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I want Mathematica to simplify my expression using MySymmetricFunction[x,y] = MySymmetricFunction[y,x], so that 

expr = MySymmetricFunction[x,y]-MySymmetricFunction[y,x];
Simplify[expr]

yields 0.

I tried

MySymmetricFunction[x_, y_] = MySymmetricFunction[y,x]

but here Mathematica assumes recursion.

Please note, that I don't want to implement an explicit version of MySymmetricFunction[x,y] as of yet.

edit:
As a follow up question:
Why does this give zero 

SetAttributes[MySymmetricFunction, Orderless]
MySymmetricFunction[x,y]-MySymmetricFunction[y,x]

But this does not:

<<FeynCalc`
SetAttributes[MySymmetricFunction, Orderless]
Simplify[MySymmetricFunction[pResonance, pRho]*FV[NucleonOut, m] - MySymmetricFunction[pRho, pResonance]*FV[pNucleonOut, m]]

edit2 Thanks to @QuantomDot for pointing out my silly spelling mistake.

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  • 3
    $\begingroup$ Could you give MySymmetricFunction the Orderless attribute? $\endgroup$
    – Carl Woll
    Mar 23, 2018 at 0:17
  • 2
    $\begingroup$ Set[MySymmetricFunction, Orderless] is a complete solution. $\endgroup$
    – evanb
    Mar 23, 2018 at 0:18
  • 3
    $\begingroup$ @evanb you mean SetAttributes. $\endgroup$
    – b3m2a1
    Mar 23, 2018 at 0:20
  • 1
    $\begingroup$ Orderless has interesting effects on pattern matching. $\endgroup$
    – rcollyer
    Mar 23, 2018 at 0:21
  • 3
    $\begingroup$ NucleonOut is not the same as pNucleonOut. $\endgroup$
    – QuantumDot
    Mar 23, 2018 at 0:33

1 Answer 1

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5
$\begingroup$ 
SetAttributes[msf, Orderless]
msf[x, y] - msf[y, x]
(* 0 *)
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1
  • $\begingroup$ Yes, this works, but please see my follow up question $\endgroup$
    – infinitezero
    Mar 23, 2018 at 0:28

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