More notes on error bounds.

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Marek Nečada 2018-09-11 12:44:15 +03:00
parent 70f1a0a67a
commit 2388ed6c99
1 changed files with 70 additions and 11 deletions

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@ -3209,7 +3209,7 @@ For the short-range part
\end_inset
, the radially varying part reads
\begin_inset Formula $f_{\eta}^{\mathrm{L}}\left(R_{pq}\right)\equiv R_{pq}^{n}\int_{\eta}^{\infty}e^{-R_{pq}^{2}\xi^{2}}e^{k^{2}/4\xi^{2}}\xi^{2n}\ud\xi$
\begin_inset Formula $f_{\eta}^{\mathrm{S}}\left(R_{pq}\right)\equiv R_{pq}^{n}\int_{\eta}^{\infty}e^{-R_{pq}^{2}\xi^{2}}e^{k^{2}/4\xi^{2}}\xi^{2n}\ud\xi$
\end_inset
and for its integral as in
@ -3222,7 +3222,7 @@ reference "eq:lsum_bound"
we have
\begin_inset Formula
\begin{eqnarray*}
B_{R_{\mathrm{s}}}\left[f_{\eta}^{\mathrm{L}}\right] & = & \int_{R_{\mathrm{s}}}^{\infty}r^{n+1}\int_{\eta}^{\infty}e^{-r^{2}\xi^{2}}e^{k^{2}/4\xi^{2}}\xi^{2n}\ud\xi\,\ud r\\
B_{R_{\mathrm{s}}}\left[f_{\eta}^{\mathrm{S}}\right] & = & \int_{R_{\mathrm{s}}}^{\infty}r^{n+1}\int_{\eta}^{\infty}e^{-r^{2}\xi^{2}}e^{k^{2}/4\xi^{2}}\xi^{2n}\ud\xi\,\ud r\\
& \le & e^{k^{2}/4\eta^{2}}\int_{R_{\mathrm{s}}}^{\infty}\int_{\eta}^{\infty}r^{n+1}e^{-r^{2}\xi^{2}}\xi^{2n}\ud\xi\,\ud r\\
& = & e^{k^{2}/4\eta^{2}}\frac{\eta^{2n+1}R_{\mathrm{s}}^{2+n}\left(E_{\frac{1}{2}-n}\left(\eta^{2}R_{\mathrm{s}}^{2}\right)-E_{-\frac{n}{2}}\left(\eta^{2}R_{\mathrm{s}}^{2}\right)\right)}{2\left(n-1\right)}\\
& = & e^{k^{2}/4\eta^{2}}\frac{\eta^{2n+1}R_{\mathrm{s}}^{2+n}\left(\left(\eta R_{\mathrm{s}}\right)^{-2n-1}\Gamma\left(n+\frac{1}{2},\eta^{2}R_{\mathrm{s}}^{2}\right)-\left(\eta R_{\mathrm{s}}\right)^{-n-2}\Gamma\left(\frac{n}{2}+1,\eta^{2}R_{\mathrm{s}}^{2}\right)\right)}{2\left(n-1\right)}\\
@ -3258,8 +3258,48 @@ Apparently, this expression is problematic for
\begin_inset Formula $_{2}F_{2}$
\end_inset
, resulting in:
\begin_inset Formula
\[
B_{R_{\mathrm{s}}}\left[f_{\eta}^{\mathrm{S}}\right]\le e^{k^{2}/4\eta^{2}}\left(\frac{\eta R}{2}{}_{2}F_{2}\left(\begin{array}{cc}
\frac{1}{2}, & \frac{1}{2}\\
\frac{3}{2}, & \frac{3}{2}
\end{array};-\eta^{2}R_{\mathrm{s}}^{2}\right)-\frac{\sqrt{\pi}}{8}\left(\gamma_{\mathrm{E}}-2\mathrm{erfc}\left(\eta R_{\mathrm{s}}\right)+2\log\left(2\eta R_{\mathrm{s}}\right)\right)\right).
\]
\end_inset
The problem is that evaluation of the
\begin_inset Formula $_{2}F_{2}$
\end_inset
for large argument is very problematic.
However, Mathematica says that the value of the right parenthesis drops
below DBL_EPSILON for
\begin_inset Formula $\eta R_{\mathrm{s}}>6$
\end_inset
.
