It appears that RICHARD J. MATHAR has written a paper on this subject  of Int(f(x)) where f(x) = (-1)^x * x^(1/x) = exp(I*Pi*x + ln(x)/x).  http://arxiv.org/PS_cache/arxiv/pdf/0912/0912.3844v2.pdf.

He develops three new acceleration methods for computing the integral.

We opened this blog with a divergent series (summed over the integers) whose sum is the MRB constant .

Now consider this divergent series (summed over the gaussian integers) evaluated with the same formula, (1-a)*(1/2)+sum((-1)^n*(n^(1/n)-a).

(The old "a" is replceced by "a +b*I.")

f2 := seq(seq(1/2*(1-a-I*b)+sum((-1)^n*(n^(1/n)-a-I*b), n = 1 .. infinity), a = 1 .. 9), b = 1 .. 9); evalf(f2)

Again, in each case the sum is the MRB constant.

Download complexmrb.mw

marvinrayburns.com

Let M = MRB constant = sum((-1)^n*(n^(1/n)-1),n =1 .. infinity), and let MRB2 = 1-2M = 0.6242807150758657595... :

Let M4 = sum((-1)^m*(m^(1/m)+M), m = 1 .. infinity) = 1/2*(M-1) and M5 = sum((-1)^m*(m^(1/m)-M), m = 1 .. infinity) = 1/2*(3*M-1).

Let M[0] = M and M[p+1] = sum((-1)^n*(n^(1/n)+M[p]),n =1 .. infinity),

then M4+M5 = N = 3*Limit(M[p],p=infinity) = -MRB2 and because MRB2 = 1-2*M, M4+M5 = 2*M-1.

Proof that M4+M5 = 2*M-1:

M = 1/2*(M4+M5+1) = 1/2*(1/2*(M-1) +1/2*(3*M-1)+1) = M

QED

Notice that M = 1/2*(M4+M5+1); which means  sum((-1)^n*(n^(1/n)-1),n =1 .. infinity) = 1/2*(sum((-1)^m*(m^(1/m)+M), m = 1 .. infinity)+sum((-1)^m*(m^(1/m)-M), m = 1 .. infinity)).

Thus M = 1/2*(sum((-1)^m*(m^(1/m)+M), m = 1 .. infinity)+sum((-1)^m*(m^(1/m)-M), m = 1 .. infinity)), where M is the MRB constant. That equation  is not true if M=1.

This question remains open: Is the MRB constant the only value for M such that M = 1/2*(sum((-1)^m*(m^(1/m)+M), m = 1 .. infinity)+sum((-1)^m*(m^(1/m)-M), m = 1 .. infinity)) and such that sum((-1)^n*(n^(1/n)-(1-2M)), n = 1 .. infinity) = 0?

If you are new to this blog and are concerned about the divergent series:

"In the case of infinite sums, Levin's u-transform is used, which has the additional effect that sums that formally diverge may return a result that can be interpreted as evaluation of the analytic extension of the series for the sum."(from Maple's help files)
 

marvinrayburns.com

In the previous messages Maple  found a value for the divergent series M4 and M5.

Wolfram Alpha, now, gives the same numeric sums, albeit issuing a warning.

Click:

Let M4 = sum((-1)^m*(m^(1/m)+M), m = 1 .. infinity) = 1/2*(M-1)

or

let M5 = sum((-1)^m*(m^(1/m)-M), m = 1 .. infinity) = 1/2*(3*M-1).

Their combined sum was N = -MRB2,  Click here for link to published description of MRB2. (If you have any additional insights please contribute to that entry.)

Let M[0] = M and M[p+1] = sum((-1)^n*(n^(1/n)+M[p]),n =1 .. infinity), then N=3*Limit(M[p],p=infinity).

In the previous two comments I showed one way in which n-cubes can have real-life applications: it is simply a matter of defining a sufficient numer of dimensions. We looked at a Tesseract with dimensions: length, width, height and time. Then we looked at a Penteract dimensions: length, width, height, time and density. Notice that they both share four dimensions.
Let's look at their length dimension. The Penteract had length of 5^(1/5) units; likewise, the Tesseract had length of 4^(1/4) units as used in http://www.marvinrayburns.com/what_is_mrb.pdf . If we were to originate both the Penteract and the Tesseract at the origin there would exist a length that corresponds to the differences of their edges. That length would be 4^(1/4)  minus 5^(1/5) or approximately 0.0344839009 units.
We could do the same with all three of the other shared dimensions. For instance, we could say both scenarios, represented by the n-cubes, started at the same time, and we simply wanted to know the difference between when one scenario and the other ended.
In my next post I would like to show why I think all scenarios in life can be compared to each other in their shared dimensions.

