March 2014 - mathfun@mailman.xmission.com

Re: [math-fun] How to find Malaysia370 missing plane.
by Warren D Smith 16 Mar '14

16 Mar '14

The picture https://dl.dropboxusercontent.com/u/3507527/malay370.png summarizes my deduction. This is based on now finding in wikipedia the timing data I needed. It was at 6deg 55min 15sec N, 103deg 34min 43sec E at 1:30am local time = 17:30 UTC. It then was detected by Malay military radar at 2:40am local time "near Pulau Perak, in the Straits of Malacca." http://my.news.yahoo.com/mh370-detected-above-malacca-straits-2-40am-062617… Based on this I estimate the center S of the second circle and time T at which plane was there, as shown in figure. If plane then flew straight at 905 km/hr then at 8:11am local time, it had to be on a circle of radius=4462km centered at S. Intersecting this circle with the satellite-pig circle, we deduce the plane's location at 8:11am Malay time as shown in figure. ---- NY Times: "Pilot Spoke to Air Controllers After Shutoff of Data System" indicates pilot spoke to air control after the data system was shut off, and did not indicate anything awry. The only reasonable explanation of this is that the pilot himself was the one whodecied to take the plane on its odyssey.

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Re: [math-fun] [CES #10484] How to find Malaysia370 missing plane.
by Warren D Smith 16 Mar '14

16 Mar '14

The NY Times right now (sunday 1pm) http://graphics8.nytimes.com/newsgraphics/2014/03/15/satellite-contact-map/… still has not managed to perform the deduction (intersecting 2 circles) I suggested. All they have deduced is the plane had to have come down somewhere within the two bent-sausage-shaped regions. In fact, we CAN deduce (subject merely to the assumption those flying the plane were not idiots) fairly exactly where on the two red circular arcs the plane was at 811pm, plus then deduce from the then-known direction of flight plus the known fuel-remaining, a small region where the plane had to have come down. Upon then finding every suitable landing strip within that small region, one can probably deduce its exact location. The NY Times still has not made that deduction. Has anybody? I hope it does not require a math PhD to tell the Great Expert Panels in the press and/or govt how to intersect two circles, but this one tried, and as far as I can tell, not a soul in the US govt or media has listened to me. But we can tell from the NYT graphic already that it did not land in Australia, and (again subject to its flyers not being idiots) therefore had to be on the more northern of the two red arcs.

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Re: [math-fun] [CES #10484] How to find Malaysia370 missing plane.
by Bill Gosper 16 Mar '14

16 Mar '14

WDS> The NY Times right now (sunday 1pm) http://graphics8.nytimes.com/newsgraphics/2014/03/15/satellite-contact-map/… still has not managed to perform the deduction (intersecting 2 circles) I suggested. All they have deduced is the plane had to have come down somewhere within the two bent-sausage-shaped regions. In fact, we CAN deduce (subject merely to the assumption those flying the plane were not idiots) fairly exactly where on the two red circular arcs the plane was at 811pm, plus then deduce from the then-known direction of flight plus the known fuel-remaining, a small region where the plane had to have come down. Upon then finding every suitable landing strip within that small region, one can probably deduce its exact location. The NY Times still has not made that deduction. Has anybody? I hope it does not require a math PhD to tell the Great Expert Panels in the press and/or govt how to intersect two circles, but this one tried, and as far as I can tell, not a soul in the US govt or media has listened to me. But we can tell from the NYT graphic already that it did not land in Australia, and (again subject to its flyers not being idiots) therefore had to be on the more northern of the two red arcs. <WDS It took NYT until the Moon Landing to retract a decades-earlier editorial ridiculing Robert Goddard for proposing space travel with rockets, when there is no air for them to push against. --rwg

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[math-fun] Solving quintics with hypergeometrics
by Bill Gosper 16 Mar '14

