Merge branch 'master' into vincent_update

main.tex

@@ -125,7 +125,6 @@ If the likelihood is Gaussian (with noise covariance $N$ and response $R$)
 \begin{align*}
 \mathcal H(d|s) \propto (d-R(s))^\dagger N^{-1}(d-R(s))
 \end{align*}
-
 or Poissonian
 \begin{align*}
 \mathcal H(d|s) \propto - \log (R(s))^\dagger d+\sum_i R(s)_i,
@@ -152,7 +151,7 @@ the course of the algorithm. In \texttt{s}, NIFTy will store the reconstruction
 of the physical field which the user wants to infer. \texttt{s} would be a
 one-dimensional array for a time-series, two-dimensional for a multi-frequency
 time-series or for a single-frequency image of the sky, three-dimensional for a
-multi-frequency image of sky, etc. To cut a long story short: be aware of the
+multi-frequency image of the sky, etc. To cut a long story short: be aware of the
 shape of \texttt{s} and \texttt{d}!
 
 \subsection*{The response}
@@ -176,7 +175,7 @@ response_out = R(np.ones(shp))
 if response_out.shape == d.shape:
     print('Yay!')
 else:
-    raise ValueError('Output of response doesn't have the correct shape.')
+    raise ValueError('Output of response does not have the correct shape.')
 \end{lstlisting}
 
 \subsection*{Derivative of response}
@@ -201,7 +200,7 @@ following shape: \footnote{There are various ways to think about derivatives and
 What needs to be implemented is a function \texttt{R\_prime(position, s0)} which
 takes the arguments \texttt{position} (which is an array of shape \texttt{s.shape}
 and determines the position at which we want to calculate the derivative) and
-the array \texttt{s0} which shall be the derivative taken of.
+the array \texttt{s0} of which the derivative shall be taken.
 \texttt{R\_prime} is nonlinear in \texttt{position} in general and linear in
 \texttt{s0}. The output of \texttt{R\_prime} is of shape \texttt{d.shape}.
 
@@ -314,7 +313,7 @@ Why is this already sufficient?
   (R(s))^\dagger d+\sum_i R(s)_i$. Implementing $R$ and stating that the data is
   Poissonian determines this form.
   \item Since $R$ is a composition of a convolution and a sampling, both of which
-    is a linear operation, $R$ itself is a linear operator.\footnote{I.e. $R(\alpha
+    are linear operations, $R$ itself is a linear operator.\footnote{I.e. $R(\alpha
     s_1+s_2) = \alpha R(s_1) + R(s_2)$.} Thus, $R' = R$ and $R'^\dagger =
   R^\dagger$. All in all, we need an implementation for $R$ and $R^\dagger$.
 \end{itemize}