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Discussion Thread: New Sum to get the MRB Constant   [#43546] / Theoretical
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The Question by Marvin Ray Burns :
2011-03-27 at 21:55GMT

Let
x1=25.65665403510586285599072933607445153794770546058072048626118195...

Then

\( \sum _{n=1}^{\infty } (-1)^n \left(n^{\frac{1}{n}}-1\right)= \sum _{n=1}^{\infty } (-1)^n \left(n^{\frac{x1}{n}}-1\right) = MRB. \)

A Response by foull :
2011-03-29 at 05:02GMT

What is definition of MRB ? (Is it Marvin Ray Burns ?)
A Response by Marvin Ray Burns :
2011-03-30 at 00:42GMT

The name MRB constant was first used by Simon Plouffe in 1999 at http://pi.lacim.uqam.ca/fra/table_fr.html to designate a certain constant I described to Steven Finch.
 \(MRB\) is a symbol for the MRB constant. Wolfram Alpha was the first to use it in 2010 at
http://www2.wolframalpha.com/input/?i=MRB+constant&t=ietb01.
A precise definition is found in MathWorld at http://mathworld.wolfram.com/MRBConstant.html, there recognize the LHS of the equation I wrote in my message, where I pointed out that "1" can be replaced by "x1" and still have the formula yield \(MRB\).
A Response by MRB :
2018-06-19 at 15:15GMT

So, let

x1=25.65665403510586285599072933607445153794770546058072048626118194900973217186212880099440071247379896...

and

x2=25.65248234360905618039719248115221022016494628424517244329671894861415553029321073438487828134403347...

then

 
\[\sum _{n=1}^{\infty } (-1)^n \left(n^{\text{x1}/n}-1\right)=MRB,\]




 
\[\sum _{n=1}^{\infty } (-1)^n \left(n^{\text{x2}/n}-1\right)=1,\]


and

\[\sum _{n=1}^{\infty } (-1)^n \left(n^{\frac{\text{x2}-\text{x1}}{n}}-1\right)+\sum _{n=1}^{\infty } (-1)^n \left(n^{\frac{\text{x1}-\text{x2}}{n}}-1\right)\simeq 0\]



 I'm here to talk, so if you have any questions just ask!
A Response by Peter :
2018-06-19 at 15:33GMT

Where/why is that useful? Seriously, seems like a waste of time.
A Response by MRB :
2018-06-19 at 15:45GMT

The MRB constant is a key fundamental constant used  for polylogarithm, L-series, and zeta variants:

 pp. 28 and 29 at

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.695.5959&rep=rep1&type=pdf .

Also found at

http://marvinrayburns.com/UniversalTOC25.pdf .



  I've got a lot more about the constant and it's place in math! Please keep up the conversation! 
A Response by MRB :
2018-06-19 at 21:44GMT

So, let

x1=25.65665403510586285599072933607445153794770546058072048626118194900973217186212880099440071247379896...,

 

x2=25.65248234360905618039719248115221022016494628424517244329671894861415553029321073438487828134403347...,



x3 =25.65761722706287108794710554508433209225296965507274176587991048009809769217563540539803575569277386...,

and

x4=25.66273322482106030200256088916637844238842909746268912899596894805459817617989507564650951482436947...



Then

 
\[\sum _{n=1}^{\infty} (-1)^n \left(n^{\text{x1}/n}-1\right)=MRB,\]




 
\[\sum _{n=1}^{\infty} (-1)^n \left(n^{\text{x2}/n}-1\right)=1,\]



\[\sum _{n=1}^{\infty} (-1)^n \left(n^{\text{x3}/n}-1\right)=0,\]


and

\[\sum _{n=1}^{\infty } (-1)^n \left(n^{\text{x4}/n}-1\right)=-1.\]



  ...All within a domain of 25.652<x<25.663 for a range of -1<f(x)<1. I wonder why the function is so sensitive to x and if we can use that to calculate digits of the MRB constant faster? I think the only hope for that would be if we found an useful x that was known to be algebraic -- other than x=1 (which we know gives MRB). A rational x in that neighborhood of 25 might even be better!


 There's more to come!
A Response by MRB :
2018-06-20 at 02:14GMT

\(\sum _{n=1}^{\infty } (-1)^n \left(n^{x/n}-1\right)\) equaled MRB for x=1 and x= 25.6566540 so the sum is not a linear function of x.

Doing a little searching, shows that
 

\[\sum _{n=1}^{\infty } (-1)^n \left(n^{x/n}-1\right)\]


has a maximum of 199.64867358621564318... at x = 22.876644538946901710...

The plot of the sums look something like this where the x's run from 22 to 26:


[xvals1.jpg]

Here is a more complete plot from x=0 to x=26 in 2600 steps of 0.01 increment size:


[xvals2.jpg]
A Response by MRB :
2018-06-20 at 21:27GMT

A good look at that second plot above shows that x1 does not speed up convergence of the MRB constant, because the smallest change in x1 produces a great error in f(x1) i.e. the MRB constant.

I still say this is more like research than a waste of time, however, because if the shape of the plot had been the other way around i.e. steep for x=1 and flat for x=x1, then x1 would have produced a faster convergence! 
A Response by The Rock :
2018-06-22 at 16:26GMT

Marvin, when you and Simon Plouffe get to the Heavenly Gates,
St-Pierre will let you both in no questions asked, just in
case you guys wanna discuss your "numbers" with him....:)
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