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The MRB constant Z will probably have several parts.

The following example is from the Maple help pages
> with(GraphTheory);
> with(SpecialGraphs);
> H := HypercubeGraph(3);
DrawGraph(H)
 
 


What I would like to do in the MRB constant z,  MRB constant z part2, and etc. is to draw a series of graphs that show the some of the geometry of the MRB constant.

See http://math-blog.com/2010/11/21/the-geometry-of-the-mrb-constant/. I would like to draw a tesseract of 4 units^4, a penteract of 5 units^5, etc and take an edge from each and line the edges up as in Diagram 3:

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As usual I'm asking for your help.

``

``

 

Download May262012.mw

 

This post can be downloaded here:  Download May202012.mw

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!

 

 

The MRB constant is evaluated by

 

with(numtheory):

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)

What are the quotients  ot the  continued fration of the sum of f(1)+f(2)+f(3)+f(4)+...

Here are the  quotients  of some partial sums.

``

cfrac(evalf(sum(f(x), x = 1 .. 2)), 'quotients')

[0, 2, 1, 1, 1, 21, 10, 4, 1, 4, 8, `...`]

(2)

cfrac(evalf(sum(f(x), x = 1 .. 3)), 'quotients')

[0, 6, 1, 2, 3, 1, 1, 2, 3, 3, 24, `...`]

(3)

cfrac(evalf(sum(f(x), x = 1 .. 4)), 'quotients')

[0, 2, 1, 2, 1, 4, 2, 1, 3, 1, 1, `...`]

(4)

cfrac(evalf(sum(f(x), x = 1 .. 5)), 'quotients')

[0, 5, 1, 99, 1, 1, 1, 6, 1, 3, 1, `...`]

(5)

cfrac(evalf(sum(f(x), x = 1 .. 6)), 'quotients')

[0, 2, 1, 6, 1, 2, 1, 2, 2, 1, 1, `...`]

(6)

cfrac(evalf(sum(f(x), x = 1 .. 7)), 'quotients')

[0, 5, 1, 1, 142, 1, 1, 1, 1, 19, 1, `...`]

(7)

cfrac(evalf(sum(f(x), x = 1 .. 8)), 'quotients')

[0, 2, 1, 47, 1, 1, 1, 1, 27, 4, 1, `...`]

(8)

cfrac(evalf(sum(f(x), x = 1 .. 9)), 'quotients')

[0, 5, 5, 3, 1, 7, 1, 1, 1, 2, 1, `...`]

(9)

cfrac(evalf(sum(f(x), x = 1 .. 100)), 'quotients')

[0, 3, 1, 1, 1, 11, 2, 2, 1, 1, 4, `...`]

(10)

cfrac(evalf(sum(f(x), x = 1 .. 200)), 'quotients')

[0, 3, 1, 2, 1, 1, 1, 11, 3, 4, 6, `...`]

(11)

cfrac(evalf(sum(f(x), x = 1 .. 400)), 'quotients')

[0, 3, 1, 3, 3, 3, 1, 18, 1, 2, 1, `...`]

(12)

cfrac(evalf(sum(f(x), x = 1 .. 800)), 'quotients')

[0, 3, 1, 3, 1, 4, 16, 14, 3, 23, 2, `...`]

(13)

cfrac(evalf(sum(f(x), x = 1 .. 1600)), 'quotients')

[0, 3, 1, 4, 7, 4, 436, 1, 1, 1, 2, `...`]

(14)

``

Here are the quotients of the  continued fration  of the sum. 

cfrac(evalf(sum(f(x), x = 1 .. infinity)), 'quotients')

[0, 3, 1, 4, 1, 1, 1, 1, 1, 9, 1, `...`]

(15)

With the exception of the leading 0, that is close to the integer squence of pi.

``evalf((65241/65251)*Pi)

3.141111191

(16)

The exponents of 2 that sum the numerator and denominator, in the following way, of that multiple of pi give rise to the integer sequences {0,1,2,3,8,16},numbers such that floor[a(n)^2 / 7] is a square, and {0,2,3,4,8,16},{0,3} union powers of 2.

evalf((2^16-2^8-2^5-2^2-2-2^0)*Pi/(2^16-2^8-2^4-2^3-2^2-2^0))

3.141111191

(17)

We can do the same thing for the first 20 quotients giving rise to the integer sequences {0,1,2,5,6,8,10,13,17,19,22,23,24,28,31} and {0,4,6,9,12, 14,15,16,18,22, 23,24,28,31}. What can be said of these sequences?

