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author | Jaron Kent-Dobias <jaron@kent-dobias.com> | 2019-08-23 17:34:05 -0400 |
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committer | Jaron Kent-Dobias <jaron@kent-dobias.com> | 2019-08-23 17:34:05 -0400 |
commit | 508e7e3b93ca48e9ac3da695c7c8fff1cd26b7c6 (patch) | |
tree | f9febb57809d194a48f0a4b60170262b059328e4 | |
parent | 3e1c9cfa3dc9aabcf66fb8f93ccdf816ee4459f7 (diff) | |
download | PRB_102_075129-508e7e3b93ca48e9ac3da695c7c8fff1cd26b7c6.tar.gz PRB_102_075129-508e7e3b93ca48e9ac3da695c7c8fff1cd26b7c6.tar.bz2 PRB_102_075129-508e7e3b93ca48e9ac3da695c7c8fff1cd26b7c6.zip |
more inverse function verbosity
-rw-r--r-- | main.tex | 3 |
1 files changed, 2 insertions, 1 deletions
@@ -313,7 +313,8 @@ which implicitly gives $\eta[\epsilon]$ and $F_\e[\epsilon]=F[\eta[\epsilon],\ep cannot be solved explicitly, we can make use of the inverse function theorem. First, denote by $\eta^{-1}[\eta]$ the inverse functional of $\eta$ implied by \eqref{eq:implicit.eta}, which gives the function $\epsilon_\X$ corresponding -to each solution of \eqref{eq:implicit.eta} it receives. Now, we use the inverse function +to each solution of \eqref{eq:implicit.eta} it receives. This we can immediately identify from \eqref{eq:implicit.eta} as $\eta^{-1}[\eta](x)=-2/b(\delta F_\o[\eta]/\delta\eta(x))$. +Now, we use the inverse function theorem to relate the functional reciprocal of the derivative of $\eta[\epsilon]$ with respect to $\epsilon_\X$ to the derivative of $\eta^{-1}[\eta]$ with respect to $\eta$, yielding \begin{equation} \begin{aligned} |