Re: [math-fun] Equation mixing integers and reals
Christian writes: << Let f(x , x , ..., x ) = k x + k x + ... + k x 1 2 n 1 1 2 2 n n where k , k , ..., k are n REAL constants 1 2 n and x , x , ..., x are n INTEGERS (positive or negative) 1 2 n If we try to have the best possible approximations of f(x , x , ... x ) = 0 1 2 n do you know good algorithms giving best values of (x , x , ..., x )?
Assume x_j = 0 for all j is not true. Let K and X be the n-dim vectors of real constants and unknown integers, respectively. Then we want to find X so that <K,X> = 0 as near as possible. The set of all values G := {<K,X> : X in Z^n} is a subgroup of the reals, so it's either discrete or dense. The discrete case is easy to handle, and only occurs for measure 0 among possible vectors K in R^n. Otherwise the set G* := {<K,X> : X in Z^n - {0}} contains elements arbitrarily close to 0, so there is no "best" value of X. I'll leave finding "good" values of X to the number theorists, once "good" is defined. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
On Tue, Sep 16, 2008 at 12:13 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Christian writes:
<< Let f(x , x , ..., x ) = k x + k x + ... + k x 1 2 n 1 1 2 2 n n
where k , k , ..., k are n REAL constants 1 2 n
and x , x , ..., x are n INTEGERS (positive or negative) 1 2 n
If we try to have the best possible approximations of
f(x , x , ... x ) = 0 1 2 n
do you know good algorithms giving best values of (x , x , ..., x )?
Assume x_j = 0 for all j is not true.
Let K and X be the n-dim vectors of real constants and unknown integers, respectively. Then we want to find X so that <K,X> = 0 as near as possible.
The set of all values G := {<K,X> : X in Z^n}
is a subgroup of the reals, so it's either discrete or dense. The discrete case is easy to handle, and only occurs for measure 0 among possible vectors K in R^n. Otherwise the set
G* := {<K,X> : X in Z^n - {0}}
contains elements arbitrarily close to 0, so there is no "best" value of X.
That's true, but one can ask for what the best value of X is given that all elements of X are bounded in absolute value by H. There is an algorithm due to Hastad, Just, Lagarias and Schnorr that is designed to answer the posed question. When looked at the right way it's very close to the L^3 lattice reduction algorithm. Victor Here's the reference: # Polynomial time algorithms for finding integer relations among real numbers , J. Hastad, B. Just, J. C. Lagarias and C. P. Schnorr, SIAM J. Computing 18 (1989), pp. 859-881. [Preliminary version in: STACS '86 , Lecture Notes in Computer Science No. 210, Springer-Verlag: New York 1986, pp. 105-118.]
I'll leave finding "good" values of X to the number theorists, once "good" is defined.
--Dan
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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I too spent time in the 70's tinkering with a naive algorithm for this problem --- just around the time that LLL came along and blew my pedestrian notions clean out of the water! I didn't know about the 1989 paper --- has it proposals less artificial to apply than LLL, which was in an essential sense discrete (for lattice basis minimisation, in the course of factoring polynomials over a finite field)? Despite trying the idea many times, I have to confess I have never actually discovered a new relation in this manner. Has anybody else out there had better luck? WFL On 9/16/08, victor miller <victorsmiller@gmail.com> wrote:
On Tue, Sep 16, 2008 at 12:13 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Christian writes:
<< Let f(x , x , ..., x ) = k x + k x + ... + k x 1 2 n 1 1 2 2 n n
where k , k , ..., k are n REAL constants 1 2 n
and x , x , ..., x are n INTEGERS (positive or negative) 1 2 n
If we try to have the best possible approximations of
f(x , x , ... x ) = 0 1 2 n
do you know good algorithms giving best values of (x , x , ..., x )?
Assume x_j = 0 for all j is not true.
Let K and X be the n-dim vectors of real constants and unknown integers, respectively. Then we want to find X so that <K,X> = 0 as near as possible.
The set of all values G := {<K,X> : X in Z^n}
is a subgroup of the reals, so it's either discrete or dense. The discrete case is easy to handle, and only occurs for measure 0 among possible vectors K in R^n. Otherwise the set
G* := {<K,X> : X in Z^n - {0}}
contains elements arbitrarily close to 0, so there is no "best" value of X.
That's true, but one can ask for what the best value of X is given that all elements of X are bounded in absolute value by H. There is an algorithm due to Hastad, Just, Lagarias and Schnorr that is designed to answer the posed question. When looked at the right way it's very close to the L^3 lattice reduction algorithm.
Victor
Here's the reference:
# Polynomial time algorithms for finding integer relations among real numbers , J. Hastad, B. Just, J. C. Lagarias and C. P. Schnorr, SIAM J. Computing 18 (1989), pp. 859-881. [Preliminary version in: STACS '86 , Lecture Notes in Computer Science No. 210, Springer-Verlag: New York 1986, pp. 105-118.]
I'll leave finding "good" values of X to the number theorists, once "good" is defined.
