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I'm getting a huge output, but it appears that almost every term should actually be zero, unless I'm misunderstanding the expression.

For example, part of the output reads

a (0.25 + 0. b) + b^2 (0. + 0./c) + b (-1. + 0. c) - 0.5 c + (b (-1. b - 0.5 c) c)/a^2

Simplify and FullSimplify won't get rid of these zero terms even with assumptions that a,b and c are real and non-zero. Suggestions ?

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    $\begingroup$ expr /. {x_Real /; x==0 -> 0} $\endgroup$
    – ssch
    Commented Oct 28, 2013 at 17:32
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    $\begingroup$ Examine the FullForm of the result. You may be surprised. Chop is useful too. $\endgroup$
    – Sjoerd C. de Vries
    Commented Oct 28, 2013 at 17:33
  • $\begingroup$ SetOptions[Simplify, TransformationFunctions :> Function[x, x //. {0. -> 0, 1. y_ :> y, -1. y_ :> y}]] and then Simplify simplifies as you want. $\endgroup$
    – Rolf Mertig
    Commented Oct 28, 2013 at 18:29
  • $\begingroup$ @RolfMertig. I think you meant to write -1. y_ :> -y $\endgroup$
    – RunnyKine
    Commented Oct 28, 2013 at 20:15
  • $\begingroup$ @RunnyKine : sure. thanx. $\endgroup$
    – Rolf Mertig
    Commented Oct 28, 2013 at 22:27

1 Answer 1

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To get what was said in the comments on record as an answer.

Chop is probably what you are looking for.

expr = 
  a (0.25 + 0. b) + b^2 (0. + 0./c) + b (-1. + 0. c) - 0.5 c + 
   (b (-1. b - 0.5 c) c)/a^2 // Chop

0.25 a - 1. b - 0.5 c + (b (-1. b - 0.5 c) c)/a^2

You might want to go further

expr = expr /. {1. x_ :> x, -1. x_ :> -x}

0.25 a - b - 0.5 c + (b (- b - 0.5 c) c)/a^2

Update

In general expressions with inexact coefficients can contain the forms 0.x, 1.x, and -1.x. None of these cause computation difficulties, but they are unattractive, and often we want to eliminate them.

With a help from Mr.Wizard, I offer the following pattern to do this task.

fixerPattern = u_ /; Sign[u] == u :> Sign[u]

With this we can clean up

expr = 
  a (0.25 + 0. b) + b^2 (0. + 0./c) + 1.b (-1. + 0. c) - 0.5 c + 
    (b (-1. b - 0.5 c) c)/a^2

without using Chop and it takes care of the instances of both 1.b and -1.b.

expr /. fixerPattern

fixup

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    $\begingroup$ I think you'll want something like /. {c_ x_ /; c == 1 :> x, c_ x_ /; c == -1 :> -x} for generality? (As you used in a more recent answer...) $\endgroup$
    – Mr.Wizard
    Commented Jun 8, 2017 at 2:17
  • $\begingroup$ @Mr.Wizard. What do you think of /. c_ x_ /; Abs[c ] == 1 :> Sign[c] x as a more concise alternative? $\endgroup$
    – m_goldberg
    Commented Jun 8, 2017 at 6:16
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    $\begingroup$ Might not that affect complex values in a strange way? A contrived example: (0.5` + 0.8660254037844386` I) /. c_ /; Abs[c] == 1 :> foo $\endgroup$
    – Mr.Wizard
    Commented Jun 8, 2017 at 6:47
  • $\begingroup$ @Mr.Wizard. Good point, What about this? /. c_Real x_ /; Abs[c] == 1 :> Sign[c] x $\endgroup$
    – m_goldberg
    Commented Jun 8, 2017 at 6:51
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    $\begingroup$ I believe we could handle all three cases (including zero) with one rule: x_ /; Sign[x] == x :> Sign[x] $\endgroup$
    – Mr.Wizard
    Commented Jun 8, 2017 at 7:08

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