Re: Zero times a variable in numerical calculations

*To*: mathgroup at smc.vnet.net*Subject*: [mg68410] Re: Zero times a variable in numerical calculations*From*: Peter Pein <petsie at dordos.net>*Date*: Fri, 4 Aug 2006 03:59:23 -0400 (EDT)*References*: <a02020013-1047-ppc-D86740659B534E4384129C036A76A845@QuarkNew.local> <easjm1$fu2$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Maurits Haverkort schrieb: > Dear Bill and others who answered my question. > > Chop[0.0 a] indeed works and returns nicely 0 however I still have some > problems when using this on Sparse Arrays. > > I define: > M = SparseArray[{{1, 1} -> a, {2, 2} -> 1.0, {3, 3} -> 1.0, {2, 3} -> b, {3, > 2} -> b, {_, _} -> 0}, {3, 3}]; > and > T = SparseArray[{{1, 1} -> 1.0, {2, 2} -> -Sqrt[0.5], {3, 3} -> Sqrt[0.5], > {2, 3} -> Sqrt[0.5], {3, 2} -> Sqrt[0.5], {_, _} -> 0}, {3, 3}]; > I calculate T.M.T This should result in a sparse array with 3 nonzero > elements. (a,1-b and 1+b on the diagonal). > However > T.M.T=SparseArray[<5>, {3, 3}] > (T.M.T)[[3,2]]=0. (0.707107 + 0.707107 b) > and > (T.M.T)[[2,3]]=0.707107 (0.707107- 0.707107 b) + 0.707107 (-0.707107 + > 0.707107 b) > > Chop works almost as > Chop[T.M.T]=SparseArray[<5>, {3, 3}] > But > Chop[T.M.T][[3,2]]=0 > However > Chop[T.M.T][[2,3]]=0.707107 (0.707107 - 0.707107 b) + 0.707107 (-0.707107 + > 0.707107 b) > > Simplify does nothing. i.e. > Simplify[T.M.T]=T.M.T > > Chop[Simplify[Normal[T.M.T]]] > does what I want, but does not work with sparse arrays and in between all > elements become large expressions so I run out of memory. > > Thanks in advance, > Maurits Haverkort > > (P.S. Mathematica 5.2 under Linux (Redhat) 1 Gb and Windows XP, 512Mb, the > matrix size I aim for is 646*646, however I curently can't evaluate a > 134*134 matrix) > > ----- Original Message ----- > From: "Bill Rowe" <readnewsciv at earthlink.net> To: mathgroup at smc.vnet.net > Subject: [mg68410] Re: Zero times a variable in numerical calculations > > > On 8/2/06 at 10:49 AM, Haverkort at ph2.uni-koeln.de (Maurits Haverkort) > wrote: > >> 2) What is the fastest way to multiply a numerical matrix times a >> half numerical half analytical sparse matrix, whereby the result is >> band diagonal. (The current result has 0. a + 0. b + 0. c + 0. d + >> 0. e + 0. f on all places where it should be zero.) > > How are you defining the sparse matrix? > > If I do > > In[17]:= > a=SparseArray[{{1, 2} -> Random[], {3, 4} -> Random[]}, > {5, 5}]; > > followed by > > In[18]:= > Normal@a > > Out[18]= > {{0, 0.49847100218212825, 0, 0, 0}, {0, 0, 0, 0, 0}, > {0, 0, 0, 0.009744974537243703, 0}, {0, 0, 0, 0, 0}, > {0, 0, 0, 0, 0}} > > > I see the zero elements are exact 0's not inexact 0's. Which means the > matrix multiplication will be evaluating 0 a etc which will evaluate to 0. > -- > To reply via email subtract one hundred and four > Hi Maurits, you might want to manipulate the array rules: m = SparseArray[{{1, 1} -> a, {2, 2} -> 1., {3, 3} -> 1., {2, 3} -> b, {3, 2} -> b, {_, _} -> 0}, {3, 3}]; t = SparseArray[{{1, 1} -> 1., {2, 2} -> -Sqrt[0.5], {3, 3} -> Sqrt[0.5], {2, 3} -> Sqrt[0.5], {3, 2} -> Sqrt[0.5], {_, _} -> 0}, {3, 3}]; (t . m . t)[[2,3]] gives the unsimplified element of the matrix. result = SparseArray[ArrayRules[t . m . t] /. HoldPattern[ix_ -> val_] :> ix -> Chop[Simplify[val]]]; result[[2,3]] gives 0 Normal[result] --> {{1.*a, 0, 0}, {0, 1.0000000000000002 - 1.0000000000000002*b, 0}, {0, 0, 1.0000000000000002 + 1.0000000000000002*b}} hth, Peter