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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!

 Have to solve and ODE in the domain of [-infnity +infinity ] via specific analytical method but due to some restrictions it could not be solved. In order to solve it, I have separated the domain into [-infinity 0 ] and [0 infinity]. So, I have to add some boundary values at x=0 to the problem. Assuming the solution of the mentioned ODE in  [-infinity 0 ] is g(x) and in [0 infinity]  is f(x), I added the boundary values of f(0)=g(0)=a and f ' (0)=b and obtained f(x...

 

 

The MRB constant is evaluated by

MRB constant T

January 13 2012 by Marvin Ray Burns 465 Maple

0

2

constants


Let c=MRB constant -1/2

MRB constant S part 2

January 08 2012 by Marvin Ray Burns 465 Maple

0

0

constants

 

 

 

``

> 

restart; Digits := 64

``

 

````Define s as the following function involving a divergent series.

> 

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

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

 (1)

``

``

 

 

``The upper limit point of the partial sums, of s is very slowly convergent.

> 

evalf(s(100))

.211329543346941069485035868216520490712148674852018130412747187

 (2)
> 

evalf(s(1000))

.191323989712141370638688981469071803275457219110707245455878532

 (3)
> 

evalf(s(10000))

.188320351076950504638897789942367214051161086517598649780487746

 (4)

 

``

Let mrb be tthe upper limit point of s as x goes to infinity.

``

> 

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

.1878596424620671202485179340542732300559030949001387861720046841

 (5)

``

``

``

 

Define f as the following function involving the divergnet series sum((-1)^n*(n^(a/n)-a), n = 1 .. infinity).NULL

``

``

> 

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

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

 (6)

``

``

 

 

``Let c be the value for a in the neighborhood of 26 such that f(a)=mrb.

c := fsolve(eval(f(x)) = mrb, x = 26)

25.71864739101744668471488151161460875040712539231550975094037406

 (7)

``

``

 

``The average of the upper and lower limit points of the partil sums of f converges much faster than the  upper limit point of the partial sums of s.

> 

evalf((sum((-1)^n*(n^(c/n)-c), n = 1 .. 100)+sum((-1)^n*(n^(c/n)-c), n = 1 .. 101))*(1/2))

.195238896203546569611605945649919224928195587923897718988014700

 (8)
> 

evalf((sum((-1)^n*(n^(c/n)-c), n = 1 .. 1000)+sum((-1)^n*(n^(c/n)-c), n = 1 .. 1001))*(1/2))

.187904922391719396683391551158554482265830937732923110694243700

 (9)
> 

evalf((sum((-1)^n*(n^(c/n)-c), n = 1 .. 10000)+sum((-1)^n*(n^(c/n)-c), n = 1 .. 10001))*(1/2))

.187860182910509428926222275077446745338505139578191116998518780

 (10)

 

Download Jan72012.mw

MRB constant S

January 02 2012 by Marvin Ray Burns 465 Maple

0

15

constants

 


NULL

NULL

Let f(c)= sum((-1)^n*(n^(c/n)-c), n = 1 .. infinity)

NULL

NULL

Then f(1) = the MRB constant:

evalf(eval(sum((-1)^n*(n^(c/n)-c), n = 1 .. infinity), c = 1)) = .1878596425NULL

NULL

NULL

What if we change the value of c and use Levin's u-transform to compute the values for the analytic extension of the sum?

Then can we find values for c such that f(c)=c?

 

evalf(eval(sum((-1)^n*(n^(c/n)-c), n = 1 .. infinity), c = -1.351776595077954)) = -1.351776595 

evalf(eval(sum((-1)^n*(n^(c/n)-c), n = 1 .. infinity), c = 7.020400867228059)) = 7.020400867

evalf(eval(sum((-1)^n*(n^(c/n)-c), n = 1 .. infinity), c = 25.58774196597964)) = 25.58880851

``

As an alalytic extension of the sum is there another value for c such that f(c) = the MRB constant? I haven't found one.

NULL

 

 

``


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MRB constant R

January 01 2012 by Marvin Ray Burns 465 Maple

0

0

constants summand

 

``

MRB constant P

September 04 2011 by Marvin Ray Burns 465 Maple

0

2

constants

 

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

 

Concerning the following divergent and convergent series, we see that

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

When can we expect Maple 15.02 to appear, to correct that major error of matrix multiplication and the plotting problem with the classic interface in particular?

     Now a new set of fundamental physical constants has been released, as of 2011 June, making the values embedded in Maple's package Scientific Constants from the preceding millennium a further step obsolescent.  I understand, however, that the values of mathematical constants pi and exp(1) are still current.

fourier coefficients

August 11 2011 by AliKhan 5 Maple 14

0

0

constants

HI,

 

I am trying to solve for fourier coefficients but can't figure out the error in the code. Need help to debug the code.

Also if possible how to find the harmonics in the curve defined in the code.

following is the website from where i got this code

http://www.mapleprimes.com/questions/121551-Fourier-Serie-And-Discrete-Fourier-Transform

fouriercoefficient_h.mw

The problem

Back in 1996 I was working for the Symbolic Computation Group at the University of Waterloo, developing algorithms and code...

 

Here is the progress made in the investigation of what I call the convergents constants:
https://oeis.org/wiki/Table_of_convergents_constants

I wonder if anyone would be interested in adding anything to it. I would like to see the convergents constants studied some in Maple to compare with my Mathematica results; my investigation is in dire need of some proof other than my...

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;

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...

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