February 2007 - mathfun@mailman.xmission.com

Re: [math-fun] sections of quadratic surfaces
by Daniel Asimov 14 Feb '07

14 Feb '07

Bill G. writes: << Some find it counterintuitive that slicing a cone gives an ellipse and not an oval. What about slicing a paraboloid of rotation? More generally how can eliminating a quadratic mumbloid vs a linear (planar) equation yield anything but a quadratic (conic)? >> What happens when a plane passes through both sheets of a hyperboloid (of two sheets) in R^3, but contains the vertex of only one sheet? (Naively it's hard to believe the intersection has the symmetry of a hyperbola.) ------------------------------------------------------------------------------------------- But to escalate the question a bit, what happens when two quadratic surfaces in R^3 intersect? It appears that all kinds of bizarre space curves are possible (that I never learned about in school). --Dan

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Re: [math-fun] Matrix Calculus
by Henry Baker 14 Feb '07

14 Feb '07

One of the coolest algorithms of all time is the use of Newton iterations to compute the "polar decomposition" of a matrix: http://www.cs.wisc.edu/graphics/Courses/838-s2002/Papers/polar-decomp.pdf I believe that this algorithm is used in graphics animation systems. Re quaternions: matrix algorithms can always be used, by simply using the matrix version of a quaternion (either real 4x4 or complex 2x2). I am not aware of any Newton iterations for quaternions that aren't inherited from real/complex numbers (and therefore assume commutativity), or are inherited from more general matrices. *** Challenge *** Are there any such Newton algorithms that work _only_ for quaternions? At 09:09 AM 2/13/2007, Fred lunnon wrote: >Recent discussion of Newton-Raphson iteration led me to >reconsider the of the iterative algorithm for reciprocal of a >scalar c --- > x <- 2 x - a x^2 >--- derived via N-R applied to the scalar function of a scalar > f(x) = c - 1/x. > >This has a matrix analogue (mentioned here a while back, >in a thread discussing the shortcomings of Gauss-Seidel >iteration) for the inverse of a matrix A --- > X <- 2 X - X A X --- >which, by a rather nice twist, can actually be employed in >inverting the Jacobian matrix required to apply Newton- >Raphson to a vector function of a vector variable! > >But what matrix calculus framework might permit the >matrix iteration to be derived in a fashion analogous >to the scalar version? In particular, is it possible to restrict >the universe of matrix functions --- as is done for functions >of a complex variable, and attempted less successfully for >a quaternion variable --- in such a way as to facilitate this? > >There is indeed a recognised topic which appears relevant >--- see on the web e.g. Wikipedia; or appendix D of > John Dattoro "Convex Optimization & Euclidean Distance >Geometry" --- >but closer iinspection reveals this to be merely a reworking >of tensor notation omitting the subscripts: in particular, >the derivative of a matrix function of a matrix variable is >a rank 4 object, in effect a Jacobian. > >Fred Lunnon

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[math-fun] Quintic equation
by Christian Boyer 14 Feb '07

14 Feb '07

With the 5th-degree equation: x(x² + a)(x² + b) = c x, a, b, c real numbers > 0 c > ab what is the fastest method to get an approximated value of x? (very few steps) Christian.

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[math-fun] Matrix Calculus
by Fred lunnon 13 Feb '07

13 Feb '07

Recent discussion of Newton-Raphson iteration led me to reconsider the of the iterative algorithm for reciprocal of a scalar c --- x <- 2 x - a x^2 --- derived via N-R applied to the scalar function of a scalar f(x) = c - 1/x. This has a matrix analogue (mentioned here a while back, in a thread discussing the shortcomings of Gauss-Seidel iteration) for the inverse of a matrix A --- X <- 2 X - X A X --- which, by a rather nice twist, can actually be employed in inverting the Jacobian matrix required to apply Newton- Raphson to a vector function of a vector variable! But what matrix calculus framework might permit the matrix iteration to be derived in a fashion analogous to the scalar version? In particular, is it possible to restrict the universe of matrix functions --- as is done for functions of a complex variable, and attempted less successfully for a quaternion variable --- in such a way as to facilitate this? There is indeed a recognised topic which appears relevant --- see on the web e.g. Wikipedia; or appendix D of John Dattoro "Convex Optimization & Euclidean Distance Geometry" --- but closer iinspection reveals this to be merely a reworking of tensor notation omitting the subscripts: in particular, the derivative of a matrix function of a matrix variable is a rank 4 object, in effect a Jacobian. Fred Lunnon

