Marvin Ray Burns

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I've been using Maple since 1997 or so.

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These are Posts that have been published by Marvin Ray Burns

MRB Constant W...

Posted: Marvin Ray Burns 550

April 14 2012

0 1

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

(1)

What are the quotients ot the continued fration of the sum of

Here are the quotients of some partial sums.

$cfrac(evalf(sum(f(x), x = 1 .. 2)), 'quotients')$

(2)

$cfrac(evalf(sum(f(x), x = 1 .. 3)), 'quotients')$

(3)

$cfrac(evalf(sum(f(x), x = 1 .. 4)), 'quotients')$

(4)

$cfrac(evalf(sum(f(x), x = 1 .. 5)), 'quotients')$

(5)

$cfrac(evalf(sum(f(x), x = 1 .. 6)), 'quotients')$

(6)

$cfrac(evalf(sum(f(x), x = 1 .. 7)), 'quotients')$

(7)

$cfrac(evalf(sum(f(x), x = 1 .. 8)), 'quotients')$

(8)

$cfrac(evalf(sum(f(x), x = 1 .. 9)), 'quotients')$

(9)

$cfrac(evalf(sum(f(x), x = 1 .. 100)), 'quotients')$

(10)

$cfrac(evalf(sum(f(x), x = 1 .. 200)), 'quotients')$

(11)

$cfrac(evalf(sum(f(x), x = 1 .. 400)), 'quotients')$

(12)

$cfrac(evalf(sum(f(x), x = 1 .. 800)), 'quotients')$

(13)

$cfrac(evalf(sum(f(x), x = 1 .. 1600)), 'quotients')$

(14)

Here are the quotients of the continued fration of the sum.

$cfrac(evalf(sum(f(x), x = 1 .. infinity)), 'quotients')$

(15)

With the exception of the leading 0, that is close to the integer squence of pi.

(16)

The exponents of 2 that sum the numerator and denominator, in the following way, of that multiple of pi give rise to the integer sequences {0,1,2,3,8,16},numbers such that floor[a(n)^2 / 7] is a square, and {0,2,3,4,8,16},{0,3} union powers of 2.

evalf((2^16-2^8-2^5-2^2-2-2^0)*Pi/(2^16-2^8-2^4-2^3-2^2-2^0))

(17)

We can do the same thing for the first 20 quotients giving rise to the integer sequences {0,1,2,5,6,8,10,13,17,19,22,23,24,28,31} and {0,4,6,9,12, 14,15,16,18,22, 23,24,28,31}. What can be said of these sequences?

$cfrac(evalf(sum(f(x), x = 1 .. infinity), 20), 20, 'quotients')$

(18)

(19)

evalf((2^31-2^28-2^24-2^23-2^22-2^19-2^17-2^13-2^10-2^8-2^6-2^5-2^2-2-2^0)*Pi/(2^31-2^28-2^24-2^23-2^22-2^18-2^16-2^15-2^14-2^12-2^9-2^6-2^4-2^0), 20)

(20)

MRB Constant V...

Posted: Marvin Ray Burns 550 Product: Maple

March 23 2012

0 2

The*MRB*constant = sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity) and sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity) = sum((-1)^n*(n^(1/n)-1), n = 2 .. infinity)

But what can we say about

(∏)(-1)^(n)*(n^(1/(n))-1)?

Maple does not evaluate it:

evalf(product((-1)^n*(n^(1/n)-1), n = 2 .. infinity))

product: Cannot show that (-1)^n*(n^(1/n)-1) has no zeros on [2,infinity] product((-1)^n*(n^(1/n)-1), n = 2 .. infinity)

(1)

And perhaps it should not because of the alternating sign;

evalf(product((-1)^n*(n^(1/n)-1), n = 2 .. 10^2))

(2)

evalf(product((-1)^n*(n^(1/n)-1), n = 2 .. 10^3))

(3)

evalf(product((-1)^n*(n^(1/n)-1), n = 2 .. 10^3+1))

(4)

Download 3232012.mw

MRB constant U...

Posted: Marvin Ray Burns 550 Product: Maple

March 18 2012

0 0

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 .