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Mathematics Mathematica Calculus

A Bunch Of MRB Constants?

Udo Krause

Udo Krause, NOrganization

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 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

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Marvin Ray Burns

Marvin Ray Burns, original investigator of the MRB constant

Posted 9 years ago

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, 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: The bottom half of which looks like the points are shifted over by one unit:

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

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 9 years ago

These appear to be the ensemble of `k`th 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 `j`th 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)^kk^(1/k)(Exp[Ij/(2k)]-1)`. This in turn is approximated, to first order, by `Ij/(2k)`. 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 `k`th roots of `k` since it was not clear what would be their role in these alternate sums.

POSTED BY: Daniel Lichtblau

Marvin Ray Burns

Marvin Ray Burns, original investigator of the MRB constant

Posted 9 years ago

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 !!!!!!!!