February 2014 - mathfun@mailman.xmission.com

Re: [math-fun] oldest stars?
by Henry Baker 12 Feb '14

12 Feb '14

Apparently, the first generation ("Population III") stars formed quite quickly & died almost as quickly (3-5 Myears). As I recall, I was surprised to find that the first stars died in less time than it took the Earth's Moon to form & get a stable orbit (perhaps 150 Myears). At 01:07 PM 2/12/2014, Charles Greathouse wrote: >> Obviously, these stars are not the oldest, since they contain iron >> that came from still-older stars -- but they come close to being first >> generation.

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[math-fun] oldest stars?
by Warren D Smith 12 Feb '14

12 Feb '14

New York Times science section today discusses very old stars, in our galaxy, found by searches for ones with very small iron content. They then claim the oldest confirmed age of a star is 13.2 Gyr, confirmed by use of uranium and thorium half-lives. http://phys.org/news98033554.html QUESTION: Huh? How can you date a star by measuring its U and Th isotope content? It seems to me you'd have to know its isotope ratios when it started, and you don't. Obviously, these stars are not the oldest, since they contain iron that came from still-older stars -- but they come close to being first generation. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] atomic clocks & gravitational time dilation
by Bill Gosper 12 Feb '14

12 Feb '14

Thank you Scott! I couldn't have managed it without profanity. Anyone living on an East Coast barrier island knows that the surfside tide floods just as the Full Moon pops over the horizon, *exactly* the opposite of those stupid bulging blob illustrations! What is it that makes water so susceptible to exogenous gravity? Lower density? Lower viscosity? Then why aren't there huge tides in the atmosphere? Gravity *gradient* and "solid" tides rule. Not all of the great sloshing ocean of truth eluded Sir Isaac. But it sure eluded most of his successors. --rwg ScottH>On 2014-02-11 15:25, Huddleston, Scott wrote: > I think this link is a typical incomplete and rather misleading > explanation of tides. It doesn't even suggest the solid tides that > earth experiences, much less hint at an explanation for them. Nor > does it suggest or begin to explain the asymmetrical ocean tides. > > It's all gravitational gradient, experienced equally by solid earth > and liquid ocean. Tidal forces stress solids and liquids equally. > The reason ocean tides are much bigger than solid earth tides: the > tidal gradient has a component normal to earth's surface (which lifts > or lowers), and a component tangent to earth's surface (which causes > water, but not the semi-rigid earth, to flow). If it were not for the > tidal-induced water flows (away from the tidal "equator" and toward > the tidal "poles"), ocean and earth tides would be almost > indistinguishable and we wouldn't notice them. > > Re symmetry of near/far tides: IIRC, the ratio of solar : lunar tidal > effects is about 4 : 9. Also, tidal effects are proportional to d^3, > where d is the earth-moon (or earth-sun) distance. > Given the much farther earth-sun distance, the 4/13 solar tidal > near/far component is pretty close to symmetrical. The much close > earth-moon distance makes a stronger tidal gradient, and (I think) is > the biggest factor in asymmetrical daily tide. I get a > back-of-envelope estimate of about 1.066 near vs. far for the 9/13 > lunar tidal effect, using .25M miles as rough earth/moon distance and > 4K miles as rough earth radius. > > I make no claim having a good explanation for K-12 level. And I > welcome any clarification or correction to any of the above. > > Thanks, > - Scott > > -----Original Message----- > From: math-fun-bounces(a)mailman.xmission.com > [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov > Sent: Tuesday, February 11, 2014 2:35 PM > To: math-fun > Subject: Re: [math-fun] atomic clocks & gravitational time dilation > > This explanation from the text Mike sugggests is the same explanation > I've long heard from reliable sources: > > ----- > * Water on the side of Earth facing the moon is pulled hardest by the > moon's gravity. This causes a bulge of water on that side of Earth. > That bulge is a high tide. > > * Earth itself is pulled harder by the moon's gravity than is the > ocean on the side of Earth opposite the moon. As a result, there is > bulge of water on the opposite side of Earth. This creates another > high tide. > > * With water bulging on two sides of Earth, there's less water left in > between. This creates low tides on the other two sides of the planet. > ----- > > But I'm curious: > > * Are the tides on the sides of the earth nearest and farthest from > the moon symmetrical? > > * If so, why (since the reasons given for those tides are different) ? > > --Dan > > > On 2014-02-11, at 2:17 PM, Mike Stay wrote: > >> On Tue, Feb 11, 2014 at 2:41 PM, Bill Gosper <billgosper(a)gmail.com> wrote: >>> PS, has anyone ever found a correct explanation of tides in a middle >>> or high school text? >> >> http://www.ck12.org/earth-science/Tides/lesson/Tides-Basic/ >> >> -- >> Mike Stay - metaweta(a)gmail.com >> http://www.cs.auckland.ac.nz/~mike >> http://reperiendi.wordpress.com

