March 2011 - mathfun@mailman.xmission.com

[math-fun] Benford
by Alex Bellos 31 Mar '11

31 Mar '11

Hi I've been playing around with Benford's Law and noticed the following curiosity. For a finite sequence of the positive integers, the probability of a number beginning with 9 is lowest when we cut off the sequence just before the reappearance of 9. So, up to 89, the chance of a number starting with a 9 is 1/89. Up to 899, the chance of a number starting with a 9 is 11/899. Up to 8999, the chance of a number starting with a 9 is 111/8999 If we carry on, this appears to be converging to 1/81, which is the lovely repeating decimal 0.0123456790123456790... My maths isnt good enough to prove this rigorously, but it seems to be true, and is delightful to me anyway! It's not in any of the Benford's literature that i have. Does anyone else have any fun insights into Benford's Law? Alex

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[math-fun] surd conjecture dead?
by Bill Gosper 30 Mar '11

30 Mar '11

Oops, I can't factor the unnested quadrinomial (3^(1/4) + sqrt(2)) * sqrt(3) + 3^(1/4) + 2 (from eta(e^( - 24 pi))) into binomial surds to rational powers. What a strange space. After so many successes, I'd've expected the "smallest" counterexample to have larger integers, or more of them. This probably means that, in the large, "hardly any" (measurespeak) polynomial surds factor. Either that or I overlooked some factors. --rwg

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[math-fun] Fw: FBI Needs Cryptanalysis Help
by Michael Beeler 30 Mar '11

30 Mar '11

I just read about this - the FBI is looking for help solving a murder case where the victim was found with two coded notes on him. They've been unable to crack the code, and now they're asking the public for help. Obviously if the FBI's experts can't figure it out my odds are pretty slim, but I thought it was an interesting situation nonetheless. Here's an article about the case with images of the two encrypted notes: http://www.fbi.gov/news/stories/2011/march/cryptanalysis_032911

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[math-fun] Succinct aequatio difficulty
by Bill Gosper 29 Mar '11

29 Mar '11

Solve the aequatio for eta(q)^8 (c86) (fancy_display:false, Solve( 16*ETA(Q)^8*ETA(Q^4)^16+ETA(Q)^16*ETA(Q^4)^8=ETA(Q^2)^24,eta(q)^8)) 24 4 24 2 12 4 8 sqrt(64 eta (q ) + eta (q )) + 8 eta (q ) (d86) [eta (q) = - --------------------------------------------, 4 4 eta (q ) 24 4 24 2 12 4 8 sqrt(64 eta (q ) + eta (q )) - 8 eta (q ) eta (q) = --------------------------------------------] 4 4 eta (q ) (Don't you wish Mathematica would let you.) By the infinite product formula, eta(positive)>0. For real q, the first rhs is strictly negative (hence the wrong one for positive q) and both roots are real. Also by the product formula, eta(negative) = negative^(1/24)*positive product. Thus the phase of eta(negative)^8 = 8 pi/24, not 0. The aequatio generally fails off the real unit interval. The magnitude is usually wrong, too. --rwg

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[math-fun] Decently geometric convergence,
by Bill Gosper 28 Mar '11

28 Mar '11

Out[270]=Product[1 + E^(-((2*n*Pi)/5)) - (2*Cos[(2*n*Pi)/5])/E^((n*Pi)/5), {n,Infinity}] -> (Sqrt[5]*E^(Pi/60)*Gamma[1/4]^2)/(2*2^(3/4)*Pi^(3/2)) Amazing inaccurate numerics: In[272]:= Timing[N[%270, 105]] During evaluation of In[272]:= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in n near {n} = {185.8931420129281672288691448025888296859652775220603165483862095513741610526450692769280980058083317102024170890843}. NIntegrate obtained 0.00004366400531989029430161228066757194803765686030072049096264985465928621657418744962590402711855355493876269474449273`115. and 7.505823825434631924585246964567939510489203332210688578001209387505086611802644511529216233982994651935733171228298`115.*^-85 for the integral and error estimates. >> Out[272]= {29.6032, 1.65371039642291406870084030182905258859277665525891059324072211312882326107919099574691792549413518551613 -> 1.65371039642291406696569291506073467057126586745207656418583564992967124414571613904819032470934233477443} It's faster and hugely more accurate just to add the terms up: In[281]:= Timing[%270[[1]] /. Product[x_, L_] :> NProduct[x, {n, \[Infinity]}, WorkingPrecision -> 106, PrecisionGoal -> 105, NProductFactors -> 691]] During evaluation of In[281]:= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in n near {n} = {861.8940068943142691044101719338395586499378157406267071093412114640734679433460889438059665960107225000717}. NIntegrate obtained 1.87009213033821895841004092891566274691306955453889921013030172021676370553701588750870721398048732949125`106.*^-189 and 2.602515810387861805818516463584710573116703810473833425997695963619028961690372962373958113185701592545913`106.*^-264 for the integral and error estimates. >> Out[281]= {26.7773, \ 1.653710396422914066965692915060734670571265867452076564185835649929671244145716139048190324709342334774} Upshot: N[%270/.Rule->Equal],n] gives False for nearly all n, discrediting the identity. --rwg

