Partial sums of the MRB constant
Obtain several digits of m from the cloud
Obtain several digits of m from the cloud
In[]:=
m=WolframAlpha["MRB constant",{{"DecimalApproximation",1},"ComputableData"},PodStates{"DecimalApproximation__More digits"}];
Brute force for 60 terms gives only 2 digits
Brute force for 60 terms gives only 2 digits
f[n_]:=(-1)^n(n^(1/n)-1);a1=(m-NSum[f[n],{n,1,60},Method"AlternatingSigns",WorkingPrecision100])
Out[]=
-0.03508058312888533476029149152413305701285431849013881336109994374863959009392347441816059869179541
Van Wijngaarden transformation for 60 terms give around 4 digits
Van Wijngaarden transformation for 60 terms give around 4 digits
f[n_]:=(-1)^n(n^(1/n)-1);g[n_]:=f[n+1];m-NSum[(g[n]+f[n])/2,{n,1,60},Method"AlternatingSigns",WorkingPrecision100]
Out[]=
-0.0002235554251167115653241529678883538450873340383873767965854185185546751063897016113494197187047609
Euler’s transform for 60 terms gives 22 digits
Euler’s transform for 60 terms gives 22 digits
N[m-Sum[Binomial[n,k-1],{n,60},{k,n+1}],30]//Timing
-(n+1)
2
k
(-1)
1/k
k
Out[]=
{0.03125,-3.72569817451617981829539904872×}
-22
10
An integral from 1 to only 6*I gives 9 digits
An integral from 1 to only 6*I gives 9 digits
m+NIntegrate(-1)Sin[Pi*x],{x,1,6I},WorkingPrecision->100//Im
1/x
x
Out[]=
1.5311525834464219175000053646095293386748535923195480774528784225182384285458621955030374600×
-9
10
Obtain several digits of m from the cloud
Obtain several digits of m from the cloud
In[]:=
m=WolframAlpha["MRB constant",{{"DecimalApproximation",1},"ComputableData"},PodStates{"DecimalApproximation__More digits","DecimalApproximation__More digits","DecimalApproximation__More digits","DecimalApproximation__More digits","DecimalApproximation__More digits"}];
m//Precision
Out[]=
2514.
The same integral from 1 to only 60*I gives 84 digits
The same integral from 1 to only 60*I gives 84 digits
m+NIntegrate(-1)Sin[Pi*x],{x,1,60I},WorkingPrecision->200//Im
1/x
x
Crandall’s First formula for 60 terms gives 111 digits
Crandall’s First formula for 60 terms gives 111 digits
c1=ActivateLimitDerivative[n][DirichletEta][x]/.n1,x1,Unevaluated[Sum]
n
(-1)
n!
In[]:=
f1[n_]:=Derivative[n][DirichletEta][n];
n
(-1)
n!
m+(c1+Sum[N[f1[n],150],{n,2,60}])
Out[]=
-4.4991140943357666465094281289401584760306×
-111
10
Burns’ program for Crandall’s First formula for 60 terms also gives 111 digits
Burns’ program for Crandall’s First formula for 60 terms also gives 111 digits
p=60;prec=4p;etaMM[m_,pr_]:=Block[{a,s,k,b,c,jj,p,q},a[j_]:=(jj=j+1;{p,q}={Log[SetPrecision[jj,prec]]^m,Power[jj,m]};p/q);{b,c,s}={-1,-d,0};Do[c=b-c;s=s+ca[k];b=(k+n)(k-n)b/((k+1)(k+1/2)),{k,0,n-1}];Return[N[s/d,pr](-1)^m]];eta1=N[EulerGammaLog[2]-Log[2]^2/2,prec];n=Floor[132/100prec];d=N[ChebyshevT[n,3],prec];MRBtest=eta1-Total[ParallelCombine[((Cos[Pi#])etaMM[#,prec]/N[Gamma[#+1],prec])&,Range[2,Floor[p]],Method->"CoarsestGrained"]];Print[N[m-MRBtest,10]];
In[]:=
m=WolframAlpha["MRB constant",{{"DecimalApproximation",1},"ComputableData"},PodStates{"DecimalApproximation__More digits","DecimalApproximation__More digits","DecimalApproximation__More digits","DecimalApproximation__More digits","DecimalApproximation__More digits","DecimalApproximation__More digits"}];
Precision[m]
Out[]=
4999.27
48/25.