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[math-fun] Ingredients: ...,water,...
by Bill Gosper 12 Feb '14

12 Feb '14

Spoiler time. rwg> Puzzle: Why does this absolutely dry and crispy snack I am munching list water as an ingredient? APG>Being edible, I imagine that it is composed of biological cells (plant? animal? fungus?), and the cytoplasm is full of water.<Adam No, these things are seriously sere, unpleasantly chewy a few hours after unsealing the package. --rwg Carr's Table Water Crackers would be a good try, but the answer is Japanese rice crackers, which are made with soy sauce, whose main ingredient, water, is dutifully listed, though totally absent. Homeopathy in reverse: instead of the water remembering, the crackers remember the water. Are there other listings of manifestly absent ingredients? "Read the label and set a healthy table." Your tax dollars at work. --rwg

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Re: [math-fun] atomic clocks & gravitational time dilation
by Bill Gosper 11 Feb '14

11 Feb '14

On 2014-02-11 08:08, Warren D Smith wrote: >> On 2/9/2014 12:40 PM, Wouter Meeussen wrote: >>> in http://www.nist.gov/pml/div688/2013_1_17_newera_atomicclocks_3.cfm >>> a atomic clock, using Y-Sr ?lattice?, is said to achieve 3E-18 accuracy & >>> stability. >>> This would allow to detect time dilation caused by a mere 4 cm altitude >>> difference. >>> >>> If guestimating correctly, this would be equivalent to the gravitational >>> effect of a 19 kg >>> mass at a horizontal distance of 10 cm. > > --The time-dilating factor is (to first order) > sqrt( 1 + 2*P/c^2 ) > where P is the (negative) gravitational potential energy per unit mass of you, > G=Newton gravit. constant, c=light speed. > > Using P = -G * (19 kg) / (10 cm) > one finds about > 1 - 1.4*10^(-25). > So you'd got that wrong by factor > million. > > To make the time-dilate factor be 1-3*10^(-18), which is the accuracy > claimed for the clock, we would want to lower the clock in Earth gravity field > by height h where > h = 3*10^(-18) * c^2 / g = 2.8 cm. > > You might want to consider the fact that the ground under your feet > rises and falls twice a day by about 50 cm due to the moon ("solid > tide"). Is not the Sun almost equally culpable? > You don't normally notice this happening. In 1961 I toured a lab with a gyro recording this while sitting on a concrete column down to bedrock. (Despite the difficulty of touring while sitting on a column.) > This would seem > to suggest that any "true" time standard would have to > be produced by a space-based clock, Earth-based clocks are inherently > too inaccurate? > Or software adjustment? Or what? The very meaning of "time" depends > where you are > on the Earth and it would seem we are reaching the limits of what one might call > "usable accuracy." Further clock improvements would seem futile. What if you want to record the close passage of a 19 kg child? --rwg PS, has anyone ever found a correct explanation of tides in a middle or high school text?