--Dan
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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WFL> I too spent time in the 70's tinkering with a naive algorithm for this problem
--- just around the time that LLL came along and blew my pedestrian notions clean out of the water!
Here is an ancient Lisp algorithm, invoked from Macsyma, telephoned ca. 1970 by Rich to Robert Maas, finding quartics for pi. It works mod 1, so there's no pi^0 term. The left column is the matrix row being printed, the 2nd column is the error, the third is the vector, the 4th is its norm, and the fifth is a figure of merit approximating the number of excess digits of agreement. (c147) round(2^105*makelist(%pi^k,k,1,4)) (d147)[127438138015862315638027235646471, 400358718177795200392018112753682, 1257764007827987814231061425713179, 3951382166942061712360453895340836] (c148) load("c:\\rwg\\climax\\munlsp.lsp")$ (c149) ?lsp("(user::precinit 105)") (d149) 72.7804 (c150) ?lsp("(user::linrel (cdr $d147))") Init of numrel has finished okay [0, 0.40909, [0, 0, 0, 1], 1, -1.77018285868794d0] [0, 0.00628, [0, 0, 1, 0], 1, 0.04390718722343d0] [0, 0.00136, [-1, -1, 2, 0], 6, 0.40674400805398d0] [0, 4.30503e-4, [-1, 2, -1, 1], 7, 0.85547926831103d0] [0, 7.30684e-5, [3, 5, 0, 3], 43, 0.67148385803703d0] [0, 9.54798e-6, [-6, 0, 5, 2], 65, 1.25846167486465d0] [0, 1.27879e-6, [-7, -12, 10, -4], 309, 0.88290059041933d0] [0, 3.62128e-7, [-18, -16, -6, -11], 737, 0.69333326841885d0] [0, 1.82782e-8, [-22, 13, -23, -5], 1207, 1.56677297182571d0] [0, 1.92441e-9, [33, 12, 48, -1], 3538, 1.61548864677868d0] This is Jim's monic quartic. But we can easily make monics by rubbing together pairs from the hereinafter. [0, 5.50442e-10, [68, 51, -11, 76], 13122, 1.02256306683229d0] [0, 1.3231e-11, [58, -28, 29, -159], 30270, 1.9160654198039d0] [0, 8.25957e-13, [-102, 154, -18, -84], 41500, 2.84671402663485d0] [0, 2.38296e-13, [-530, 10, -647, -131], 716770, 0.91211003918297d0] [0, 2.69997e-14, [741, 497, -525, -838], 1773959, 1.07075195229049d0] [0, 8.85595e-16, [-568, -561, -2006, 645], 5077406, 1.64147564053187d0] ... (c160) (makelist(%pi^k,k,1,4),[ [58, -28, 29, -159].%%,%%. [68, 51, -11, 76]])$ (c161) bfloat(%,33) (d161) [ - 1.46829999999999867690017321923245b4, 7.77900000000055043781524460966016b3] (c162) cfexpand(cf(159/76)) 159 23 (d162) ( ) 76 11 (c163) expand(d160.[11,23]) (d163) - pi^4 + 66 * pi^3 + 865 * pi^2 + 2202 * pi (c164) dfloat(%) (d164) 17404.0000000128d0
Despite trying the idea many times, I have to confess I have never actually discovered a new relation in this manner. Has anybody else out there had better luck?
WFL
Sure, that's how I find those valuations of eta'(e^-(pi n/d)). And I think it's how Simon found his nth-bit-of-pi series. And although they're not new, it's the easiest way to simplify products of gammas of rationals. (Look for relations among the logs.) And here's a neat anecdote: Maas was doing adaptive data compression and needed to solve y-2 3-2y 1-y y 1-x x x-3 (1-x) x (1-y) y = 1 = (1-x) x (1-y) y . Curious whether the roots were algebraically independent, we fed them to Rich's recognizer and they turned out to be independently algebraic! Cubics, in fact, which we were then able to exhibit by rubbing the transcendentals together. I need also to nag that the powerseries analog of this process is a great way to uncover functional relationships. E.g., to find the ODE for Bessel_J[3], (c165) sqfr(tlinrel(makelist(('diff(bessel_j[3](z),z,k), %%=taylor(apply_nouns(%%),z,0,17)),k,0,2))) (d165) 2 2 d d 5971968000 (z (--- (bessel_j (z))) + z (-- (bessel_j (z))) 2 3 dz 3 dz 2 13 + (z - 9) bessel_j (z))/z = 0 3 where 17 is any sufficiently large degree, whose size does not affect the answer. (Not even that z^13.) With a little more trouble, one can find the ODE for Bessel_J[n].
On 9/16/08, victor miller <victorsmiller@gmail.com> wrote:
On Tue, Sep 16, 2008 at 12:13 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Christian writes:
<< Let f(x , x , ..., x ) = k x + k x + ... + k x 1 2 n 1 1 2 2 n n
where k , k , ..., k are n REAL constants 1 2 n
and x , x , ..., x are n INTEGERS (positive or negative) 1 2 n
If we try to have the best possible approximations of
f(x , x , ... x ) = 0 1 2 n
do you know good algorithms giving best values of (x , x , ..., x )?
