September 2015 - mathfun@mailman.xmission.com

[math-fun] ATT: infinity ~ 2^34
by Henry Baker 17 Sep '15

17 Sep '15

"Unlimited" = infinity ~ 2^34. I guess they checked this on a 32-bit computer... http://www.theregister.co.uk/2015/09/17/att_fined_unlimited_data_plan/

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[math-fun] Wei Dai: advanced civilizations dump their waste heat into black holes? Not quite.
by Warren D Smith 17 Sep '15

17 Sep '15

Here is the text from http://www.weidai.com/black-holes.txt ... (It's a brilliant idea?): Advanced civilizations probably have extensive cooling needs. Computing and communication equipment both work better at lower temperatures. A cooler computer means a faster computer with lower energy needs, and a cooler transceiver has lower thermal noise. Since these equipment cannot operate with perfect efficiency, they will need to eliminate waste heat. It's not too difficult to cool a system down to the temperature of the cosmic background radiation. All you need to do is build a radiator in interstellar space with a very large surface area, and connect it with the system you're trying to cool with some high thermal-conductance material. However, even at the cosmic background temperature of T=3K, erasing a bit still costs a minimum of k*T*ln 2 = 2.87e-23 J. What is needed is a way to efficiently cool a system down to near absolute zero. I think the only way to do it is with black holes. Black holes emit Hawking radiation at a temperature of T=h*c^3/(16*pi*G*M). With the mass of the sun, the temperature of a black hole would be about 10^-8 K. At this temperature, erasing a bit costs only about 10^-31 J. If you build an insulating shell outside the event horizon of a black hole, everything inside the shell would eventually cool down to the temperature of the black hole. However, it would not be necessary to build a complete shell around a black hole in order to take advantage of its low temperature. For example you can simply point the radiators of your black hole orbiter toward the black hole and insulate the side facing away from the black hole. If it's true that the only efficient way to cool material down to near absolute zero is with black holes, we should expect all sufficiently advanced civilizations to live near them. However this prediction may be difficult to test since they would have virtually no radiation signatures. ----end. WDS comment: the trouble with Wei Dai's idea is: black holes are small. If you only are willing to live near them, then you miss out on an enormous fraction of all real estate. Also, at the present moment in the history of the universe and of galaxies, there is plenty of energy available from, e.g, stars, unfused hydrogen, etc. I happen to believe black holes will become very important in the future for "life," but right now it seems like other places seem more fun. And even if they are not, then one still would have to ask: why refuse to exploit the ecological niche of all the non-black-hole real estate in those galaxies? In view of this, the recent astronomy study I posted about, claiming failure to see any galaxies that had been taken over by ultra-competent life, is, if valid, still rather meaningful. Also, another problem. Wei Dai is saying a 6 solar mass black hole is a heat sink at 10^(-8) kelvin, fine for radiating your waste heat into so you can now, e.g, perform computations at energy cost of around kB*T per bit, where T=10^(-8) kelvin and kB=Boltzmann constant. As opposed to, if we radiated into the universe at 3 kelvin. A huge win. Sounds great at first, but a minute later, you realize that's totally lame. The trouble is the RATE at which you can radiate waste heat from a 10 nanokelvin radiator, is pathetically small. Power = Area*sigmaSB*T^4 where sigmaSB=Stefan Boltzmann constant. If we use area=4*10^9 square meters (total horizon area) and T=10 nanokelvins we get power=2*10^(-30) watts. This would enable you to perform Area*(sigmaSB/kB)*T^3 bit operations per second if each cost kB*T, i.e. about 16 per second. Whoo whee. Yah, that's some super intelligent & competent lifeform there. Unfortunately, at the present time it would be massively outcompeted by other lifeforms using a different approach.