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Re: [math-fun] OK kids, what's going on here?-)
by Bill Gosper 11 Feb '14

11 Feb '14

(Followup on a puzzle to my usual suspects that I hope amuses other funsters) On Sat, Feb 8, 2014 at 5:30 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > FullSimplify[ > Sum[x^(4*n - 3)/(4*n - 3) + x^(4*n - 1)/(4*n - 1) - x^(2*n)/(2*n), > {n,∞}] /. x -> 1 > > Out[448]= Log[2] > > Sum[x^(4*n - 3)/(4*n - 3) + x^(4*n - 1)/(4*n - 1) - x^(2*n)/(2*n) /. > x -> 1, {n,∞}] > > Out[445]= Log[8]/2 > --Bill > I assume nobody answered because it was embarrassingly obvious. This is a rearrangement of the series ln(1+x) = x - x^2/2 + x^3/3 - ... which becomes illegal when the series becomes conditionally convergent at x = 1, ln 2 = 1 -1/2 + 1/3 - ... . Why isn't this a counterexample to http://en.wikipedia.org/wiki/Abel%27s_theorem ? In any case, the infinite permutation 1, 3, 2, 5, 7, 4, 9, 11, 6, 13, 15, 8, 17, 19, 10, ... where we take two odds for every even, is a fun instrument for torturing Mathematica. First, let's make some terms to torture FindSequenceFunction. Starting with the evenly odd and oddly odd ones: In[23]:= Transpose[Partition[Range[1, 19, 2], 2]] Out[23]= {{1, 5, 9, 13, 17}, {3, 7, 11, 15, 19}} Tack on the even ones and riffle: In[25]:= Flatten[Transpose[Append[%23, 2*Range[5]]]] Out[25]= {1, 3, 2, 5, 7, 4, 9, 11, 6, 13, 15, 8, 17, 19, 10} Too bad Riffle fails here: In[63]:= Riffle[%[[1]], %[[2]], 2*Range[5]] During evaluation of In[63]:= Riffle::rspec: The third argument {2,4,6,8,10} should be a positive integer or a list with three integers. Out[63]= Riffle[{1, 5, 9, 13, 17}, {3, 7, 11, 15, 19}, {2, 4, 6, 8, 10}] Pressing on, In[26]:= FindSequenceFunction[%25, n] Out[26]= DifferenceRoot[ Function[{\[FormalY], \[FormalN]}, {-30 - 20 \[FormalN] + (-3 - 2 \[FormalN]) \[FormalY][\[FormalN]] + 6 \[FormalY][1 + \[FormalN]] + 6 \[FormalY][2 + \[FormalN]] + (3 + 2 \[FormalN]) \[FormalY][ 3 + \[FormalN]] == 0, \[FormalY][1] == 1, \[FormalY][2] == 3, \[FormalY][3] == 2}]][n] constructs for itself a recurrence it can't solve. Is it even an an answer? In[27]:= Table[%, {n, 14, 22}] Out[27]= {19, 10, 21, 23, 12, 25, 27, 14, 29} Sure enough, at least Mathematica understands it. Let's find a nicer formula by trisecting the sequence using In[28]:= FindSequenceFunction[{0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1}, n] Out[28]= Mod[1 + 2 n^2, 3] Actually, I expected In[39]:= Assuming[n \[Element] Integers, FullSimplify[Sum[E^(2*I*\[Pi]*k*n/3), {k, 3}]/3]] Out[39]= 1/3 (1 + 2 (-1)^n Cos[n π/3]) In[40]:= Table[%, {n, 8}] Out[40]= {0, 0, 1, 0, 0, 1, 0, 0} Mathematica seems completely unable to show the equivalence In[42]:= Assuming[n \[Element] Integers, FullSimplify[%39-%28]] Out[42]= 1/3 (1 + 2 (-1)^n Cos[(n \[Pi])/3] - 3 Mod[1 + 2 n^2, 3]) OK, so find me one counterexample: In[44]:= FindInstance[ 1 + 2 (-1)^n Cos[(n \[Pi])/3] != 3 Mod[1 + 2 n^2, 3], n, Integers, 1] Out[44]= {} Of course, it can't. But doesn't {}, without the usual disclaimer about how there might be examples beyond the ken of FindInstance, indicate conviction that there are none? Now let's make our sequence function using my triggy trisector because it will probably simplify more. We could introduce six(!) arbitrary functions by trisecting, e.g., {foo[1], bar[2], 2, foo[4], bar[5], 4,...} etc, but it's unlikely we could later choose those functions with greater simplification than plain old linear interpolation: In[64]:= InterpolatingPolynomial[{{3, 2}, {6, 4}}, n] // Expand Out[64]= (2 n)/3 In[65]:= Table[%, {n, 6}] Out[65]= {2/3, 4/3, 2, 8/3, 10/3, 4} In[66]:= InterpolatingPolynomial[{{1, 1}, {4, 5}}, n] // Expand Out[66]= -(1/3) + (4 n)/3 In[67]:= Table[%, {n, 6}] Out[67]= {1, 7/3, 11/3, 5, 19/3, 23/3} Combining (rather manually) all three: In[49]:= FullSimplify[%64*%39 + ((n + 1)/3*4 - 1)*(%39 /. n -> n + 1) + ((n + 2)/3*4 - 3)*(%39 /. n -> n + 2), n \[Element] Integers] gives our sequence function Out[49]= 2/9 (5 n - 2 (-1)^n n Cos[(n \[Pi])/3] + (-1)^n Sqrt[3] Sin[(n \[Pi])/3]) In[50]:= Table[%, {n, 15}] Out[50]= {1, 3, 2, 5, 7, 4, 9, 11, 6, 13, 15, 8, 17, 19, 10} Let's turn this into an actual function In[51]:= Function @@ {%49 /. n -> #} Out[51]= 2/ 9 ((-1)^#1 Sqrt[3] Sin[(\[Pi] #1)/3] + 5 #1 - 2 (-1)^#1 Cos[(\[Pi] #1)/3] #1) & and see if it satisfies that strange-looking recurrence from FindSequenceFunction: In[52]:= %26[[0, 1]][%, n] Out[52]= {-30 - 20 n + 2/9 (-3 - 2 n) (5 n - 2 (-1)^n n Cos[(n \[Pi])/3] + (-1)^n Sqrt[3] Sin[(n \[Pi])/3]) + 4/3 (5 (1 + n) - 2 (-1)^(1 + n) (1 + n) Cos[1/3 (1 + n) \[Pi]] + (-1)^(1 + n) Sqrt[3] Sin[1/3 (1 + n) \[Pi]]) + 4/3 (5 (2 + n) - 2 (-1)^(2 + n) (2 + n) Cos[1/3 (2 + n) \[Pi]] + (-1)^(2 + n) Sqrt[3] Sin[1/3 (2 + n) \[Pi]]) + 2/9 (3 + 2 n) (5 (3 + n) - 2 (-1)^(3 + n) (3 + n) Cos[1/3 (3 + n) \[Pi]] + (-1)^(3 + n) Sqrt[3] Sin[1/3 (3 + n) \[Pi]]) == 0, True, True, True} In[53]:= FullSimplify[%] Out[53]= {True, True, True, True} We