Shane Flynn edited Absolute Formulation.tex  over 9 years ago

Commit id: f62cd51bf66ef901e4658895ab39a5c85cc43294

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\begin{equation}  GFE_a=\frac{d}{dt}\sum [\gamma_i(t)k_a(t)]  \end{equation}  We also know that $k_a\equiv \frac{d}{dt}[\ln S_i]$ and that $\gamma_i=\frac{S_i}{\sum[S_i]}$. With these definition we can determine the time derivative of the average rate as follows.   \begin{equation}  \frac{d}{dt}\sum [\gamma_i(t)k_a(t)]=\sum[\frac{d}{dt}[\gamma_i(t)k_a(t)]]  \end{equation}  Now applying the product rule and distributing the sum we find the following.   \begin{equation}  GFE_a=\sum[\gamma_i\frac{d}{dt}[k_a]]+\sum[k_a\frac{d}{dt}[\frac{S_i}{\sum[S_i]}]]  \end{equation}  The second term can be evaluated and simplified using the product rule for the derivative as follows.  \begin{equation}  GFE_a=\sum[\gamma_i\frac{d}{dt}[k_a]]+\sum[k_a(S_i\frac{d}{dt}[(\Sigma[S_i])^{-1}]+(\Sigma[S_i])^{-1}\frac{d}{dt}[S_i])]  \end{equation}  Simplifying further we find  \begin{equation}  GFE_a=\sum[\gamma_i\frac{d}{dt}[k_a]]+\sum[k_a(\frac{-S_i}{(\Sigma S_i)^2}\frac{d}{dt}[(\Sigma S_i)]+(\Sigma S_i)^{-1}\frac{d}{dt}[S_i])]