I have looked here which describes how to define operators. I was wondering whether there was a way to assign thes operators to special characters? eg Let $\odot:=(a+b)(ab),$ so 4\[CircleDot]3 would yield 84? It is really a stylistic / display issue - of course the same could be achieved with cd[a_,b_]:=(a+b) a b, implemented with cd[a,b].
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1$\begingroup$ Take a look at this question, see if my answer there might help you along: How to assign symbols to functions $\endgroup$– MarcoBCommented Jun 19, 2015 at 12:32
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$\begingroup$ @MarcoB I have taken a look at the link, but usure on syntax when it comes to applying it to the example above $\endgroup$– martinCommented Jun 19, 2015 at 12:38
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1$\begingroup$ I think Karsten took care of that below :-) $\endgroup$– MarcoBCommented Jun 19, 2015 at 12:46
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1 Answer
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4
CircleDot[a_, b_] := (a + b) a b
Now
4⊙3
84
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$\begingroup$ ha! I didn't think it would be that simple - I thought the operator would have to preceed the arguments - thank you! :) $\endgroup$– martinCommented Jun 19, 2015 at 12:47
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3$\begingroup$ @martin
4⊙3//FullFormmay shed some light on your confusion. Some operators are infix, meaning⊙is the same as~CircleDot~and we know, thata~CircleDot~bis the same asCircleDot[a,b]. But in fact, you could just as easily set like so:a_⊙b_:=(a + b) a b, MMA would interpret this just like the full form that Karsten showed in this answer. $\endgroup$– LLlAMnYPCommented Jun 19, 2015 at 13:06 -
$\begingroup$ @LLlAMnYP ah, I see - much like
~Join~$\endgroup$– martinCommented Jun 19, 2015 at 13:07 -
1$\begingroup$ @martin precisely. When in doubt, always check the
FullForm:) $\endgroup$– LLlAMnYPCommented Jun 19, 2015 at 13:08