### 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 $R$) \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} ... ...
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