PSLQ for dilogarithm identities

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imported>Pythagoras0님의 2012년 10월 29일 (월) 10:09 판
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introduction

 

 

Implement the PSLQ algorithm first.

 

  1.  
    PSLQ[inx_List, prec_] :=
     Block[
      {
       x,
       n = Length[inx],
       \[Gamma] = 2/Sqrt[3],
       A, B, H, D, Dinv, t, i, j, k, l, iter,
       \[Alpha], \[Beta], \[Lambda], \[Delta], r, R
       },
      (*Initialize*)
      x = N[inx /Sqrt[inx . inx], prec];
      s = Sqrt[MapIndexed[Plus @@ Drop[x^2, First[#2] - 1] &, x]];
      A = B = IdentityMatrix[n];
      H = Table[Which[
         i > j, (-xi*xj)/(sj*sj + 1),
         i == j, si + 1/si,
         i < j, 0
         ], {i, 1, n}, {j, 1, n - 1}];
      (* Reduce H *)
      t = HermiteReduce[H];
      D = First[t];
      Dinv = Inverse[D];
      (*Update*)
      H = Last[t]; x = x.Dinv; A = D.A; B = B.Dinv;
      For[iter = 0, iter < $IterationLimit, ++iter,
       (* Step One *)
       r = MaxIndex[MapIndexed[\[Gamma]^First[#2] Abs[#1] &, Tr[H, List]]];
       If[r < n - 1, \[Alpha] = Hr, r; \[Beta] = 
         Hr + 1, r; \[Lambda] = Hr + 1, r + 1; \[Delta] = 
         Sqrt[\[Beta]^2 + \[Lambda]^2]];
       R = IdentityMatrix[n]; t = Rr; Rr = Rr + 1
       Rr + 1 = t;
       x = x.R; H = R.H; A = R.A; B = B.R;
       (* Step Two *)
       If[r < n - 1,
        H = H.Table[
           Which[
            i == r && j == r, \[Beta]/\[Delta],
            i == r && j == r + 1, -\[Lambda]/\[Delta],
            i == r + 1 && j == r, \[Lambda]/\[Delta],
            i == r + 1 && j == r + 1, \[Beta]/\[Delta],
            i == j && j != r || i == j && j != r + 1, 1,
            True, 0],
           {i, 1, n - 1}, {j, 1, n - 1}]
        ];
       (* Step Three *)
       t = HermiteReduce[H];
       D = First[t];
       Dinv = Inverse[D];
       (*Update*)
       H = Last[t]; x = x.Dinv; A = D.A; B = B.Dinv;
       (* Step Four *)
       If[Min[Abs[Union[x, Tr[H, List]]]] <= 10^(-prec + 5), Break[]]
       ];(*Main Iteraton*)
      Return[Transpose[B][[MaxIndex[-Abs[x]]]]]
      ]

 

Then find a relation.

 

  1. Clear[a]
    f[x_] := x^3 + x - 1
    Solve[f[x] == 0, x]
    N[%]
    a := -(2/(3 (9 + Sqrt[93])))^(1/3) + (1/2 (9 + Sqrt[93]))^(1/3)/3^(2/3)
    N[a, 20]
    L[x_] := PolyLog[2, x] + 1/2 Log[x] Log[1 - x]
  2. L[1] := Pi^2/6
  3. (* choose a expected height of the dilogarithm identity *)
  4. deg:=6
    S := Table[L[a^i],{i,0,deg}]
    N[S, 100]
    PSLQ[N[S, 100], 1000]
  5. N[N[S, 50].%, 50]

 

 

I found 

\(-2L(1)+2L(\alpha)+2L(\alpha^{2})-L(\alpha^{4})=0\) or

\(2L(\alpha)+2L(\alpha^{2})-L(\alpha^{4})=\frac{\pi^2}{3}\)

 

 

 

 

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