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MathGroup Archive 2007

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Replacement according to the pattern

  • To: mathgroup at smc.vnet.net
  • Subject: [mg77526] Replacement according to the pattern
  • From: tomuf at seznam.cz
  • Date: Tue, 12 Jun 2007 01:18:31 -0400 (EDT)

Hello,
I wanted to parse a big expression and according to some rules
integrate it term by term. For example, I wanted to replace symbol

HPL[{1, 1, 0}, u]

where HPL(list,variable) is function from package, with (note that
this simplified example has no mathematical sense)

HPL[{1, 1, 0}, u] - D[HPL[{1, 1, 0}, u], u]

i.e. that function minus its derivative with respect to 'u'. I have
loaded package which allows me to differentiate HPL. This works fine.
But when I try to generalize this to handle all lists like {1,1,0} at
the same time so that I don't have to type the rule for all
combinations, the replacement doesn't work well. When I type:

in = HPL[{1, 1, 0}, u]
in /. HPL[{1, 1, 0}, u] -> HPL[{1, 1, 0}, u] - D[HPL[{1, 1, 0}, u], u]
in /. HPL[a_, u] -> HPL[a, u] - D[HPL[a, u], u]

the third line produces different result than the second line (the
difference is in the term involving derivative). Does anyone know what
could be wrong? Thanks in advance

Best regards,
Tomas Prochazka



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