## 495 Reputation

14 years, 169 days

I've been using Maple since 1997 or so.

## For some -.38<x<-.18 (x^x^x^x) has...

For some -.38 < x < -.18, (x^x^x^x) has a large magnitude. (It has real and imaginary parts far from 0 in either the positive or negative direction.) Then x^(x^x^x^x) is very close to 0.

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## Levin's u-transform...

evalf(sum) uses Levin's u-transform to compute the value for the analytic extension of the sum.

I came accross it in the following discussion:

http://www.mapleprimes.com/posts/35778-MRB-Constant-F

## Thank you Markiyan.The answer to my...

Thank you Markiyan.

The answer to my question,"As an alalytic extension of the sum is there another value for c such that f(c) = the MRB constant?" is yes. (It is 25.71864....)

 (1)

## f Has a maximium...

I see; f has a maximium near (22,202)

## Let f(c)=...

Let f(c)=

I asked, "As an alalytic extension of the sum is there another value for c such that f(c) = the MRB constant?" i.e. 0.187859...

It appears that the slope of the graph of f is positive throughout, never equals 0, and never allows f(c) to equal 0.187859 again.

showing, interestingly, that

=

However, if we let g(c)=

then

=

## not necessarily f(c)=MRB...

Markiyan, It looks like you were showing me where f(c)=c, not necessarily  f(c)=MRB. I wonder if our solution of f(c)=c near c=26 is valid. I mean why would the graph of f have a singularity near 26?

## sum(n^(1/n)-1, n = 1 .. infinity) diverg...

sum(n^(1/n)-1, n = 1 .. infinity) diverges because n^(1/n) - 1 > exp(1/n) - 1 for n >= 3 and

limit((exp(1/n) - 1)/(1/n), n = infinity)=1.

Therefore, sum(exp(1/n)-1, n = 1 .. infinity) diverges and so does sum(n^(1/n)-1, n = 1 .. infinity).

## I believe the previous table simply show...

I believe the previous table simply shows some "maxing out" of the software instead of showing convergence of the limit(sum(n^(1/n)-1, n = 1 .. N), N = infinity).

marvinrayburns.com

## The limit(sum(n^(1/n...

From the following emperical evidence alone it appears that the limit(sum(n^(1/n)-1, n = 1 .. N), N = infinity)

just might converge.

Here is a table of limits

x    limit(sum(n^(1/n)-1,n=1..N),N=10^x)

1      3.1511647323226333

2     11.452637624960385

3     24.818755742523738

4     43.39893188518323

5     67.26154566874983

6     96.42261224741904

7     130.8850394120835

8     170.649287381816

9     215.7154228449

10   266.08345507

11   321.7533848

12   382.725223

13   448.99902

14   520.57333

15   597.45191

16   679.63721

17   767.0216

18   829.7285

19   829.727

20   829.809

21   829.71

22   829.71

23   829.71

24   829.71

25   829.71

26   829.71

27   829.71

28   829.71

The table is from Mathematica using the following code:

Table[{x,NSum[n^(1/n)-1,{n,1,10^x}, Method

->{"EulerMaclaurin", Method->{"NIntegrate", "MaxRecursion"->20, Method->"DoubleExponential"}}]},{x,28}]

## My record high precision computation of ...

My record high precision computation of the MRB constant is now exactly 300,000 digits to the right of the decimal point.

They were computed from Sat 8 Oct 2011 23:50:40 to Sat 5 Nov 2011 19:53:42.

This run was 0.5766 seconds per digit slower than the 299,998 digit computation even though it used 16GB physical DDR3 RAM on the same machine. The working precision and accuracy goal combination were maximized for exactly 300,000 digits, and the result was automatically saved as a file instead of just being displayed on the front end. Windows reserved a total of 63 GB of working memory of which at 52 GB were recorded being used.

## The partial sums of g and s do have a fe...

We did let

g:=x-> ;

f:=x->;

and

s:=x->;

[Comment being edited.]

marvinrayburns.com

## beautiful how the maximum...

We did let s:=x->.

I think it is beautiful how the maximums of the partial sums of s(x) at n= 5, 7, 9, and 11 together with the sum are so separated! I think I will have more to say about the partial sums of s, f and g and how they compare with each other this weekend.

plot([sum((-1)^n*(n^(x/n)-x), n = 1 .. 5), sum((-1)^n*(n^(x/n)-x), n = 1 .. 7), sum((-1)^n*(n^(x/n)-x), n = 1 .. 9), sum((-1)^n*(n^(x/n)-x), n = 1 .. 11), sum((-1)^n*(n^(x/n)-x), n = 1 .. infinity)], x = 1 .. 26, y = -150 .. 250)

## A General Form...

My goal here was to identify the best general form to categorize the MRB constant under. Of course it falls under g(x,d)=sum((-1)^n* (n^(x/n)-d)  ,n=1..infinity)  where g(1,1)=MRB constant. However, since I now have the form of Laurent series,  h(z,c,a(n)) =sum((z^n-c)*a(n) ,n=1..infinity), I got thinking that perhaps the MRB constant can best be categorized under the unnamed function f(z,x,d)= sum(z^n*(n^(x/n)-d) ,n=1..infinity); then  f(-1,1,1)=MRB constant.

Any thoughts are appreciated.

## Normality in Base 10...

Here you can see for yourself how normal the first 5,000 digits of the MRB constant are. However, that doesn't mean that the constant is either normal or irrational. It is fun, however, to see its digits being analyzed so quickly!