f(n)=(-1)^n* n^(1/n)

THEOREM MRBK 4.0

When n is in the set of integers the derivative of f is exactly I*Pi*f+(1-ln(n))*f/n^2

Proof:

THEOREM MRBK 2.0 says the Imaginary part is Pi*f.

THEOREM MRBK 3.0 says the real part is (1-ln(n))*f/n^2.

A complex number is the sum of its imaginary part and its real part.

A reminder: If a function is differentiable at some point c of its domain, then it is also continuous at c. However here we extend the notion of differentiability to be valid for individual points on the real number line, specifically positive integers.

f(n)=(-1)^n* n^(1/n)

THEOREM MRBK 3.0

When n is in the set of integers the derivative of f has a real part of exactly (1-ln(n))/n^2*f.

Proof:
> simplify(`assuming`([Re(diff(f(n), n)-(1-ln(n))*f(n)/n^2)], [n::posint]));
0

You are correct Alec.  I'm sorry that I let the misreading of my own notation lead me astray.

 COROLLAY MRBK 1.1a

 When n is in the set of integers the derivative of f has an imaginary part of Pi*f.

If a function is differentiable at some point c of its domain, then it is also continuous at c. However here we extend the notion of differentiability to be valid for individual points on the real number line, specifically positive integers.

DEFINITIONS

f(n)=(-1)^n* n^(1/n)

g(n)= n^(1/n)

k(n)' =d/dn k(n)

∈ means in.

n  In the set of integers is n ∈ {1,2,3,...}.

Re(f(n)) is the real part of f(n)

THEROM MRBK 1.0                                                                                   .

  Re(((-1)^n*n^(1/n))')=(-1)^n*(n^(1/n))' | n ∈ {1,2,3,...},

The real part of the instantaneous change of f relative to n are the same as the instantaneous change of g relative to n.

Proof:

restart

simplify(`assuming`([Re(diff((-1)^n*n^(1/n), n))-(-1)^n*(diff(n^(1/n), n))], [n::posint]))

0

 COROLLAY MRBK 1.1

 When n is in the set of integers the derivative of f  has no imaginary part.

Since the MRB constant is an alternating sum of all integers to their own roots, f(n)=(-1)^n* n^(1/n); a thorough understanding of the changes in f, as n changes, is important. So this will be covered in a future blog MRB constant K.

Concerning Re(int((-1)^n*n^(1/n), n = exp(-1) .. exp(1))) or the integral over cos(pi*x)x^(1/x) between 1/E and E, 

According to the Euler–Maclaurin formula, the odd nth derivatives mentioned in my last post, which are rich with cancelation and approximation at the greater endpoint, e, do simplify the integral of concern. However our definite integral has a lesser endpoint of 1/e that possess very little cancelation within its higher derivatives. Thus I conclude that the nearness of the value of this integral to the value of the MRB constant is just coincidence. However the wealth of cancelation I demonstrated in my previous post will probably help me in writing a future blog.

The value of the integral of concern is found in the online version of integer sequences under the A-number A177218.

For more on the Euler–Maclaurin formula see http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula.

 marvinrayburns.com

For f(x)=(-1)^x*x^(1/x) I said there was a critical point at x=e.

At x=e:

n        real part of the nth deriviative

1        -Pi^1 * e^(1/e) * sin(e*Pi) within 0

2        -Pi^2 * e^(1/e) *cos(e*Pi) within a remainder of 0.0455474...

3        Pi^3*e^(1/e)*sin(e*Pi) within a remainder of 0.440864...

4        Pi^4*e^(1/e)*cos(e*Pi) within a remainder of -3.830448090...

5       -Pi^5*e^(1/e)*sin(e*Pi) within a remainder of -6.35268...

6       -Pi^6*e^(1/e)*cos(e*Pi) within a remainder of  100.9196638... , where Pi^7>3020.

7      Pi^7*e^(1/e)*sin(e*Pi) within a remainder of -51.137422... , where Pi^8>9488

I asked, "Is there a connection between the MRB constant and MKB(E^-1,E) ?"

One person wrote me back and suggested that,

"The two constants are different without doubt. Generally speaking
there may be some approximate mapping with the Euler-Maclaurin formula
to convert the sum and the integral with favorite cancellation because
the derivative involves the logarithm, [ d/dn of n^(1/n) is
n^(1/n)*(1-log n)/n^2 ] but beyond this approximation this is just
coincidence."

On the other hand, I was trying to work on a connection by means of the following:

At n=1/2 and n=5/2  the real parts of (-1)^n*n^(1/n) are 0.

eval((-1)^n*n^(1/n), n = 5/2) = 0+I*(5/2)^(2/5).

eval((-1)^n*n^(1/n), n = 1/2) = 0+I*(1/4).

