, at least for me, well before reaching the boiling point. > Why?? > --rwg How do I stop idiot GMail from linebreaking? IBM punch cards were 80 column.

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Re: [math-fun] Balancing chemical equations by hand
by Henry Baker 26 Apr '11

26 Apr '11

Doesn't there also have to be some notion of "atomic" (gcd?) reaction, otherwise you couldn't separate 2 H2 + 3 O2 + C <-> 2 H2O + CO2 into its constituent reactions 2 H2 + 2 O2 <-> 2 H2O + energy and C + O2 <-> CO2 + energy At 11:29 AM 4/25/2011, mcintosh(a)unam.mx wrote: >> There must be more to it than just determinants & linear algebra, >> because you aren't guaranteed to get integers. Or do you simply >> multiply by the (integer-valued) determinant to clear the fractions? >> >> Is there any chance that you might end up doing "integer linear >> programming", which is considerably harder than linear programming >>(because it subsumes hard problems like the knapsack problem) ? > >Well, I didn't pursue it that far. But since the solution would be >rational, presumably a multiple would suffice. > >In truth, there is a problem of reaction kinetics involved, an there >is an implicit assumption of stable compounds; otherwise the reaction >could just as well be reversible. (hence your second question) > >But this is all simple introductory chemistry. And I was mostly >delighted that I had found a rational scheme for understanding those >reaction equations. > >-hvm

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[math-fun] prices these days
by quad 26 Apr '11

26 Apr '11

How much did this cost 50 years ago? http://i.imgur.com/qwocC.jpg

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[math-fun] Do sweat it
by Bill Gosper 26 Apr '11

26 Apr '11

Sprinkle some water on the horizontal inside surface of the opened glass door of a room-temperature toaster oven. Close the door, and wait half a minute for the larger drops to run down. Press TOAST. What happens to the remaining drops? I would have expected them to evaporate. Instead, they run down in decreasing order of size, at least for me, well before reaching the boiling point. Why?? --rwg

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Re: [math-fun] Balancing chemical equations by hand
by Henry Baker 25 Apr '11

25 Apr '11

There must be more to it than just determinants & linear algebra, because you aren't guaranteed to get integers. Or do you simply multiply by the (integer-valued) determinant to clear the fractions? Is there any chance that you might end up doing "integer linear programming", which is considerably harder than linear programming (because it subsumes hard problems like the knapsack problem) ? At 08:04 PM 4/24/2011, mcintosh(a)unam.mx wrote: >> Can anyone come up with a nice way to balance chemical equations >> manually? >> >> All the web seems to offer is either vague "fiddle around until it >> works" or the nuclear option "translate into a simultaneous linear >> system and solve". > >As an undergraduate, I discovered how to balance chemical reactions >using determinants. Teacher response: This is MY course and we will >do it MY way! Actually I had added conservation of charge to >conservation of mass because redox equations were otherwise ambiguous. > >I wonder of thete is really any alternatice to conservation equations >and linear algebra, even if most textbook equations can be solved by >inspection. Of course the algebra part is rely mentioned in a chemistry >course. > >-hvm

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[math-fun] US scientists get glimpse of antihelium
by Henry Baker 25 Apr '11

25 Apr '11

FYI -- So it sounds like there could still be anti-galaxies out there... http://www.guardian.co.uk/science/2011/apr/24/antihelium-antimatter-brookha… US scientists get glimpse of antihelium Heaviest particles of antimatter seen in a lab survive for about 10 billionths of a second before crashing into collider's detector Ian Sample, science correspondent guardian.co.uk, Sunday 24 April 2011 17.59 BST They were gone as soon as they appeared, but for a fleeting moment they were the heaviest particles of antimatter a laboratory has seen. Scientists in the US produced a clutch of antihelium particles, the antimatter equivalents of the helium nucleus, after smashing gold ions together nearly 1bn times at close to the speed of light. The discovery of antihelium at the Relativistic Heavy Ion Collider at Brookhaven national laboratory in New York will aid the search for exotic phenomena in the distant universe, including antimatter versions of stars and even galaxies. Antimatter looks and behaves like normal matter but has one crucial difference: particles of antimatter have an equal and opposite charge to those that make up the world around us. When antimatter meets matter, the two annihilate one another, leaving nothing but a burst of energy. Researchers at the US laboratory recorded 18 antihelium particles that survived for about 10 billionths of a second before they crashed into the collider's detector and vanished in the tiniest of fireballs. "Antihelium is stable, so if it doesn't encounter anything it will survive forever," said Aihong Tang, a physicist at the laboratory. "Unless there is a major breakthrough in accelerator technology, this will be the heaviest antimatter made for decades to come." Antihelium is the heaviest breed of antimatter created by scientists, with each particle is roughly 10 million billion times lighter than a grain of sand. The next heaviest that is stable is antilithium, but this is so rare the Brookhaven collider would have to run for thousands of years to detect just one particle. Antimatter is central to one of the greatest mysteries of our existence. Equal amounts of matter and antimatter were created in the Big Bang and should have destroyed each other in one enormous cosmic explosion. But for reasons unknown, only normal matter seems to have survived to make up all we know in the visible universe. Particles of antimatter were first discovered in 1932 when a US researcher found antielectrons, or positrons, among the debris of cosmic ray collisions. Cosmic rays are highly charged streams of particles that can span vast stretches of space. Paul Dirac, the British physicist who predicted antimatter, speculated that some regions of the universe might be home to entire galaxies made of antimatter. The latest research, published in the journal Nature, is a benchmark for space-based experiments that will hunt for antimatter elsewhere in the universe. Next week, the penultimate mission of the space shuttle will deliver a $2bn instrument called the alpha magnetic spectrometer (AMS) to the International Space Station. From there, it will scour space for signs of heavenly bodies made from antimatter. "Collisions among cosmic rays near Earth can produce antimatter, but the odds of these collisions producing an intact antihelium nucleus are so vanishingly small that finding even one would strongly suggest that it had drifted to Earth from a distant region of the universe dominated by antimatter," said Hans Georg Ritter at the University of California, Berkeley. "Antimatter doesn't look any different from ordinary matter, but AMS finding just one antihelium nucleus would suggest that some of the galaxies we see are antimatter galaxies."

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[math-fun] periodic nag
by Bill Gosper 25 Apr '11

25 Apr '11

Can anybody prove: Sum[k/n*Binomial[-n, n - k]*Fibonacci[k + 1], {k, 1, n}] == 1 ? I was almost surprised when it defeated Julian last nite. --rwg

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