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The MRB Constant-A

Posted: February 05 2007 by Marvin Ray Burns 465

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This blog is an extension of MRB Constant post (at that time called a blog). It was written in December of 2006 as a part of the MRB constant post. I seperateded it and MRB constant B to try to make the original post easier to read.

On May 5 and 6, 2012 I added some old content that was once lost. However, some of the replaced content might be in the wrong place.

Some Plot Qualities Part 1 As I post these plot qualities of the MRB Constant. Any input as to triviality will be welcome.

Beginning in this post, we will consider the graph of the terms of f1, which terms we treat as a continuous function g1.

g1 Makes the terms of f1(ie. the absolute convergent form of the MRB Constant)
Within special ranges, the sequence, g1, has an interesting spiral design that finds itself a center at the origin.

A Spiral Made from the Terms of MRB

g1 Makes the terms of f1(ie. the absolute convergent form of the MRB Constant)]
Within special ranges, the sequence, g1, has an interesting spiral design that finds itself a center at the origin.

>	g1:=x->(-1)^x*(x^(1/x)-1):

>	f1:=sum(g1(x),x=1..infinity):

>	with(plots):

Warning, the name changecoords has been redefined

>	for n from -1 to 4 do plot([Re(g1(x)),Im(g1(x)),x=0..1/Pi+n*f1]) od; for n from 2 to 10 do plot([Re(g1(x)),Im(g1(x)),x=2^n..2^(n+1)]) od;

There also is great quasi - symmetric beauty in the polar plot of g1(x) +n*f1

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There also is great quasi - symmetric beauty in the polar plot of g1(x) +n*f1

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Some Plot Qualities Part 2 In part 1, g1 with 0 < x < 1 had a helical from 0-i to 0+0i. Not shown but indicated, g1 with 1 < x < 2 (from 0+0i to 0.414+0i) transitions from a helix to a spiral. This post will look into this transition and uncover a couple of facts that we might use later.

> g1:=x->(-1)^x*(x^(1/x)-1):

> for x from 1 by 0.1 to 2 do ;printf("x=%a->%a
",x,evalf(g1(x))) od;

x=1->0.
x=1.1->-.8608027175e-1-.2796917575e-1*I
x=1.2->-.1327468570-.9644623704e-1*I
x=1.3->-.1314441248-.1809173169*I
x=1.4->-.8395312828e-1-.2583811608*I
x=1.5->-.3103706970*I
x=1.6->.1055143292-.3247397140*I
x=1.7->.2153281519-.2963737753*I
x=1.8->.3124221289-.2269879633*I
x=1.9->.3822154489-.1241893276*I
x=2.0->.414213562

> with(plottools):
extra := circle([0.1,-0.15], 0.2, color=red):

Warning, the name arrow has been redefined

> main:=plot([Re(g1(t)),Im(g1(t)),t=1..2],color=blue):

> with(plots):display({main,extra},axes=box,title="g1 compared to a cirlce");

Warning, the name arrow has been redefined

Maple Plot

The area enclosed by the graph under the x-axis is:

> with(VectorCalculus):

> SetCoordinates( cartesian[t,y] ):

> field:=VectorField( ):

> LineInt( field , Path( , t=1..2 ));evalf(%);

The local minimum imaginary value occurs just counterclockwise of Pi/2.

> for x from 1.5 by 0.01 to 1.6 do ;printf("x=%a->%a
",x,evalf(g1(x))) od;

x=1.5->-.3103706970*I
x=1.51->.9856555762e-2-.3136406899*I
x=1.52->.1991376631e-1-.3165202517*I
x=1.53->.3015454619e-1-.3190017777*I
x=1.54->.4056163262e-1-.3210783834*I
x=1.55->.5111761269e-1-.3227439046*I
x=1.56->.6180495057e-1-.3239928971*I
x=1.57->.7260601377e-1-.3248206334*I
x=1.58->.8350309906e-1-.3252230984*I
x=1.59->.9447845960e-1-.3251969910*I
x=1.60->.1055143292-.3247397140*I

> for x from 1.58 by 0.001 to 1.59 do ;printf("x=%a->%a
",x,evalf(g1(x))) od;

x=1.58->.8350309906e-1-.3252230984*I
x=1.581->.8459740596e-1-.3252398223*I
x=1.582->.8569247753e-1-.3252522559*I
x=1.583->.8678829669e-1-.3252603987*I
x=1.584->.8788484512e-1-.3252642457*I
x=1.585->.8898210519e-1-.3252637951*I
x=1.586->.9008005928e-1-.3252590447*I
x=1.587->.9117868963e-1-.3252499918*I
x=1.588->.9227797830e-1-.3252366335*I
x=1.589->.9337790748e-1-.3252189671*I
x=1.590->.9447845960e-1-.3251969910*I

So the local min is approximately 0.3252643, with a corresponding real value of approx. 0.087884.