Since (-1)^n*n^(1/n) is continuous from 1/e to e,

int((-1)^n*n^(1/n), n = exp(-1) .. exp(1)) = int((-1)^n*n^(1/n), n = exp(-1) .. 1/2) + int((-1)^n*n^(1/n), n = 1/2 .. 5/2) + int((-1)^n*n^(1/n), n = 5/2 .. exp(1)).

So,

Re(int((-1)^n*n^(1/n), n = exp(-1) .. exp(1))) = Re(int((-1)^n*n^(1/n), n = exp(-1) .. 1/2)) + Re(int((-1)^n*n^(1/n), n = 1/2 .. 5/2)) + Re(int((-1)^n*n^(1/n), n = 5/2 .. exp(1))).

Or approximately,

0.18777903132304277043 = 0.0032282119429439541474 + 0.28848088137392251193 + (-.10393006199382369565)

But I don't see anywhere to go from there.

I started calling this constant the MKB constant because at first I thought that its numerical value equaled the MRB constant's +1/2 which would be very remarkable by establishing such a connection between a matching integral and sum. So now that we know that is false, the question remains as to what precise value must be added to the MRB constant to equal the MKB constant?  Since we know a very precise value of the MRB constant and it is  a magnitude, or so, easier to compute, it would suffice us to know an equally precise value of the MKB constant, and then we would naturally know what their difference is to an equally precise value. It was my hope that we could look at their difference and recognize an non trivial combination of constants that was equal to it, but as more digits to their difference has been revealed, it seems unlikely that such a combination exists.

Here is the best approximation of their difference that I know of.

MRB constant                 0.187859642462067120249

MKB constant                 0.68765236892769436980

their difference               0.499792726465627249551

I computed the value of the MRB constant in Maple with the simple Maple command Digits := 21; evalf(sum((-1)^n*(n^(1/n)-1), n = 1 .. infinity)), while the value for the MKB constant was computed by Richard J. Mather and published in http://arxiv.org/abs/0912.3844 .

I don't think Mather ever called the second constant the MKB constant; he called it M1.

Here is how I came up with the name MKB constant. Since the MRB constant = ,

I thouht if I cahnge "sum" to "int", integrate insteand of summing, what do we have? .

The second expression is so closely related to the first! Since the first is named by my initials, I named the second by my wife's initials.

The MRB constant is Sloane's A037077 and the MKB constant is Sloane's A157852.

Acer, while reviewing this blog, I noticed you said,

"I couldn't follow what the summations mean when the index ranges from 1 to sqrt(3). I had thought that the dummy index n was representing positive integers. What does this mean,
sum((-1)^n*n^(1/n),n = 1 .. 27*6^(1/2))));"

I had defined a function, f:=x->sum((-1)^n*(n^(1/n)),n=1..x).

When I plotted f Maple appeared to give me the plot of a continuous function. I was curious to see if Maple had some way of evaluating f at non-integer value of x.

This reply is a little late. Sorry.

I call that value, the integral over (-1)^x*x^(1/x) between 1 and infinity, the MKB constant.

 It is Sloane's A157852.

Concerning the unlikely approximation to the MRB constant, MKB(e^-1,e) = Re(int((-1)^n*n^(1/n), n = e^-1..e)) = 0.1877790313...,

one note to myself and anyone else working on this.

n=E Gives a maximum for f in the function f(n)=Abs((-1)^n*n^(1/n)) in the range (0, oo).

That critical point is (e, e^(1/e)).

Another note:

f(n)=abs((-1)^n*n^(1/n)).

n = 1/e,  f = (1/e)^(e).

There is the very efficient approximation to Pi of 22/7;

but a better approximation, to the MRB constant, is 3/16.

Compare 22/7 -  3.141596 = 0.00126114

with          0.1878596 - 3/16 = 0.0003596. 

marvinrayburns.com

  =  , where m is the MRB constant which is, : See the previous messages.

Levin's u-transform assigns values to divergent series.

marvinrayburns.com

The point of my laat message was,
S=sum((-1)^k (k^(1/k) - a),k=1..infinity)=1/2*(a+2m-1) where m is the MRB constant.
So I conclude as follows:
S=1/2*(a+2m-1).
Thus,
a=2S+(1-2m) where 1-2m is MRB2, the solution for a in sum((-1)^k (k^(1/k) - a),k=1..infinity)=0.
So, like m is achieved from sum((-1)^k (k^(1/k) - a),k=1..infinity) with a=1, any value for S may be achieved from S=sum((-1)^k (k^(1/k) - a),k=1..infinity); the value for a needed is simply 2S+MRB2.

In these series the sum is the numerical value which is arrived at according to principles of analysis.