Closed lmoffatt closed 3 weeks ago
Now we can calculate the variations against $\Delta\beta_{i}^{i+1}$
A_{i,i+1} \approx \Phi \left (1,\Delta \beta_{i}) \cdot ( L_{i+1}-L_{i}), {\Delta \beta_{i}}^2 \cdot( vL_i + vL_{i+1})
\right)
\frac{ \partial A_{i,i+1} } {\partial \Delta \beta_{i}}
\approx
\mathcal{N} \left (1,\Delta \beta_{i} \cdot ( L_{i+ 1}-L_{i}),\Delta \beta_{i}^2 \cdot( vL_{i} + vL_{i+1}) \right)
\cdot
-\frac
{(\Delta \beta_{i}) \cdot \frac{\partial L_{i+1}}{\partial \beta_{i+1}} +( L_{i+1}-L_{i})}
{|\Delta \beta_{i}| \cdot \sqrt{vL_i + vL_{i+1}}}
-\frac
{1-\Delta \beta_{i}\cdot (L_{i+1}-L_{i}) }
{\Delta \beta_{i}^2\cdot \sqrt{(vL_i+vL_{i+1})}}
\cdot 2 \cdot \frac {\Delta \beta_{i}}{|\Delta \beta_{i}|}
that is
\frac
{( L_{i}-L_{i+1})-\Delta \beta_{i} \cdot \frac{\partial L_{i+1}}{\partial \beta_{i+1}} -\frac{2}{\Delta \beta_{i}} }
{|\Delta \beta_{i}|\cdot \sqrt{(vL_i+vL_{i+1})}}
\mathcal{N_p}(\Delta \beta_i) \equiv \mathcal{N} \left (1,\Delta \beta_{i}^{i+1} \cdot ( L_{i+ 1}-L_{i}),{\Delta \beta_{i}^{i+1}}^2 \cdot( vL_{i} + vL_{i+1}) \right)
\frac{ \partial A_i^{i+1} } {\partial \beta_{i}}
\approx
-\mathcal{N_p}(\Delta \beta_i^{i+1})
\cdot
\frac
{( L_{i+1}-L_{i})-\Delta \beta_{i} ^{i+1}
\cdot
\frac {\partial L_{i}}{\partial \beta_{i}} -\frac{2}{\Delta \beta_{i}^{i+1}} }
{|\Delta \beta_{i}^{i+1}|\cdot \sqrt{(vL_i+vL_{i+1})}}
\frac{ \partial A_i^{i+1} } {\partial \beta_{i+1}}
\approx
\mathcal{N_p}(\Delta \beta_i^{i+1})
\cdot
\frac
{( L_{i+1}-L_{i})-\Delta \beta_{i}^{i+1}
\cdot
\frac {\partial L_{i+1}}{\partial \beta_{i+1}} -\frac{2}{\Delta \beta_{i}^{i+1}} }
{|\Delta \beta_{i}^{i+1}|\cdot \sqrt{(vL_i+vL_{i+1})}}
\exp(s_i) =\frac {1}{\beta_{i}} -\frac {1}{\beta_{i+1} }
\exp(s_i) +\frac {1}{\beta_{i+1} }=\frac {1}{\beta_{i}}
\beta_{i} = \frac {1}{\frac {1}{\beta_{i+1} }+\exp(s)}
\beta_{i+1} = \frac {1}{\frac {1}{\beta_{i} }-\exp(s)}
\frac{\partial \beta_{i} }{ \partial s_{i+1}^i} =-\frac {exp(s)}{(\exp(s)+\frac {1}{\beta_{i+1} })^2} =-\exp(s)\cdot \beta_i ^2
= -(\frac {\beta_i ^2}{\beta_{i}} -\frac {\beta_i ^2}{\beta_{i+1} })
\frac{\partial \beta_{i} }{ \partial s_{i+1}^i} = \frac {\beta_i ^2}{\beta_{i+1} }-\beta_i
\frac{\partial \beta_{i+1} }{ \partial s_{i+1}^i} =\frac {-\exp(s)}{(-\exp(s)+\frac {1}{\beta_{i} })^2} =-(\frac {1}{\beta_{i}} -\frac {1}{\beta_{i+1} })\cdot \beta_{i+1} ^2
\frac{\partial \beta_{i+1} }{ \partial s_{i+1}^i} =\beta_{i+1} -\frac {\beta_{i+1} ^2}{\beta_{i}}
Now the expression that we want to make zero is $A{i+1}^{i+2} - A{i}^{i+1}$ by optimizing $s_i^{i+1} $ so, we differentiate:
\frac{ \partial (A_{i+1}^{i+2} - A_{i}^{i+1}) } {\partial s_i^{i+1}}
= \frac{ \partial A_{i+1}^{i+2} } {\partial s_i^{i+1}} -\frac{ \partial A_{i}^{i+1} } {\partial s_i^{i+1}}
\frac{ \partial (A_{i+1}^{i+2} - A_{i}^{i+1}) } {\partial s_i^{i+1}}
= \frac{ \partial A_{i+1}^{i+2} } {\partial \beta_{i+1}} \cdot \frac{\partial \beta_{i+1}} {\partial s_i^{i+1}} -\frac{ \partial A_{i}^{i+1} } {\partial \beta_{i+1}} \cdot \frac{\partial \beta_{i+1}} {\partial s_i^{i+1}}
\frac{ \partial (A_{i+1}^{i+2} - A_{i}^{i+1}) } {\partial s_i^{i+1}}
= \frac{ \partial A_{i+1}^{i+2} } {\partial \beta_{i+1}} \cdot \beta_{i+1}^2 -\frac{ \partial A_{i}^{i+1} } {\partial \beta_{i+1}} \cdot \beta_{i+1}^2
then our algorithm step is
s_i^{i+1} (t+1) =s_i^{i+1} (t) + \frac{\delta} {\frac{ \partial (A_{i}^{i+1} -A_{i+1}^{i+2} ) } {\partial s_i^{i+1}} }\cdot(A_i^{i+1}-A_{i+1}^{i+2})
most of the equations in this issue are wrong. i will post a new issue with an appropiate derivation
Instead of parameterizing the betas as temperatures and its logarithm, lets try to formulate the acceptance in terms of betas, logL and its variance.
If we approximate the jump probability by the integral of a normal distribution it can be shown that