## 495 Reputation

14 years, 169 days

I've been using Maple since 1997 or so.

## The red seqment in the above mess is not...

The red seqment in the above mess is not in the right place neither should it be along the y-axis.

## The following mess is a two dimentional ...

The following mess is a two dimensional projection of the segment, square, cube, tesseract, penteract and hexeract lined up not quite according to Diagram 3. The red segment should  represent 1-sqrt(2)+3^(1/3)+4^(1/4)-5^(1/5)+6^(1/6), but I'm not sure if I kept track of its possition altogether correctly.Furthermore, it is not necessarily to scale.

Here is the workseet the n-cubes came from. (I tried to erase the yellow and black colors.)

## Digits:=20Let c be the MRB constant.Let...

`Digits:=20`
`Let c be the MRB constant.`
`Let V be the figure eight hyperbolic volume.`
`c := .1878596424620671202485179340542732300559030949001387`
`V := 2.0298832128193072500424051085490405`
`(15*(V+667))/(839*V+2605)  = 2.329452296051860367745`
`evalf(c+Pi-1)              = 2.329452296051860358712`
`Let p be the prime constant.`
`p := .4146825098511116602481096221543077083657742381379`
`(9509-169*p)/(50*(18*p+23)) = 6.196711067333514470392`
`evalf(c+7*(4-Pi))           = 6.19671106733351445101`
` `
`Digits := 53`
`c := .187859642462067120248517934054273230055903094900138786172`
`evalf(17Pi-642614815285663014792765/12074864494484015995117-c) = -1.*10^(-51)`

## I see how to draw the tessera...

I see how to draw the tesseract and penteract in 2 or 3 dimensions. The graph gets pretty crowded for the penteract. I don't see how I can use an individual edge of graphs of the hexeract and so on. I wonder if the default volume of an n-cube graph in Maple is n units. (That would be convenient.)

with(GraphTheory): with(SpecialGraphs): H4 := HypercubeGraph(4): H5 := HypercubeGraph(5):
DrawGraph([H4, H5], style = spring, dimension = 2)
DrawGraph([H4, H5], style = spring, dimension = 3)

## Digits := 23Let c be the MRB constantc...

Digits := 23

Let c be the MRB constant

c := .187859642462067120248517934054273230055903094900139

Let C be Catalan's Constant.

C := .9159655941772190150546035149323841107741493742816721

evalf(-(119/19)*C-115+(479561360045/59753497176)*Pi+(525/76)*Pi^2+(102/19)*Pi*ln(2)+(353/76)*Pi*ln(3)-2*c) = -8.*10^(-21)

Digits:=24

c := .187859642462067120248517934054273230055903094900139

evalf((2/7163)*(-90155*e-29388+37921*e^2)/(e*Pi)-c) = -5.19*10^(-22)

evalf(((1325323718/907073065)*Pi-4)/Pi-c) = -2.6*10^(-22)

## Again let c be the MRB constant an...

Digits:=22

Again let c be the MRB constant and this time let b be the probability that at least two people in a room of 23 share birthday.

b:=.5072972343239854072254172283370325002359718452929878
c:= 0.187859642462067120248517934054273230055903094900139

(6200 b-239)/(2 (1550 b+437))-1 =0.187859642462067125421

(5 (1860 b+127))/(2 (1550 b+437))-2 =0.187859642462067125421

(5 (4960 b+1001))/(2 (1550 b+437))-7 = 0.187859642462067125421

0.187859642462067125421-c = 5.1725*10^(-18)

Also

evalf(23212425209 /3144920904*Pi-23 - c) = 2.*10^(-20)

and

evalf(20561436223 /2375900782*Pi-27 - c) = 3.*10^(-20)

Also

evalf((-216+53*Pi^(1/2)+363*Pi+1526*Pi^(3/2)+1062*Pi^2)/(92*Pi)-69-c) = -3.*10^(-20)

and

evalf(1/890*(115850*Zeta(3)+10010*Zeta(5)-1785*Pi^2-979*Pi^4)-41-c) = -5.*10^(-20)

and

evalf((6/11)*Pi*arccosh(292945824/1726159)^2-58-c) = 6.*10^(-20)

marvinrayburns.com

## Below we have approximations involving t...

Below we have approximations involving the MRB constant. The MRB constant plus a fraction is saved as P while a combination of another constant is saved as Q. We then subtract Q from P and always have a very small result.

Let

Let c be the MRB constant, 0.1878596424620671202485179340542732300559030949001387.

P:= c+7/47

Let G be Graham's Biggest Little Hexagon area.

