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| -rw-r--r-- | 2-point.tex | 67 |
1 files changed, 22 insertions, 45 deletions
diff --git a/2-point.tex b/2-point.tex index e8a7a57..f0fec59 100644 --- a/2-point.tex +++ b/2-point.tex @@ -529,17 +529,17 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case, \begin{align} &\log\det \begin{bmatrix} - C^{00}&iR^{00}&C^{01}&iR^{01}&X_1\\ - iR^{00}&D^{00}&iR^{10}&D^{01}&\hat X_1\\ - C^{01})^T&iR^{10})^T&C^{11}&iR^{11}&X_2\\ - iR^{01})^T&D^{10})^T&iR^{11}&D^{11}&\hat X_2\\ - X_1)^T&\hat X_1)^T&X_2)^T&\hat X_2)^T&A + C^{00}&iR^{00}&C^{01}&iR^{01}&X_0\\ + iR^{00}&D^{00}&iR^{10}&D^{01}&\hat X_0\\ + (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}&X_1\\ + (iR^{01})^T&(D^{10})^T&iR^{11}&D^{11}&\hat X_1\\ + X_0^T&\hat X_0^T&X_1^T&\hat X_1^T&A \end{bmatrix}\\ &=\log\det\left( A- \begin{bmatrix} - X_1\\\hat X_1\\X_2\\\hat X_2 - \end{bmatrix})^T + X_0\\\hat X_0\\X_1\\\hat X_1 + \end{bmatrix}^T \begin{bmatrix} C^{00}&iR^{00}&C^{01}&iR^{01}\\ iR^{00}&D^{00}&iR^{10}&D^{01}\\ @@ -547,11 +547,18 @@ Because the matrices $C^{00}$, $R^{00}$, and $D^{00}$ are diagonal in this case, (iR^{01})^T&(D^{10})^T&iR^{11}&D^{11}\\ \end{bmatrix}^{-1} \begin{bmatrix} - X_1\\\hat X_1\\X_2\\\hat X_2 + X_0\\\hat X_0\\X_1\\\hat X_1 \end{bmatrix} \right) \end{align} \begin{equation} + \begin{bmatrix} + C^{00}&iR^{00}&C^{01}&iR^{01}\\ + iR^{00}&D^{00}&iR^{10}&D^{01}\\ + (C^{01})^T&(iR^{10})^T&C^{11}&iR^{11}\\ + (iR^{01})^T&(D^{10})^T&iR^{11}&D^{11}\\ + \end{bmatrix}^{-1} + = \begin{bmatrix} A & B \\ C & D @@ -745,7 +752,7 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are && \hat X_1 =\begin{bmatrix} - 0&\cdots&0\\ + \hat x_1^0&\cdots&\hat x_1^0\\ \hat x_1^1&\cdots&\hat x_1^1\\ \vdots&\ddots&\vdots\\ \hat x_1^1&\cdots&\hat x_1^1 @@ -753,51 +760,21 @@ which is \emph{not} a hierarchical matrix! But, all these new hat variables are \end{align} \[ - x_0^2\tilde d^{00}_\mathrm d-\hat x_0^2\tilde c^{00}_\mathrm d+2x\hat x\tilde r^{00}_\mathrm d - -2 - \begin{bmatrix} - x_0q\tilde d^{00}_\mathrm d+\hat x_0q\tilde r^{00}_\mathrm d+x_0r_{10}\tilde r^{00}_\mathrm d-\hat x_0r_{10}\tilde c^{00}_\mathrm d - \\ - i(x_0(r_{01}\tilde d^{00}_\mathrm d-d_{01}\tilde d^{00}_\mathrm d) - +\hat x_0(d_{01}\tilde c^{00}_\mathrm d+r_{01}\tilde r^{00}_\mathrm d)) - \end{bmatrix}^T - \begin{bmatrix} - C^{11}&iR^{11}\\iR^{11}&D^{11} - \end{bmatrix}^{-1} - \begin{bmatrix} - X_1\\i\hat X_1 - \end{bmatrix} -\] -\begin{align} - \hat c^{11}=\sum_{i}^n\sum_{j=2}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}D^{11}]_{ij}\\ - \hat r^{11}=\sum_{i}^n\sum_{j=2}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}R^{11}]_{ij}\\ - \hat d^{11}=\sum_{i}^n\sum_{j=2}^n[(C^{11}D^{11}+R^{11}R^{11})^{-1}D^{11}]_{ij} -\end{align} -\[ - \begin{aligned} - &(\tilde d^{00}_\mathrm d\hat d^{11}q+\tilde r^{00}_\mathrm d\hat d^{11}r_{10}-\tilde r^{00}_\mathrm d\hat r^{11}d_{01}+\tilde d^{00}_\mathrm d\hat r^{11}r_{01})x_0x_1 - +(\tilde r^{00}_\mathrm d\hat d^{11}q-\tilde c^{00}_\mathrm d\hat d^{11}r_{10}+\tilde c^{00}_\mathrm d\hat r^{11}d_{01}+\tilde r^{00}_\mathrm d\hat r^{11}r_{01})\hat x_0x_1 \\ - &+(\tilde r^{00}_\mathrm d\hat c^{11}d_{01}-\tilde d^{00}_\mathrm d\hat c^{11}r_{01}+\tilde d^{00}_\mathrm d\hat r^{11}q+\tilde r^{00}_\mathrm d\hat r^{11}r_{10})x_0\hat x_1 - -(\tilde c^{00}_\mathrm d\hat c^{11}d_{01}+\tilde r^{00}_\mathrm d\hat c^{11}r_{01}-\tilde r^{00}_\mathrm d\hat c^{11}q+\tilde c^{00}_\mathrm d\hat r^{11}r_{10})\hat x_0\hat x_1 - \end{aligned} -\] -all a constant $\ell\times\ell$ matrix. - -\[ \log\det(A-c)=\log\det A-\frac{c}{\sum_{i=0}^k(a_{i+1}-a_i)x_{i+1}} \] where $a_{k+1}=1$ and $x_{k+1}=1$. So the basic form of the action is (for replica symmetric $A$) \[ - \frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\beta\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TB\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}-\frac1{1-a_0}\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix}^TC\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1\end{bmatrix} + \frac14\beta^2f''(1)(1-a_0^2)+\frac12\log(1-a_0)+\frac12\frac{a_0}{1-a_0}+\frac12\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix}^T\left(\beta B-\frac1{1-a_0}C\right)\begin{bmatrix}x_0\\\hat x_0\\x_1\\\hat x_1^1\\\hat x_1^0\end{bmatrix} \] for \[ B=\begin{bmatrix} - \hat\beta_0f''(q)+r_{10}f'''(q)&f''(q)&0&0\\ - f''(q)&0&0&0\\ - 0&0&-\hat\beta_1f''(q^{11}_0)-r^{11}_0f'''(q^{11}_0)&-f''(q_0^{11})\\ - 0&0&-f''(q_0^{11})&0 + \hat\beta_0f''(q)+r_{10}f'''(q)&f''(q)&0&0&0\\ + f''(q)&0&0&0&0\\ + 0&0&-\hat\beta_1f''(q^{11}_0)-r^{11}_0f'''(q^{11}_0)&-f''(q_0^{11})&0\\ + 0&0&-f''(q_0^{11})&0&0\\ + 0&0&0&0&0 \end{bmatrix} \] Use $X$ for the big vector. Then |