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

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<h5>introduction</h5>
 
<h5>introduction</h5>
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* [http://pythagoras0.springnote.com/pages/4682477 체비셰프 다항식]<br><math>U_n(x)^2=1+U_{n-1}(x)U_{n+1}(x)</math><br>
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* [http://pythagoras0.springnote.com/pages/6904713 정다각형의 대각선의 길이]<br><math>r_i^2=1+r_{i-1}r_{i+1}, 1\leq i \leq n-3</math><br>
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* Question : for what values of <math>r_1=x</math>, is the recurrence <math>r_i^2=1+r_{i-1}r_{i+1}</math> periodic? (<math>r_0=1</math>)
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# A := RecurrenceTable[{a[n] a[n - 2] + 1 == a[n - 1]^2, a[1] == x,<br>    a[2] == y}, a, {n, 10}]<br> Simplify[A]
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* Laurent phenomenon is true
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* total positivity is broken
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* 정오각형의 경우
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* <math>r_i^2=1+r_{i-1}r_{i+1}</math>, <math>r_0=1,r_3=1</math>
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* 3가지 점화식의 해가 존재
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* <math>\{1,-1,0,1\}</math>, <math>\{1,\frac{-\sqrt{5}+1}{2},\frac{-\sqrt{5}+1}{2},1 \}</math> , <math>\{1,\frac{\sqrt{5}+1}{2},\frac{\sqrt{5}+1}{2},1 \}</math>
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# A := RecurrenceTable[{a[n] a[n - 2] + 1 == a[n - 1]^2, a[1] == 1,<br>    a[2] == 2 y}, a, {n, 10}]<br> Simplify[A]<br> NSolve[-4 y + 8 y^3 == 1, y]<br> {1, 2 y, -1 + 4 y^2, -4 y + 8 y^3,<br>    1 - 12 y^2 +<br>     16 y^4} /. {{y -> -0.5`}, {y -> -0.30901699437494745`}, {y -><br>      0.8090169943749475`}} // TableForm
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<h5>total positivity</h5>
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* <math>r_{i-1}r_{i+1}=r_i^2+1</math>
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# A := RecurrenceTable[{a[n] a[n - 2] - 1 == a[n - 1]^2, a[1] == x,<br>    a[2] == y}, a, {n, 10}]<br> Simplify[A]
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* [[rank 2 cluster algebra examples]]
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<h5>relation to 5-term relation</h5>
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* [http://pythagoras0.springnote.com/pages/5956565 5항 관계식 (5-term relation)]<br><math>1-x_{i}=x_{i-1}x_{i+1}</math><br>
  
 
 
 
 
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<h5>related items</h5>
 
<h5>related items</h5>
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* [[cluster algebra]]
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* [[Nahm's equation]][[cluster algebra and dilogarithm identities|]]
  
 
* [[3 central charge of CFT, L-values, volume of threefolds and dilogarithm]]
 
* [[3 central charge of CFT, L-values, volume of threefolds and dilogarithm]]

2011년 3월 12일 (토) 09:30 판

introduction

 

  1. A := RecurrenceTable[{a[n] a[n - 2] + 1 == a[n - 1]^2, a[1] == x,
       a[2] == y}, a, {n, 10}]
    Simplify[A]
  • Laurent phenomenon is true
  • total positivity is broken
  • 정오각형의 경우
  • \(r_i^2=1+r_{i-1}r_{i+1}\), \(r_0=1,r_3=1\)
  • 3가지 점화식의 해가 존재
  • \(\{1,-1,0,1\}\), \(\{1,\frac{-\sqrt{5}+1}{2},\frac{-\sqrt{5}+1}{2},1 \}\) , \(\{1,\frac{\sqrt{5}+1}{2},\frac{\sqrt{5}+1}{2},1 \}\)

 

  1. A := RecurrenceTable[{a[n] a[n - 2] + 1 == a[n - 1]^2, a[1] == 1,
       a[2] == 2 y}, a, {n, 10}]
    Simplify[A]
    NSolve[-4 y + 8 y^3 == 1, y]
    {1, 2 y, -1 + 4 y^2, -4 y + 8 y^3,
       1 - 12 y^2 +
        16 y^4} /. {{y -> -0.5`}, {y -> -0.30901699437494745`}, {y ->
         0.8090169943749475`}} // TableForm

 

 

 

total positivity
  • \(r_{i-1}r_{i+1}=r_i^2+1\)
  1. A := RecurrenceTable[{a[n] a[n - 2] - 1 == a[n - 1]^2, a[1] == x,
       a[2] == y}, a, {n, 10}]
    Simplify[A]

 

 

relation to 5-term relation

 

 

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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