Again, consider the diagram at http://www.marvinrayburns.com/what_is_mrb.pdf.
It starts with a Line Segment of 1 linear unit.
Next, Square of 2 square units.
Next, Cube of 3 cubic units.
Next, Tesseract of 4 units^4.
Now that we can picture those 4 dimensions, let's move on to the
Penteract of 5 units^5.
Penteract of 5 units^5 has 80 edges of size 5^(1/5), approximately 1.37972966... units.
The penteract of 5 units^5 is equivalent to a particle with
density that changes from "0." to 1.3... pounds per cubic inch that moves 1.3... inches to the right, 1.3... inches backwards, 1,3...inches up for 1,3... seconds.
Perhaps, you can now understand that all hypercubes can represent real life
scenarios, given we include enough detail about the situation.
Next, I will try to show what happens when you add one edge of a hypercube to
one edge of another hypercube.
 

Consider the diagram at http://www.marvinrayburns.com/what_is_mrb.pdf.
It starts with a Line Segment of 1 linear unit.
Next, Square of 2 square units.
Next, Cube of 3 cubic units.
Those need no explanation.
After those, however, there is the Tesseract of 4 units^4. Let me give you an
example of that. 4^(1/4) = sqrt(2) approximately 1.414.
So a Tesseract of 4 units^4 has (32) edges each with length approximately 1.414.
We often call time the fourth dimension; so a Tesseract of 4 units^4 can
represent the following. A particle moves 1.4... inches to the right, 1.4...
inches backward, 1,4... inches up for 1,4... seconds.
I'll stop there for now. In my next message I will try to explain a Penteract of
5 units^5.
 

Someone I respect has suggested that "MRB2" should not be considered a new constant,  rather, it should be considered simply as an offshoot of the MRB constant.. I suppose it is a matter of taste.

So then, M is the MRB constant:

1-2M is the solution for a in the following two expressions.

solve(limit(sum((-1)^m*(m^(1/m)-a), m = 1 .. 2*N),N=infinity) = M,a)

 and  when we reverse the order of a and m^(1/m) we still get M

solve(limit(sum((-1)^m*(a-m^(1/m)), m = 1 .. 2*N-1),N=infinity) = M,a).

In contrast,

1 is the solution for a in the following two expressions.

solve(limit(sum((-1)^m*(m^(1/m)-a), m = 1 .. 2*N),N=infinity) = M,a)

 and  when we reverse the order of a and m^(1/m) we get negative M

solve(limit(sum((-1)^m*(a-m^(1/m)), m = 1 .. 2*N-1),N=infinity) = -1*M,a).

MRB2 will be used in my next blog,

MRB constant G

.

Here is the formula for MRB2 in Maple:

Digits := 50; fsolve(sum((-1)^j*(j^(1/j)-n), j = 1 .. infinity) = 0, n), giving 0.6242807150758657595029641318914535398881938101997;

and in Mathematica:

Block[{$MaxExtraPrecision = 10000},

 FindRoot[NSum[(-1)^j*(j^(1/j) - n), {j, Infinity}], {n, 0.6},

  WorkingPrecision -> 50]], giving {n -> 0.62428071507608096085825341581276859415083890780807}.

There is a huge discrepancy between the two results. But they agree that MRB2 approximately = 0.624280715076.

However, according to Richard Mathar's trivial solution, let M=the MRB constant and MRB2=1-2M=0.62428071507586575950296413189145353988819381019972242765599

063182104553687067957259340669113378500619231530828748396187753725981048

15467391221413022076317583253267478367727945237472412 ....

Previously we found a exact form for the generating function, M[p+1] = sum((-1)^n*(n^(1/n)+M[p]), n = 1 .. infinity), which was -1/3MRB2 where MRB2 is 1-2M.

As before, M is the MRB constant and Maple will be using the Levin's u-transform which can assign values to divergent series. ( Euler also found a value to the, divergent, first general infinite series:, as seen in http://www.math.dartmouth.edu/~euler/docs/translations/E352.pdf .)