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Some Plot Qualities Part 3 In this post we want to get a slightly bigger picture of the building up of the MRB Constant. It seems we stumble accross the value of Pi/8.

The first few partial sums

(From Terms to Sums)

In our last post we made some measurements of a special part of the graph of the terms of f1, which terms we treated as a continuous function g1.

>	g1:=x->(-1)^x*(x^(1/x)-1):

>	main:=plot([Re(g1(t)),Im(g1(t)),t=1..2],color=blue):

>	with(plots):

Warning, the name changecoords has been redefined

>	display(main,title="a special part of g1");

The area enclosed by the graph under the x-axis was 0.07790843494

The local minimum imaginary value occurred just counterclockwise of Pi/2, approximately 0.3252643, with a corresponding real value of approx. 0.087884.

Those numbers may come in handy later, but for now lets refocus on f1, the summing terms.
So next we will be plotting the partial sums of the MRB Constant in the complex field.

>	f1:=x->sum((-1)^n*(n^(1/n)-1),n=1..x):

>	for a from 2 to 3 do plot([Re(f1(x)),Im(f1(x)),x=1..a]) od;

It appears that we have two nearly- identiacal semi-elliptical curves, one in the second quadrant and one in the fourth. Approximations for elliptical fits are given here:

>	main:=plot([Re(f1(x)),Im(f1(x)),x=1..3],color=blue): extra_neg:=plot(sqrt(.25-(x+1/2)^2),x=-1..0,color=red): extra_pos:=plot(-sqrt(.25-(x-0.91)^2),x=-0..1.5,color=green): display([main,extra_neg,extra_pos]);

Hence the size of each curve is given by:

>	int(sqrt(.25-(x+1/2)^2),x=-1..0);

>	int(-sqrt(.25-(x-0.91)^2),x=-0..1.5):abs(Re(%));

Further plotting shows that the plot of curves rotates clockwise very slowly

but each curve remains at the same size of Pi/8 or at least approx. 0.392699.

>	evalf(Pi/8);

>	for a from 2 to 3 do plot([Re(f1(x)),Im(f1(x)),x=1..5^a],axes=none) od;

If we turn the plot about 45 degrees to the left we would have an, or several similar, ellipsis.

Now to say the summing of the terms of the MRB Constant forms an elliptical field invites great opportunity for further exploration.
This is a good time to try to re-read,

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Some Plot Qualities Part 4 SO I say, HELP! Here, I'm asking for some help. We just began to touch upon the graphs of the partial sums of the MRB Constant and seemingly got a closed form number for some area representing the partial sums. However, as you will see, I felt we left out something important when we moved on so quickly into partial sums, when so many questions were left unanswered about the continuous function.

Taking following sums, on to infinity, is a way of determinig the value of the MRB Constant.

>	evalf(sum((-1)^n*(n^(1/n)-1),n=1..10));

>	evalf(sum((-1)^n*(n^(1/n)-1),n=1..100));

>	evalf(sum((-1)^n*(n^(1/n)-1),n=1..1000));

>	evalf(sum((-1)^n*(n^(1/n)-1),n=1..10000));

Here it would be convenient to have some form returned for f1 that would not take an infinite number of terms to compute. Maple is not quick to give us a closed form formula for it.

>	f1(x):=sum((-1)^n*(n^(1/n)-1),n=1..x);

f1(x) := (1/2)*(-1)^(x+1)+1/2+sum((-1)^n*n^(1/n), n = 1 .. x)

Notwithstanding, in our last post we considered the complex graph of f1, when turned to the left, is one ellipse with area of Pi/8. This is true
looking at the graph as a 3d object projected on to a 2d plane.

One reason for Maple's reluctance to give us a quick closed form formula is that we are summing a discreet series, not a continuous function.

When we look at f1 in the complex, we may use it as a continuous function.