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Re: [math-fun] proportion of bernoulli numbers that are = 1/6.
by N. J. A. Sloane 13 Feb '07

13 Feb '07

Simon said: Subject : The proportion of Bernoulli numbers that are = to 1/6. I made a calculation for the numbers that are = 1/6, I went up to B(5285000). Me: did you mean those that have DENOMINATOR 6 ? If so, see A051222 in the OEIS, which has references to the work of H. Cohen and G. Tenenbaum. Neil

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[math-fun] Comment re Allan Weshler's 1996 posting to the math-fun mailing list.
by RAMSEYKk2＠aol.com 13 Feb '07

13 Feb '07

Sloane's ONline encyclopedia of integer sequences" lists Allan Weshler's J-Determinant sequence as sequence A022344. Until I rediscovered this sequence and added my own comments no information on this sequence was given except a reference to "Posting to math-fun mailing list dated December 4, 1996". Was this your mailing list? If so members may be interested in my following conjectures: Background: Every integer based Fibonacci sequence ultimately appears as a row, n, of Wtthoffs array (Sloane's sequence no A035513) and can be characterized by their J-Determinant as given by Allan Weshler's J-Determinant sequence (A022344 where a(n) = (the absolute value of F(n,i-1)*F(n,i+1) - F(n,i)^2, which is invariant with i varying. Also, in 1961 Horadam, Amer. Math. Monthly 68(1961) pp 751-753) showed that for any Fibonacci sequence, F, that the triple (2F(n+1)*F(n+2), F(n)*F(n+3),2F(n+1)*F(n+2) + F(n)^2) is a Pythagorean number triple (a,b,c) for all n, i.e. a^2 + b^2 = c^2. Conjecture 1: Let any two integer based Fibonacci sequences, F1 and F2, be sequences appearing in rows "n" and "m" respectively and thus have characteristic values, a(n) and a(m) respectively. Then there is a third Fibonacci sequence that is defined by F3(i) = F1(i)*F2(j+1) - F1(i+1)*F2(j) where j is a constant. This third Fibonacci sequence appears in a row, p, of Wythoffs array and has the characteristic value a(p) = a(n)*a(m). Conjecture 2: For every Fibonacci sequence having coprime adjacent terms and appearing in a row (n) of Wythoff's array, the "c" values of Horadam's Pythagorean triples, 2F(n+1)*F(n+2) -F(n)^2), where the c values are reduced by dividing through by the greatest common divisor, form the bijection of a Fibonacci sequence having the characteristic value of (a(n))^2. Anyone interested in this may refer to my further comments to sequences A035513, A022344 in Sloane's online encyclopedia and my own sequence (Sloane's A127561) which is a tabulation of terms, a(n) from A022344, as a table where the row and column positions dictate the starting values of a Fibonacci sequence having the characteristic value a(n), and which has very interesting prime modulus properties.

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[math-fun] proportion of bernoulli numbers that are = 1/6.
by Simon Plouffe 12 Feb '07

12 Feb '07

Hello everybody, Subject : The proportion of Bernoulli numbers that are = to 1/6. I made a calculation for the numbers that are = 1/6, I went up to B(5285000). - The proportion is NOT 1/6, after B(5285000) the ratio is .152616880782547821315725 and still slowly decreasing. It is hard to tell where it goes, 0.15 ? 1/7 ? I don't know, your guess is certainly as good as mine. any idea ? I believe somebody did such a calculation in the past, Wagstaff ? SimonPlouffe

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[math-fun] more pi methods
by R. William Gosper 10 Feb '07

