From 37c8960783c4c152de44e1aa7e1f69b4ee712d99 Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Fri, 2 Sep 2022 12:05:03 +0200 Subject: kugel: remove Eq. for a more consistent style --- buch/papers/kugel/spherical-harmonics.tex | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) (limited to 'buch/papers') 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) \\ -- cgit v1.2.1