>	for a from 1 to 4 do plot([Re(f1(x)),Im(f1(x)),x=10^a+1..10^a+7]) od;

>	f1:=x->sum((-1)^n*(n^(1/n)-1),n=1..x);

f1 := proc (x) options operator, arrow; sum((-1)^n*(n^(1/n)-1), n = 1 .. x) end proc

f1(9);

Since we are concerning ourselves with a continuous function, we may replace the discreet sum sign with the continuous sum sign.

>	g1:=n->(-1)^n*(n^(1/n)-1):

>	for a from 2 to 10 do evalf(int((-1)^n*(n^(1/n)-1),n=1..a)) od;

>	evalf(int((-1)^n*(n^(1/n)-1),n=1..100));

>	abs(0.7073779165e-1-0.6238188574e-1*I);

A maping of area of iside the ellipses as x gets large

   x                          area                                        absolute area
n=1..x                Re          Im                                  sqt(Re^2+Im^2)

10 0.06911897789- 0.1297273908*I 0.1469919352

100 0.07073779165-0.06238188574*I 0.09431508276

1,000           0.07064725368- 0.1492489408*I       0.1651250459

2,000           0.07064091220- 0.1813553542*I       0.1946276008

3,000           0.07063861072- 0.2015317773*I       0.2135529690

4,000           0.07063740145- 0.2164757753*I       0.2277090331

5,000           0.07063665010- 0.2284274612*I       0.2390996474

6,000           0.07063613556- 0.2384266367*I       0.2486699112

7,000           0.07063599534- 0.2422468231*I       0.2523350296

8,000           0.07063570845- 0.2498342235*I       0.2596276999

9,000           0.07063536222- 0.2601027200*I       0.2695232445

10,000          0.07063508194- 0.2693926669*I       0.2784990553

>	re:={[10,0.06911897789],[100,0.07073779165],[1000,0.07064725368],[2000,0.07064091220],[3000,0.07063861072],[4000,0.07063740145],[5000,0.07063665010],[6000,0.07063613556],[7000,0.07063599534],[8000,0.07063570845],[9000,0.07063536222],[10000,0.07063508194]}:

>	im:={[10,0.1297273908],[100,0.06238188574],[1000,0.1492489408],[2000,0.1813553542],[3000,0.2015317773],[4000,0.2164757753],[5000,0.2284274612],[6000,0.2384266367],[7000,0.2422468231],[8000,0.2498342235],[9000,0.2601027200],[10000,0.2693926669]}:

>	ab:={[10,0.1469919352],[100,0.09431508276],[1000, 0.1651250459],[2000,0.1946276008],[3000,0.2135529690],[4000,0.2277090331],[5000,0.2390996474],[6000,0.2486699112],[7000,0.2523350296],[8000,0.2596276999],[9000,0.2695232445],[10000,0.2784990553]}:

>	with(plots):

>	Real:=pointplot(re,color=black):Imaginary:=pointplot(im,color=blue,style=LINE):Absolute:=pointplot(ab,color=red,style=LINE):display(Real,Imaginary,Absolute);

Question: What is the..

>	limit (Re(int((-1)^n*(n^(1/n)-1),n=1..x)),x=infinity);

limit(Re(int((-1)^n*(n^(1/n)-1), n = 1 .. x)), x = infinity)

>	limit (Im(int((-1)^n*(n^(1/n)-1),n=1..x)),x=infinity);

limit(Im(int((-1)^n*(n^(1/n)-1), n = 1 .. x)), x = infinity)

>	limit (Abs(int((-1)^n*(n^(1/n)-1),n=1..x)),x=infinity);

limit(Abs(int((-1)^n*(n^(1/n)-1), n = 1 .. x)), x = infinity)

Can anyone help here?

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By the post Some Plot Qualities Part 3, I believe the absolutve value(int(f1),n=infinty) = Pi/8~0.39....

Can anyone help here?

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Some Plot Qualities Part 5 In our last post, Some Plot Qualities Part 4, we changed a series that makes up the MRB Constant, (1-1)-(sqrt(2)-1)+(3^(1/3)-1)-(4^(1/4)-1)+..., into a continuous function, int((-1)^n*(n^(1/n)-1),n=1..infinity). Since the continuous function has complex values, we asked about the absolute value of the reals and imaginaries. We hope to find a closed formula (here it means, one that has a finite number terms of elementary operators and functions) to the integral. It is then hoped for that the integral function will tell us more about the discreet sum function. It is then hoped that more knowledge about the MRB Constant will give us more understanding of the integers. Finally, it is hoped that more knowledge about the integers will give us a greater understanding of the universe in a transcendental way. (Here, transcendental means philosophy independent of human experience of phenomena but within the range of knowledge.) Well, let's get started.