16 Mar '14

On Sat, Mar 15, 2014 at 2:41 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > > On Wed, Mar 12, 2014 at 5:24 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > >> WDS> >> >> On 3/12/14, Warren D Smith <warren.wds(a)gmail.com> wrote: >> >> Of course 4F3(x) and indeed any pFq(x) with p<=q+1, is defined for >> all complex x except for a set of measure zero, via analytic continuation. >> And this particular 4F3 is of course an algebraic function which makes >> that obvious. >> >> >> --and >> http://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQ/ >> gives explicit formulas such as integral representations. >> Pade and Borel summation should also work. >> <wds >> >> For more ideas, this ancient Macsyma Newsletter: >> http://gosper.org/newsrope.pdf >> >> (which, of course, isn't restricted to quintics.) >> >> >> --rwg >> >> Mma's InverseSeries should be able to do some of this. Then >> >> FindSequenceFunction of a bleepload of terms might give you some >> >> interesting identities. >> >> >> Sadly, no. Mathematica's series stuff is still decades behind Macsyma's. > Expansions in negative powers of (n+1/2) (like for the fixed points of > Tan) are illegal. > SeriesCoefficient[Log[n!], {n, \[Infinity], k}] can't recite Stirling's > formula. > FindSequenceFunction doesn't know Bernoullis. > But it did find > Log[Binomial[2*n, n]] == 2*n*Log[2] + > (1/2)*Log[1/(n*Pi)] + > Sum[((-2 + 2^(1 - k))*n^(1 - k)*BernoulliB[k])/ > ((-1 + k)*k), {k, 2, Infinity}] > after I divided out (and later restored) the Bernoullis. > It won't even let you tell it the coefficients: > In[340]:= Series[Sum[a[k]*x^k, {k, \[Infinity]}], {x, 0, 4}] > > Out[340]= \!$ > \*UnderoverscriptBox[\(\[Sum]$, $k$, $\[Infinity]$]$ > \*SuperscriptBox[\(x$, $k$]\ a[k]\)\) > > OtOH, FindSequenceFunction miraculously fit the crazy coefficients > of the root near 1 (for small y) of x^5-x = y: > {1, 1/4, -(5/32), 5/32, -(385/2048), 1/4, -(23205/ > 65536), 2145/4096, -(6705765/8388608), 5/4, -(534734915/ > 268435456), 3380195/1048576, -(90759684405/17179869184), 35/4,...], > > producing a huge pile of Gammas that Macsyma simplified to > > Out[376]= 1 + y/4 + Sum[(5*(-1)^(1 + n)*y^n*Gamma[(5/4)*(-1 + n)])/ > (4*Gamma[(1/4)*(-1 + n)]*Gamma[1 + n]), > {n, 2, Infinity}] > > on which Series malfunctions: > In[377]:= Series[%, {y, 0, 9}] > > Out[377]= SeriesData[y, 0, {1, 1/4}, 0, 10, 1] + > Sum[(5*(-1)^(1 + n)*y^n*Gamma[(5/4)*(-1 + n)])/ > (4*Gamma[(1/4)*(-1 + n)]*Gamma[1 + n]), > {n, 2, Infinity}] > > In[378]:= %/.\[Infinity]->9 > Out[378]= 1+y/4-(5 Gamma[5/4] y^2)/(8 Gamma[1/4])+(5 y^3)/32-(5 > Gamma[15/4] y^4)/(96 Gamma[3/4])+y^5/4-(Gamma[25/4] y^6)/(576 > Gamma[5/4])+(2145 y^7)/4096-(Gamma[35/4] y^8)/(32256 Gamma[7/4])+(5 > y^9)/4+O[y]^10 > In[380]:= FullSimplify[%377^5-%377/.\[Infinity]->11] > Out[380]= y+O[y]^10 > > I.e.x^5-x=y. > --rwg > I was thinking of sending the Mma developers the > > http://gosper.org/newsrope.pdf URL. > > But it seems I already did years ago. > > This actually gives all five roots--the original one -3 + 4 n Gamma[-4 + 5 n] y Sum[-(-------------------------), {n, ∞}] Gamma[n] Gamma[-2 + 4 n] and the other four, k n 1 5 n n I Gamma[-(-) + ---] y 1 4 4 - Sum[-------------------------, {n, 0, ∞}] 4 3 n Gamma[- + -] Gamma[1 + n] 4 4 -------------------------------------------------- k I k ∈ 0..3 all five based on the five roots of x^5-x=0 . (I:=√-1) Test that they satisfy the symmetric functions: In[506]:= FullSimplify[Series[Product[t-ReleaseHold[%/.∞->29],{k,4}]* ReleaseHold[t-%%/.∞->8],{y,0,29}]] Out[506]= (-t+t^5)-y+O[y]^30 --rwg

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[math-fun] Finding missing Malaysia370 plane using trigonometry.
by Warren D Smith 15 Mar '14

