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A Bunch Of MRB Constants?

Posted 9 years ago

The terms of the series

(* first 100 terms of the series:*)
m100 = Table[N[(-1)^k (k^(1/k) - 1), 200], {k, 1, 100}];

whose partial sums give raise the the MRB constant (see for example in this community Record breaking direct summation of MRB constant ) contain the k-th root of k.

As always the question about all the other roots in the complex plane comes up:

Remove[burnsStrip]
burnsStrip[k_?NumericQ, w_?NumericQ] := 
     Block[{m = RotationTransform[k, {-1, 0}]},
      Polygon[{m[{-5/2, -w}], m[{1/2, -w}], m[{1/2, w}], m[{-5/2, w}]}]
      ]

    Graphics[{{Green, burnsStrip[0, 1/40]}, {Yellow, 
       Sequence @@ (burnsStrip[#, 
            1/40] & /@ (Degree {143.8, 120, 90, 60, 36.2}))}, 
      Table[Point /@ 
        Evaluate[
         ReIm[(-1)^k  ((Flatten[Evaluate[Block[{x}, Solve[x^k == k, x]]]])[[All, 2]] - 1)]], {k, 1, 217}]}, 
     Frame -> True, AspectRatio -> 3/4.7, PlotRange -> All, 
     PlotLabel -> "A Bunch Of Burns Constants"]

giving

enter image description here

The terms for the MRB constant are in the green strip in the interval (-1/2,1/2). It is special, as the graphics shows clearly.

But is it sure, that the partial sums of terms on straight lines through -1 + 0 I or on straight lines through 1 + 0 I do all go to zero? Those (if any) which do not are satellites of the MRB constant.

POSTED BY: Udo Krause
6 Replies

Using Degree (Degree Table[x, {x, 0, 172.5, 7.5}]) we get a good alignment on several of the points of a "concentric circle" for a special set of k. Enter

Remove[burnsStrip]
burnsStrip[k_?NumericQ, w_?NumericQ] := 
 Block[{m = RotationTransform[k, {-1, 0}]}, 
  Polygon[{m[{-5/2, -w}], m[{1/2, -w}], m[{1/2, w}], m[{-5/2, w}]}]]

Graphics[{{Green, burnsStrip[0, 1/40]}, {Yellow, 
   Sequence @@ (burnsStrip[#, 1/40] & /@
      (Degree Table[x, {x, 0, 172.5, 7.5}]))},
  Table[k = 2 (3 n^3); 
   Point /@ 
    Evaluate[
     ReIm[(-1)^
        k ((Flatten[Evaluate[Block[{x}, Solve[x^k == k, x]]]])[[All, 
           2]] - 1)]], {n, 1, 5}]}, Frame -> True, AspectRatio -> 3/3,
  PlotRange -> All, PlotLabel -> "Let the sunshine in!"]

and get,

enter image description here

For more detail and more accuracy in the top half, try the code,

Remove[burnsStrip]
burnsStrip[k_?NumericQ, w_?NumericQ] := 
 Block[{m = RotationTransform[k, {-1, 0}]}, 
  Polygon[{m[{-5/2, -w/4}], m[{1/2, -w/4}], m[{1/2, w/4}], 
    m[{-5/2, w}/4]}]]

Graphics[{{Green, burnsStrip[0, 1/40]}, {Yellow, 
   Sequence @@ (burnsStrip[#, 1/40] & /@
      (Degree Table[x, {x, 0, 180, 7.5/8}]))},
  Table[k = 2 (3 n^3); 
   Point /@ 
    Evaluate[
     ReIm[(-1)^
        k ((Flatten[Evaluate[Block[{x}, Solve[x^k == k, x]]]])[[All, 
           2]] - 1)]], {n, 1, 4}]}, Frame -> True, AspectRatio -> 3/3,
  PlotRange -> All, PlotLabel -> "Let the sunshine in!", 
 ImageSize -> 2000]

for what a portion of which looks like the following:

enter image description here

The bottom half of which looks like the points are shifted over by one unit: enter image description here

POSTED BY: Marvin Ray Burns

These appear to be the ensemble of kth roots of k for k from 1 onward. As the pictures pretty much show, the set (-1)^k*(rootsofk-1) will approach the two translates of the unit circle in the complex plane, by +-1 along the x axis.

From here it is not clear to me which are to be selected in forming variants on the MRB evaluation. One possibility is to fix an integer j, and in a given summation always use the jth root counting counterclockwise starting (j=0, that is) at the principle root on the positive x axis (and when j is larger in magnitude than k, take j modulo k). So for j=0 we recover the usual MRB summation.

What happens for other values of j? It is not difficult to show that for fixed j and k sufficiently large, these terms approach the usual MRB summands and the difference is (-1)^k*k^(1/k)(Exp[I*j/(2*k)]-1). This in turn is approximated, to first order, by I*j/(2*k). So the tails of the modified and standard MRB sums will start to differ by roughly I*j/2 times the tail of the standard alternating harmonic series.

I'm not sure whether this tells us much about what these variant summations look like. Mostly I was proposing a possible use for the different kth roots of k since it was not clear what would be their role in these alternate sums.

POSTED BY: Daniel Lichtblau

I think an important thought we can take home about the points, or "terms of the partial sums of the many satellites (on straight lines through -1 + 0 I or on straight lines through 1 + 0 I) of the MRB constant," or roots of x^k=k is that for every k there are a finite number of unique ones! or at least so it seems !!!!!!!!

POSTED BY: Marvin Ray Burns

The points given by specific values of k follow the pattern shown in the following snippet, were k=13 is used for a clear example.

enter image description here

Here is the graph with just the points from k=13:

enter image description here

Here are the points with k=4 and their graph: (I think there are less unique points because with k=4 you have points made from sin of multiples of 2 Pi, which give repeated points.)

{{Point[{-1 - Sqrt[2], 0}], Point[{-1, -Sqrt[2]}], 
  Point[{-1, Sqrt[2]}], Point[{-1 + Sqrt[2], 0}]}}

enter image description here

k =( An odd number ^n) gives points that make up the RHS unit circle (or one close to it) with concentric circles of less and less points, while points that make up the LHS unit circle (or one close to it) with concentric circles of less and less points are given by k =( an even number ^n):

enter image description here

A remarkable quality of the points is that the sum of all k=(positive integer, b)^(positive integer, n), is simply b^n !!!!!!!

Table[k = b^n; 
  N[Total[Flatten[
     Evaluate[
      ReIm[(-1)^
         k ((Flatten[Evaluate[Block[{x}, Solve[x^k == k, x]]]])[[All, 
            2]] - 1)]]]]], {n, 2, 7}, {b, 2, 7}] // TableForm

gives

A huge list of all such points

followed by

enter image description here

POSTED BY: Marvin Ray Burns

: Vitaliy Kaurov et al, I looked at the OP and tried to understand what was happening. I could still use some more insight into it! For instance, what happens when you sum the terms of each specific "satellite" of the MRB? And what is their sum in terms of the MRB? And why are the terms of each specific "satellite" follow the path given by the mentioned angle i.e.. Degree {143.8, 120, 90, 60, 36.2}. ? And why are those points of the "satellites" outside that negative unit circle, and not inside?

I will see what insight others can bring to the table, here, and probably build upon those insights. Thank you @Udo Krause for opening up the door to this previously undiscovered relationship!

POSTED BY: Marvin Ray Burns

Very interesting @Udo Krause, I am sure @Marvin Ray Burns would love to take a look.

POSTED BY: Vitaliy Kaurov
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