Commit 53d1c9b2 authored by Philipp Arras's avatar Philipp Arras
Browse files

Merge branch 'master' into vincent_update

parents fc53aac7 269a8536
Pipeline #79997 passed with stage
in 32 seconds
......@@ -125,7 +125,6 @@ If the likelihood is Gaussian (with noise covariance $N$ and response $R$)
\mathcal H(d|s) \propto (d-R(s))^\dagger N^{-1}(d-R(s))
or Poissonian
\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:
raise ValueError('Output of response doesn't have the correct shape.')
raise ValueError('Output of response does not have the correct shape.')
\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 $R^\dagger$.
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