Previously, we inquired into the following three problems.

>	limit (Re(int((-1)^n*(n^(1/n)-1),n=1..x)),x=infinity):

>	limit (Im(int((-1)^n*(n^(1/n)-1),n=1..x)),x=infinity):

>	limit (Abs(int((-1)^n*(n^(1/n)-1),n=1..x)),x=infinity):

I have a hunch that the answer to the third one was directly related the post Plot Qualities 3 and was specifically Pi/8.

To take that as an conclusion, however, seems a little premature.

Lets see if we can find a solution for the third problem.

First lets see if it is likely that a solution exist.

Consider that the graph of the third function just might go on to infinity as x does.

Before we try to find a value for its limit we had better be sure it exists.

We will define the third problem and call the solution "abb."

>	f1:=x->(-1)^x*(x^(1/x)-1): rea:=inf->Re(int(f1(x),x=1..inf)): imi:=inf->Im(int(f1(x),x=1..inf)): abb:=inf->sqrt(rea(inf)^2+imi(inf)^2);

abb Starts out like this:

>	plot(abb(x),x=2..4);

>	plot(abb(x),x=1..4);plot(abb(x),x=5..20);

plot(abb(x),x=21..31);

As x gets much larger, the wave is somewhat dampend and from a distance the graph looks more like this:

>	ab:=[[10,0.1469919352],[100,0.09431508276],[1000, 0.1651250459],[2000,0.1946276008],[3000,0.2135529690],[4000,0.2277090331],[5000,0.2390996474],[6000,0.2486699112],[7000,0.2523350296],[8000,0.2596276999],[9000,0.2695232445],[10000,0.2784990553]]:

>	with(plots):pointplot(ab,color=red,style=LINE);

I am stumped on how to prove that the series represented by abb does converge.

But it is fair to think it has a maximum value

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We have not found the closed formula for the integral. We are, however, quite sure there is an exact value for it. Will we find what we want? Or will we be forever stumped?

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Some Plot Qualities Part 6 Several hours ago, we were lost in the land of I. Good news, we didn't stay lost in Limbo too long. We did get the answer.

restart;

We may deduce that the derivative of abb is

>	dbb:=exp(PixI)*exp(ln(x)/x);

The integral, abb, continuously sums the real and imaginary values of .

We will simply see what happens as is summed. What variables will change the total of the summing?

Here we will look at two variables. n will be intervals of n/Pi. y will be multiples of 1 term summed

>	s:=0:for n from 0.1 by .05 to 4 do for y from 1 to 10^3 do x:=n/Pi: s:=s+abs(dbb)od; printf("using multiples of %a /Pi and %a terms gives us a sum of %f\n",n,y,evalf(Re(s))) od:

using multiples of .1 /Pi and 1001 terms gives us a sum of 0.000000
using multiples of .15 /Pi and 1001 terms gives us a sum of 0.000000
using multiples of .20 /Pi and 1001 terms gives us a sum of 0.000000
using multiples of .25 /Pi and 1001 terms gives us a sum of 0.000000

using multiples of .30 /Pi and 1001 terms gives us a sum of 0.000000
using multiples of .35 /Pi and 1001 terms gives us a sum of 0.000003
using multiples of .40 /Pi and 1001 terms gives us a sum of 0.000096
using multiples of .45 /Pi and 1001 terms gives us a sum of 0.001379

using multiples of .50 /Pi and 1001 terms gives us a sum of 0.011037
using multiples of .55 /Pi and 1001 terms gives us a sum of 0.058597
using multiples of .60 /Pi and 1001 terms gives us a sum of 0.230519
using multiples of .65 /Pi and 1001 terms gives us a sum of 0.723619

using multiples of .70 /Pi and 1001 terms gives us a sum of 1.908291
using multiples of .75 /Pi and 1001 terms gives us a sum of 4.386862
using multiples of .80 /Pi and 1001 terms gives us a sum of 9.033430

using multiples of .85 /Pi and 1001 terms gives us a sum of 17.007340
using multiples of .90 /Pi and 1001 terms gives us a sum of 29.739315
using multiples of .95 /Pi and 1001 terms gives us a sum of 48.894603