Out[]=
1.92
Burns’ program for Crandall’s First formula for 960 terms gives 2871 digits
Burns’ program for Crandall’s First formula for 960 terms gives 2871 digits
p=960;prec=4p;etaMM[m_,pr_]:=Block[{a,s,k,b,c,jj,p,q},a[j_]:=(jj=j+1;{p,q}={Log[SetPrecision[jj,prec]]^m,Power[jj,m]};p/q);{b,c,s}={-1,-d,0};Do[c=b-c;s=s+ca[k];b=(k+n)(k-n)b/((k+1)(k+1/2)),{k,0,n-1}];Return[N[s/d,pr](-1)^m]];eta1=N[EulerGammaLog[2]-Log[2]^2/2,prec];n=Floor[132/100prec];d=N[ChebyshevT[n,3],prec];MRBtest=eta1-Total[ParallelCombine[((Cos[Pi#])etaMM[#,prec]/N[Gamma[#+1],prec])&,Range[2,Floor[p]],Method->"CoarsestGrained"]];Print[N[m-MRBtest,10]];
Obtain even more digits of m from the cloud
Obtain even more digits of m from the cloud
m//Clear;CloudGet["https://www.wolframcloud.com/obj/ed331fd0-6500-46bf-bb3b-c32faadec459"];m//Precision
Out[]=
10100.
Burns’ second program for Crandall’s First formula for 960 terms also gives 2871 digits
Burns’ second program for Crandall’s First formula for 960 terms also gives 2871 digits
1000(1.+M)/2/960
Out[]=
25
48
terms=960;prec=terms126/20;(**Numberofrequireddecimals.*.*)ClearSystemCache[];T0=SessionTime[];expM[pre_]:=Module{lg,a,d,s,k,bb,c,end,iprec,xvals,x,pc,cores=16(*=4*numberofphysicalcores*),tsize=2^7,chunksize,start=1,ll,ctab,pr=Floor[1.0002pre]},chunksize=cores*tsize;(*AbovewecomputedtheneededratiotomatchCrandall'sconvergentrate,andmisitselfthebestchoiceoffractionalvaluetoaddto1.toarivethereexactly.*)n=Floor(1.`+m)pr;end=Ceiling[n/chunksize];Print["Using ",pr," decimals of precision, "];d=ChebyshevT[n,3];{b,c,s}={SetPrecision[-1,1.1*n],-d,0};iprec=pr2^6;Do[xvals=Flatten[ParallelTable[Table[ll=start+j*tsize+l;lg=Log[ll]/(ll);x=N[E^(lg),iprec];pc=iprec;While[pc<pr,pc=Min[4pc,pr];x=SetPrecision[x,pc];xll=x^ll;z=(ll-xll)/xll;t=2ll-1;t2=t^2;x=x*(1+SetPrecision[4.5,pc](ll-1)/t2+(ll+1)z/(2llt)-SetPrecision[13.5,2pc]ll(ll-1)/(3llt2+t^3z))];x-lg,{l,0,tsize-1}],{j,0,cores-1},Method->"EvaluationsPerKernel"->16]];ctab=ParallelTable[Table[c=b-c;ll=start+l-2;b*=2(ll+n)(ll-n)/((ll+1)(2ll+1));c,{l,chunksize}],Method->"EvaluationsPerKernel"->16];s+=ctab.(xvals-1);start+=chunksize;st=SessionTime[]-T0;kc=k*chunksize;,{k,0,end-1}];N[-s/d,pr];t2=Timing[MRBeta2toinf=expM[prec];];Print[terms," terms gave ",-Log10[N[m-MRBeta2toinf+Log[2]^2/2-EulerGammaLog[2],200]]//Re//Floor," digits."]
25
48
Burns’ second program for Crandall’s First formula for 1000 terms gives 2990 digits
Burns’ second program for Crandall’s First formula for 1000 terms gives 2990 digits
The above-used integral from 1 to 1000*I gives around 523 digits
The above-used integral from 1 to 1000*I gives around 523 digits