March 2011 - mathfun@mailman.xmission.com

Re: [math-fun] What is "geometric" about a geometric progression?
by Henry Baker 03 Mar '11

03 Mar '11

I programmed an interactive version of Robert's "means" diagram in Firefox javascript, which Robert put on his website here: http://www.symbo1ics.com/files/means.htm I'm not proud of the code, but this was a good excuse to learn how to program in Javascript with JSXGraph. --- BTW, you should be aware that one of the coolest features of Javascript is _full closures_ (as found in Common Lisp & Scheme), so you can use function-returning functions to your heart's content, and even play with "Curried" functions. At 07:55 PM 3/2/2011, quad wrote: >On Wed, Mar 2, 2011 at 6:36 PM, Gareth McCaughan ><gareth.mccaughan(a)pobox.com> wrote: >> OED doesn't have much to say about "geometric[al] mean", regrettably. >> The entry for "mean" has a citation from about 1450 concerning the >> "meene proporcionalle", which is clearly the same thing as the GM >> (the OED entry for "mean proportional" quotes the same thing and >> seems to think it might mean something other than the GM, but I've >> no idea what else they think it could be). There's nothing for >> "arithmetic[al] mean" or "geometric[al] mean" until much later. >> >> So it's inconclusive, but seems to me like weak evidence against the >> theory that "geometric mean" came *before* "geometric progression". > >With regards to the geometric mean, I thought you all might be >interested in a construction I just made of four means. See the PDF >file here: > ><http://www.symbo1ics.com/files/means.pdf> > > Key >==================================================== >Light Blue : Arbitrary Value : x >Dark Blue : Arbitrary Value : y >Dark Green : Arithmetic Mean : (x+y)/2 >Light Green : Harmonic Mean : 2/(1/x + 1/y) >Red : Geometric Mean : sqrt(x*y) >Orange : Root Mean Square : sqrt[(x^2 + y^2)/2] >==================================================== > >The constructions should be easy to infer from the diagram. :) > >This does make for some nice identities, somewhat interesting methods >of computation, and easy derivation of inequalities. > >-Robert

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[math-fun] What is "geometric" about a geometric progression?
by Andy Latto 03 Mar '11

03 Mar '11

Why is a geometric progression called geometric? Wikipedia's answer is "A geometric progression gains its geometric character from the fact that the areas of two geometrically similar plane figures are in "duplicate" ratio to their corresponding sides; Given two squares whose sides have the ratio 2:3, then their areas will have the ratio 4:9; we can write this as 4:6:9 and notice that the ratios 4:6 and 6:9 both equal 2:3; so by using the side ratio 2:3 "in duplicate" we obtain the ratio 4:9 of the areas, and the sequence 4, 6, 9 is a geometric sequence with common ratio 3/2. Furthermore, the volumes of two similar solid figures are in "triplicate" ratio of their corresponding sides. Similar to with the squares, you can take two cubes whose side ratio is 2:5. Their volume ratio is 8:125, which can be obtained as 8:20:50:125, the original ratio 2:5 "in triplicate", yielding a geometric sequence with common ratio 5/2." but there's no citation given, and I'm not sure I believe that. The terminology is older than higher-dimensional geometry, I think, so there would be no reason to think of a progression of more than 4 terms as being "geometric". Any evidence for wikipedia's theory? Any alternate theories? Andy

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Re: [math-fun] Boy do we need a good radical simplifier
by Bill Gosper 03 Mar '11

03 Mar '11

Joerg> * Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> [Feb 11. 2011 09:45]: > [...] > Out[250]= > (1 - Sqrt[5] + 3 Sqrt[-2 + Sqrt[5]] + (-2 + Sqrt[5])^(3/2))/(2 Sqrt[2]) > > which canNOT be done by working outward from the innermost. Your (1 - sqrt(5) + 3*sqrt(-2+sqrt(5)) + (-2+sqrt(5))^(3/2))/(2*sqrt(2)) = 1/2*(sqrt(sqrt(5)-1) - sqrt(3-sqrt(5))) = 0.1188769458... (sqrt of 5th singular value) Yeth, LambdaStar[x]:=Sqrt[ModularLambda[Sqrt[-x]]] is the xth singular value. For the three fifthsth singular value, MathWorld misprints a doubly nested radical containing a 23 that should be 27, which upon correction denests completely to LambdaStar[3/5] = (1 - Sqrt[3] + 3 Sqrt[5] + Sqrt[15])/(8 Sqrt[2]). Bob Baillie>in mathematica 8: c = (1 + Sqrt[3] + Sqrt[2] 3^(3/4))^(1/3)/(Sqrt[3] - 1)^(1/6); FullSimplify[c] = (2*(7 + 3*Sqrt[3] + Sqrt[72 + 42*Sqrt[3]]))^(1/6) which has even fewer radicals than the one given just below. of course, one can always ask, what does "simplest" mean? bob --- For most purposes, nesting depth trumps radical count. Regrettably, FullSimplify just minimizes LeafCount. I haven't tried it in Mma 8, but in 7, tedious penalization of nesting via custom definitions of ComplexityFunction led me to conclude that Mma does not know how to denest. Perhaps there is something infeasible about Landau's 18yr old algorithm. --rwg

