kugel: remove Eq. for a more consistent styleHEADmaster

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) \\