using multiples of 1.00 /Pi and 1001 terms gives us a sum of 76.320297
using multiples of 1.05 /Pi and 1001 terms gives us a sum of 113.984544
using multiples of 1.10 /Pi and 1001 terms gives us a sum of 163.914423

using multiples of 1.15 /Pi and 1001 terms gives us a sum of 228.137593
using multiples of 1.20 /Pi and 1001 terms gives us a sum of 308.631072
using multiples of 1.25 /Pi and 1001 terms gives us a sum of 407.278897

using multiples of 1.30 /Pi and 1001 terms gives us a sum of 525.839225
using multiples of 1.35 /Pi and 1001 terms gives us a sum of 665.920525
using multiples of 1.40 /Pi and 1001 terms gives us a sum of 828.966034

using multiples of 1.45 /Pi and 1001 terms gives us a sum of 1016.245323
using multiples of 1.50 /Pi and 1001 terms gives us a sum of 1228.851761
using multiples of 1.55 /Pi and 1001 terms gives us a sum of 1467.704651

using multiples of 1.60 /Pi and 1001 terms gives us a sum of 1733.554952
using multiples of 1.65 /Pi and 1001 terms gives us a sum of 2026.993603
using multiples of 1.70 /Pi and 1001 terms gives us a sum of 2348.461637

using multiples of 1.75 /Pi and 1001 terms gives us a sum of 2698.261407
using multiples of 1.80 /Pi and 1001 terms gives us a sum of 3076.568387
using multiples of 1.85 /Pi and 1001 terms gives us a sum of 3483.443128

using multiples of 1.90 /Pi and 1001 terms gives us a sum of 3918.843060
using multiples of 1.95 /Pi and 1001 terms gives us a sum of 4382.633895

using multiples of 2.00 /Pi and 1001 terms gives us a sum of 4874.600493
using multiples of 2.05 /Pi and 1001 terms gives us a sum of 5394.457063

using multiples of 2.10 /Pi and 1001 terms gives us a sum of 5941.856652
using multiples of 2.15 /Pi and 1001 terms gives us a sum of 6516.399885

using multiples of 2.20 /Pi and 1001 terms gives us a sum of 7117.642959
using multiples of 2.25 /Pi and 1001 terms gives us a sum of 7745.104899

using multiples of 2.30 /Pi and 1001 terms gives us a sum of 8398.274112
using multiples of 2.35 /Pi and 1001 terms gives us a sum of 9076.614264

using multiples of 2.40 /Pi and 1001 terms gives us a sum of 9779.569532
using multiples of 2.45 /Pi and 1001 terms gives us a sum of 10506.569270

using multiples of 2.50 /Pi and 1001 terms gives us a sum of 11257.032130
using multiples of 2.55 /Pi and 1001 terms gives us a sum of 12030.369710

using multiples of 2.60 /Pi and 1001 terms gives us a sum of 12825.989740
using multiples of 2.65 /Pi and 1001 terms gives us a sum of 13643.298810

using multiples of 2.70 /Pi and 1001 terms gives us a sum of 14481.704830
using multiples of 2.75 /Pi and 1001 terms gives us a sum of 15340.619080

using multiples of 2.80 /Pi and 1001 terms gives us a sum of 16219.458010

using multiples of 2.85 /Pi and 1001 terms gives us a sum of 17117.644730
using multiples of 2.90 /Pi and 1001 terms gives us a sum of 18034.610290

using multiples of 2.95 /Pi and 1001 terms gives us a sum of 18969.794770
using multiples of 3.00 /Pi and 1001 terms gives us a sum of 19922.648150

using multiples of 3.05 /Pi and 1001 terms gives us a sum of 20892.631040
using multiples of 3.10 /Pi and 1001 terms gives us a sum of 21879.215250

