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Next: X2X translation formula Up: M2X, X2X, X2L in Previous: M2X, X2X, X2L in

   
M2X translation formula

Using (3.22), (4.2), (B.1), (B.2) and the following identity:

\begin{eqnarray*}\partial_{\pm}e^{i (\lambda_k/d) ((\overrightarrow{O\mbox{\bold...
... + (\overrightarrow{O\mbox{\boldmath$ x $ }})_2 \sin \alpha_j)},
\end{eqnarray*}


we obtain
$\displaystyle {\sum_{n=0}^{\infty}\sum_{m=-n}^{n}\overline{S_{n,m}}(\overrightarrow{O\mbox{\boldmath$ x $ }_1})
M_{n,m}(O)}$
    $\displaystyle =
\sum_{n=0}^{\infty}\left(\sum_{m=0}^{n} \overline{S_{n,m}}(\ove...
...overline{S_{n,-m}}(\overrightarrow{O\mbox{\boldmath$ x $ }}) M_{n,-m}(O)\right)$  
    $\displaystyle =\sum_{n=0}^{\infty}\left(
\sum_{m=0}^{n}(-1)^n \partial_{-}^{m} ...
...}{\vert\overrightarrow{O\mbox{\boldmath$ x $ }}\vert}\right)M_{n,-m}(O)
\right)$  
    $\displaystyle =\sum_{n=0}^{\infty}
\sum_{m=0}^{n}\sum_{k=1}^{s}\sum_{j=1}^{M(k)}\frac{\omega_k}{M(k)d}
(-i)^m (\lambda_k/d)^n e^{-i m \alpha_j(k)} M_{n,m}(O)$  
    $\displaystyle \qquad\qquad\qquad \times~ e^{-(\lambda_k/d) ((\overrightarrow{O\...
...\alpha_j(k) - i (\overrightarrow{O\mbox{\boldmath$ x $ }})_2 \sin \alpha_j(k))}$  
    $\displaystyle +\sum_{n=0}^{\infty}
\sum_{m=1}^{n}\sum_{k=1}^{s}\sum_{j=1}^{M(k)}\frac{\omega_k}{M(k)d}
(i)^m (\lambda_k/d)^n e^{i m \alpha_j(k)} M_{n,-m}(O)$  
    $\displaystyle \qquad\qquad\qquad \times~ e^{-(\lambda_k/d) ((\overrightarrow{O\...
...\alpha_j(k) - i (\overrightarrow{O\mbox{\boldmath$ x $ }})_2 \sin \alpha_j(k))}$  
    $\displaystyle =\sum_{k=1}^{s}\sum_{j=1}^{M(k)}
\frac{\omega_k}{M(k)d}\sum_{n=0}^{\infty}\sum_{m=-n}^{n}
(-i)^m (\lambda_k/d)^n e^{-i m \alpha_j} M_{n,m}(O)$  
    $\displaystyle \qquad\qquad\qquad \times~ e^{-(\lambda_k/d) ((\overrightarrow{O\...
...\alpha_j(k) - i (\overrightarrow{O\mbox{\boldmath$ x $ }})_2 \sin \alpha_j(k))}$  
    $\displaystyle =\sum_{k=1}^{s}\sum_{j=1}^{M(k)}X(k,j;O)
e^{-(\lambda_k/d) ((\ove...
...\alpha_j(k) - i (\overrightarrow{O\mbox{\boldmath$ x $ }})_2 \sin \alpha_j(k))}$  

where X(k,j;O) is the coefficient of the exponential expansion centred at O defined as
 
$\displaystyle X(k,j;O)=\frac{\omega_k}{M(k)d}\sum_{n=0}^{\infty}\sum_{m=-n}^{n}
(-i)^m (\lambda_k/d)^n e^{-i m \alpha_j(k)} M_{n,m}(O).$     (J.1)

Exchanging an order of the double sum in (J.1) (See Fig.J.1) we rewrite (J.1) as
 
$\displaystyle X(k,j;O)=\frac{\omega_k}{M(k)d}\sum_{m=-\infty}^{\infty}(-i)^m e^{-i m
\alpha_j(k)} \sum_{n=\vert m\vert}^{\infty} (\lambda_k/d)^n M_{n,m}(O).$     (J.2)


  
Figure J.1: Exchange of indices
\begin{figure}
\begin{center}
\leavevmode
\epsfile{file=FIG/fig4.1c.eps,scale=0.6}
\end{center}\end{figure}

Next we estimate the cost for M2X translation. Noting (J.2) and truncating the infinite series (3.22) with p terms, the cost can be roughly estimated as follows:

\begin{eqnarray*}\mbox{Cost} &=& \sum_{k=1}^{s}\sum_{m=-p}^p \sum_{n=\vert m\ver...
...C_0sp^2) + O(C_1pS\mbox{exp}) \approx O(p^3) + O(p^3) = O(p^3),
\end{eqnarray*}


where C0 and C1 are the computational cost related to $(\lambda_k/d)^n M_{n,m}(O)$ and $(-i)^m e^{-i m \alpha_j(k)}$ in (J.2), respectively and C2 is the negligible cost.
next up previous contents
Next: X2X translation formula Up: M2X, X2X, X2L in Previous: M2X, X2X, X2L in
Ken-ichi Yoshida
2001-07-28