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Marvin Ray Burns

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19 years, 256 days
Contact Marvin Ray Burns

 I've been using Maple since 1997 or so.

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Some Numeric Qualities Part 7 and moving...

Marvin Ray Burns 550
January 01 2007

We have been using a process called mit to manufacture the various family members of the MRB Constant. Continuing in that mode we will use it again to compute m, or the MRB Constant.

 



> 

 

> 

restart:
> 

mit := proc (it) global git, m; git := proc (n); (-1)^n*(n^(1/n)-it) end proc; m := sum(git(n),n = 1 .. infinity) end proc;

mit := proc (it) global git, m; git := proc (n) (-1)^n*(n^(1/n)-it) end proc; m := sum(git(n), n = 1 .. infinity) end proc

> 

m:=mit(1):

 This is important for understanding the following, the MRB Constant=mit(1)=g1, mit(0)=g0,  mit(2)=g(2), etc

 


Previously when asked:

 i.    What real numbers to their own roots equal mit(1)?

ii.    What real number do we get when we raise mit(1) to the power of itself an infinate number of times?

We discovered:

i. and ii. have the same answer.

> 

r:=fsolve(x^(1/x)=mit(1),x);

r := .4619214402

.

 

Before I show you the following trick,  in one sentance, let's review.

 

 m Is the alternating sums of all integers to their roots; r is one number to its root that equals m.

 

Now we go a step further to the area under the curve of r to all roots. And find out -- that area is nearly rational!

> 

int(r^(x/r),x=1..infinity);

.1123522026

> 

4466/39750.;

.1123522013

> 

 

> 

 

> 

 

> 

 

 

 



Download 565_Jan_1.mw


As we move on, this would be a good time to introduce a process that will give us desired approximations to the MRB Constant. I give this to you with the following caveat. Do to the internal structure of Maple the last 2 to 4 digits may not always be correct. However, you can use this process to compute many thousands of digits of MRB. I hold the present world record of 40,000 digits. With little or no fine tuning of the process, you might be able to beat my record.

e:=100:

restart:

mrb:=proc(e)
local a,b,c,d,s,k,n,m:
if e < 100 then n:=31+e;Digits:=31+e else
n:=131*round(e/100);Digits:=131*round(e/100) end if:
a:=array(0..n-1):
a[0]:=1:
for m from 1 to n-1 do
a[m]:=(m+1)^(1/(m+1)) od:
d:=(3+sqrt(8))^n:
d:=(d+1/d)/2:
b:=-1:
c:=-d:
s:=0:
for k from 0 to n-1 do c:=b-c:
b:=(k+n)*(k-n)*b/((k+1/2)*(k+1)):
s:=s+c*a[k] od:
Digits:=e:
print(evalf(1/2-s/d))
end proc:

for a from 2 to 20 do mrb(a) od:

mrb(100);

mrb(1000);

 

 

 

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My Desire Download 565_Jan 1.mw
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!!!!banengineer!!! 10 min...

Marvin Ray Burns 550
December 31 2006
Are you available right now? I have a few ideas. write to author I've got about 10 min My Desire

Things are slow right now...

Marvin Ray Burns 550
December 31 2006
I too have had problems with Maple 10. I had to use the worksheet mode exclusively, until my license ran out. They ought to fix these things for us with regular internet updates. My Desire

The Title should have read Definite Inte...

Marvin Ray Burns 550
December 24 2006
My Desire

EE!...

Marvin Ray Burns 550
December 19 2006
I never thought Pi was that straight.

Spirit of JacquesC’s Reply and f1's Conv...

Marvin Ray Burns 550
December 19 2006

In the spirit of JacquesC’s reply, since f1:=x->sum((-1)^n*(n^(1/n)-1),n=1..x); is an absolutely converging series. The work of crunching out the digits of f1 qualifies for the following methods of (accelerated) convergence; however, practically speaking, only a few really are applicable. http://numbers.computation.free.fr/Constants/Miscellaneous/seriesacceleration.html More convenient are two other methods that I know of. One is mentioned in Let.doc

Since f1 is a partially alpha-convex series(after about the third or fourth term), the methods in Download 565_4148474.pdfapply but are hard to understand.

On the other hand, since f1 is a reasonably well-behaved function with a(n)->0 as n->infinity, the algorithms found in http://projecteuclid.org/Dienst/UI/1.0/Summarize/euclid.em/1046889587 are easier to use. Attached, also, is a worksheet that demonstates the absolute convergence of f1. Download 565_mrb dec 17-2.mw
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n-Dimensional Cubes

Previously, we read the link QA-1.doc.
Then we looked at the worksheets describing functions f and f1;
we began to investigate how the summing of their terms can produce a value that represents the value of the MRB Constant.
In this worksheet we will use a pair of sequence of terms s and s1;
we will use them to stack n-dimensional cubes on a 2D coordinate plane in a way that also represents the value of the MRB Constant.

> 

restart;
> 

with(plots):

Warning, the name changecoords has been redefined

> 

s:=n->(-1)^n*n^(1/n):s1:=n->(-1)^n*(n^(1/n)-1):alias(p=pointplot):inf:=2^10:

 

 

First we have a stacking of n-dimensional cubes:

the sequence, s, centered at the y-axis, and afterward the series, s, stacked according to QA-1.doc.

> 

p({seq([s(n),n],n=1..inf)});
p({seq([sum(s(u),u=1..n),n],n=1..inf)});

 

 

Next, we have a stacking of n-dimensional cubes:

the sequence, s1, centered at the y-axis, afterward  the series, s1, stacked according to QA-1.doc.

