See A068214/A068215. - Neil On Sat, Feb 25, 2012 at 7:05 PM, Warren Smith <warren.wds@gmail.com> wrote:

I just saw this wikipedia article: http://en.wikipedia.org/wiki/Borwein_integral

Truly, the Gods are against us.

-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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-- Dear Friends, I will soon be retiring from AT&T. New coordinates: Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com

Hello, there are many exotic examples like that. Look at this one, the sequence A007699, a(n)= {nearest integer to } a(n-1)^2 /a(n-2). the sequence is : 10, 219, 4796, 105030, 2300104, 50371117, 1103102046, ... the ratio of a(n+1)/a(n) -->> 21.89949541893233438052532024370859294946 6190194047646611617481379446934015013... quite rapidly. By testing that number for algebraicity, one quickly finds that it is one of the real roots of 4 3 2 11 x - 18 x + 3 x - 22 x + 1 , the root can be calculated explicitely but it is quite ugly. BUT this is FALSE, the exact number, the ratio of a(n+1)/a(n) to high precision is : 21.89949541893233438052532024370859294946619019404764661\ 1617481379446934015013222134632935910397445152107249\ 9761349032053359109441276427149926005650693640825950\ 1335437931357406880506038288883182166121707555714731\ 7300725684058890389637589630893096102492114259366306\ 8931460123024765848662425871358354508021664249361199\ 6944188439509332536170556143293330518711985939602460\ 9252047620048792957280978376910436207360514815505096\ ... Is NOT a simple algebraic the root of the <apparent> equation is valid up to 0.11357748460267988639402531902 * 10^(-1878). As far as I know, nobody really knows WHY it behaves like that, those Pisot sequences are quite bizarre in my humble opinion. There are others, like A010900, that one too is exotic enough. E(4,13). A good idea would be to test against any asymptotics the sequence (21.8994954189323343805253202437085929494661)^n to see how far it is from the real sequence. I discussed that topic with David Boyd (UBC) a while ago and the problem with these animals is that we could have a situation where the exponents (or order) of the recurrence relation could be bery big. That means that we can have some approximations but the real thing is far more complicated. ps : I added this comment on the edit page of the sequence. PS2 : this is one example where whatever sophistication we now have with programs like PSLQ, LLL, Pari and all that, we still can find examples where it fails to give a definite answer. best regards, Simon Plouffe

From: Warren Smith <warren.wds@gmail.com> To: math-fun@mailman.xmission.com Sent: Saturday, February 25, 2012 4:05 PM Subject: [math-fun] Mathematical hell

I just saw this wikipedia article:   http://en.wikipedia.org/wiki/Borwein_integral

Truly, the Gods are against us.

-- Warren D. Smith http://RangeVoting.org  <-- add your endorsement (by clicking "endorse" as 1st step)

Let sinc(x) = sin(πx)/(πx).  Let rect(x) = 1 for |x| < 1/2, otherwise 0.  Define Fourier transform to have kernel exp(+-2πitx).  Then rect(t) and sinc(x) are a Fourier transform pair.  The convolution of two functions f and g is defined as (f*g)(x) = int(f(x-y) g(y), y=-inf..inf) = = int(f(y) g(x-y), y=-inf..inf).  With F denoting the Fourier transform operation, the convolution theorem states that F(f*g) = F(f) F(g), F(fg) = F(f)*F(g).  This shows that convolution is commutative and associative.  Also, under scaling, let g(x) = f(x/a).  Then F(g)(t) = |a| F(f)(at).  In particular, for a > 0, sinc(x/a) and (a rect(ax)) are Fourier transform pairs. We have 1 = rect(0) = F(sinc)(0) = int(sinc(x), x=-inf..inf).  Next, taking a to be positive, int(sinc(x) sinc(x/a), x=-inf..inf) = F[sinc(x) sinc(x/a)](t=0) =[[rect(t)] * [a rect(at)]](t=0). In taking the convolution the rect(at) nibbles away at each of the edges of rect(t) by amount a/2.  As long as a < 1. the nibbling doesn't reach t = 0, and the integral equals 1.  Similarly, int(sinc(x) sinc(x/a) sinc(x/b) sinc(x/c) ... , x=-inf..inf) = [[rect(t)] * [a rect(at)] * [b rect(bt)] * [c rect(ct)] * ...](t=0). Again the nibbles fail to reach 0, exactly when a + b + c + ... < 1.  By continuity, we can say that the integral equals 1 when a + b + c + ... <= 1.  But if the sum is greater than 1, the convolutions will have eaten their way to the center, and the integral is less than 1. The same trick applies to any band-limited function, one whose Fourier transform has compact support.  For example BesselJ(0,x) and 1/sqrt(1 - t^2) for |t| < 1, other wise 0, are a Fourier transform pair (modulo various factors of π).   --  Gene