Assume x_j = 0 for all j is not true.
Let K and X be the n-dim vectors of real constants and unknown integers, respectively. Then we want to find X so that <K,X> = 0 as near as possible.
The set of all values G := {<K,X> : X in Z^n}
is a subgroup of the reals, so it's either discrete or dense. The discrete case is easy to handle, and only occurs for measure 0 among possible vectors K in R^n. Otherwise the set
G* := {<K,X> : X in Z^n - {0}}
contains elements arbitrarily close to 0, so there is no "best" value of X.
That's true, but one can ask for what the best value of X is given that all elements of X are bounded in absolute value by H. There is an algorithm due to Hastad, Just, Lagarias and Schnorr that is designed to answer the posed question. When looked at the right way it's very close to the L^3 lattice reduction algorithm.
Victor
Here's the reference:
# Polynomial time algorithms for finding integer relations among real numbers , J. Hastad, B. Just, J. C. Lagarias and C. P. Schnorr, SIAM J. Computing 18 (1989), pp. 859-881. [Preliminary version in: STACS '86 , Lecture Notes in Computer Science No. 210, Springer-Verlag: New York 1986, pp. 105-118.]
I'll leave finding "good" values of X to the number theorists, once "good" is defined.
--Dan
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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Due to fading sensitivity of my "C" key, I inadvertently pasted a Macsyma notebook output section (instead of text) into a Mathematica notebook, and it worked! (I.e., it rendered a bitmap image.) So anyone with an interest in theta functions can now sneak a peek at http://gosper.org/thetpak.html. --rwg Minor correction to couple wk old mail: rwg>The formula I gave [ 2 i n pi -------- i g(n/d) pi d ----------- sqrt(- log(q)) eta(e q) 12 2 pi limit ------------------------------- = e sqrt(----), q -> 1 2 d pi ----------- 2 6 d log(q) e 1 1 g(r) := if integerp(r) then r else - g(---------) - ------------------- mod(r, 1) 2 denom (r) mod(r, 1) + floor(r) + 3 . ]
is good only for 2 |n| < d, n/=0, which is barely enough, with
2 %pi - -------- 6 log(q) eta(q) %e sqrt(- log(q)) -> sqrt(2 %pi),
to describe eta(e^(i t pi)) for rational t.
Actually, we need 2 |n| <= d, which fortunately obtains.
rwg> Actually, we need 2 |n| <= d, which fortunately obtains.
Actually just -d < 2 n <= d. The limit formula works for all (reduced) n/d with the tau flavor of eta (period 24), but the q = e^(2 i pi tau) flavor introduces this artificial period of 1. Those of you deterred by Macsyma's ugly ASCII displays of my recent eta evaluations are invited to browse http://gosper.org/etavals.html . It's already obsolete, and will likely get updated. Note that many of the special values of thetas can be written as etas, and thus we now have special values of theta derivatives, e.g., pi - ------- sqrt(3) (d105) theta''(0, e ) = 3 1/3 6 1 2 sqrt(3) Gamma (-) 7/8 3/2 1 3 3 Gamma (-) (---------------------- + 1) 3 3 32 pi - --------------------------------------------- 2/3 2 2 pi (c106) dfloat(%) (d106) - 1.32688190926034d0 = - 1.32688190926035d0 (Despite it's simple appearance, this was a tedious derivation, but I think the process can be automated, especially if nasty algebraic expressions regularly disappear as they did here. The automation of the underlying eta' evaluations is a little iffy. They're really just numerical conjectures.) --rwg
[...]
Those of you deterred by Macsyma's ugly ASCII displays of my recent eta evaluations are invited to browse http://gosper.org/etavals.html . It's already obsolete, and will likely get updated.
E.g., with a bunch of identities as follow: For positive integers a,b,c, let p{a,b,c} := the lowest degree (trivariate) polynomial equation satisfied by eta(q^a),eta(q^b), and eta(q^c). E.g., 8 16 16 8 24 p{1,2,4} = (16 h1 h4 + h1 h4 = h2 ), 3 9 6 6 9 3 12 p{1,3,9} = (27 h1 h9 + 9 h1 h9 + h1 h9 = h3 ), where h2 := eta(q^2), etc. (Maybe someone can tell us why these relations always seem to exist.) If q -> q^d, we have p{a,b,c} <=> p{da,db,dc} and can thus scale by common divisors. With resultants, we can also combine p{a,b,c} p{a,b,d} = p{b,c,d}. E.g., p{1,2,4} p{2,4,8} = p{1,2,8} 24 8 32 24 16 24 24 24 16 = (1048576 h1 h2 h8 + 131072 h1 h2 h8 + 4864 h1 h2 h8 48 16 24 32 8 48 8 8 64 + 16 h1 h8 + 48 h1 h2 h8 + h1 h2 h8 = h2 ), and similarly for p{1,4,8}. So now we can get p{2^a,2^b,2^c} and p{3^a,3^b,3^c}. But row reduction on series can, *in principle*, find *any* p{a,b,c}, e.g., 6 8 12 12 8 6 24 2 p{1,2,5} = (- 3125 h1 h2 h5 - 250 h1 h2 h5 + 256 h2 h5 24 2 18 8 + h1 h5 = h1 h2 ).