10 Feb '07

(Thought funsters might enjoy eavesdropping.) Jason, I'm still occupied by your pi question. I believe we've already rejected montecarloing or integrating Buffon's needle. How about integrating the area under sqrt(1-x^2)? Besides the numerous numerical methods, one can expand via the binomial thm and termwise integrate. Expanding at x=0 gives the slow series i 1/2 inf (- 1) ( ) ==== i %pi \ --- = > ------------, 4 / 2 i + 1 ==== i = 0 but expanding at x=1 gives one bit/term: i 1/2 inf (- 1) ( ) ==== i %pi \ --- = > ------------------. 4 / i - 3/2 ==== (2 i + 3) 2 i = 0 Better yet, only integrate from sqrt(3)/2 to 1: inf ==== i + 3/2 \ (2 - sqrt(3)) 1/2 3 sqrt(3) %pi = 12 > -------------------- ( ) + --------- / i i 2 === (2 i + 3) (- 4) i = 0 gives > 1 decimal digit/term (ratio = .06699). Expansion at points other than 0 and 1 produces a powerseries whose coefficients are themselves sums. But these obey a three term recurrence, so the double sum reduces to a 4x4 matrix product. Strangely, integrate(expansion at 1/sqrt(2),x,0,1) produces a divergent series. How about inf ==== 2 \ 1 %pi > -- = ---- ? / 2 6 ==== n n = 1 (I have an Euleroid derivation of this in the form of an annotated Macsyma notebook, starting with the atan x series.) Here's a neat one: the average value of gcd([k sqrt(2)],[k sqrt(3)])^-2 should approach pi^2/15, for large intervals of k. ([]:=floor) Too bad you don't like logs. I have *really* magic ways to get pi/ln 2 and (pi/ln 2)^2, the latter using a random bit stream. Heck, pi is basically log(-1). Ah, how about (d5) t(n) := if n = 0 then .2d1 else (t(n - 1), 2 %% ---------------------------------) %% (--, sqrt(%% + 1) + sqrt(1 - %%)) n 2 (c6) t(24) (d6) 3.14159265358979d0 or (d7) s(n) := if n = 0 then .4d1 else (s(n - 1), 2 %% --------------------) 2 %% 1 + sqrt(1 + ------) n + 1 4 (c8) s(25) (d8) 3.14159265358979d0 ? --rwg POINT SOURCE PROSECUTION

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[math-fun] Exact formula for this series?
by Mike Stay 10 Feb '07

10 Feb '07

0.38490017945975 0.29101243968677 0.21666469449485 0.15928582101343 0.11593604442994 0.08373599185262 0.06012913678590 0.04299237598573 This is the magnitude of the nonzero entries in log(G_n)/(i pi), where G_n is the unitary matrix implementing Grover's algorithm for an n-qubit quantum computer. More specifically, T_1 = |1 1| / sqrt(2) |1 -1| T_n = T_1 tensor T_{n-1} S_{n,k} = the 2^n x 2^n identity matrix except entry (k,k)=-1 Grover's algorithm is then G_n = T_n S_{n,1} T_n S_{n,m} where m is the marked state. They decrease roughly like sqrt(2) at each step, but not exactly. Does anyone know how I could figure out a closed form for the terms, or a reference if someone else already has? -- Mike Stay metaweta(a)gmail.com http://math.ucr.edu/~mike

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[math-fun] New proof of Euclid's Theorem
by Paul R. Pudaite 10 Feb '07

10 Feb '07

In the December 2006 issue of Am. Math. Monthly , Filip Saidak presents "A New Proof of Euclid's Theorem." Here's a concise rendition: Let f(x) = x^2 + x, let g[0] be a positive integer, and for n = 1, 2, 3, ..., let g[n] = f(g[n-1]). Then g[n] has at least n different prime factors. I omit the proof for those who might enjoy working it out for themselves. The case of g[0] = 1 seems somewhat canonical. I'm curious about which primes appear as factors in the sequence of g[n] (2, 6, 42, 1807, ...). This sequence begins: 2, 3, 7, 13, 43, 73, 139, 181, 547, 607, 1033, 1171, ... Paul

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