didn't even need n \[Element] Integers. Now we can use the sequence function %49 to (illegally) rearrange ln 2 = 1 -1/2 + 1/3 - ... In[55]:= -(-1)^k/k /. k -> %49; In[56]:= Sum[%, {n,∞}] Out[56]=Sum[-((9*(-1)^((2/9)*(5*n - 2*(-1)^n*n*Cos[(n*Pi)/3] + (-1)^n*Sqrt[3]*Sin[(n*Pi)/3])))/ (2*(5*n - 2*(-1)^n*n*Cos[(n*Pi)/3] + (-1)^n*Sqrt[3]*Sin[(n*Pi)/3]))), {n, Infinity}] Unsurprisingly, Sum boggles. We can suppress all the period 3 trickiness by grouping terms three at a time. Because Sum uses inconsistent limit semantics, we need start at n=0 vs n=1: In[59]:= shift[%56, 1] Out[59]=Sum[(9*(-1)^((1/9)*(1 + 10*n + 4*(-1)^n*(1 + n)* Cos[(1/3)*(1 + n)*Pi] - 2*(-1)^n*Sqrt[3]* Sin[(1/3)*(1 + n)*Pi])))/ (2*(5 + 5*n + 2*(-1)^n*(1 + n)*Cos[(1/3)*(1 + n)*Pi] - (-1)^n*Sqrt[3]*Sin[(1/3)*(1 + n)*Pi])), {n, 0, Infinity}] My functions shift and triple are at the bottom. In[60]:= triple[%] Out[60]= Sum[(9*(-1)^((1/9)*(1 + 30*n + 4*(-1)^(3*n)*(1 + 3*n)* Cos[(1/3)*(1 + 3*n)*Pi] - 2*(-1)^(3*n)*Sqrt[3]* Sin[(1/3)*(1 + 3*n)*Pi])))/ (2*(5 + 15*n + 2*(-1)^(3*n)*(1 + 3*n)* Cos[(1/3)*(1 + 3*n)*Pi] - (-1)^(3*n)*Sqrt[3]* Sin[(1/3)*(1 + 3*n)*Pi])) + (9*(-1)^((1/9)*(1 + 10*(1 + 3*n) + 4*(-1)^(1 + 3*n)* (2 + 3*n)*Cos[(1/3)*(2 + 3*n)*Pi] - 2*(-1)^(1 + 3*n)*Sqrt[3]*Sin[(1/3)*(2 + 3*n)*Pi])))/ (2*(5 + 5*(1 + 3*n) + 2*(-1)^(1 + 3*n)*(2 + 3*n)* Cos[(1/3)*(2 + 3*n)*Pi] - (-1)^(1 + 3*n)*Sqrt[3]* Sin[(1/3)*(2 + 3*n)*Pi])) + (9*(-1)^((1/9)*(1 + 10*(2 + 3*n) + 4*(-1)^(2 + 3*n)* (3 + 3*n)*Cos[(1/3)*(3 + 3*n)*Pi] - 2*(-1)^(2 + 3*n)*Sqrt[3]*Sin[(1/3)*(3 + 3*n)*Pi])))/ (2*(5 + 5*(2 + 3*n) + 2*(-1)^(2 + 3*n)*(3 + 3*n)* Cos[(1/3)*(3 + 3*n)*Pi] - (-1)^(2 + 3*n)*Sqrt[3]* Sin[(1/3)*(3 + 3*n)*Pi])), {n, 0, Infinity}] This took over an hour to give up trying to simplify, when all it needed was to simplify the summand first! In[61]:= MapAt[FullSimplify[#, n \[Element] Integers] &, %, 1] Out[61]= Log[8]/2 Remember that? --rwg shift[xp_, s_] := xp /. (cf : ContinuedFractionK | cfk | mprd | MProd | Product | Sum)[ ab__, {v_, L_: 1, h_}] :> cf @@ Append[Simplify[{ab} /. v -> v + s], {v, L - s, h - s}] pair[xp_] := xp /. (op : mprd | MProd | Product | Sum)[ab__, {v_, L_: 1, h_}] :> op[(op /. mprd | MProd -> Dot)[ab /. v -> 2*v + ¢, {¢, 0, 1}], {v, L/2, (h - 1)/2}] triple[xp_] := xp /. (op : mprd | MProd | Product | Sum)[ab__, {v_, L_: 1, h_}] :> op[(op /. mprd | MProd -> Dot)[ab /. v -> 3*v + ¢, {¢, 0, 2}], {v, L/3, (h - 2)/3}] I should have called them bisect and trisect.

