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authorNao Pross <np@0hm.ch>2022-09-02 12:05:03 +0200
committerNao Pross <np@0hm.ch>2022-09-02 12:05:31 +0200
commit37c8960783c4c152de44e1aa7e1f69b4ee712d99 (patch)
tree039ac4faa8559611063345d4520732e1ae6d5007
parentkugel: Start reviewing long proofs (diff)
downloadSeminarSpezielleFunktionen-master.tar.gz
SeminarSpezielleFunktionen-master.zip
kugel: remove Eq. for a more consistent styleHEADmaster
Diffstat (limited to '')
-rw-r--r--buch/papers/kugel/spherical-harmonics.tex8
1 files changed, 4 insertions, 4 deletions
diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex
index 76e3c91..404e151 100644
--- a/buch/papers/kugel/spherical-harmonics.tex
+++ b/buch/papers/kugel/spherical-harmonics.tex
@@ -772,7 +772,7 @@ computational cost lower by a factor of six \cite{davari_new_2013}.
\frac{d^mP_{n+1}}{dz^m} - \frac{d^mP_{n-1}}{dz^m}
= (2n+1)\frac{d^{m-1}P_n}{dz^{m-1}}.
\end{equation}
- Then, using eq. \eqref{kugel:eq:rec_3} in eq. \eqref{kugel:eq:rec_1}, we will
+ Then, using \eqref{kugel:eq:rec_3} in \eqref{kugel:eq:rec_1}, we will
have
\begin{equation}
\label{kugel:eq:rec_4}
@@ -783,7 +783,7 @@ computational cost lower by a factor of six \cite{davari_new_2013}.
+ n\frac{d^m P_{n-1}}{dz^m}=0.
\end{equation}
Finally, multiplying both sides by $(1-z^2)^{\frac{m}{2}}$ and simplifying the
- expression, we can rewrite eq. \eqref{kugel:eq:rec_4} in terms of $P^m_n(z)$,
+ expression, we can rewrite \eqref{kugel:eq:rec_4} in terms of $P^m_n(z)$,
namely
\begin{equation*}
(n+1-m)P^m_{n+1}(z)-(2n+1)zP^m_n(z)+(m+n)P^m_{n-1}(z)=0,
@@ -824,7 +824,7 @@ computational cost lower by a factor of six \cite{davari_new_2013}.
\end{proof}
\begin{proof}[Proof of \eqref{kugel:eqn:rec-leg-3}]
- To derive this expression, we can multiply eq. \eqref{kugel:eq:rec_3} by
+ To derive this expression, we can multiply \eqref{kugel:eq:rec_3} by
$(1-z^2)^{\frac{m}{2}}$ and, as always, we could express it in terms of
$P^m_n(z)$:
\begin{equation*}
@@ -849,7 +849,7 @@ computational cost lower by a factor of six \cite{davari_new_2013}.
(m+n)P^m_{n-1}(z) + (n-m+1)P^m_{n+1}(z)
\right] = P^{m+1}_n(z) + [ n(n+1)-m(m-1) ]P^{m-1}_n(z).
\end{equation*}
- Rewriting then $P^{m-1}_n(z)$ using eq.\eqref{kugel:eq:helper}, we will have
+ Rewriting then $P^{m-1}_n(z)$ using \eqref{kugel:eq:helper}, we will have
\begin{align*}
\frac{2m}{(2n+1)\sqrt{1-z^2}}
&\left[ (m+n)P^m_{n-1}(z) + (n-m+1)P^m_{n+1}(z) \right] = P^{m+1}_n(z) \\