From this you may enjoy figuring out how to get p{1,5,25}. Also notice that you can't get p{1,2,5} from p{1,2,4} and p{1,5,25}. So now we have p{2^a 5^b, 2^c 5^d, 2^e 5^f}. But I said we have p{a,b,c} from row reduction, no? *In practice*, row reduction is more strenuous than resultants, and seems limited to exponents of 50 or so, bad news because it's the only source of new primes. (Well, there's always numeric lattice reduction.) But resultants are strenuous, too! A couple of days ago Mathematica disgorged a resultant for p{4,6,9}, a geometric progression with a presumably simple polynomial, but got > 2000 terms, with thousand-digit coefficients, and square-free! Needless to say, the factorization is still in progress.
The Jacobi imaginary transformation provides one more operation, the "flip" involution: /p{a,b,c} = p{bc,ca,ab}, which is at least a nonstrenuous shortcut. The shapes of these polynomials are surprising. p{1,2,7} (deg 76) has 12 terms; p{1,3,7} (deg 26) has only 8. p{1,P,P^2} has only one P term for P=2,3,and 5, but 8 3 4 2 2 4 3 4 p{1,7,49} = (h7 - 49 h1 h49 h7 - 35 h1 h49 h7 - 7 h1 h49 h7 7 2 6 3 5 4 4 - 343 h1 h49 - 343 h1 h49 - 147 h1 h49 - 49 h1 h49 5 3 6 2 7 - 21 h1 h49 - 7 h1 h49 = h1 h49). p{1,5,6} has 247 terms, starting with 6 36 288 267901468048997825931274052676501085034012492775816057012130948209391921577066496 h1 h5 h6 . So far I only have primes through 11, which is not to say p{49,77,121} ! However, Michael Somos mentions in a paper at somos@harary.math.georgetown.edu "my database of thousands of eta-product identities which I update frequently and is available upon request", so he might have lots of primes. OtOH, his paper is about relating >>3 etas, which may also be true of much of his collection. --rwg
rwg>A couple of days ago Mathematica disgorged a resultant for p{4,6,9}, a geometric progression with a presumably simple polynomial, but got > 2000 terms, with thousand-digit coefficients, and square-free! Needless to say, the factorization is still in progress. Joke's on me. Saddam XP forced me to terminate the factorization (and then pay for the bullet) when "an initialization failed", without even letting me look in the window. But I was then able to find the various factors of p{3,6,9}/p{2,3,4}, and in a complete reversal of tradition, the answer turned out to be the *largest*, p{4,6,9} = 4096*eta(q^6)^1296 - <<878>> + 296975826110677186501810694400299649164810446749391313181000047331065773408181565393812635137798905081215065695126808741771033488065035979225233690130907136*eta(q^4)^288*eta(q^6)^144*eta(q^9)^864. rwg> The shapes of these polynomials are surprising. --rwg PS, Greg Whitehead found an unintended but very legitimate solution to Thane's 80% filled Arnold Dozenegger. The 82% model remains unsolved (except by "codesigner" Emma Cohen), despite a furious and prolonged attack by Cheny Xu. (Who keeps sending perplexing photos of bogus solutions that I find physically unfeasible.) There are presumably no unintended solutions.
Here's some more on Greg Whitehead's solution to the 80% Dozenegger http://thaneplambeck.typepad.com/thaneplambeck/ Scroll down past the blackboard photo -- but only if you want to see a photo of the solution ! Thane On Wed, Sep 24, 2008 at 11:49 PM, <rwg@sdf.lonestar.org> wrote:
rwg>A couple of days ago Mathematica disgorged a resultant for p{4,6,9}, a geometric progression with a presumably simple polynomial, but got > 2000 terms, with thousand-digit coefficients, and square-free! Needless to say, the factorization is still in progress.
Joke's on me. Saddam XP forced me to terminate the factorization (and then pay for the bullet) when "an initialization failed", without even letting me look in the window. But I was then able to find the various factors of p{3,6,9}/p{2,3,4}, and in a complete reversal of tradition, the answer turned out to be the *largest*,
p{4,6,9} = 4096*eta(q^6)^1296 - <<878>> + 296975826110677186501810694400299649164810446749391313181000047331065773408181565393812635137798905081215065695126808741771033488065035979225233690130907136*eta(q^4)^288*eta(q^6)^144*eta(q^9)^864.
rwg> The shapes of these polynomials are surprising. --rwg PS, Greg Whitehead found an unintended but very legitimate solution to Thane's 80% filled Arnold Dozenegger. The 82% model remains unsolved (except by "codesigner" Emma Cohen), despite a furious and prolonged attack by Cheny Xu. (Who keeps sending perplexing photos of bogus solutions that I find physically unfeasible.) There are presumably no unintended solutions.