Now let's just concentrate on the nested part of the generating function, namley, sum((-1)^m*(m^(1/m)+M), m = 1 .. infinity), and the like summation, sum((-1)^m*(m^(1/m)-M), m = 1 .. infinity).

Through numeric evaluation it becomes rather clear that

sum((-1)^m*(m^(1/m)+M), m = 1 .. infinity) = 1/2*(M-1)

and

sum((-1)^m*(m^(1/m)-M), m = 1 .. infinity) = 1/2*(3*M-1),

which can be shown in Maple as follows.

Digits:=70:

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

M4:=sum((-1)^m*(m^(1/m)+evalf(M)), m = 1 .. infinity):evalf(M4-1/2*(M-1)); M5:=sum((-1)^m*(m^(1/m)-evalf(M)), m = 1 .. infinity):evalf(M5-1/2*(3*M-1));

Thier combined sum is -1+2*M = -(1-2M) = -MRB2, as seen by

evalf(M4+M5-2*M+1).

Will links to old version of Maple Primes postings still work after we switch to the new version? I have posted a few links to various messages; will I have to replace all of them with new URLs? Moreover, will the present day Google search engine results that go to Maple Primes messages still work?

Let M = MRB constant = sum((-1)^n*(n^(1/n)-1),n =1 .. infinity), and let MRB2 = 1-2M = 0.6242807150758657595... :

The zero for sum((-1)^n*(n^(1/n)-a),n = 1 .. infinity) is a = MRB2.

Let M4 = sum((-1)^m*(m^(1/m)+M), m = 1 .. infinity) = 1/2*(M-1),

and

let M5 = sum((-1)^m*(m^(1/m)-M), m = 1 .. infinity) = 1/2*(3*M-1).

Let N = M4 + M5 = -1+2*M = -(1-2M) = -MRB2.

Let M[0] = M and M[p+1] = sum((-1)^n*(n^(1/n)+M[p]),n =1 .. infinity), then Limit(M[p],p=infinity) = -1/3MRB2 = 1/3*N

In the case of infinite sums, Levin's u-transform is used, which has the additional effect that sums that formally diverge may return a result that can be interpreted as evaluation of the analytic extension of the series for the sum.

Remember, in the case of infinite sums, Levin's u-transform is used by Maple, which has the additional effect that sums that formally diverge may return a result that can be interpreted as evaluation of the analytic extension of the series for the sum.

Since 1-2M = -1*(2M-1), we can let M=MRB constant=sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity) and  MRB2=1-2M=0.62428071507586575950296413189145353988819381019972242765599...,

so the zero for sum((-1)^n*(n^(1/n)-a), n = 1 .. infinity) is MRB2,

and let M[0]=sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity) and M[p+1] = sum((-1)^n*(n^(1/n)+M[p]), n = 1 .. infinity), then Limit(M[p],p=infinity) = -1/3MRB2.

In the blog MRB Constant F we looked at the sequences of series of the form sum((-1)^n*(n^(1/n)-a), n = 1 .. infinity) and found an a, that Maple says is a zero for sum((-1)^n*(n^(1/n)-a), n = 1 .. infinity), namely 1-2*MRB constant.

Now lets look at what happens if we let a=M where M=sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity), the MRB constant, and apply the generating function, M[p+1] = sum((-1)^m*(m^(1/m)+M[p]), m = 1 .. infinity) for several iterations.

To begin to see the hyper-geometric application of each step of this iteration follow the link at MRB Constant E.

In Maple, Digits := 30; M := evalf(sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity)); for a to 60 do M := evalf(sum((-1)^m*(m^(1/m)+M), m = 1 .. infinity)); print(evalf(M)) end do; (giving an end result of -.208093571691955...). According to Wolfram Alpha, -0.208093571691955253167654710630484513296064603399907475885... is 1/3(2*MRB constant-1).

So it appears that the infinite iteration of M[0]=sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity) and M[p+1] = sum((-1)^n*(n^(1/n)+M[p]), n = 1 .. infinity) gives M[p=infinity] = 1/3(2*MRB constant-1).

Note from ?evalf,Sum help-page: In the case of infinite sums, Levin's u-transform is used, which has the additional effect that sums that formally diverge may return a result that can be interpreted as evaluation of the analytic extension of the series for the sum.