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[math-fun] The sound of mathematical truth: DRUP, DRUP, DRUP ....
by Thane Plambeck 11 Feb '14

11 Feb '14

"The negative witness, that is, the DRUP unsatisfiability certificate, is probably one of longest proofs of a non-trivial mathematical result ever produced. Its gigantic size is comparable, for example, with the size of the whole Wikipedia, so one may have doubts about to which degree this can be accepted as a proof of a mathematical statement." http://gowers.wordpress.com/2014/02/11/recent-news-concerning-the-erdos-dis… -- Thane Plambeck tplambeck(a)gmail.com http://counterwave.com/

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[math-fun] Drum Corps does 3d rotating tetrahedron
by W. Edwin Clark 11 Feb '14

11 Feb '14

Is this cool or not? http://24.media.tumblr.com/5eb47fccd6b4c363a9076527c85a7375/tumblr_ms0xa2EG…

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[math-fun] atomic clocks & gravitational time dilation
by Warren D Smith 11 Feb '14

11 Feb '14

> On 2/9/2014 12:40 PM, Wouter Meeussen wrote: >> in http://www.nist.gov/pml/div688/2013_1_17_newera_atomicclocks_3.cfm >> a atomic clock, using Y-Sr ?lattice?, is said to achieve 3E-18 accuracy & >> stability. >> This would allow to detect time dilation caused by a mere 4 cm altitude >> difference. >> >> If guestimating correctly, this would be equivalent to the gravitational >> effect of a 19 kg >> mass at a horizontal distance of 10 cm. --The time-dilating factor is (to first order) sqrt( 1 + 2*P/c^2 ) where P is the (negative) gravitational potential energy per unit mass of you, G=Newton gravit. constant, c=light speed. Using P = -G * (19 kg) / (10 cm) one finds about 1 - 1.4*10^(-25). So you'd got that wrong by factor > million. To make the time-dilate factor be 1-3*10^(-18), which is the accuracy claimed for the clock, we would want to lower the clock in Earth gravity field by height h where h = 3*10^(-18) * c^2 / g = 2.8 cm. You might want to consider the fact that the ground under your feet rises and falls twice a day by about 50 cm due to the moon ("solid tide"). You don't normally notice this happening. This would seem to suggest that any "true" time standard would have to be produced by a space-based clock, Earth-based clocks are inherently too inaccurate? Or software adjustment? Or what? The very meaning of "time" depends where you are on the Earth and it would seem we are reaching the limits of what one might call "usable accuracy." Further clock improvements would seem futile.

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[math-fun] Can these AES instructions be used for anything else?
by Henry Baker 11 Feb '14

11 Feb '14

FYI -- Are these specialized AES instructions useful for any other interesting computations? http://software.intel.com/en-us/articles/intel-advanced-encryption-standard… IntelÂ® Advanced Encryption Standard (AES) Instructions Set - Rev 3.01 Submitted by Shay Gueron (Intel) on Thu, 08/02/2012 - 14:11 Introduction IntelÂ® AES instructions are a new set of instructions available beginning with the all new 2010 IntelÂ® CoreÂ processor family based on the 32nm IntelÂ® microarchitecture codename Westmere. These instructions enable fast and secure data encryption and decryption, using the Advanced Encryption Standard (AES) which is defined by FIPS Publication number 197. Since AES is currently the dominant block cipher, and it is used in various protocols, the new instructions are valuable for a wide range of applications. The architecture consists of six instructions that offer full hardware support for AES. Four instructions support the AES encryption and decryption, and other two instructions support the AES key expansion. The AES instructions have the flexibility to support all usages of AES, including all standard key lengths, standard modes of operation, and even some nonstandard or future variants. They offer a significant increase in performance compared to the current pure-software implementations. Beyond improving performance, the AES instructions provide important security benefits. By running in data-independent time and not using tables, they help in eliminating the major timing and cache-based attacks that threaten table-based software implementations of AES. In addition, they make AES simple to implement, with reduced code size, which helps reducing the risk of inadvertent introduction of security flaws, such as difficult-to-detect side channel leaks. This paper gives an overview of the AES algorithm and Intel's new AES instructions. It provides guidelines and demonstrations for using these instructions to write secure and high performance AES implementations. This version of the paper also provides a high performance library for implementing AES in the ECB/CBC/CTR modes, and discloses for the first time, the measured performance numbers. [Revisions history: Rev. 1.0 in 4/2008; Rev. 2.0 in 4/2009; Rev. 3.0 in 5/2010; Rev. 3.01 in 9/2012] Download Article http://download-software.intel.com/sites/default/files/article/165683/aes-w…

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