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Re: [math-fun] Surprisingly simple sin pi/2n
by Bill Gosper 03 Mar '11

03 Mar '11

Axel>I guess you will trash the following (Maple) lame way: > sin(Pi/42); convert(%, exp); expand(%); -1/2*I*(-1)^(1/42)-1/2*I*(-1)^(41/42) Yes, that's what happens when you heed Mma's message: "Developer`TrigToRadicals::obs: Developer`TrigToRadicals has been superseded by ToRadicals, and is now obsolete. It will not be included in future versions of Mathematica. " I can't find this interesting (and endangered?) functionality elsewhere in Mma. The Mapleoid answer seems an elegant demonstration that sin pi/n is always expressible in radicals. How can we articulate our dissatisfaction with it? Simply that the order of the roots is needlessly high, and we'll actually trade greater nesting depth to lower it? Apparently Developer`TrigToRadicals somehow exploits a theorem(?) that the Galois group of the minimal polynomial of trig(pi/n) is always cyclic. --rwg

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Re: [math-fun] Surprisingly simple sin pi/2n
by Axel Vogt 03 Mar '11

03 Mar '11

I guess you will trash the following (Maple) lame way: sin(Pi/42); convert(%, exp); expand(%); -1/2*I*(-1)^(1/42)-1/2*I*(-1)^(41/42)

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[math-fun] Weird arcsin pi formulas
by Bill Gosper 03 Mar '11

03 Mar '11

ArcSin[(-1)^(1/4)/2^(3/4)] == Pi/8 + (1/2)*I*Log[1 + Sqrt[2]] (Compute just the real part?) Pi/2 == ArcSin[13/85] + ArcSin[5/13] + ArcSin[8/17] + ArcSin[33/65] (Obviously inferior to ArcSin[1/2].) Pi/2 == ArcSin[73/2665] + ArcSin[21/221] + ArcSin[13/85] + ArcSin[9/41] + ArcSin[7/25] + 2*ArcSin[5/13] Pi/2 == 4*ArcSin[191/18241] + 20*ArcSin[187/17485] - 4*ArcSin[157/12325] + 9*ArcSin[99/4901] + 8*ArcSin[93/4325] + 20*ArcSin[164/6725] + 8*ArcSin[160/6401] - ArcSin[140/4901] + 12*ArcSin[136/4625] --rwg asin(x)*asin(y) = sqrt(%pi)* inf ==== 2 2 2(n + 1) 2(n + 1) 2(n + 1) \ n! ((x sqrt(1-y ) + sqrt(1-x ) y) - y - x ) > ------------------------------------------------------------------ / 1 ==== 4 (n + 1) (n + -)! n = 0 2

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[math-fun] Fwd: Question from journalist re maze solving
by Paul Reiners 03 Mar '11

03 Mar '11

Anyone interested in talking to a journalist, Jon Cartwright, on a recent development in maze solving? See his email below. I'm not really qualified to comment on this. I'm including Jon on the Cc line. Paul ---------- Forwarded message ---------- From: Jon Cartwright <jon.a.cartwright(a)googlemail.com> Date: Wed, Mar 2, 2011 at 11:22 AM Subject: Question from journalist re maze solving To: paul.reiners(a)gmail.com Dear Paul, I am a UK-based freelance journalist writing on behalf of ScienceNOW, the leading science new source based in Washington, DC. I was wondering if you can help me. I am considering writing a news story about a preprint recently submitted to Cornell University's arXiv ( http://arxiv.org/pdf/1103.0021v1) by Massimiliano Di Ventra of the University of California in San Diego and Yuriy Pershin of the University of South Carolina in Columbia. Di Ventra and Pershin have come up with a new way to solve mazes with an electronic network of "memristors" — devices like resistors but with memory, which were first created in 2008 (see, e.g., http://physicsworld.com/cws/article/news/34013) The researchers say the method could work a lot faster than previous methods to solve mazes, and could have applications in various fields from traffic optimisation to psychology. It seems fascinating, so I'm looking for the perspective of other scientists who have working in the field. Since I believe you have done some work in the past on mazes, I was hoping you might be willing to offer your opinion. If you can spare a few minutes, would you be able to tell me if you think it is an interesting development, and if so why? I appreciate you may be busy, so for whatever input you can offer I would be truly grateful. My (preliminary) deadline is Friday morning, UK time. Let me know what you think. Best wishes, Jon ---------------------------------------- Jon Cartwright UK freelance journalist office +44 (0) 117 373 9345 mobile +44 (0) 7866 721 555 jon.a.cartwright(a)gmail.com http://jcartwright.co.uk/ ----------------------------------------

