Rational Padé Replacement of the MRB Residue Sine Factor

This notebook tests how much of the remarkable MRB residue approximation survives after replacing the trigonometric factor
sin(π/m)
by a very low-order rational Padé approximation.
The residue-magnitude family is
A(m)
2
π
-1/m
m
sin
π
m
.
Using the Padé $[3/2]$ approximation
sinz≈
z-
7
60
3
z
1+
1
20
2
z
,
the
m8
term becomes the completely trig-free expression
3840-7
2
π
12
3/8
2
(1280+
2
π
)
,
which differs from the MRB constant by about
-1.3343
-7
10
.
That approximation is too crude at
m8
to preserve the exceptional
-8
10
agreement of the exact residue term. However, the situation changes dramatically at the second greedy index,
m
2
105719004.
Here
π/
m
2
is extremely small, so the same Padé $[3/2]$ approximation to sine is extraordinarily accurate. Keeping the exact, trig-free first term
A(8)
2-
2
3/8
2
π
and replacing only the second sine by its Padé approximation gives
A(8)-
A
rat
(105719004)-
C
MRB
2.58394457818020
-17
10
.
Thus even the simple Padé $[3/2]$ rational approximation is sufficient to preserve the roughly
-17
10
accuracy of the second greedy MRB approximation. The experiment suggests that rational approximations to the trigonometric factor become increasingly effective as the greedy indices grow, because the argument
π/m
rapidly approaches zero. A third index gives even an even better approximation.
In[]:=
ClearAll["Global`*"];​​​​(*High-precisionMRBconstant*)​​mrbN=0.187859642462067120248517934054273230055903094900138786`58;​​​​(*Pade[3/2]approximationtoSin[z]*)​​sinp[z_]=PadeApproximant[Sin[z],{z,0,{3,2}}];​​​​(*Residue-familyapproximationobtainedfromthatPadeform*)​​ar[x_]=(2/Pi)x^(-1/x)sinp[Pi/x];​​​​(*Greedyindices*)​​m2=105719004;​​m3=77401040908104410;​​​​(*Pade[3/2]approximationatm=8*)​​pade8=(3840-7Pi^2)/(122^(3/8)(1280+Pi^2));​​​​(*Exact,trig-freeA(8)*)​​exact8=Sqrt[2-Sqrt[2]]/(2^(3/8)Pi);​​​​(*Firstrow*)​​middle8=HoldForm[q-CMRB]/.q->pade8;​​​​(*Secondrow*)​​middle2=HoldForm[-((2(60m2^2-7Pi^2)m2^-(1/m2+1))/(3(20m2^2+Pi^2)))-CMRB+Sqrt[2-Sqrt[2]]/(2^(3/8)Pi)];​​​​(*Thirdrow*)​​middle3=HoldForm[-((2(60m2^2-7Pi^2)m2^-(1/m2+1))/(3(20m2^2+Pi^2)))-((2(60m3^2-7Pi^2)m3^-(1/m3+1))/(3(20m3^2+Pi^2)))-CMRB+Sqrt[2-Sqrt[2]]/(2^(3/8)Pi)];​​​​TableForm[{{TraditionalForm[sinp[z]],TraditionalForm[middle8],ScientificForm[N[pade8-mrbN,30],15]},{TraditionalForm[sinp[z]],TraditionalForm[middle2],ScientificForm[N[-((2(60m2^2-7Pi^2)m2^-(1/m2+1))/(3(20m2^2+Pi^2)))-mrbN+exact8,40],15]},{TraditionalForm[sinp[z]],TraditionalForm[middle3],ScientificForm[N[-((2(60m2^2-7Pi^2)m2^-(1/m2+1))/(3(20m2^2+Pi^2)))-((2(60m3^2-7Pi^2)m3^-(1/m3+1))/(3(20m3^2+Pi^2)))-mrbN+exact8,55],15]}},TableHeadings->{{"m = 8","m2 = 105719004","m3 = 77401040908104410"},{"Padé formula for Sin[z]","Formula","Approximation - MRB"}}]
Out[]//TableForm=
Padé formula for Sin[z]
Formula
Approximation - MRB
m = 8
z-
7
3
z
60
2
z
20
+1
3840-7
2
π
12
3/8
2
(1280+
2
π
)
-CMRB
-1.33431249843590×
-7
10
m2 = 105719004
z-
7
3
z
60
2
z
20
+1
-
2(60
2
m2
-7
2
π
)
-
1
m2
+1
m2
3(20
2
m2
+
2
π
)
-CMRB+
2-
2
3/8
2
π
2.58394457818020×
-17
10
m3 = 77401040908104410
z-
7
3
z
60
2
z
20
+1
-
2(60
2
m2
-7
2
π
)
-
1
m2
+1
m2
3(20
2
m2
+
2
π
)
-
2(60
2
m3
-7
2
π
)
-
1
m3
+1
m3
3(20
2
m3
+
2
π
)
-CMRB+
2-
2
3/8
2
π
8.04970519014935×
-35
10