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jy:results [2018/07/25 02:05] – [7.24.18 Inverse Problem] yilingjy:results [2019/09/16 09:16] – yiling
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 +====== 2019-09-16 :Case N1: long tail antiferromagentic ======
 +
 +
 +
 ====== 2018-07-2 :The solutions to the inverse Ising problem  ====== ====== 2018-07-2 :The solutions to the inverse Ising problem  ======
  
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 ====== 7.19.18 Inverse Problem  ====== ====== 7.19.18 Inverse Problem  ======
  
-The mean field theory (nMF): +{{page>dynstat:mean-field}}
- +
-$\mathbf{J}^\text{nMF}=-\mathbf{C}^{-1}, (i \neq j) $, +
- +
-$h_{i}^\text{nMF}=\tanh^{-1}\left< s_{i} \right>-\sum_{j=1}^{N} J{ij}^\text{nMF}\left< s_{j} \right> $ . +
- +
- +
-The independent-pair theory (IP): +
- +
-$J_{ij}^\text{pair}=0.25*\ln\left[ \frac{(1+m_{i}+m_{j}+C_{ij}^{*}) (1-m_{i}-m_{j}+C_{ij}^{*})} {(1-m_{i}+m_{j}-C_{ij}^{*})(1+m_{i}-m_{j}-C_{ij}^{*})} \right]$, with $C_{ij}^{*}=C_{ij}+m_{i}m_{j} , (i \neq j)$ +
- +
-$h_{i}^\text{pair}=0.5*\ln\left( \frac{1+m_{i}} {1-m_{i}} \right)-\sum_{j=1}^{N} J_{ij}^\text{pair} m_{j}$. +
- +
- +
-The Thouless-Anderson-Palmer theory (TAP): +
- +
-$\mathbf{C}^{-1}+J_{ij}^\text{TAP}+2(J_{ij}^\text{TAP})^{2}m_{i}m_{j}=0 , (i \neq j)$ +
- +
-$h_{i}^\text{TAP}=h_{i}^\text{nMF}-m_{i}\sum_{i=1}^{N} (J_{ij}^\text{TAP})^{2} (1-m_{j})^{2}$. +
- +
- +
-The Sessak and Monasson theory (SM): +
- +
-$J_{ij}^\text{SM}=J_{ij}^\text{nMF}+J_{ij}^\text{pair}-\frac{C_{ij}}{(1-m_{i}^{2})(1-m_{j}^{2})-C_{ij}^{2}} , (i \neq j)$ +
- +
-$h_{i}^\text{SM}=0.5*\ln\left( \frac{1+m_{i}} {1-m_{i}} \right)-\sum_{j=1}^{N} J_{ij}^\text{SM} m_{j}$.+
  
 The coefficient of $R^{2}$ is: The coefficient of $R^{2}$ is:
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-====== 7.24.18 Inverse Problem  ======+======== 7.24.18 Inverse Problem  ========
  
 ^ h:n(0.0,0.07) & J:n(0.0,0.07), n=200 at T=1   ^ $\sqrt{200}$ ^ ^ h:n(0.0,0.07) & J:n(0.0,0.07), n=200 at T=1   ^ $\sqrt{200}$ ^
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 | h:n(0.0,0.22) & J:n(0.0,0.22), n=20  at T=1   | $\sqrt{20} $ | | h:n(0.0,0.22) & J:n(0.0,0.22), n=20  at T=1   | $\sqrt{20} $ |
  
- 
-We use the normal distribution of h and J to get the coefficients of $R^{2}$ and the RMSerror. 
  
 We set the mean equal to zero and the variance of normal distribution of h and J  We set the mean equal to zero and the variance of normal distribution of h and J 
 from $\frac{1}{\sqrt{200}}$: $\frac{1}{\sqrt{150}}$: $\frac{1}{\sqrt{100}}$:$\frac{1}{\sqrt{50}}$:$\frac{1}{\sqrt{20}} \approx 0.07 : 0.08 : 0.1 : 0.14 : 0.22 $. from $\frac{1}{\sqrt{200}}$: $\frac{1}{\sqrt{150}}$: $\frac{1}{\sqrt{100}}$:$\frac{1}{\sqrt{50}}$:$\frac{1}{\sqrt{20}} \approx 0.07 : 0.08 : 0.1 : 0.14 : 0.22 $.
 +
 +We use the normal distribution of h and J to get the coefficients of $R^{2}$ and the RMSerror at T=1.
  
  
 {{jy:RMSt1.png?400}} {{jy:RMSt1.png?400}}
 {{jy:RR.png?400}} {{jy:RR.png?400}}
 +
 +{{jy:hrms.png?400}}
 +{{jy:hRR.png?400}}
  
  
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 {{jy:h20_0.22t1.png}} {{jy:h20_0.22t1.png}}
 {{jy:j20_0.22t1.png}} {{jy:j20_0.22t1.png}}
 +
 +====== 7.30.18 Inverse Problem data ======
 +{{jy:j78.png}}
 +{{jy:ha78.png?450}}
 +{{jy:ja78.png?450}}
 +
 +{{jy:j781.png}}
 +{{jy:h78a1.png?450}}
 +{{jy:j78a1.png?450}}
 +