15 Mar '14

CNN's idiotic news on the missing Malaysia370 plane said a satellite ping timing deduced that at 8:11pm, the plane had to be on a certain circle drawn on the globe. What CNN did NOT say was, if we ASSUME the plane flew on a straight path (i.e. geodesic great circle) at its most-efficient miles/gallon speed (which of course it would because the plotters were not idiotic) starting from a known place S at a known time T (both of which we pretty much know) then we can DEDUCE EXACTLY WHERE on that circle, the plane was a 8:11pm. You follow? There are two circles, one from the satellite, the other centered at S and of radius V*T where V is the 777's most efficient speed. The intersection of BOTH these circles is: two points. Find those two points, and you'll know where the plane was at 8:11 and you'll also know (under our straightness assumption) which direction it was going. Follow that direction out and intersect with all suitable landing strips, and you've found it. PS. Could anybody relay this message to, like, the appropriate government or whoever bodies? I tried to send it to them myself, but my experience whenever I try to contact the press or government about anything ever, is they always ignore me either forever or at least for weeks. I see no reason this should be any different.

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[math-fun] Dodgson's determinant by concentration
by Bill Gosper 15 Mar '14

15 Mar '14

Besides Warren's proposals, Dodgson's method is valuable theoretically, helping Rich prove my Somos addition formulæ http://arxiv.org/abs/math/0703470 and Zeilberger prove the famously challenging Alternating Sign Matrix Theorem (A005130), the "Stirling's formula" for which is Out[264]= (2^(5/12 - 2*n^2)*3^(-(7/36) + (3*n^2)/2)* E^((1/3)*HoldForm[Derivative[1][Zeta][-1]])*Pi^(1/3))/ (n^(5/36)*Gamma[1/3]^(2/3)) (the HoldForm to fend off that vexatious Glaisher symbol). In[265]:= Table[N[ReleaseHold[%]], {n, 9}] // InputForm Out[265]//InputForm= {1.0063254118710128, 2.003523267231662, 7.0056223910285915, 42.01915917750558, 429.12582410098327, 7437.518404899576, 218380.8077275304, 1.085146545456063*^7, 9.119184824937415*^8} the actual sequence being 1, 1, 2, 7, 42, 429, 7436, 218348, 10850216, 911835460, 129534272700, Adamchik may have already worked this out someplace, in spite of the painful obstacles thrown up by Mathematica. OEIS could probably use more such asymptotic formulæ. --rwg

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[math-fun] Hyperstupormetric
by Bill Gosper 15 Mar '14

15 Mar '14

First of all, why, to save your life, don't the minus signs cancel in In[444]:= Simplify[FullSimplify[FunctionExpand[ -Sum[-((Gamma[-(1/4) + (5*n)/4]*y[n])/ (4*Gamma[3/4 + n/4]*Gamma[1 + n])), {n, 0, Infinity}]]]] Out[444]= -Sum[-((Gamma[-(1/4) + (5*n)/4]*y[n])/ (4*Gamma[3/4 + n/4]*Gamma[1 + n])), {n, 0, Infinity}] Why don't they cancel automatically? Now take two terms of the series In[445]:= % /. ∞-> 1 Out[445]= (Gamma[-(1/4)] y[0])/(4 Gamma[3/4]) + y[1]/4 observing that the constant (n=0) term is -1 if y[n_]:=y^n In[446]:= FunctionExpand[% /. y[n_] -> y^n] Out[446]= -1 + y/4 Now impose y[n_]:=y^n on the whole sum: In[447]:= %%% /. y[n_] -> y^n Out[447]= (1/(32 Gamma[3/4]))(8 Gamma[-(1/4)] + 8 Gamma[-(1/4)] HypergeometricPFQ[{-(1/20), 3/20, 7/20, 11/20}, {1/ 4, 1/2, 3/4}, (3125 y^4)/256] + 8 y Gamma[3/ 4] HypergeometricPFQ[{1/5, 2/5, 3/5, 4/5}, {1/2, 3/4, 5/4}, ( 3125 y^4)/256] + 5 y^2 Gamma[3/ 4] HypergeometricPFQ[{9/20, 13/20, 17/20, 21/20}, {3/4, 5/4, 3/ 2}, (3125 y^4)/256] + 5 y^3 Gamma[3/ 4] HypergeometricPFQ[{7/10, 9/10, 11/10, 13/10}, {5/4, 3/2, 7/ 4}, (3125 y^4)/256]) This outrageously stupid and unhelpful canonicalization is what drove me to use y[n] when I meant y^n in the first place. This should never be automatic-- either put it on (overloaded) FunctionExpand, or provide ToHypergeometric, which surely is already present internally. And yet in %445 you won't canonicalize Gamma[-1/4]/Gamma[3/4], which would actually be helpful. But the most wonderful thing about the preceding eructile dysfunction is that it's wrong! In[448]:= FunctionExpand[% /. y -> 0] Out[448]= -2 should be -1 ! --Bill Gosper In[61]:= $LicenseID Out[61]= "L3290-7570"