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-- Thane Plambeck tplambeck@gmail.com http://www.plambeck.org/ehome.htm
rwg>So far I only have primes through 11, Foo, I just remembered how I used to get these on a timeshared, 1 meg Macsyma over the Arpanet. For etas, you want to avoid nonconstant coefficients in the reduction process, and all you really need is the ECHELON function. So 0=-16777216*h13^2*h2^72-196608*h1^24*h13^2*h2^48+302875106592253*h1^22*h13^28*h2^24+605750213184506*h1^24*h13^26*h2^24+582452128062025*h1^26*h13^24*h2^24+351263437231252*h1^28*h13^22*h2^24+146681435327336*h1^30*h13^20*h2^24+44326807379140*h1^32*h13^18*h2^24+9858921494006*h1^34*h13^16*h2^24+1610126946220*h1^36*h13^14*h2^24+189124030238*h1^38*h13^12*h2^24+15274994020*h1^40*h13^10*h2^24+778915592*h1^42*h13^8*h2^24+21099988*h1^44*h13^6*h2^24+196885*h1^46*h13^4*h2^24-22*h1^48*h13^2*h2^24+h1^50*h2^24-h1^72*h13^2 2 72 24 2 48 0 = - 16777216 h13 h2 - 196608 h1 h13 h2 22 28 24 24 26 24 + 302875106592253 h1 h13 h2 + 605750213184506 h1 h13 h2 26 24 24 28 22 24 + 582452128062025 h1 h13 h2 + 351263437231252 h1 h13 h2 30 20 24 32 18 24 + 146681435327336 h1 h13 h2 + 44326807379140 h1 h13 h2 34 16 24 36 14 24 + 9858921494006 h1 h13 h2 + 1610126946220 h1 h13 h2 38 12 24 40 10 24 + 189124030238 h1 h13 h2 + 15274994020 h1 h13 h2 42 8 24 44 6 24 + 778915592 h1 h13 h2 + 21099988 h1 h13 h2 46 4 24 48 2 24 50 24 72 2 + 196885 h1 h13 h2 - 22 h1 h13 h2 + h1 h2 - h1 h13 This is, of course, a completely nonrigorous and nearly mindless process, and somewhere there's probably a well developed theory of these things. --rwg
rwg>For etas, you want to avoid nonconstant coefficients in
the reduction process, and all you really need is the ECHELON function. And more than an I286 worth of VM. So for 17, moved to MMa. Which has no Echelon function! So, just use RowReduce. What can it cost? A factor of two? Try a few thousand! With RowReduce, it couldn't even do the 13 case! So I had to bleeping write my own Echelon:
4294967296*h17^2*h2^96-1648529244160*h1^12*h17^6*h2^80+67108864*h1^24*h17^2*h2^72-7771775074107392*h1^18*h17^16*h2^64+137158277070848*h1^24*h17^10*h2^64-9110093824*h1^30*h17^4*h2^64-19318702080*h1^36*h17^6*h2^56-5071143952847839744*h1^24*h17^26*h2^48-94961376686749696*h1^30*h17^20*h2^48-922310028528640*h1^36*h17^14*h2^48-2119493040128*h1^42*h17^8*h2^48+138752*h1^48*h17^2*h2^48-60716992766464*h1^42*h17^16*h2^40+1071549039616*h1^48*h17^10*h2^40-71172608*h1^54*h17^4*h2^40-827240261886336764177*h1^30*h17^36*h2^32-28624230515098157930*h1^36*h17^30*h2^32-437549300159550511*h1^42*h17^24*h2^32-3052451941032780*h1^48*h17^18*h2^32-7818231011807*h1^54*h17^12*h2^32+2834555350*h1^60*h17^6*h2^32-h1^66*h2^32-19809156065811874*h1^48*h17^26*h2^24-370942877682616*h1^54*h17^20*h2^24-3602773548940*h1^60*h17^14*h2^24-8279269688*h1^66*h17^8*h2^24+30*h1^72*h17^2*h2^24-118587876497*h1^66*h17^16*h2^16+2092869218*h1^72*h17^10*h2^16-139009*h1^78*h17^4*h2^16-98260*h1^84*h17^6*h2^8+h1^96*h17^2 2 96 12 6 80 4294967296 h17 h2 - 1648529244160 h1 h17 h2 24 2 72 18 16 64 + 67108864 h1 h17 h2 - 7771775074107392 h1 h17 h2 24 10 64 30 4 64 + 137158277070848 h1 h17 h2 - 9110093824 h1 h17 h2 36 6 56 24 26 48 - 19318702080 h1 h17 h2 - 5071143952847839744 h1 h17 h2 30 20 48 36 14 48 - 94961376686749696 h1 h17 h2 - 922310028528640 h1 h17 h2 42 8 48 48 2 48 - 2119493040128 h1 h17 h2 + 138752 h1 h17 h2 42 16 40 48 10 40 - 60716992766464 h1 h17 h2 + 1071549039616 h1 h17 h2 54 4 40 30 36 32 - 71172608 h1 h17 h2 - 827240261886336764177 h1 h17 h2 36 30 32 - 28624230515098157930 h1 h17 h2 42 24 32 - 437549300159550511 h1 h17 h2 48 18 32 54 12 32 - 3052451941032780 h1 h17 h2 - 7818231011807 h1 h17 h2 60 6 32 66 32 + 2834555350 h1 h17 h2 - h1 h2 48 26 24 54 20 24 - 19809156065811874 h1 h17 h2 - 370942877682616 h1 h17 h2 60 14 24 66 8 24 - 3602773548940 h1 h17 h2 - 8279269688 h1 h17 h2 72 2 24 66 16 16 + 30 h1 h17 h2 - 118587876497 h1 h17 h2 72 10 16 78 4 16 + 2092869218 h1 h17 h2 - 139009 h1 h17 h2 84 6 8 96 2 - 98260 h1 h17 h2 + h1 h17 There is very probably a simpler p{a,b,17}, but the Resultants are so enormous that it is probably easier to Echelon for it de novo. --rwg PS, there's probably nothing wrong with RowReduce--it's just the wrong tool for this job.
