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Sep 20, 2013 at 17:58 vote accept jlperla
Sep 19, 2013 at 21:08 answer added Jens timeline score: 2
Sep 19, 2013 at 20:47 comment added jlperla I checked the other links. They look related, but I can't figure out how to apply them to my problem.
Sep 19, 2013 at 19:43 history edited jlperla CC BY-SA 3.0
Added a few reasons why I want to use rules.
Sep 19, 2013 at 19:31 comment added jlperla See my edits to the question. It definitely solves the exact problem, but I left off part of the motivation for using rules.
Sep 19, 2013 at 19:19 comment added Mr.Wizard @Blackbird Alright then.
Sep 19, 2013 at 19:17 comment added Pankaj Sejwal @Mr.Wizard : You have a permanent go ahead from my side always :)
Sep 19, 2013 at 19:09 comment added Mr.Wizard @Blackbird I went ahead and posted it. If you choose to post I will delete mine.
Sep 19, 2013 at 19:09 answer added Mr.Wizard timeline score: 3
Sep 19, 2013 at 19:03 comment added Mr.Wizard Related, possible duplicates: (7876), (6089), (27230) -- have you looked at these?
Sep 19, 2013 at 18:57 comment added Pankaj Sejwal Instead of using Replace simply use function f[c_] := a c, you will get same result.
Sep 19, 2013 at 18:50 comment added jlperla That does it for this example. The problem is that in reality I am applying this substitution to pretty complicated expressions where I want to write out the expression properly and then do the substitution. (e.g. something like exp = D[f[c],c] + a c / (D[f[c],c,c]^2); exp //. fsub; That approach would force me to re-write the entire expression with the substitutions embedded throughout.
Sep 19, 2013 at 18:47 comment added Pankaj Sejwal May be this D[f[c] //. fsub, c] //. fsub.Raise it to power or anything else to see if its done or not.
Sep 19, 2013 at 18:28 history asked jlperla CC BY-SA 3.0