Marvin Ray Burns

 I've been using Maple since 1997 or so.

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These are Posts that have been published by Marvin Ray Burns

As some of you know, I'm hoping to, some day, find a closed form expression for the MRB constant.

 Here is my latest little nugget.

Let x=MRB constant.

(1-604*x)/(28+209*x) = log(x) with an error< ...

 Here is a example of a cesaro sum used when a series fails to converge in the usual sense of a sum.

sum((-1)^n*(n^(1/n)-a), n = 1 .. infinity) has a cesaro sum of 1/2*(a+2MRB constant-1).


We are given that                            S=  sum((-1)^n*(n^(1/n)-a), n = 1 .. infinity) .

Expanding the infinite series we get         S=(a-1)+(2^(1/2)-a)+(a-3^(1/3))+(4^(1/4)-a)+... .

Collecting the a's and the surds we see that S=a-a+a-a+...+(-1+2^(1/2)-3^(1/3)+4^(1/4)...) .

By Grandi's series  we know that             S=1/2*a+(-1+2^(1/2)-3^(1/3)+4^(1/4)...) .

Collecting the infinite series we get        S=1/2*a+ sum((-1)^n*(n^(1/n)), n = 1 .. infinity).

Which can be shown to be                     S=1/2*a+ sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity)-1/2 .

Thus, by factoring out the 1/2, we get           S=1/2*(a+2 sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity)-1) .

Therefore,                                   S=1/2*(a+2MRB constant-1) .


In order to give symbolic results for that familly of sums Maple shouldmake this identify an integral part of maple in future versions!

Marvin Ray Burns

Original investigator of the MRB constant.


The MRB constant is defined at

On about Dec 31, 1998 I computed 1 digit of the MRB constant with my TI-92's, by adding 1-sqrt(2)+3^(1/3)-4^(1/4) as far as I could. That first digit by the way is just 0.

On Jan 11, 1999 I computed 3 digits of the MRB constant with the Inverse Symbolic Calculator.

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