ran out of memory, but I got {1,3,19} (48 terms, degree 74) if anybody wants to see it. I'm being stupid. All I need is an Echelon that jettisons previous rows as it goes. I can also find out how big {1,2,19} is, if anybody's curious. --rwg
With the MatrixRank function and a finite modulus I can detect relations from matrices too large to directly Echelon. So far, I can't even smell a p{a,b,23}. Because it's 24-1? Or the nearly infinite 69/3? rwg>(Maybe someone can tell us why these relations always seem to exist.) And if. Joerg Arndt>You need formula 31.1-10a on p.628 of
http://www.jjj.de/fxt/#fxtbook (note my eta(q) is defined as prod(n=1,infty, 1-q^n)) Joerg later explained (privately?) that he dropped the q^(1/24) because he was just using eta to generate partitions. If Ramanujan had done that, he and Rademacher wouldn't have found their miraculous partitions formula. --rwg While seeking postprandial excercise last nite, I was led, Newtonlike, to the discovery of a new cause for tort litigation: Turning out the parking lot lights while your boomerang is in the air.
On Tue, Sep 30, 2008 at 2:48 AM, <rwg@sdf.lonestar.org> wrote:
While seeking postprandial excercise last nite, I was led, Newtonlike, to the discovery of a new cause for tort litigation: Turning out the parking lot lights while your boomerang is in the air.
Yikes! Was it a soft plastic one like an Aerobie, or was it a real wooden one? -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
theta[3](%i*sqrt(3)*%pi/4,%e^-(sqrt(3)*%pi)) = 3^(1/8)*(-4*sqrt(6)+7*sqrt(3)+8*sqrt(2)-9)^(1/8)*%e^(sqrt(3)*%pi/16)*gamma(1/3)^(3/2)/(2*2^(17/48)*%pi) i sqrt(3) pi - sqrt(3) pi theta (------------, e ) 3 4 sqrt(3) pi ---------- 1/8 1/8 16 = 3 (- 4 sqrt(6) + 7 sqrt(3) + 8 sqrt(2) - 9) e 3/2 1 65/48 Gamma (-)/(2 pi) ~ 1.066114065917882492374666215948483823267149247604 3
On Tue, Sep 30, 2008 at 2:48 AM, <rwg@sdf.lonestar.org> wrote:
While seeking postprandial excercise last nite, I was led, Newtonlike, to the discovery of a new cause for tort litigation: Turning out the parking lot lights while your boomerang is in the air.
Yikes! Was it a soft plastic one like an Aerobie, or was it a real wooden one? -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com
Aerobie A30, probably the only US outdoor boomerang safe enough to get product liability insurance. For nearly 20 years. I wonder if the barratry industry has managed to zap an apple tree owner for falling fruit. "I cannot tell a lie. It was to save us from the Burgesses." Actually, that day's most liable product was (naturally) mainland Chinese. I often sit in bed polishing plastic puzzle parts with microfiber cloths from TaP Plastics, who recently switched from a Korean to Chinese source. Beware: The latter give off clouds of microscopic lint. The next morning, it took two hours to tear six intractable knots from my hair. --rwg EMISSION SIMONIES (Carbon credits?)
rwg>The (mostly impressive) Mma 6.0 doc gives, under applications of EllipticF,
"Parametrization of a mylar balloon (two flat sheets of plastic sewn together at their circumference and then inflated):", followed by three definitions and a call to ParametricPlot3D that makes an "M&M". Inelastic mylar would crinkle at the equator, so the true solution would be more like two shallow paper cupcake molds.
