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L2L translation formula

From (3.123), we have
 
$\displaystyle \overline{{\cal I}}_n^m(k,\overrightarrow{\mbox{\boldmath$ x $ }_...
...l^{-m-m'}(k,\overrightarrow{\mbox{\boldmath$ x $ }_0\mbox{\boldmath$ x $ }_1}).$     (I.3)

Substituting (I.3) into (3.130), we obtain

\begin{eqnarray*}\lefteqn{\sum_{n=0}^{\infty}\sum_{m=-n}^{n} (2n+1)
\frac{\parti...
...box{\boldmath$ x $ }})
L_{n}^{m}(k,\mbox{\boldmath$ x $ }_1),\\
\end{eqnarray*}


where $L_n^m(k,\mbox{\boldmath$\space x $ }_1)$ is the coefficient of the local expansion at $\mbox{\boldmath$\space x $ }_1$ given by

\begin{eqnarray*}L_n^m(k,\mbox{\boldmath$ x $ }_1)=\sum_{n'=0}^{\infty}\sum_{m'=...
...ox{\boldmath$ x $ }_1}) L_{n'}^{m'}(k,\mbox{\boldmath$ x $ }_0).
\end{eqnarray*}




Ken-ichi Yoshida
2001-07-28