On Tue, Jul 8, 2014 at 11:43 AM, Bill Gosper <billgosper@gmail.com> wrote:
---------- Forwarded message ---------- From: Julian Ziegler Hunts <julianj.zh@gmail.com> Date: Tue, Jul 8, 2014 at 9:37 AM Subject: Re: [math-fun] Thue-Morse sums To: Bill Gosper <billgosper@gmail.com>
then S0 = -3 S1 and S3 = (S1 - 3 S2)/2.
Nope. The convergence is great; 2^18 terms gets you >30 digits and these are only accurate to about 10^-5.
Julian
YIKES! I must've been had by FindIntegerNullVector. Or just general Mma numerics.
Now just a dipole moment: False false Alarm! NeilB & I just reconfirmed the original calculations purporting those simple relations. Then Julian turned around and *proved* them! ------------------- Looking for more linear relations reveals the pattern, which quickly leads to a proof. Let a[k]=Sum[morse[n]/(n+k),{n,0,∞ }]. Then a[k]=a[k+1]+2a[2k]+2a[2k+1]. This comes from combining the morse[2n] and morse[2n+1] terms of a[2k]+a[2k+1]; they give morse[n]*(1/(2n+2k)-1/(2n+1+2k)+1/(2n+2k+1)-1/(2n+1+2k+1))=morse[n]*1/2*(1/(n+k)-1/(n+k+1)) (this even works for n=k=0; we just need to interpret 1/0 as 0). In fact this works for a[k,c,m]=Sum[morse[n]*(n+k)^c,{n,0,2^m-1}], and shows that a[k,c,m]-a[k+1,c,m]=(a[2k,c,m+1]+a[2k+1,c,m+1])/2^c. Are there more linear relations? Would coalescing terms twice give nothing, a new linear relation, or a combination of old ones? --------------------------- Unfortunately, he punted the notebook containing the bad derivations, but thinks they were his fault vs Mathematica's. Note here spontaneous reduction to machine precision:
[Chop] Even more blatant: In[224]:= morse /@ Range[0, 15] Out[224]= {1, -1, -1, 1, -1, 1, 1, -1, -1, 1, 1, -1, 1, -1, -1, 1} In[223]:= Sum[morse[n]/(1`69 - 1 + n), {n, 2^10 - 1}] Out[223]= -1.19628326432525379989311583225439076264934483334544909189104057071473 In[222]:= Sum[morse[n]/(1`69 - 1 + n), {n, 2^20 - 1}] Out[222]= -1.19628 (Taking several minutes to destroy nearly all precision.) I should have regarded these results with greater suspicion. No wonder
people believed that I believed that ISC delusion. At any rate, the convergence seems to *accelerate*, like a confluent hypergeometric series.
Julian, seeing only my remarks: You mean more than a constant number of bits/term? Yes, it does, for some nice functions like the ones we're dealing with. Breaking the sum into pairs is the same as changing f(n) to g(n)=f(2n)-f(2n+1), which for the functions we're dealing with is about a constant times f'(n), so by doing this repeatedly we get convergence better than c_k*(n/2^k)^-k, c_k some sequence of constants, i.e. better than k bits/term but with a bad starting estimate on accuracy. The same pairing-up also easily shows that for f a polynomial the sum is 0.
--rwg
Numerics. Treacherous as always.
On Mon, Jul 7, 2014 at 10:33 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2014-07-07 00:39, Bill Gosper wrote:
[Adding subject; fixing bug.]
Whoa, how are you getting these? Convergence is absent or useless.
Holy crapoli, I completely missed the point of this Flanelle business.
(Wasn't he the guy who helped Dumbledore make a Philosopher's Stone?)
WDS>
h(n) = (-1)^(parity of bit-sum of binary representation of n)
f(x) S=SUM[ f(n) * h(n), for n=0..2^k-1 with k large]
ln(x+1) S=-log(2)/2 ln(x+2) S = -0.1379330125 ln(x+3) -0.07070756527 x^j 0 for any fixed integer j>=0 1/(x+1) 0.398761088108
[http://isc.carma.newcastle.edu.au/advancedCalc identifies this as
3^(1/6)/Zeta[3]^(5/237)/3 . I'm surprised you didn't notice.] http://arewomenhuman.me/wp-content/uploads/2012/10/trolls.jpeg
1/(x+2) 0.1049709156499 sqrt(x) -0.63407426 1/sqrt(x+1) 0.1983140804979
It appears that the cases f(x):=1/(x+k) (various k) are simply related. If
S0:=SUM[h(n)/n, for n=1..2^k-1 with k large] ~ -1.19628326432525643722229,
S1:=SUM[h(n)/(n+1), for n=0..2^k-1 with k large] ~ 0.398761088108
S2:=SUM[h(n)/(n+2), for n=0..2^k-1 with k large] ~ 0.1049709156499
S3:=SUM[h(n)/(n+3), for n=0..2^k-1 with k large] ~ 0.0419241705
then S0 = -3 S1 and
S3 = (S1 - 3 S2)/2 .
--rwg
After a bit of poking around, NeilB and I share Warren's deep curiosity over Flanelle's biographical details. And sorry about accidentally trolling Mr. Goucher. --rwg