cfrac(evalf(sum(f(x), x = 1 .. infinity), 20), 20, 'quotients')``

[0, 3, 1, 4, 1, 1, 1, 1, 1, 9, 1, 3, 1, 2, 1, 1, 1, 5, 1, 3, 11, `...`]

(18)

evalf((1849023129/1849306543)*Pi, 20)

3.1411111913121115131

(19)

````

evalf((2^31-2^28-2^24-2^23-2^22-2^19-2^17-2^13-2^10-2^8-2^6-2^5-2^2-2-2^0)*Pi/(2^31-2^28-2^24-2^23-2^22-2^18-2^16-2^15-2^14-2^12-2^9-2^6-2^4-2^0), 20)

3.1411111913121115131

(20)

``


 

MRB Constant V

March 23 2012 by Marvin Ray Burns 465 Maple

The*MRB*constant = sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity) and sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity) = sum((-1)^n*(n^(1/n)-1), n = 2 .. infinity)

But what can we say about

 (∏)(-1)^(n)*(n^(1/(n))-1)?

``

``

Maple does not evaluate it:

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

product: Cannot show that (-1)^n*(n^(1/n)-1) has no zeros on [2,infinity] product((-1)^n*(n^(1/n)-1), n = 2 .. infinity)

(1)

And perhaps it should not because of the alternating sign;

evalf(product((-1)^n*(n^(1/n)-1), n = 2 .. 10^2))

-0.3908773173e-101

(2)

evalf(product((-1)^n*(n^(1/n)-1), n = 2 .. 10^3))

-0.7676360791e-1799

(3)

evalf(product((-1)^n*(n^(1/n)-1), n = 2 .. 10^3+1))

0.5316437097e-1801

(4)

``

 

Download 3232012.mw

If you use all the convergents of the simple continued fraction of the MRB constant as the terms of a generalized continued fraction, then likewise use the new convergents in another generalized continued fraction, and so on... you arrive at 0.5557531....  For more on this process see https://oeis.org/wiki/Convergents_constant .

Hi,

I've figure out several ways to accomplish this, using a series of commands, but is it possible using only dsolve to solve a first-order ivp involving two constants: the constant of integration and a constant of proportionality. For example, the type that arise using Newton's basic law of heating/cooling:

y'(t) = k*(y(t)-10), y(0)=70, y(1/2)=50

I welcome all ideas, especially if there is something easy I'm missing, thanks! 

Dear All,

 

I have a question related to Modulated Markov Rate Process (MMRP). Please see MMRP and here. For more information see this

In this kind of markov chain it is necessary to have ON and OFF state that separte by a exponential distribution. Also the number of items as mentioned in the figures has a...

Since -1 = i^2 I thought that there could be some meaning behind "alternating" series that instead of beginning with (-1)^n begin with (a+b*i)^n, with real coefficients, for abs(a)<1 and abs(b)<1. I'm not sure but it seems that such series are absolutely convergent, because (a+b*i)^n -> 0+0I as n->infinity, hence the term utterly diminishing series instead of alternating series.

As an example,
Where sum((-1)^n*(n^(1/n)-1),n=1..infinity)= 0.187859642462067120248... ,

Hello!

I'm trying to solve numerically an ODE system with piecewise. And this piecewise is very important for this task.

This system describes behavior of a pulley with friction. There are some constants: m, c, g, mu and J. Values of this constants are not important.

> sys := m*a(t) = piecewise(a(t) < a0, F0*time, a(t) >= a0, 0), v(t) = diff(x(t), t), a(t) = diff(v(t), t);
> m := 5; F0 := 10; a0 := 5;
> initialconditions := x(0) = 0, v(0) = 0;

 

 

 The first few convergents from the continued fraction expansion of the MRB constant are 0,1/5,3/16,31/165,34/181 and 65/346. If you were to use those convergents as terms of a generalized continued fraction, it would represent, approximately,

 

0+1/(0.20+1/(0.1875+1/(0.1878787+1/(0.187845+1/0.187861))))=1.83346...

Regards,

I have a very large equation which has an arctan(x,y). I need to be able to extract the arguments x,y and assign them in some variables.

I have tried the solution given here.

Unfortunately, that solution only works for constants, not equations.

For example, if I use the proc given in there with arctan(10,11) it works. But if I use something like arctan...

 The best definition for the MRB constant that I know of is found at

http://mathworld.wolfram.com/MRBConstant.html.

 

Up until...

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