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Re: [math-fun] Advanced civilizations do not exist?
by Henry Baker 16 Sep '15

16 Sep '15

Here's a research plan for the NSA: 1. Build a quantum computer; in parallel with 2. Find a small (but not too small) black hole to cool it with. (3. Then move it to Bluffdale?) There, we found the money to "invest" in a huge space program... At 12:58 PM 9/16/2015, Mike Stay wrote: >Anything advanced enough to colonize the galaxy will radiate waste heat into a black hole. >http://www.weidai.com/black-holes.txt > >Advanced civilizations probably have extensive cooling needs. Computing >and communication equipment both work better at lower temperatures. A >cooler computer means a faster computer with lower energy needs, and a >cooler transceiver has lower thermal noise. Since these equipment cannot >operate with perfect efficiency, they will need to eliminate waste heat. > >It's not too difficult to cool a system down to the temperature of the >cosmic background radiation. All you need to do is build a radiator in >interstellar space with a very large surface area, and connect it with the >system you're trying to cool with some high thermal-conductance material. > >However, even at the cosmic background temperature of T=3K, erasing a bit >still costs a minimum of k*T*ln 2 = 2.87e-23 J. What is needed is a way to >efficiently cool a system down to near absolute zero. I think the only way >to do it is with black holes. > >Black holes emit Hawking radiation at a temperature of >T=h*c^3/(16*pi*G*M). With the mass of the sun, the temperature of a black >hole would be about 10^-8 K. At this temperature, erasing a bit costs only >about 10^-31 J. If you build an insulating shell outside the event horizon >of a black hole, everything inside the shell would eventually cool down to >the temperature of the black hole. However, it would not be necessary to >build a complete shell around a black hole in order to take advantage of >its low temperature. For example you can simply point the radiators of >your black hole orbiter toward the black hole and insulate the side facing >away from the black hole. > >If it's true that the only efficient way to cool material down to near >absolute zero is with black holes, we should expect all sufficiently >advanced civilizations to live near them. However this prediction >may be difficult to test since they would have virtually no radiation >signatures.

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[math-fun] Advanced civilizations do not exist?
by Warren D Smith 16 Sep '15

16 Sep '15

https://www.astron.nl/about-astron/press-public/news/werp/good-night-sleep-… tried to find galaxies that had been commandeered by ultra-competent energy-using life forms. They failed. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Eigenvalues and Asimov's group-metric problem
by Warren D Smith 16 Sep '15

16 Sep '15

> If F is the complex numbers, using dist(Id,M) is some suitable function of > the eigenvalues of M, such as dist=SUM |log(eigenvalue)|, > perhaps works? --In order for this distance to define a "metric" we would need the triangle inequality lemma that the product AB of two matrices, has F(AB)=SUM |log(eigenvalues of AB)| <= F(A) + F(B). Is this lemma valid? Apparently we need to use a branch cut for log(z) on the negative z axis. For FINITE matrix groups, the eigenvalues always are roots of unity since every group element has finite order, and then I think the lemma is true because of viewing it as a statement about the angular effects of rotation matrices (?).

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[math-fun] Reference for: sum of 1/(2k - 1) is not an integer
by Robert Baillie 16 Sep '15

16 Sep '15

In a paper I'm working on, I want to claim that for n > 1, Sum[1/(1k - 1), {k, 1, n}] is not an integer. The result is peripheral to what the paper is about, so I hesitate to spend space proving it there. I'd like to just cite a source for the claim. Can anyone give a reference to where this claim is proved? Thanks, Bob

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Re: [math-fun] Calendar puzzle
by Keith F. Lynch 16 Sep '15

