ExploringCrandall'sFirstFormulafortheMRBConstantCMRB=S=∑m>=1m(-1)m!(m)η(m):SplittingtheSeriesandIntegralRepresentations
ExploringCrandall'sFirstFormulafortheMRBConstantCMRB=S=(m):SplittingtheSeriesandIntegralRepresentations
∑
m
>=
1m
(-1)
m!
(m)
η
AccordingtotheWolframDocumentation:istypicallyusedtosolveproblemsforwhichnoexactsolutioncanbefoundortogetsimpleranswersforcomputation,comparisonandinterpretation.Insuchcases,anasymptoticapproximationoftengivesenoughinformationforsimplifyingorsolvingapplicationproblems.
Let m be the MRB constant, the sum of -1.
1/k
k
exp(πk)
SInce a exact solution for the MRB constant in terms of elementary functions seems illusive,
In[]:=
S=-1
∞
∑
k=1
1/k
k
exp(πk)
Out[]=
∞
∑
k=1
-kπ
1
k
k
We have the following nontrivial relations:
Asymptotic[-1,k∞]==(theAsymptoticterm)
1/k
k
log(k)
k
In[]:=
Asymptotic[-1,k∞]==
1/k
k
log(k)
k
Out[]=
True
In[]:=
eta1=Sum,{k,1,Infinity}
log(k)
kexp(πk)
Out[]=
EulerGammaLog[2]-
2
Log[2]
2
Surprisingly, of which, means
∞
∑
k=1
1/k
k
exp(πk)
∞
∑
k=1
1/k
k
log(k)
k
exp(πk)
∞
∑
k=1
1/k
k
exp(πk)
Inessence,thisequalityshowsthattheoriginalseriescanbesplitintoanasymptoticcomponentof-1==eta1,andaremainder--1==(m),andthesecomponentstogetherreconstructtheoriginalseries,convergingtotheMRBconstant,(m)=S.
1/k
k
exp(πk)
∞
∑
k=1
1/k
k
log(k)
k
exp(πk)
∑
m
>
1m
(-1)
m!
(m)
η
∑
m
>=
1m
(-1)
m!
(m)
η
shown next
In[]:=
Neta1+--1==S//Chop
∞
∑
k=1
1/k
k
Log[k]
k
Exp[πk]
Out[]=
True
Concerning an analog of the above sum
Concerning an analog of the above sum
ThefollowingadditioncodeevaluatsafunctionthatinvolvestheMeijerGfunctionandcomparingittoanumericalintegralanalogoftheMRBconstant(theanalog).Let'sbreakdownwhat'shappens:1.FunctionDefinition:2.TableEvaluation: .3.ComplexResults:4.MagnitudeCalculation:
• The function is defined using the function. The Meijer G-function is a very general function that includes many special functions as special cases. It is used here in a complex configuration.
• The function is then used in a sum that depends on a complex power of (the imaginary unit) to equal the analog.
• Two tables are computed. Each table evaluates the difference between a sum involving and a numerical integral.
• The first table evaluates the sum from to x and compares it to an integral with the integrand .
•
The second table evaluates the sum from
to
x
and compares it to an integral with the integrand
• The results of the table evaluations are complex numbers, showing both real and imaginary parts. The use of ensures these are numerical approximations with a specified precision.
• Following each table, the function is applied to the results to obtain the magnitudes of the complex numbers. This gives a sense of the size of the differences.
| a 1 a n a n+1 a p b 1 b m b m+1 b q m,n G p,q
|
M1
In[]:=
f[n_]:=MeijerG[{{},Table[1,{n+1}]},{Prepend[Table[0,n+1],-n+1],{}},-π];Table[N[Sum[(I/Pi)^(1-n)*f[n],{n,1,x}],x+6]-NIntegrate[(-1),{n,1,xI},WorkingPrecision->x+6],{x,1,11}]
n
(-1)
1/n
n
Out[]=
{-0.0060083+0.0352107,-0.00100558+3.55×,-0.000093587-0.000070483,-5.7199×-7.6050×,-2.6165×-5.5902×,-8.972×-3.2745×,-1.988×-1.6159×,6.8×-6.888×,3.07×-2.651×,2.75×-9.89×,1.29×-2.50×}
-6
10
-6
10
-6
10
-7
10
-7
10
-9
10
-8
10
-10
10
-9
10
-13
10
-11
10
-13
10
-12
10
-14
10
-14
10
-15
10
-15
10
In[]:=
Abs[%]
Out[]=
{0.0357197,0.00100559,0.000117160,9.5159×,6.1722×,3.3952×,1.6281×,6.889×,2.669×,1.027×,2.81×}
-6
10
-7
10
-8
10
-9
10
-11
10
-12
10
-13
10
-15
10
M2