"PSLQ for dilogarithm identities"의 두 판 사이의 차이

수학노트
둘러보기로 가기 검색하러 가기
imported>Pythagoras0
잔글 (찾아 바꾸기 – “<h5 (.*)">” 문자열을 “==” 문자열로)
imported>Pythagoras0
 
(같은 사용자의 중간 판 7개는 보이지 않습니다)
1번째 줄: 1번째 줄:
 
==introduction==
 
==introduction==
 +
* Implement the [[PSLQ algorithm]] first.
 +
* I found
  
−
 
+
<math>-2L(1)+2L(\alpha)+2L(\alpha^{2})-L(\alpha^{4})=0</math> or
  
−
 
+
<math>2L(\alpha)+2L(\alpha^{2})-L(\alpha^{4})=\frac{\pi^2}{3}</math>
  
−
Implement the [[PSLQ algorithm]] first.
+
  
−
 
 
−
 
−
#   <br> PSLQ[inx_List, prec_] :=<br>  Block[<br>   {<br>    x,<br>    n = Length[inx],<br>    \[Gamma] = 2/Sqrt[3],<br>    A, B, H, D, Dinv, t, i, j, k, l, iter,<br>    \[Alpha], \[Beta], \[Lambda], \[Delta], r, R<br>    },<br>   (*Initialize*)<br>   x = N[inx /Sqrt[inx . inx], prec];<br>   s = Sqrt[MapIndexed[Plus @@ Drop[x^2, First[#2] - 1] &, x]];<br>   A = B = IdentityMatrix[n];<br>   H = Table[Which[<br>      i > j, (-x[[i]]*x[[j]])/(s[[j]]*s[[j + 1]]),<br>      i == j, s[[i + 1]]/s[[i]],<br>      i < j, 0<br>      ], {i, 1, n}, {j, 1, n - 1}];<br>   (* Reduce H *)<br>   t = HermiteReduce[H];<br>   D = First[t];<br>   Dinv = Inverse[D];<br>   (*Update*)<br>   H = Last[t]; x = x.Dinv; A = D.A; B = B.Dinv;<br>   For[iter = 0, iter < $IterationLimit, ++iter,<br>    (* Step One *)<br>    r = MaxIndex[MapIndexed[\[Gamma]^First[#2] Abs[#1] &, Tr[H, List]]];<br>    If[r < n - 1, \[Alpha] = H[[r, r]]; \[Beta] = <br>      H[[r + 1, r]]; \[Lambda] = H[[r + 1, r + 1]]; \[Delta] = <br>      Sqrt[\[Beta]^2 + \[Lambda]^2]];<br>    R = IdentityMatrix[n]; t = R[[r]]; R[[r]] = R[[r + 1]]; <br>    R[[r + 1]] = t;<br>    x = x.R; H = R.H; A = R.A; B = B.R;<br>    (* Step Two *)<br>    If[r < n - 1,<br>     H = H.Table[<br>        Which[<br>         i == r && j == r, \[Beta]/\[Delta],<br>         i == r && j == r + 1, -\[Lambda]/\[Delta],<br>         i == r + 1 && j == r, \[Lambda]/\[Delta],<br>         i == r + 1 && j == r + 1, \[Beta]/\[Delta],<br>         i == j && j != r || i == j && j != r + 1, 1,<br>         True, 0],<br>        {i, 1, n - 1}, {j, 1, n - 1}]<br>     ];<br>    (* Step Three *)<br>    t = HermiteReduce[H];<br>    D = First[t];<br>    Dinv = Inverse[D];<br>    (*Update*)<br>    H = Last[t]; x = x.Dinv; A = D.A; B = B.Dinv;<br>    (* Step Four *)<br>    If[Min[Abs[Union[x, Tr[H, List]]]] <= 10^(-prec + 5), Break[]]<br>    ];(*Main Iteraton*)<br>   Return[Transpose[B][[MaxIndex[-Abs[x]]]]]<br>   ]<br>
 
−
 
−
 
 
−
 
−
Then find a relation.
 