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[math-fun] log(eta)ish surprise
by Bill Gosper 03 Mar '11

03 Mar '11

In the Mittag-Leffler section, W&W derive an identity equivalent to (1/(x^3*Pi)) - (Pi/(x*(Cosh[x*Pi] - Cos[x*Pi]))) == Sum[(((-1)^k*Csch[k*Pi])/(x^2 + 2*k*x + 2*k^2)), {k, -Infinity, -1}] + Sum[(((-1)^k*Csch[k*Pi])/(x^2 + 2*k*x + 2*k^2)), {k, 1, Infinity] where the two sums just mean sum k/=0. Note the similarity of Csch[k Pi] to Simon's recent summands, especially if you use y/(y^2 - 1) = y/(y - 1) - y^2/(y^2 - 1) . So this sum involves e^(-k pi)/(1-e^(-k pi)) and same(2 k), but with a complex degree of freedom! Analyzing the derivation might enable generalization to e^(-k pi/2)/(1-e^(-k pi/2)), etc. --rwg

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[math-fun] Surprisingly simple sin pi/2n
by Bill Gosper 03 Mar '11

03 Mar '11

Well hidden and threatening to disappear, Developer`TrigToRadicals seems able to do all trig pi/n. n = 34 gives a ridiculous mess that Corey's denester cracks, but into unmanageably large pieces. Some Macsymas know that Sin[Pi/34] ==((-Sqrt[Sqrt[17] + 3]*Sqrt[4*Sqrt[17] - 2*Sqrt[34 - 2*Sqrt[17]]] + Sqrt[34 - 2*Sqrt[17]] + Sqrt[17] - 1)/16) But I'm amazed at #==Apart[Simplify[Developer`TrigToRadicals[#]]]&@Sin[Pi/13/2] Sin[Pi/26] == (1/12)*(-1 + Sqrt[13]) + (1/12)*I*(I + Sqrt[3])* ((1/2)*(26 - 5*Sqrt[13] - 3*I*Sqrt[39]))^(1/3) + ((1 + I*Sqrt[3])*(-13 + Sqrt[13]))/ (12*2^(2/3)*(26 - 5*Sqrt[13] - 3*I*Sqrt[39])^(1/3)) . Also, the denester and I managed to bash 42 down to Sin[Pi/42] == ((2 + 6*I*Sqrt[3])^(1/3)* (3*I - 5*Sqrt[3] + Sqrt[7] - I*Sqrt[21]))/(24*2^(2/3)*7^(1/6)) - (1/6)*I*((1/2)*(-14*I + Sqrt[7] + 3*I*Sqrt[21]))^ (1/3) + (1/6)*I*((1/2)*(14*I + Sqrt[7] + 3*I*Sqrt[21]))^(1/3) + (1/12)*(1 + Sqrt[21]) which is better than I got with sin(pi/6 - pi/7). --rwg

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Re: [math-fun] eta triples again
by Bill Gosper 03 Mar '11

03 Mar '11

Joerg>A collection of (currently) 5416 such identities by Michael Somos is available at http://eta.math.georgetown.edu/ Enjoy! * Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> [Oct 01. 2010 09:35]: > [...] These identities seem to fail for complex values. E.g., for the simplest, good old aequatio satis abstrusa, In[318]:= Equal[0, 16*(DedekindEta[-((Log[q]*I)/(2*Pi))])^8*(DedekindEta[-((Log[q^4]* I)/(2*Pi))])^16 + (DedekindEta[-((Log[q]*I)/(2* Pi))])^16*(DedekindEta[-((Log[q^4]*I)/(2* Pi))])^8 - (DedekindEta[-((Log[q^2]*I)/(2*Pi))])^24] In[319]:= %318 /. q -> Exp[-\[Pi]*(I + Sqrt[5])] Out[319]= False (*yet*) In[320]:= %318 /. q -> Exp[-\[Pi]*(+Sqrt[5])] Out[320]= 0 == -DedekindEta[I Sqrt[5]]^24 + DedekindEta[(I Sqrt[5])/2]^16 DedekindEta[2 I Sqrt[5]]^8 + 16 DedekindEta[(I Sqrt[5])/2]^8 DedekindEta[2 I Sqrt[5]]^16 Can they be restated to hold more broadly? --rwg BtW, DedekindEta[1/2 + (I Sqrt[5])/2] == ((-1)^(1/24) Sqrt[ Gamma[1/20] Gamma[9/20]])/(2 5^(5/16) GoldenRatio^(1/8) \[Pi]^(3/4))

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