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

If you use all the convergents of the simple continued fraction of the MRB constant as the terms of a generalized continued fraction, then likewise use the new convergents in another generalized continued fraction, and so on... you arrive at 0.5557531....  For more on this process see https://oeis.org/wiki/Convergents_constant .


Let c=MRB constant -1/2

 

 

 

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restart; Digits := 64

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````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)

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``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)

 

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

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

.1878596424620671202485179340542732300559030949001387861720046841

(5)

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

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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)

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``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)

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``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

 


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

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Then f(1) = the MRB constant:

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

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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

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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.

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Download jan022012.mw

 

 

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