−
 
−
 
 
−
 
−
#  Clear[a]<br> f[x_] := x^3 + x - 1<br> Solve[f[x] == 0, x]<br> N[%]<br> a := -(2/(3 (9 + Sqrt[93])))^(1/3) + (1/2 (9 + Sqrt[93]))^(1/3)/3^(2/3)<br> N[a, 20]<br> L[x_] := PolyLog[2, x] + 1/2 Log[x] Log[1 - x]<br>
 
−
#  L[1] := Pi^2/6<br>
 
−
#  (* choose a expected height of the dilogarithm identity *)<br>
 
−
#  deg:=6<br> S := Table[L[a^i],{i,0,deg}]<br> N[S, 100]<br> PSLQ[N[S, 100], 1000]<br>
 
−
#  N[N[S, 50].%, 50]<br>
 
−
 
−
 
 
−
 
−
 
 
−
 
−
I found 
 
−
 
−
<math>-2L(1)+2L(\alpha)+2L(\alpha^{2})-L(\alpha^{4})=0</math> or
 
−
 
−
<math>2L(\alpha)+2L(\alpha^{2})-L(\alpha^{4})=\frac{\pi^2}{3}</math><br>
 
−
 
−
 
 
−
 
−
 
 
−
 
−
 
 
−
 
−
 
 
−
 
−
==history==
 
−
 
−
* http://www.google.com/search?hl=en&tbs=tl:1&q=
 
−
 
−
 
 
−
 
−
 
 
  
 
==related items==
 
==related items==
  
−
* [[PSLQ algorithm]]<br>
+
* [[PSLQ algorithm]]
−
* [[Slater 34]]<br>
+
* [[Slater 34]]
−
 
 
−
 
 
−
 
 
−
 
 
−
 
 
−
==encyclopedia==
 
−
 
 
−
* http://en.wikipedia.org/wiki/
 
−
* http://www.scholarpedia.org/
 
−
* http://www.proofwiki.org/wiki/
 
−
* Princeton companion to mathematics([[2910610/attachments/2250873|Companion_to_Mathematics.pdf]])
 
−
 
 
−
 
 
−
 
 
−
 
 
−
 
 
−
==books==
 
−
 
 
−
 
 
−
 
 
−
* [[2010년 books and articles]]<br>
 
−
* http://gigapedia.info/1/
 
−
* http://gigapedia.info/1/
 
−
* http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
 
−
 
 
−
[[4909919|4909919]]
 
−
 
 
−
 
 
−
 
 
−
 
 
−
 
 
−
==articles==
 
−
 
 
−
 
 
−
 
 
−
* http://www.ams.org/mathscinet
 
−
* [http://www.zentralblatt-math.org/zmath/en/ ]http://www.zentralblatt-math.org/zmath/en/
 
−
* http://arxiv.org/
 
−
* http://www.pdf-search.org/
 
−
* http://pythagoras0.springnote.com/
 
−
* http://math.berkeley.edu/~reb/papers/index.html
 
−
* http://dx.doi.org/
 
−
 
 
−
 
 
−
 
 
−
 
 
−
 
 
−
==question and answers(Math Overflow)==
 
−
 
 
−
* http://mathoverflow.net/search?q=
 
−
* http://mathoverflow.net/search?q=
 
−
 
 
−
 
 
−
 
 
−
 
 
−
 
 
−
==blogs==
 
−
 
 
−
*  구글 블로그 검색<br>
 
−
**  http://blogsearch.google.com/blogsearch?q=<br>
 
−
** http://blogsearch.google.com/blogsearch?q=
 
−
* http://ncatlab.org/nlab/show/HomePage
 
−
 
 
−
 
 
−
 
 
−
 
 
−
 
 
−
==experts on the field==
 
  
−
* http://arxiv.org/
 
  
−
 
+
==computational resource==
 +
* https://docs.google.com/file/d/0B8XXo8Tve1cxNW0ta2hCUVJPdmc/edit
  
−
 
 
  
−
==links==
 
  
−
* [http://detexify.kirelabs.org/classify.html Detexify2 - LaTeX symbol classifier]
+
[[분류:개인노트]]
−
* [http://pythagoras0.springnote.com/pages/1947378 수식표현 안내]
+
[[Category:계산]]
−
* [http://www.research.att.com/~njas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
+
[[Category:quantum dimensions]]
−
* http://functions.wolfram.com/
+
[[분류:dilogarithm]]
−
[[분류:개인노트]][[분류:개인노트]]
+
[[분류:migrate]]

2020년 11월 12일 (목) 22:53 기준 최신판

introduction

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

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



related items


computational resource

원본 주소 "https://wiki.mathnt.net/index.php?title=PSLQ_for_dilogarithm_identities&oldid=34982"