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[math-fun] (no subject)
by Bill Gosper 15 Mar '14

15 Mar '14

WDS> On 3/12/14, Warren D Smith <warren.wds(a)gmail.com> wrote: Of course 4F3(x) and indeed any pFq(x) with p<=q+1, is defined for all complex x except for a set of measure zero, via analytic continuation. And this particular 4F3 is of course an algebraic function which makes that obvious. --and http://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQ/ gives explicit formulas such as integral representations. Pade and Borel summation should also work. <wds For more ideas, this ancient Macsyma Newsletter: http://gosper.org/newsrope.pdf (which, of course, isn't restricted to quintics.) --rwg Mma's InverseSeries should be able to do some of this. Then FindSequenceFunction of a bleepload of terms might give you some interesting identities.

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Re: [math-fun] calculating pi with puzzle pieces
by Fred Lunnon 14 Mar '14

14 Mar '14

Then just fix up a few relays to save reading all those inputs, and maybe an industrial steam-engine to do the heavy-lifting ... hey presto, a difference engine! WFL On 3/14/14, Steve Witham <sw(a)tiac.net> wrote: > Thinking about calculating pi... if only there were a way to do > calculations by hand that was absolutely error-proof, so one wouldn't > die having calculated ten thousand digits, only to have an error in > digit 7195 detected after one's death. > > Well, you could calculate with appropriately-designed puzzle pieces. > Think of the basic cell in a multiplication > > total factor A > input & carry input > ^ > / \ > + + > factor B | | factor B > output | | input > + + > \ / > factor A v total output > & carry > output > > In decimal, there are 10,000 combinations of the four input digits > (of which 9100 are possible). So, just make molds or cutting dies > for all those combinations, with appropriate outputs of course. Then > you need a frame, and pieces for the edge conditions. On the input > and output edges, you could put raised type and print the thing when > done. > > In fact you could design a set of pieces around the sides, or maybe > wires, turning shafts or sliders, to convey all the inputs and > outputs to a small area for printing. > > If you broke each digit into two pairs of binary-coded-decimal (BCD) > bits, you would need four little pieces out of an alphabet of only > 64, inside each cell. Then you would need to recode the result as > BCD, but there's an issue there. > > The bit-pairs for factor A need to be interleaved with the bit-pairs > for the carry, in order to get one of each to each upper-right little > piece. I can imagine either using signal-crossing pieces, or > handling the output of all combinations of (factor A, carry, total) > with single pieces out of a set of 1000. > > But, a better way to distribute the factor inputs would be as > crisscrossing bars lying under the pieces. Then there would still be > 64 types of little piece, but only 100 BCD recoder shapes. > > > --Steve > "Did you check your calculations?" > "I have my boards, right here!" > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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[math-fun] avoiding divisions
by Warren D Smith 14 Mar '14

14 Mar '14

> From: Henry Baker <hbaker1(a)pipeline.com> > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] avoidance of division (in English) > Message-ID: <E1WOSHp-0007oE-Bd(a)elasmtp-scoter.atl.sa.earthlink.net> > Content-Type: text/plain; charset="us-ascii" > > The practical problem is that putting off division typically > blows up the sizes of the eventual numerators & denominators, > so that the total number of bit operations actually increases. --no... Any algorithm involving +,-,*,/ can be done with rationals in such a way that only one division per output is needed, thus postponing the divisions until one at end. But that is not what Strassen & I were speaking of. Strassen was ENTIRELY eliminating divisions, not postponing or reducing to one. Example problem: given a matrix compute its determinant (or the coefficients of its characteristic polynomial.) These are normally done by algorithms involving divisions. But algorithms involving NO divisions are possible. By combining Strassen's trick with some other tricks by Kaltofen & Villard, and some updating, I can show that these things may be computed for NxN matrix with O(N^2.6945) ring-arithmetic operations. Meanwhile the best determinant and charpoly algorithms I know which do use divisions, take O(N^2.3727) field-arithmetic operations. It would be very interesting if you could decrease either of these magic exponents... or if you could prove a gap between the division-using and not complexities... -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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