using multiples of 3.15 /Pi and 1001 terms gives us a sum of 22881.884240
using multiples of 3.20 /Pi and 1001 terms gives us a sum of 23900.133460
using multiples of 3.25 /Pi and 1001 terms gives us a sum of 24933.470590
using multiples of 3.30 /Pi and 1001 terms gives us a sum of 25981.415720
using multiples of 3.35 /Pi and 1001 terms gives us a sum of 27043.501410
using multiples of 3.40 /Pi and 1001 terms gives us a sum of 28119.272730
using multiples of 3.45 /Pi and 1001 terms gives us a sum of 29208.287230
using multiples of 3.50 /Pi and 1001 terms gives us a sum of 30310.114840
using multiples of 3.55 /Pi and 1001 terms gives us a sum of 31424.337800
using multiples of 3.60 /Pi and 1001 terms gives us a sum of 32550.550460
using multiples of 3.65 /Pi and 1001 terms gives us a sum of 33688.359150
using multiples of 3.70 /Pi and 1001 terms gives us a sum of 34837.381940
using multiples of 3.75 /Pi and 1001 terms gives us a sum of 35997.248450
using multiples of 3.80 /Pi and 1001 terms gives us a sum of 37167.599590
using multiples of 3.85 /Pi and 1001 terms gives us a sum of 38348.087350
using multiples of 3.90 /Pi and 1001 terms gives us a sum of 39538.374530
using multiples of 3.95 /Pi and 1001 terms gives us a sum of 40738.134430
using multiples of 4.00 /Pi and 1001 terms gives us a sum of 41947.050670

Is our conclusion, there is a specific real value for the integral, abb; we just can not say what it will be?

Nonetheless, the smaller interval we use, the smaller the sum.
It would seem then, when we use the "continuous sum over infinitesimal intervals" operator (integration) the sum is a value less than 1.

or

In the following does Maple correctly sum infinitesimal amounts of F(x)? Look at the sudden jump in estimated area from x->5816to x->5817; this is just wrong!

>	x:='x':plot({Re(dbb),Im(dbb)},x=1..10);evalf(abs(int(dbb,x=1..10)));

>	x:='x':plot({Re(dbb),Im(dbb)},x=1..100);evalf(abs(int(dbb,x=1..100)));

>	x:='x':plot({Re(dbb),Im(dbb)},x=1..100);evalf(abs(int(dbb,x=1..100)));

>	x:='x':plot({Re(dbb),Im(dbb)},x=1..5000);evalf(abs(int(dbb,x=1..5816)));

>	x:='x':plot({Re(dbb),Im(dbb)},x=1..6000);evalf(abs(int(dbb,x=1..5817)));

>	x:='x':plot({Re(dbb),Im(dbb)},x=1..infinity);evalf(Abs(int(dbb,x=1..infinity)));

Abs(int(exp(I*Pi*x)*exp(ln(x)/x), x = 1 .. infinity))

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Marvin Ray Burns

Some Plot Qualities Part 7 and moving on We must not be too harsh on Maple; I knew ahead of time about the danger we were getting into. It is an error I discovered first in Mathematica. That program is actually so picky that that error shows up about 50 times before it shows in Maple once. (Here picky means that program is more sensitive to possible singularities -- even when they actually do not exist -- there is just a theoretical chance there could be discontinuities.) It really should be fixed in both programs.

Another Critical point of g1

restart:

>	g1:=x->(-1)^x*(x^(1/x)-1):g2(x):=diff(g1(x),x);

g2(x) := I*(-1)^x*Pi*(x^(1/x)-1)+(-1)^x*x^(1/x)*(-ln(x)/x^2+1/x^2)

>	plot(abs(g1(x)),x=1..5);

>	x:='x':plot(abs(g2(x)),x=1..2,title="A derivative of g1 from 1 to 2");plot(abs(g2(x)),x=1..5,title="A derivative of g1 from 1 to 5");

In Some Plot Qualities Part 3 we saw something weird happed in the graph of g1 between x=1 and x=2.
Well, something else happens between 2 and 3.
That is the x-value of the constant e;
it is a maximum absolute value of the derivative of g1.
The MRB constant is the alternating sum of integers raised to their own roots. In the summing of the terms of the MRB constant, no term is more critical than the 3rd.
We often call the MRB Constant by m. m~0.18785964.
We will se the greatest disparity between m and a partial sum of m happens when the third term is added.

>	m:=0.18785964:

>	evalf(abs(m-(1^(1/1))));

>	evalf(abs(m-(1^(1/1)-2^(1/2))));

>	evalf(abs(m-(1^(1/1)-2^(1/2)+3^(1/3))));

>	evalf(abs(m-(1^(1/1)-2^(1/2)+3^(1/3)-4^(1/4))));

It also turns out that that third partial sum has a coincidental connection to m. If you divided m by the third sum, you get an approximation, as close to m, that would otherwise take nearly 1,000 terms to get.