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[math-fun] linear algebra theorem
by Jean Gallier 28 Mar '11

28 Mar '11

The answer is yes! If H and K are two hermitian matrices that are PSD, then HK can be diagonalized and has positive eigenvalues. This follows from the following proposition by Denis Serre (Matrices Theory and Applications, GTM 216, Proposition 6.1): Prop: If H and K are two Hermitian matrices with H positive symmetric definite, then HK is diagonalizable with real eigenvalues. Furthermore, the number of positive eigenvalues of HK is equal to that of K. Proof. Since H is hermitian, it can be diagonalized w.r.t a unitary matrix U and its eigenvalues are real: H = U^* diag(mu_1, ..., mu_n) U. Because H is PSD, mu_i > 0 so if we let S = U^* diag(sqrt(mu_1), ..., sqrt(mu_n)) U, we have S^2 = H and S is also Hermitian PSD. Observe that HK = SS K SS^{-1} = S(SK S) S^{-1} = S(S^* K S) S^{-1}. Therefore, HK and S^*KS have the same eigenvalues and since S^* K S is hermitian, it can be diagonalized and it has real eigenvalues, and so the same holds for HK. The number of positive eigenvalues of S^*KS is the largest dimension of a subspace of complex^n on which the hermitian form x |-> x^* S^* K S x is positive definite, which means that S\complex^n is the subspace on which the hermitian form y |-> y^*Ky is positive definite. Since S is nonsingular, this max dimension is the same. Best, -- Jean

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[math-fun] A linear algebra theorem?
by Eugene Salamin 28 Mar '11

28 Mar '11

Is this a theorem? Let A and B be real symmetric matrices all of whose eigenvalues are positive. Then all eigenvalues of AB are positive. -- Gene

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[math-fun] EllipticK[1/3]?
by Bill Gosper 28 Mar '11

28 Mar '11

(Question from Julian.) Apparently, K(k) has a closed form at and only at "singular values", i.e. when K(1-k)/K(k) = sqrt(rational), (periodic CF). There's no period in the first 10^4 terms of K(2/3)/K(1/3), but, e.g., EllipticK[4 Sqrt[2] (-1 + Sqrt[2])^2] -> ((2 + Sqrt[2]) Pi^(3/2))/(4 Gamma[3/4]^2) EllipticK[(-1 + Sqrt[2])^4] == ((2 + Sqrt[2]) Pi^(3/2))/(8 Gamma[3/4]^2) EllipticK[(-Sqrt[2] + Sqrt[3])^4/(-1 + Sqrt[2])^4] == (3^(3/4) (-1 + Sqrt[2]) (1 + Sqrt[3]) Gamma[1/3]^3)/( 16 2^(5/6) (-Sqrt[2] + Sqrt[3]) Pi) EllipticK[( 16 Sqrt[2] (-Sqrt[2] + Sqrt[3])^2)/((-1 + Sqrt[2])^2 (1 + Sqrt[ 3])^3)] == (3^(1/4) (-1 + Sqrt[2]) (1 + Sqrt[3]) Gamma[1/3]^3)/( 8 2^(5/6) (-Sqrt[2] + Sqrt[3]) Pi) EllipticK[(-1 + 2^(1/4))^4/(1 + 2^(1/4))^4] == ((-1 + Sqrt[ 2])^2 Pi^(3/2))/(8 Sqrt[2] (-1 + 2^(1/4))^2 Gamma[3/4]^2) EllipticK[(8 2^(1/4) (1 + Sqrt[2]))/(1 + 2^(1/4))^4] == ((1 + 2^(1/4))^2 Pi^(3/2))/(2 Sqrt[2] Gamma[3/4]^2) I don't recall seeing the last four (which are equivalent in pairs). I have to check if K(1-k)/K(k) = sqrt(rational1) + i sqrt(rational2) can also work. Also, that unlikely conjecture that denestable <-> factorable into binomials has survived a minor complexes test: DedekindEta[-((I*((I*Pi)/2 - Pi/4))/(2*Pi))] == (Gamma[1/4]*Sqrt[Sqrt[2] + 1]* I^(13/24))/(2^(25/32)*(Sqrt[2] - 1)^(1/8)*Sqrt[2^(1/4)*I - 1]*Pi^(3/4)) (A virtue of binomial factorization is that it tends to minimize the most "deeply rooted" subexpressions, and we know how unwelcome radicalism is around here.) --rwg

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Re: [math-fun] perpendicular planes
by Dan Asimov 27 Mar '11

27 Mar '11

<< Is the D. Asimov in the Sloane et al paper our own D.A.? just curious Yes, he is I. --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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Re: [math-fun] perpendicular planes
by Dan Asimov 27 Mar '11

27 Mar '11

Gene wrote: << From: Dan Asimov <dasimov(a)earthlink.net> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Sun, March 27, 2011 12:20:32 AM Subject: Re: [math-fun] perpendicular planes If two lines are given parametrically by {u0 + su} and {v0 + tv} for (s,t) in R^2, vectors u0, v0, and unit vectors u, v, then the distance between the lines is just (setting w0 = u0-v0) |w0 * u x v|, or am I missing something? --Dan ________________________________________________________________________________________ Let's divide that expression by |u x v|. >> Yes, let's. --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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