"Y-system and functional dilogarithm identities"의 두 판 사이의 차이

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12번째 줄: 12번째 줄:
 
* is this also an example of a cluster variable?
 
* is this also an example of a cluster variable?
 
* [[asymptotic analysis of basic hypergeometric series]]
 
* [[asymptotic analysis of basic hypergeometric series]]
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# f[{x_, y_, z_, w_}] := Simplify[(x - z)/(x - w)*(y - w)/(y - z)]<br> A := Permutations[{0, 1, w, z}]<br> Table[Limit[f[A[[i]]], w -> \[Infinity]], {i, 24}]<br> B := Subsets[{0, x*y, 1, y, z}, {4}]<br> g[i_] := Table[<br>   Limit[f[n], z -> \[Infinity]], {n, Permutations[B[[i]]]}]<br> Table[f[B[[i]]], {i, 1, 5}]<br> Table[g[i], {i, 5}]
  
 
 
 
 
20번째 줄: 24번째 줄:
  
 
* [[rank 2 cluster algebra]]
 
* [[rank 2 cluster algebra]]
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<math>y_{m-1}y_{m+1}=y_m+1</math>
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Start with two variables <math>y_1,y_2</math>.
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<math>y_3y_1=y_2+1</math>. so <math>y_3=\frac{y_2+1}{y_1}</math>
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<math>y_2y_4=y_3+1 </math>implies <math>y_4=\frac{y_3+1}{y_2}=\frac{y_1+y_2+1}{y_1y_2}</math>
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<math>y_3y_5=y_4+1</math> implies <math>y_5=\frac{y_4+1}{y_3}= \frac{y_1+1}{y_2}</math> we are getting Laurent polynomials
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<math>y_4y_6=y_5</math> implies <math>y_6=\frac{y_5+1}{y_4}= \frac{\frac{y_1+1}{y_2}+1}{\frac{y_1+y_2+1}{y_1y_2}}=\frac{y_1(y_1+1)+y_1y_2}{y_1+y_2+1}=y_1</math>
  
 
 
 
 

2011년 2월 25일 (금) 12:25 판

introduction

 

 

five-term relation of dilogarithm

 

  1. f[{x_, y_, z_, w_}] := Simplify[(x - z)/(x - w)*(y - w)/(y - z)]
    A := Permutations[{0, 1, w, z}]
    Table[Limit[f[Ai], w -> \[Infinity]], {i, 24}]
    B := Subsets[{0, x*y, 1, y, z}, {4}]
    g[i_] := Table[
      Limit[f[n], z -> \[Infinity]], {n, Permutations[Bi]}]
    Table[f[Bi], {i, 1, 5}]
    Table[g[i], {i, 5}]

 

 

rank 2 example

\(y_{m-1}y_{m+1}=y_m+1\)

Start with two variables \(y_1,y_2\).

\(y_3y_1=y_2+1\). so \(y_3=\frac{y_2+1}{y_1}\)

\(y_2y_4=y_3+1 \)implies \(y_4=\frac{y_3+1}{y_2}=\frac{y_1+y_2+1}{y_1y_2}\)

\(y_3y_5=y_4+1\) implies \(y_5=\frac{y_4+1}{y_3}= \frac{y_1+1}{y_2}\) we are getting Laurent polynomials

\(y_4y_6=y_5\) implies \(y_6=\frac{y_5+1}{y_4}= \frac{\frac{y_1+1}{y_2}+1}{\frac{y_1+y_2+1}{y_1y_2}}=\frac{y_1(y_1+1)+y_1y_2}{y_1+y_2+1}=y_1\)

 

 

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