16 Sep '15

James Buddenhagen <jbuddenh(a)gmail.com> wrote: > Well, I get errors when I type that in. We must be on different > operating systems. The cal command should work on all Unix, Linux, and Mac OS X command lines. Are any other command line interfaces (e.g. VMS, MS/DOS, ITS) still in use? Anyhow, the problem isn't actually about the actual cal command, but about an idealized version of it. "Adam P. Goucher" <apgoucher(a)gmx.com> wrote: > The Gregorian calendar has a period of 400 years, so cal is a > function of (year mod 400). True, but years can have the same calendar even if they don't differ by a multiple of 400. There are, after all, only 14 different calendars in our system. > This reduces the puzzle to a finite check. I don't see how that follows. Indeed, it turns out there are an infinite number of solutions. I suppose one could argue that they all follow one of a finite number of distinct patterns. Tom Duff <td(a)pixar.com> wrote: > Sometime around 4140 AD, the Gregorian rule will lose > synchronization with the equinoxes and an extra Feb 29 will > be needed. So the result depends on what fix humanity adopts. Not that soon, given that it's designed, not to track all equinoxes and solstices equally, but only to track the vernal equinox. And it really is very good at doing so. In the very long run, those average out to the same, but not over a mere 3000 years or so. Anyhow, by the time the predictable precession of the equinoxes makes a difference, the less predictable irregular slowing of the Earth's rotation will make a bigger difference. Perhaps you underestimate what our descendants will be capable of in another thousand years. They could resolve the Earth's slowing rotation by transporting lots of mass to higher latitudes, which will reduce its moment of inertia, meaning a higher rotation rate for the same angular momentum. And they could build lots of tidal power stations and run them in reverse to increase the angular momentum. With these two processes they could restore Earth's rotation rate to what it was when it was equivalent to atomic time (officially 1900, but actually closer to 1850) and keep it there indefinitely. Anyhow, I solved it for the simplest case, assuming the Gregorian calendar had been in use since the year 1 and will remain in use until the year 1 quintillion (10^18). 1 2 3 4 5 6 7 8 9 18 31 50 62 74 93 99 181 187 313 319 501 507 626 745 933 939 995 1810 1877 3130 3197 5014 5070 6265 7455 9333 9339 9395 9953 9959 18103 18773 18779 31975 50146 62654 74553 74559 93333 93339 93395 93953 93959 99531 99598 181038 187731 187798 319755 626541 626547 745534 745590 933333 933339 933395 933953 933959 939531 939598 995311 995981 995987 1877314 3197558 6265415 6265471 7455342 7455900 7455906 9333333 9333339 9333395 9333953 9333959 9339531 9339598 9395311 9395981 9395987 9953114 9959818 9959874 18773145 31975586 62654158 62654711 74553421 74553427 74559066 93333333 93333339 93333395 93333953 93333959 93339531 93339598 93395311 93395981 93395987 93953114 93959818 93959874 99531142 99598189 99598742 187731455 319755862 626541589 626547111 745534210 745534277 745590666 933333333 933333339 933333395 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The command "cal -R 1/1/1 <year>" for any year before 10000 should show the Gregorian calendar for that year. (Except the year 1, which has a bug.) Here is my very short C program which generated the above. Have fun figuring out how it works. #include <stdio.h> int main() { long long i,j,k,p,y,c[400],h[1000]; c[0] = 7; i = 2; for (j=1;j<400;j++) { c[j] = (c[j-1] + i) % 7; i = 1; if ((j%4 == 0) && (j%100 != 0)) { c[j] += 7; i = 2; } } for (j=0;j<9;j++) h[j] = j+1; p = 9; y = 0; for (j=0;(j<p) && (y < 1000000000000000000);j++) { for (k=0;k<10;k++) { y = h[j]*10 + k; if (c[y%400] == c[h[j]%400]) { h[p++] = y; printf("%lld\n",y); } } } return(0); }

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[math-fun] Calendar puzzle
by Keith F. Lynch 15 Sep '15

15 Sep '15

I just attempted to type "cal 2015" at the command line prompt to view this year's calendar. But I mistyped it as "cal 201". However, it gave exactly the same result. So I tried taking it a step further, with "cal 20". Not the same. Oh well. For which years would you get the same result if you left off any number of digits from the right? Assume the "cal" command allowed years of any length, rather than being limited to four-digit years. What about leaving digits off the left? What about deleting any subset of digits whatsoever? (You must leave at least one digit, since "cal" with no argument shows just the current month. And the last remaining digit can't be 0 since there was no year 0.) (Note that "cal" switches to the Julian calendar before 1752. Feel free to disregard this to make the problem more elegant, but please say so if you do so. Thanks.)

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[math-fun] Doomsday argument
by Warren D Smith 15 Sep '15

15 Sep '15

> From: Charles Greathouse <charles.greathouse(a)case.edu> > The doomsday argument suggests considering oneself as a random sample from > the total number of humans to be born. Estimating the number of humans born > so far at 100 billion and the number of annual births at ~0.13 billion per > year (assuming stabilization around 10 billion, if current trends continue) > a 95% confidence interval suggests that humans will be around at least 20 > more years but not more than 30,000 years. Adjust assumptions to suit. --That argument yields very different conclusions if we assume that future "humans" (which by the way might not be considered human) will have much longer lifespans. For example I think cybernetic "life" organisms will become important and their lifespans could be comparable to the lifespans of stars. In that case, assuming you are near-median chronologically does NOT at all imply that intelligent life will end in the next 100K years.

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[math-fun] Series convergence test(?)
by Bill Gosper 13 Sep '15

13 Sep '15

Iff Sum[a[k]-a[2k-1]-a[2k]] -> 0. True? Named? --rwg

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