Hence it might make sense to take a rougher estimate using (for
\end_layout
\begin_layout Standard
Also the expression for
\begin_inset Formula $n\ne1$
\end_inset
decreases very fast, so as long as the value of
\begin_inset Formula $e^{k^{2}/4\eta^{2}}$
\end_inset
is reasonably low, there should not be much trouble.
\end_layout
\begin_layout Standard
\begin_inset Note Note
status open
\begin_layout Plain Layout
Maybe it might make sense to take a rougher estimate using (for
\begin_inset Formula $n=1$
\end_inset
@ -3285,8 +3325,7 @@ symmetric
\begin_inset Formula $R_{\mathrm{s}}\leftrightarrow\eta$
\end_inset
, so we can write either TODO; dammit, I should implement the hypergeometric
fn instead.
, so we can write either
\begin_inset Formula
\[
B_{R_{\mathrm{s}}}\left[f_{\eta}^{\mathrm{L}}\right]\le e^{k^{2}/4\eta^{2}}\int_{R_{\mathrm{s}}}^{\infty}\int_{\eta}^{\infty}e^{-r^{2}\xi^{2}}r^{2}\xi^{2}\ud\xi\,\ud r
@ -3295,6 +3334,11 @@ B_{R_{\mathrm{s}}}\left[f_{\eta}^{\mathrm{L}}\right]\le e^{k^{2}/4\eta^{2}}\int_
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Subsubsection
@ -3334,7 +3378,22 @@ For
\begin{eqnarray*}
\left(\beta_{pq}/k\right)^{n-2j}\Gamma_{j,pq}\left(\gamma_{pq}\right)^{2j-1} & = & \left(\beta_{pq}/k\right)^{n-2j}\Gamma\left(\frac{1}{2}-j,\frac{\beta_{pq}^{2}-k^{2}}{4\eta^{2}}\right)\left(\frac{\beta_{pq}^{2}}{k^{2}}-1\right)^{j-\frac{1}{2}}\\
& \le & \left(\beta_{pq}/k\right)^{n-2j}\left(\frac{\beta_{pq}^{2}-k^{2}}{4\eta^{2}}\right)^{-j-\frac{1}{2}}e^{-\frac{\beta_{pq}^{2}-k^{2}}{4\eta^{2}}}\left(\frac{\beta_{pq}^{2}}{k^{2}}-1\right)^{j-\frac{1}{2}}\\
& & TODO
& = & \left(2\eta\right)^{2j+1}e^{-\frac{\beta_{pq}^{2}-k^{2}}{4\eta^{2}}}k^{-n-1}\beta_{pq}^{n-2j}\left(\frac{\beta_{pq}^{2}}{k^{2}}-1\right)^{-1}\\
& = & e^{-\frac{\beta_{pq}^{2}-k^{2}}{4\eta^{2}}}\left(\frac{\beta_{pq}}{k}\right)^{n}\frac{2\eta}{k}\left(\frac{2\eta}{\beta_{pq}}\right)^{2j}\left(\frac{\beta_{pq}^{2}}{k^{2}}-1\right)^{-1}.
\end{eqnarray*}
\end_inset
The only diverging factor here is apparently
\begin_inset Formula $\left(\beta_{pq}/k\right)^{n}$
\end_inset
; Mathematica and [DMLF] say
\begin_inset Formula
\begin{eqnarray*}
\int_{B_{\mathrm{s}}}^{\infty}e^{-\frac{\beta^{2}}{4\eta^{2}}}\beta^{n}\beta\ud\beta & = & \frac{B_{\mathrm{s}}^{n+2}}{2}E_{-\frac{n}{2}}\left(\frac{B_{\mathrm{s}}^{2}}{4\eta^{2}}\right)\\
& = & \frac{B_{\mathrm{s}}^{n+2}}{2}\left(\frac{B_{\mathrm{s}}^{2}}{4\eta^{2}}\right)^{-1-\frac{n}{2}}\Gamma\left(1+\frac{n}{2},\frac{B_{\mathrm{s}}^{2}}{4\eta^{2}}\right)\\
& = & \frac{\left(2\eta\right)^{n+2}}{2}\Gamma\left(1+\frac{n}{2},\frac{B_{\mathrm{s}}^{2}}{4\eta^{2}}\right).
\end{eqnarray*}
\end_inset