How can I create a function F that works like this:
x = 2 f1 = x^2 + 3 f2=x+5 F[f1] x^2 + 3 = 2^2 + 3 = 7
F[f2] x + 5 = 2 + 5 = 7
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2 
1 Answer
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2 First you will need to make your definitions with SetDelayed (:=) rather than Set (=):
x := 2 f1 := x^2 + 3 f2 := x + 5 You can get a list of all evaluation steps using Trace or TracePrint. To get only the steps that transform the entire expression use the Option TraceDepth -> 1, and you can format with Row
Row[Trace[f1, TraceDepth -> 1], "="] f1 = x^2+3 = 4+3 = 7
This is not exactly what you asked for but I hope it is close enough to help. The additional steps are accessible with Trace but I could not think of a simple and robust way to use them and I can't spend more time on this right now. For reference:
Trace[f1] {f1,x^2+3,{{x,2},2^2,4},4+3,7}
- $\begingroup$ I think your answer might be closer to what the OP is asking for if you added a definition for his
Falong the lines of:SetAttributes[F, HoldFirst]; F[var_Symbol] := Row[Trace[var, TraceDepth -> 1], "="]-- providing, of course, that you think this would be robust enough to meet the OP needs. $\endgroup$m_goldberg– m_goldberg2013-11-28 02:15:53 +00:00Commented Nov 28, 2013 at 2:15 - $\begingroup$ @m_goldberg Good point. user10833: that is indeed the way to package an expression such as this as a function. I did not do that as I did not feel this was complete, e.g. there is no
2^2+3step shown. $\endgroup$Mr.Wizard– Mr.Wizard2013-11-28 02:24:58 +00:00Commented Nov 28, 2013 at 2:24
xinx^2 + 3. User: is this correct? $\endgroup$