No, that can't be right either. More efficient would be for the inward creases to be cusps and the outward ones rounded. But then the distance from the equator to a pole along a groove would be less than along a ridge, and the only obvious sink for the extra length would be the grooves wastefully curving inward past vertical. Real mylar balloons solve this problem by having the grooves change to ridges on the way up, which I'm happy to have predicted before confirming at the supermarket. But on what scale are these grooves and ridges? Don't they depend on the thickness of the mylar? I.e., wouldn't thinner mylar permit a greater number of smaller puckers? Might it be that the grooves in a 200' diameter "satelloon" (hopefully bearing a giant Happy Face and the legend, "Eat This, PG&E") be no deeper than a one footer? (I wonder if the Echo designers considered this presumably cheaper alternative before somehow figuring out how to make that moby sphere.) If the wrinkles don't scale, then they're infinitesimal in mathematically ideal mylar, and the Mathematica plot is (probably) right! Ideal mylar would be this strange, infinitely shrinkable but finitely stretchable material, as if composed of a triangular webwork of microscopic keychains. --rwg PS: Researching an accessible shape for p{eta(q^a),eta(q^b),eta(q^23)}, I guessed that p{eta(q^3),eta(q^5),eta(q^11)} might be nice because 11-3 and 11-5 both divide 24. After a one day Resultant and a three day Factor, 1.03 MB, 1091 terms, degree 256, smallest coefficient = 3^81.
prod((q^(2*2^-n)+2*cos(theta/2^n)*q^2^-n+1)/4,n,1,inf) = (q^2-2*cos(theta)*q+1)/(log(q)^2+theta^2) - n - n 2 2 theta 2 inf q + 2 cos(-----) q + 1 /===\ n 2 | | 2 q - 2 cos(theta) q + 1 | | -------------------------------- = ----------------------- | | 4 2 2 n = 1 log (q) + theta --rwg
For odd d, 'limit(theta[3](0,%e^(2*%i*%pi*n/d)*r)*sqrt(-log(r)),r,1) = sqrt(%pi/d)*%e^(%i*%pi*(-2*g(4*n/d)+5*g(2*n/d)-2*g(n/d))/(12*d)) 2 i pi n -------- d limit theta (0, e r) sqrt(- log(r)) = r -> 1 3 4 n 2 n i pi (- 2 g(---) + 5 g(---) - 2 g(n/d)) d d --------------------------------------- pi 12 d sqrt(--) e d Where, as with the eta limits, g(r) := if integerp(r) then r else (floor(r),denom(r)*(%%+3)-(g(1/(r-%%))+1/denom(r))/(r-%%)) g(r) := if integerp(r) then r else (floor(r), 1 1 g(------) + -------- r - %% denom(r) denom(r) (%% + 3) - --------------------), r - %% a n(ot obviously) integer-valued function. The exponent expression -2*g(4*n/d)+5*g(2*n/d)-2*g(n/d) 4 n 2 n n - 2 g(---) + 5 g(---) - 2 g(-) d d d appears to be 0 mod 6, period n for fixed n, period d for fixed d, but otherwise no simpler than g. Even d looks not much harder. (Betcha you reparsed!) --rwg PS, how can a natural boundary grow only like 1/sqrt(-log q)? The other thetas are *much* wilder.
For odd d, 'limit(theta[3](0,%e^(2*%i*%pi*n/d)*r)*sqrt(-log(r)),r,1) = sqrt(%pi/d)*%e^(%i*%pi*(-2*g(4*n/d)+5*g(2*n/d)-2*g(n/d))/(12*d))
2 i pi n -------- d limit theta (0, e r) sqrt(- log(r)) = r -> 1 3
4 n 2 n i pi (- 2 g(---) + 5 g(---) - 2 g(n/d)) d d --------------------------------------- pi 12 d sqrt(--) e d
Where, as with the eta limits, g(r) := if integerp(r) then r else (floor(r),denom(r)*(%%+3)-(g(1/(r-%%))+1/denom(r))/(r-%%))
g(r) := if integerp(r) then r else (floor(r),
1 1 g(------) + -------- r - %% denom(r) denom(r) (%% + 3) - --------------------), r - %% a n(ot obviously) integer-valued function.
The exponent expression -2*g(4*n/d)+5*g(2*n/d)-2*g(n/d)
4 n 2 n n - 2 g(---) + 5 g(---) - 2 g(-) d d d
appears to be 0 mod 6, period n
2n
for fixed n, period d for fixed d, but otherwise no simpler than g. Even d looks not much harder. (Betcha you reparsed!) --rwg PS, how can a natural boundary grow only like 1/sqrt(-log q)? The other thetas are *much* wilder.
This question is braindead thrice over. Theta_4 goes "wildly" to 0 similarly to the oddly even d case of theta_3. Theta_2 (odd d) is similar to theta_3: theta[2](0,%e^(2*%i*%pi*n/d)*r)*sqrt(-log(r)) = sqrt(%pi/d)*%e^(%i*%pi*(2*g(4*n/d)-g(2*n/d))/(12*d)) 2 i n pi -------- d theta (0, e r) sqrt(- log(r)) = 2 4 n 2 n i (2 g(---) - g(---)) pi d d ------------------------ 12 d pi e sqrt(--) d The poles are at irrational points. Duh. PS, almost exactly a week after buying it, Greg Whitehead has become the first solver of the 82% Arnold Dozenegger other than Emma Cohen, who actually took more than a week to rediscover pretty nearly her own design, despite her abundant free time as a Cal Tech student.
