Marvin Ray Burns

 I've been using Maple since 1997 or so.

MaplePrimes Activity


These are replies submitted by Marvin Ray Burns

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 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.)

 

with(GraphTheory):

DrawGraph([H1, H2, H3, H4, H5, H6], style = spring, dimension = 2)

 

Download MRB_in_parts.mw

 

 

 

 

 

 

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 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)

Download May26b2012.mw

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)

 

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 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 constant N Part 3.

 

Digits := 20:

c := evalf(sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity))

.18785964246206712025

(1)

Again let b be (1-2 MRB constant)^-1

b := 1/(1-2*c)

1.6018434909982361492

(2)

``

Then we have the divergent series``

````

evalf(sum((-1)^n*(-b*n^(1/n)-b), n = 1 .. infinity))

1.3009217454991180746

(3)

That is MRB constant times b plus 1

 

c*b+1

1.3009217454991180746

(4)

 

NULL

We also have the convergent series

evalf(sum((-1)^n*(b*n^(1/n)-b), n = 1 .. infinity))

.30092174549911807459

(5)

NULL

That is obviously MRB constant times b:``

c*b

.30092174549911807460

(6)

 

NULL

That means that we have the following equations.

 

 

evalf(sum((-1)^n*(-b*n^(1/n)-b), n = 1 .. infinity)-(sum((-1)^n*(b*n^(1/n)-b), n = 1 .. infinity)))

1.0000000000000000000

(7)

NULL

``

evalf(sum((-1)^n*(-b*n^(1/n)-b-b*n^(1/n)+b), n = 1 .. infinity))

1.0000000000000000000

(8)

``

evalf(sum((-1)^n*(-b*n^(1/n)-b*n^(1/n)), n = 1 .. infinity))

1.0000000000000000000

(9)

``

evalf(sum((-1)^n*(-2*b*n^(1/n)), n = 1 .. infinity))

1.0000000000000000000

(10)

evalf(-2*(sum((-1)^n*b*n^(1/n), n = 1 .. infinity)))

1.0000000000000000000

(11)

evalf(sum((-1)^n*b*n^(1/n), n = 1 .. infinity))

-.50000000000000000001

(12)

 

 

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

evalf(sum(-(-1)^n*b*n^(1/n), n = 1 .. infinity))

.50000000000000000001

(13)

``

evalf(-b*(sum((-1)^n*n^(1/n), n = 1 .. infinity)))

.50000000000000000000

(14)

But sum((-1)^n*n^(1/n), n = 1 .. infinity)ia a divergent series whoes Cesàro summation is MRB constant -1/2

evalf(sum((-1)^n*n^(1/n), n = 1 .. infinity))-c+1/2

0.

(15)

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

-b*(c-1/2)

.50000000000000000000

(16)

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

b*(1/2-c)

.50000000000000000000

(17)

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

b*c-(b-1)*(1/2)

0.

(18)

``

``

``

``

 

Download May122012d.mw

 

solve(sum((-1)^n*(a*n^(1/n)-a), n = 1 .. infinity) = x, a)

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."

solve(sum((-1)^n*(a*n^(1/n)-a), n = 1 .. infinity) = x, a)

x/(sum((-1)^n*n^(1/n), n = 1 .. infinity)-(sum((-1)^n, n = 1 .. infinity)))

(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? 

Download cincodemayo2012.mw

 

 

 

 

For a few x that I have tried, a is solved for in sum((-1)^n*(a*n^(1/n)-a), n = 1 .. infinity)=x  when a= x/MRB constant .

Maple confirms this by the following:

 



solve(sum((-1)^n*(a*n^(1/n)-a), n = 1 .. infinity) = x, a)

x/(sum((-1)^n*n^(1/n), n = 1 .. infinity)-(sum((-1)^n, n = 1 .. infinity)))

(1)

evalf(sum((-1)^n*n^(1/n), n = 1 .. infinity)-(sum((-1)^n, n = 1 .. infinity)))

.1878596425

(2)

evalf(solve(sum((-1)^n*(a*n^(1/n)-a), n = 1 .. infinity) = x, a))

5.323123086*x

(3)

``



Download april272012.mw

@icegood

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.

See http://www.mapleprimes.com/posts/95240-Computing-Digits-Of-The-MRB-Constant

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

``

Let C be the MRB constant:

C := evalf(sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity))

.1878596425

(1)

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

evalf(sum((-1)^n*(n^(1/n)-1), n = 1 .. 10000)-C)

0.460663e-3

(2)

Define g:

g := proc (x) options operator, arrow; sum((-1)^n*(n^(1/n)-1), n = 1 .. x) end proc

proc (x) options operator, arrow; sum((-1)^n*(n^(1/n)-1), n = 1 .. x) end proc

(3)

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

NULL

(1/10000)*evalf(sum(g(x), x = 1 .. 10000))-C

-0.76848e-5

(4)

NULL

 

``

 

Download mran.mw


Previously we had the function,

f := proc (x) options operator, arrow; sum((-1)^n*(n^(1/n)-1), n = x .. infinity) end proc

proc (x) options operator, arrow; sum((-1)^n*(n^(1/n)-1), n = x .. infinity) end proc

(1)

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

``

NULL

evalf(sum(f(x), x = 1 .. infinity), 20)

.26163115389725031288

(2)

It is a less apealing multple of pi:

evalf((92951968/1116140855)*Pi, 18)

.261631153897250313

(3)

``


Download 4162012.mw

 

 

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:

plot([Im(x^(x^(x^x))), Re(x^(x^(x^x)))], x = -.4 .. -.3326)

 

plot([Im(x^(x^(x^x))), Re(x^(x^(x^x))), x = -.40 .. -.39])

 

 

plot([Im(x^(x^(x^x))), Re(x^(x^(x^x))), x = -.40 .. -.385])

 

 

plot([Im(x^(x^(x^x))), Re(x^(x^(x^x))), x = -.39 .. -.38])

 

 

plot([Im(x^(x^(x^x))), Re(x^(x^(x^x))), x = -.38 .. -.37])

 

 

plot([Im(x^(x^(x^x))), Re(x^(x^(x^x))), x = -.37 .. -.36])

 

 

plot([Im(x^(x^(x^x))), Re(x^(x^(x^x))), x = -.36 .. -.35])

 

plot([Im(x^(x^(x^x))), Re(x^(x^(x^x))), x = -.36 .. -.335])

 

plot([Im(x^(x^(x^x))), Re(x^(x^(x^x))), x = -.35 .. -.3326])

 

``

 

Download 1212012b.mw

 

 

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

plot(abs(x^(x^(x^x))), x = -.4 .. -.3325)

 

plot(abs(x^(x^(x^x))), x = -.34 .. -.3325)

 

plot(abs(x^(x^(x^x))), x = -.336 .. -.334)

 

``

 

Download 1212012c.mw

 

 

 

marvinrayburns.com

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