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[ewald] Dudopráce 

Former-commit-id: a72c2f10d8ff5f02afb177eb7b64728599d8b951
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Marek Nečada 2017-08-17 21:07:41 +00:00
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@ -1021,6 +1021,103 @@ now the Stirling number of the 2nd kind
\begin_inset Formula $\kappa>t$ \begin_inset Formula $\kappa>t$
\end_inset \end_inset
.
\end_layout
\begin_layout Plain Layout
What about the gamma fn on the left? Using DLMF 5.5.5, which says
\begin_inset Formula $Γ(2z)=\pi^{-1/2}2^{2z-1}\text{Γ}(z)\text{Γ}(z+\frac{1}{2})$
\end_inset
we have
\begin_inset Formula
\[
\text{Γ}\left(2-q+n\right)=\frac{2^{1-q+n}}{\sqrt{\pi}}\text{Γ}\left(\frac{2-q+n}{2}\right)\text{Γ}\left(\frac{3-q+n}{2}\right),
\]
\end_inset
so
\begin_inset Formula
\begin{eqnarray*}
\pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & = & \frac{\kor{\text{Γ}\left(2-q+n\right)}}{\kor{2^{n}}k_{0}^{q}}\kor{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\kappa!\left(-1\right)^{\kappa}\left(\frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\kor{\text{Γ}\left(\frac{3-q+n}{2}\right)}\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s}{t}\begin{Bmatrix}t\\
\kappa
\end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s-t}-\frac{\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\kor{\text{Γ}\left(\frac{2-q+n}{2}\right)}\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s+1}{t}\begin{Bmatrix}t\\
\kappa
\end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right)\\
& = & \frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\kappa!\left(-1\right)^{\kappa}\left(\frac{\kor{\text{Γ}\left(\frac{2-q+n}{2}\right)\left(\frac{2-q+n}{2}\right)_{s}}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s}{t}\begin{Bmatrix}t\\
\kappa
\end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s-t}-\frac{\kor{\text{Γ}\left(\frac{3-q+n}{2}\right)\left(\frac{3-q+n}{2}\right)_{s}}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s+1}{t}\begin{Bmatrix}t\\
\kappa
\end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right)\\
\mbox{(D5.2.5)} & = & \frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\kappa!\left(-1\right)^{\kappa}\left(\frac{\text{Γ}\left(\frac{2-q+n}{2}+s\right)\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s}{t}\begin{Bmatrix}t\\
\kappa
\end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s-t}-\frac{\text{Γ}\left(\frac{3-q+n}{2}+s\right)\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s+1}{t}\begin{Bmatrix}t\\
\kappa
\end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right)
\end{eqnarray*}
\end_inset
The two terms have to be treated fifferently depending on whether q
\begin_inset Formula $q+n$
\end_inset
is even or odd.
\end_layout
\begin_layout Plain Layout
First, assume that
\begin_inset Formula $q+n$
\end_inset
is even, so the left term has gamma functions and pochhammer symbols with
integer arguments, while the right one has half-integer arguments.
As
\begin_inset Formula $n$
\end_inset
is non-negative and
\begin_inset Formula $q$
\end_inset
is positive,
\begin_inset Formula $\frac{q+n}{2}$
\end_inset
is positive, and the Pochhammer symbol
\begin_inset Formula $\left(\frac{2-q-n}{2}\right)_{s}=0$
\end_inset
if
\begin_inset Formula $s\ge\frac{q+n}{2}$
\end_inset
, which transforms the sum over
\begin_inset Formula $s$
\end_inset
to a finite sum for the left term.
However, there still remain divergent terms if
\begin_inset Formula $\frac{2-q+n}{2}+s\le0$
\end_inset
(let's handle this later; maybe D15.8.6–7 may be then be useful)! Now we
need to perform some transformations of variables to make the other sum
finite as well
\end_layout
\begin_layout Plain Layout
Pár kroků zpět:
\begin_inset Formula
\begin{eqnarray*}
\pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\kor{\text{Γ}\left(2-q+n\right)}}{\kor{2^{n}}k_{0}^{q}}\kor{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\left(\sigma c-ik_{0}\right)^{2s}\times\left(\underbrace{\frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\kor{\text{Γ}\left(\frac{3-q+n}{2}\right)}\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}}_{\equiv c_{q,n,s}}-\underbrace{\frac{\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\kor{\text{Γ}\left(\frac{2-q+n}{2}\right)}\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}}_{č_{q,n,s}}\frac{\left(\sigma c-ik_{0}\right)}{k}\right)\\
& = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\left(\sigma c-ik_{0}\right)^{2s}\times\left(\frac{\text{Γ}\left(\frac{2-q+n}{2}\right)\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}-\frac{\text{Γ}\left(\frac{3-q+n}{2}\right)\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\frac{\left(\sigma c-ik_{0}\right)}{k}\right)
\end{eqnarray*}
\end_inset
\end_layout \end_layout
@ -1313,7 +1410,7 @@ where the spherical Hankel transform
2) 2)
\begin_inset Formula \begin_inset Formula
\[ \[
\bsht lg(k)\equiv\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right). \bsht lg(k)\equiv\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).
\] \]
\end_inset \end_inset
@ -1323,7 +1420,7 @@ Using this convention, the inverse spherical Hankel transform is given by
3) 3)
\begin_inset Formula \begin_inset Formula
\[ \[
g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\,k^{2}\bsht lg(k)j_{l}(k), g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\, k^{2}\bsht lg(k)j_{l}(k),
\] \]
\end_inset \end_inset
@ -1336,7 +1433,7 @@ so it is not unitary.
An unitary convention would look like this: An unitary convention would look like this:
\begin_inset Formula \begin_inset Formula
\begin{equation} \begin{equation}
\usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition} \usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition}
\end{equation} \end{equation}
\end_inset \end_inset
@ -1390,7 +1487,7 @@ where the Hankel transform of order
is defined as is defined as
\begin_inset Formula \begin_inset Formula
\begin{equation} \begin{equation}
\pht mg\left(k\right)=\int_{0}^{\infty}\ud r\,g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition} \pht mg\left(k\right)=\int_{0}^{\infty}\ud r\, g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}
\end{equation} \end{equation}
\end_inset \end_inset