G:=0.6749814429301047036884958318514002889802977322780266

Q:=(79-940*G)/(3530*G-4032)

p-Q= -3.3803*10^(-18)

P := c+9/46

Let Q be the positive root of 41309050*x^3-2330137

Q:=0.3835118163751105434087

P-Q=5.51007*10^(-17)

P:=c+46/47

Let p be the plastic constant.

p:=1.3247179572447460259609088544780973407344040569017333

Q:=-5*(37*p-2196)/(7000*p-71)

p-Q=6.97*10^(-19)

P:=c+35/48

e:=evalf(exp(1))

Q:= -17(e-5)*(5*e-6)/(-751+215*e+66*e^2)

P-Q= -1.710*10^(-19)

P:=c+40/49

Let r be the rabbit constant

r:=0.7098034428612913146417873994445755970125022057678605

Q:=6*(167*r+1060)/(25*r+7024)

P-Q= -1.20*10^(-19)

P:=c+29/51

Let Pg be plouffe's gamma constant.

Pg:=0.1475836176504332741754010762247405259511345238869178

Q:=-20*(303Pg-40)/(178Pg-151)

P-Q=-1.12336*10^(-17)

P:=c+15/17

Let F be the Fransén-Robinson Constant,

F:=2.8077702420285193652215011865577729323080859209301982

Q:=(20882-2007*F)/(1050 + 4700*F)

P-Q= -1.739*10^(-18)

P:=c+37/52

Let QRS be the QRS constant.

QRS:=0.6054436571967327494789228424472074752208994969563226

Q:=(3970*QRS+83)/(6053*QRS-900)

P-Q=1.7719*10^(-18)

P:=c+43/52

Let f1 be the first Foias constant.

f1:=2.2931662874118610315080282912508058643722572903271212

Q:=(3881-220*f1)/(100*f1+3098)

P-Q= -3.670*10^(-18)

P:=c+16/53

Let Pa be Plouffe's A constant

Pa:=0.1591549430918953357688837633725143620344596457404564

Q:=4*(5800Pa-773)/(6000Pa+271)

P-Q=4.6241*10^(-18)

## This post is continued in the MRB consta...

This post is continued in the MRB constant N Part 3.

## ...

 (1)

Again let b be

 (2)

Then we have the divergent series

 (3)

That is MRB constant times b plus 1

 >
 (4)

We also have the convergent series

 (5)

That is obviously MRB constant times b:

 >
 (6)

That means that we have the following equations.

 (7)

 (8)

 (9)

 (10)

 (11)

 (12)

Finally, as we saw in the original post, we have the following.

 (13)

 (14)

But ia a divergent series whoes Cesàro summation is MRB constant -1/2

 (15)

So -b times (MRB constant -1/2) =1/2:

 (16)

Thus b times (1/2-MRB constant) =1/2:

 (17)

Therefore b*MRB constant=(b-1)*1/2:

 (18)

## > ...

 >

It only requires the distributive law in reverse to solve for a. But Maple adds to the work by applying the distrbutive law twice (once for a and once for (-1)^n) and the "associative law for convergent infinite series."

 (1)

Is it just me, or do you see a convergent infinite series in that denominator?

In other words, is the series that is made from the sum of those two divergent series absolutely convergent?

## For a few x that I have tried a is solve...

For a few x that I have tried, a is solved for in =x  when a=  .

Maple confirms this by the following:

 (1)

 (2)

 (3)

## All Convergent Series?...

I wonder if the Cesàro mean is faster than the partial sums to converge for all convergent series? I also wonder if it is true that  Levin's u-transform is better than all the Cesàro transforms for computing the MRB constant is it indeed better for summing all convergent series?

About  Levin's u-transform Mathematica (using Nsum[,"AlternatingSigns"]) sums the MRB constant a magnitude of times faster than Maple(using evalf(sum)); Mathematica might use a diffferent method.

## The Chezaro mean seems to converge ...

For this series the Cesàro mean seems to converge faster than the partial sums.

Let C be the MRB constant:

 (1)

Subtract C from the partial sum of the first 10,000 terms:

 (2)

Define g:

 (3)

Subtract C from the Cesàro mean of the first 10,000 terms and you get a number with a much smaller magnitude.

 (4)

## Previously we had the funct...

 (1)

Consider just the sum of f(x) for x from 1 to infinity.

 (2)

It is a less apealing multple of pi:

 (3)

## x^x^x^x...

The MRB constant -1/2 = -.3121...  Here is what the real and imaginary parts of x^x^x^x look like as it goes to a maximum of maqnitude near x=-.335 in a plot and some parametric plots:

Here we see the local maximim magnitude is approximately3.24*10^24 at x= -.335: