Re: Numerically computing partial derivatives
- To: mathgroup at smc.vnet.net
- Subject: [mg47962] Re: Numerically computing partial derivatives
- From: "Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk>
- Date: Tue, 4 May 2004 01:08:44 -0400 (EDT)
- References: <c72d7k$jk$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
I think you should be able to make a lot of progress analytically. I attach an extract from one of my papers in which I compute the first derivative of a quantity that is closely related to yours. The main tricks to use are log det = trace log, the Baker-Hausdorff identity for expanding logs of products of matrices, trace(commutator)=0, etc. I presume these sorts of tricks can be used to compute the Hessian as well. Select from (******** to *********) and paste into Mathematica. Steve Luttrell (************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.0' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info at wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 4550, 141]*) (*NotebookOutlinePosition[ 5349, 170]*) (* CellTagsIndexPosition[ 5274, 164]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell[TextData[{ StyleBox["Differentiating", FontSlant->"Italic"], " log det" }], "Title"], Cell[TextData[{ "We differentiate the logarithm of the determinant of a \ matrix-valued quantity. We use this in order to differentiate ", ButtonBox["equation", ButtonData:>"Eq:LowerBoundIntegrated", ButtonStyle->"Hyperlink"], " (", "XXX", "), so we present the derivation using an appropriate notation." }], "Text"], Cell[BoxData[ FormBox[GridBox[{ {\(Step\ 1\), \(\[Delta]\ log[ det(\[Pi]\ \[Sigma]\^\[Prime])]\), \(\(=\)\(\(-log[ det(\[Pi](\(\[Sigma]\^\[Prime]\)\^\(-1\) + \(\ \[Delta]\[Sigma]\^\[Prime]\)\^\(-1\)))]\) + log[det(\[Pi]\ \(\[Sigma]\^\[Prime]\)\^\(-1\))]\)\)}, {\(Step\ 2\), " ", \(\(=\)\(\(-tr[ log(\(\(\[Sigma]\^\[Prime]\)\^\(-1\)\)( 1 + \(\[Sigma]\^\[Prime]\) \(\[Delta]\[Sigma]\ \^\[Prime]\)\^\(-1\)))]\) + tr[log(\(\[Sigma]\^\[Prime]\)\^\(-1\))]\)\)}, {\(Step\ 3\), " ", \(\(=\)\(-tr[ log(1 + \(\[Sigma]\^\[Prime]\) \(\[Delta]\[Sigma]\^\ \[Prime]\)\^\(-1\))]\)\)}, {\(Step\ 4\), " ", \(\(\[TildeEqual]\)\(-\(\(tr[\(\[Sigma]\^\[Prime]\) \ \(\[Delta]\[Sigma]\^\[Prime]\)\^\(-1\)]\)\(.\)\)\)\)} }], TraditionalForm]], "NumberedEquation", TextAlignment->AlignmentMarker, GridBoxOptions->{ColumnAlignments->{Left}}], Cell["\<\ We justify the various stages of this manipulation as follows:\ \>", "Text"], Cell[TextData[{ "Step 1. Matrix invert ", Cell[BoxData[ \(TraditionalForm\`\[Sigma]\^\[Prime]\)]], ", which introduces a minus sign outside the logarithm function. In \ order to calculate the derivative, write the difference that results \ from changing ", Cell[BoxData[ \(TraditionalForm\`\(\[Sigma]\^\[Prime]\)\^\(-1\)\)]], " infinitesimally." }], "Text"], Cell[TextData[{ "Step 2. Use the identity ", Cell[BoxData[ \(TraditionalForm\`log[det(X)] = tr[log(X)]\)]], "." }], "Text"], Cell[TextData[{ "Step 3. Use the identity ", Cell[BoxData[ \(TraditionalForm\`log(X\ Y) = log(X) + log(Y) + \((commutator\ terms\ from\ the\ Baker - Hausdorff\ identity)\)\)]], " to obtain ", Cell[BoxData[ \(TraditionalForm\`tr[log(X\ Y)] = tr[log(X)] + tr[log(Y)]\)]], " which causes a pair of terms to cancel, leaving only the \ infinitesimal part. Note that the trace of any commutator is zero." }], "Text"], Cell[TextData[{ "Step 4. 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