> 

p({seq([s1(n),n],n=1..inf)});
p({seq([sum(s1(u),u=1..n),n],n=1..inf)});

> 

 

As the graph goes up -- i.e. as y goes to infinity -- the plateau is removed and it is replaced with a peak.
We see that the sides of the two stacks are shifted from their position in the graph of s to other positions in s1.
Graphically, this peak makes the series' -- i.e. s1 and accordingly f1 -- absolutely converging.
Since we now have a peak, the top left of the map of the series s is shifted by the absolute value of the MRB Constant.

 

 

 

 

 

Download 565_mrb_dec_17-3.mw

One Member of the Family is an Absolutel...

Marvin Ray Burns 550
December 12 2006

I show in the worksheet, mrbgraphs a, that the constant of concern (the MRB Constant) is indeed a limiting point at Infinity for one member of the family. Since all members have equal partial sums at even intervals, they all converge to the mrb in the following sense.



Last update, Firday Dec 17, 7:17PM

> 

restart;

 

 

 

> 

f1:=x->sum((-1)^n*(n^(1/n)-1),n=1..x);

f1 := proc (x) options operator, arrow; sum((-1)^n*(n^(1/n)-1), n = 1 .. x) end proc

Technically, as x goes to infinity, f1(x) does converge. n^(1/n)->0 as n->infinity. Therefore, the function that describes the local maximums of f:

> 

limit(sum((-1)^n*n^(1/n), n = 1 .. 2*Inf), Inf = infinity)

aslo converges.

Because the function that describes the local maximums of f converges and because the limit(x^1/x,x=infinity)=1,  the function that describes the local minimums of f:

 

> 

limit(sum((-1)^n*n^(1/n), n = 1 .. 2*Inf+1), Inf = infinity)

also converges.

> 

 

> 

 limit(x^1/x,x=ininity);

1

As for the plots I was concerned about in mrbgraphs 1, I can focus solely upon f1 since we know it converges. I will look at the real and imaginary parts and the absolute value of f1.

> 

with(plots):

Warning, the name changecoords has been redefined

> 

Inf:=20:
 P0(Inf):=plot(Re(f1(x)),x=1..Inf,color=green,tickmarks=[21,3],style=point):
 P1(Inf):=plot(Im(f1(x)),x=1..Inf,color=blue,tickmarks=[21,3],style=point):
 P2(Inf):=plot(abs(f1(x)),x=1..Inf,color=red,tickmarks=[21,3],thickness=3):
 display([P0(Inf),P1(Inf),P2(Inf)]);
> 

 

> 

 

From the red graph, it appears that abs(f1) has local minimums (mins) at every odd value of x and the value of a min is getting closer to the value of the next odd value as the value of x grows without bound .

By definition, the red graph, the absolute value of f1 is simply sqrt((-1/2*sin(n*Pi))^2+Re(f1(n))^2); in maple's natural simplification: (I say nothing esle about that.)

> 

ff1:=sqrt((-1/2*sin(n*Pi))^2+Re(f1(n))^2);
> 

 

ff1 := (1/2)*(sin(n*Pi)^2+4*(1/2+(1/2)*Re((-1)^(n+1)+2*(sum((-1)^n*n^(1/n), n = 1 .. n))))^2)^(1/2)

Looking further along the x-axis of  graph we see that those mins head twoard the MRB Constant, here called f .

> 

 

> 

f:=sum((-1)^n*(n^(1/n)-1),n=1..infinity):
Inf:=200:

  mrb:=plot(f,x=120..Inf,color=black):
  P0(Inf):=plot(Re(f1(x)),x=120..Inf,color=green):
  P1(Inf):=plot(Im(f1(x)),x=120..Inf,color=blue):
  P2(Inf):=plot(abs(f1(x)),x=120..Inf,color=red,thickness=2):

#3 graphs zooming in on the MRB:
   display([mrb,P0(Inf),P1(Inf),P2(Inf)]);
   display([mrb,P0(Inf),P1(Inf),P2(Inf)],view=[120..200,evalf(f-1/2^2)..evalf(f+1/2^2)]);
   display([mrb,P0(Inf),P1(Inf),P2(Inf)],view=[180..200,evalf(f-1/2^3)..evalf(f+1/2^3)]);

> 

 

 



Download 565_mrbgraphs_a.mw


acer, you wrote:>I couldn't...

Marvin Ray Burns 550
December 12 2006
acer, you wrote: >I couldn't follow what the summations mean when the index ranges from 1 to sqrt(3). I had >thought that the dummy index n was representing positive integers. What does this mean, >sum((-1)^n*n^(1/n),n = 1 .. 27*6^(1/2)))); It is important to know, there I am just exploring to see what value is represented by the graphs that I intend to research next. You wrote: >But isn't that just the same as the limits of any subsequences of the function coloured >green itself, as x goes to infinity? What do the red and blue curves add, for illustrating >the behaviour of the green curve? Yes you are right in that it is the top of the green graph. However, as I demonstrated in the above table, the top value graphed in green is shared (every other term) by the red and blue. That sharig of values is because, as you correcly wrote, >But that's just because when x is any even positive integer all three of F1, F2, and F3 >simplify immediately to sum((-1)^n*n^(1/n),n = 1 .. infinity); The point of the post is not that it is hard to understand. The point is, my books failed to tell me that there exists rich value in breaking up some non-converging infinite series into two series that do converge. As far as my point is concerned, I only have scratched the surface. I see (and foresee) that in all disciplines, where there exists a model that has non-converging alternating sums, much progress can be made in breaking up the model into series that do indeed converge. I hope this helps, From the study bench of Marvin Ray Burns (317) 371-6571 P.O. Box 19785 Indianapolis, IN 46219

Same with 10.0...

Marvin Ray Burns 550
December 07 2006
13 months or so ago I tried 10.0 on my Dual core VIIV with XP. It had the same symptoms.
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