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May
16
accepted Applying ReplaceAll to I (imaginary unit)
May
15
comment Applying ReplaceAll to I (imaginary unit)
@SjoerdC.deVries: I understand, thank you for clarifying this to me. It would be nice if this could be stressed in the Documentation.
May
15
revised Applying ReplaceAll to I (imaginary unit)
typo
May
14
comment Applying ReplaceAll to I (imaginary unit)
@SjoerdC.deVries: Did you read the other posts? What is then, in your opinion, the mistake? I mean, it's clear that the behavior is inconsistent. But since I'm not the designer of the called function I have no clue what is going on there.
May
14
revised Applying ReplaceAll to I (imaginary unit)
the behavior of ReplaceAll is inconsistent or at least confusing
May
13
comment Applying ReplaceAll to I (imaginary unit)
Is is difficult to add a reverse procedure?
May
13
comment Applying ReplaceAll to I (imaginary unit)
aah, now I understand, I thought it's necessary to have it in FullForm before replacement, thank you!
May
13
comment Applying ReplaceAll to I (imaginary unit)
Of course, I understand. However, if I precede in my calculation I have a mixture of "normal form" expressions and FullForm and this generates problems when evaluating scalar products like FullForm[{1, 2, 3}].{1, 2, 3}
May
13
comment Applying ReplaceAll to I (imaginary unit)
Let us continue this discussion in chat.
May
13
comment Applying ReplaceAll to I (imaginary unit)
@MichaelE2: Both replacements have the same form so I thought: Since it works for x it has to work for "I" too. As I understand Bill, the issue is solved with FullForm[]. However, I thought that InputForm[] will reverse the latter after the replacement. This seems not the case, still searching for the solution...
May
13
awarded  Commentator
May
13
comment Applying ReplaceAll to I (imaginary unit)
Thank you! Can you tell me how to revert FullForm[] afterwards?
May
13
comment Applying ReplaceAll to I (imaginary unit)
@MichaelE2: I get the proper transformation from Exp[]-> E^ BUT Exp[-a I] /. I -> -I E^(-I a) however: Exp[-a x] /. x -> -x E^(x a) is evaluated correctly
May
13
comment Applying ReplaceAll to I (imaginary unit)
@MichaelE2: Well, seems strange but this is what I get as output.
May
13
comment Applying ReplaceAll to I (imaginary unit)
@MichaelE2: For me the replacement works for E^I, too, but not for Exp[I]
May
13
asked Applying ReplaceAll to I (imaginary unit)
Jul
5
awarded  Popular Question
Jul
2
awarded  Curious
May
26
awarded  Yearling
May
26
asked Bug in ListVectorPlot?