---------- 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. Note here spontaneous reduction to machine precision: In[172]:= Precision /@ Take[%168, -4] Out[172]= {26.7284, MachinePrecision, MachinePrecision, MachinePrecision} In[168]:= %152 =.; Table[ Sum[morse[n]/(1`69 - 1 + n), {n, 2^(k - 1), 2^k - 1}], {k, 23}] Out[168]= \ {-1.000000000000000000000000000000000000000000000000000000000000000000\ 000, -0.16666666666666666666666666666666666666666666666666666666666666\ 6666667, -0.\ 026190476190476190476190476190476190476190476190476190476190476190476,\ -0.003149628149628149628149628149628149628149628149628149628149628149\ 6281, -0.0002619192183025316892818376272195153227023190373683308580086\ 696571249, \ -0.0000140961597127378916942716179595142349011573826289517744899239499\ 297, -4.68535396826140722399442374299172711851286747074109655227506937\ 39*10^-7, \ -9.29696753332452313450490608814161365712389608058600043630547620*10^-\ 9, -1.0739453689565117144058603338727124796070485046345754188182264*\ 10^-10, -7.\ 08627180236246614574305619137829547332797328422000406663275*10^-13, \ -2.631884639466089306668070775152803160494886428701003929236*10^-15, \ -5.438339487510447306781452686083673103623574531677729791*10^-18, \ -6.192989936483548192430918684852680607679431272493048*10^-21, \ -3.8561058372588959715408154984974065907955402095482*10^-24, \ -1.3041154300564990475170368702167108513355356894*10^-27, \ -2.381829871124105427498092722592385259369002*10^-31, \ -2.337561278904883240397854728219244957183*10^-35, \ -1.22733321882602548663709666820289370*10^-39, \ -3.434100513325279736346293287953*10^-44, \ -5.10267262676374843629844283*10^-49, -4.01376*10^-54, \ -1.66664*10^-59, -3.64371*10^-65} Even weirder: spontaneous *creation* of "precision": In[167]:= Table[ Sum[morse[n]/(1`69 + n), {n, 2^(k - 1), 2^k - 1}], {k, 25}] Out[167]= \ {-0.500000000000000000000000000000000000000000000000000000000000000000\ 000, -0.08333333333333333333333333333333333333333333333333333333333333\ 3333333, -0.\ 015476190476190476190476190476190476190476190476190476190476190476190,\ -0.002207514707514707514707514707514707514707514707514707514707514707\ 5147, -0.0002091375027541893674392190938372057340187376836883901987123\ 870639319, \ -0.0000122946980614332692270376487316405594406332942110185551582173466\ 668, -4.32195428826170511217542239637665018410757461892500010625794694\ 03*10^-7, \ -8.87301646666058245644516124261223306314074651021882186455064548*10^-\ 9, -1.0458099643631916758928638937741904903086993472013174399559272*\ 10^-10, -6.\ 98143049429755311075516385788269917584660543240898536451685*10^-13, \ -2.610180763779562442726545899627011713573506161849931176559*10^-15, \ -5.413598355752984663980985076812050357066558893858646547*10^-18, \ -6.177575942247773207802885950958692670828387637048574*10^-21, \ -3.8508912806285963424755514484965618347262775126888*10^-24, \ -1.3031628850933154851149881302056671832958127403*10^-27, \ -2.380894944793706582745607332625954939245743*10^-31, \ -2.337070373089340522029095103932355659111*10^-35, \ -1.22719585888756635716102476624359666*10^-39, \ -3.433896409631194067248804676899*10^-44, \ -5.10251207751970643813210988*10^-49, \ -4.0136977684427557028670456513860079925490740717925430787290285399384\ 8*10^-54, \ -1.6666230725159058889654514475037587573071924772089658199531293682217\ 2*10^-59, \ -3.6436950646124655027220392184731237917460692833546825636890121781519\ 5*10^-65, \ -4.1844261203426874910214038879337676295476794340326942203659352222110\ 4*10^-71, \ -2.5187326323755771625453269689899173317142745156407402958430970838190\ 2*10^-77} 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. --rwg 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
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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
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Bill Gosper