Recall (from a couple of years) that a "biroller" is the intersection of two perpendicular circular cylinders and takes an intriguing choatic path if rolled diagonally down a shallow, inclined (bathtub-like) trough. A "triroller" is the intersection of three perpendicular cylinders, with perhaps slightly less interesting rolling behavior. In either case, the roller can only change direction if it momentarily rests on the intersection of two ridges. Now consider the intersection of the four cones circumscribing a regular tetrahedron. (http://gosper.org/tetraroller1.gif, http://gosper.org/tetraroller2.gif). Instead of tipping on one of the triple points of ridges, it chooses left or right as it rolls up an initially nonexistent ridge. I'm not sure what would be the most entertaining surface, but these might make an even more hellish alternative to Gilbert & Sullivan's "elliptical billiard balls". The intersection ridges are in fact elliptical, and the projection along an altitude of the tetrahedron reveals quite probably a Reuleaux triangle (Wankel). The vector from the center to one of the triple points is exactly -3/5 times the vector to the opposite vertex. --rwg
Is there a complex version of this, where the quadratic in q is split into the product of two factors 1 +- q e^(i theta)? If so, there might be other polynomials available as numerators. Rich ------------- Quoting rwg@sdf.lonestar.org:
prod((q^(2*2^-n)+2*cos(theta/2^n)*q^2^-n+1)/4,n,1,inf) = (q^2-2*cos(theta)*q+1)/(log(q)^2+theta^2)
- n - n 2 2 theta 2 inf q + 2 cos(-----) q + 1 /===\ n 2 | | 2 q - 2 cos(theta) q + 1 | | -------------------------------- = ----------------------- | | 4 2 2 n = 1 log (q) + theta
--rwg
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------ Dan writes: << so there is no "best" value of X. I'll leave finding "good" values of X to the number theorists, once "good" is defined.
------ Victor writes: << That's true, but one can ask for what the best value of X is given that all elements of X are bounded in absolute value by H. There is an algorithm due to Hastad, Just, Lagarias and Schnorr that is designed to answer the posed question. When looked at the right way it's very close to the L^3 lattice reduction algorithm.
--------------------- Many thanks Victor for this reference: its title looks exactly what I am searching. You are right in your answer to Dan. "Good" means using smallest integers abs(x_i) < H allowing an asked precision. For example: k_1 = pi k_2 = e Using the continued fraction algorithm on pi/e, we will find good x_1 and x_2: k_1*x_1 + k_2*x_2 ~ 0 ------------------------------ e*7 + pi*(-6) = 0.1784 e*15 + pi*(-13) = -0.0664 e*37 + pi*(-32) = 0.0454 e*52 + pi*(-45) = -0.0210 e*141 + pi*(-122) = 0.0034 ... Now, what would be solutions with: k_1 = pi k_2 = e k_3 = sqrt(2) k_1*x_1 + k_2*x_2 + k_3*x_3 ~ 0 --------------------------------- ??? ??? ??? Probably the method explained in the paper will be able to produce some "good" values. Christian.
Hello, Concerning the general problem of finding an integer relation. Here is an algorithm for 3 values, that is : if we are looking for aX + bY + Z ? 0 where a,b are 2 real numbers and X,Y,Z three integers. do {| a - b |} = c a <- b b <- c od { } is the fractional part. If we iterate this, after a certain amount of iterations we find 3 integers where the equation is near 0. I could generalize this example to more ; Here is the program in Maple : pseudoLLL:=proc(s) local a, b, c, d, nn, k,i, aa, ss, g, h,m,z; aa:=s: nn:=nops(aa): k:=[seq(frac(abs(aa[i]-aa[i+1])),i=1..nn-1),frac(abs(aa[1]-aa[nn]))]; ss:=evalf(k): g:=sort([seq([convert(1+ss[j],string),j],j=1..nn)],lexorder[2]): h:=[seq(g[i][3],i=1..nn)]: m:=[seq(k[h[z]],z=1..nn)]: return(m); end: This will work with a vector of real numbers up to a certain precision. I could make it to work with Maple 11, but they changed the syntax and the calling sequence again for the sort function (#@!$!@#($&!). For n=3 there, it can be worked out this way : L3:=proc(x, y, z) local a, b, c, d, k1, k2, k3, ll; a := x; b := y; c := z; k1 := frac(abs(a - b)); k2 := frac(abs(b - c)); k3 := frac(abs(a - c)); ll := sort([k1, k2, k3]); a := ll[1]; b := ll[2]; c := ll[3]; RETURN(a, b, c) end: Simon Plouffe
participants (9)
-
Christian Boyer -
Dan Asimov -
Fred lunnon -
Mike Stay -
rcs＠xmission.com -
rwg＠sdf.lonestar.org -
Simon Plouffe -
Thane Plambeck -
victor miller