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For example: Can we declare a function that works on pairs? Like $f(\{x,y\})$ or define an operation on pairs like $\{x,y\}\circ \{X,Y\}$?

I tried to look at f@@ but it doesn't seems to be what I want. I could declare a function such as

f[x_, y_, X_, Y_] := {x \[Theta][y] X, y Y}

And then compute

f @@ {a, b, c, d}

But this is not too "natural". Given what I know, If I want to compute, for example $\{x,y\} \circ \{x',y'\} \circ \{x'',y''\}$, I'd have to make a set $\{x,y,x',y'\}$, use f@@ which would yield another set of two elements, and now to compute the remaining operation, I'd have to construct a set with $\{i,j,x'',y''\}$ and then apply f@@ again. This is rather cumbersome.I'd like to construct an operation where I could really type $\{x,y\} \circ \{x',y'\} \circ \{x'',y''\}$ and it would work.

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1 Answer 1

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Edit

SetAttributes[CircleDot, {Flat, OneIdentity}]; CircleDot[{x_, y_}, {a_, b_}] := {x ** θ[y] ** a, y ** b}; {g1,h1}⊙{g2, h2}⊙{g3, h3}

{g1 ** θ[h1] ** g2 ** θ[h1 ** h2] ** g3, h1 ** h2 ** h3}

Update

SetAttributes[CircleDot, {Orderless, Flat, OneIdentity}]; CircleDot[{x_, y_}, {a_, b_}] := {x*θ[y]*a, y*b} {g1, h1}⊙{g2, h2}⊙{g3, h3}

{g1 g2 g3 θ[h2] θ[h1 h2], h1 h2 h3}

SetAttributes[ CircleDot, {NumericFunction, Orderless, Flat, OneIdentity}]; CircleDot[{x_, y_}, {a_, b_}] := {a*x, b*y}; {x, y}⊙{a, b}⊙{m, n}

{a m x, b n y}

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    $\begingroup$ This is really nice! I am doing the following and it seems to be assuming that things associate and commute. Can we stop that? $\endgroup$ Commented Mar 12, 2021 at 1:13
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    $\begingroup$ @BillyRubina What is $\theta$? You can edit you question and add the new code about $\theta$ $\endgroup$ Commented Mar 12, 2021 at 2:22
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    $\begingroup$ It is a function but I'd like to leave it unevaluated in the form $\theta(x)$. The trouble is that Mathematica changes the order, I don't know why. $\endgroup$ Commented Mar 12, 2021 at 2:29
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    $\begingroup$ @ SetAttributes[CircleDot, {Orderless, Flat, OneIdentity}]; CircleDot[{x_, y_}, {a_, b_}] := {x*θ[y]*a, y*b} {g1, h1}⊙{g2, h2}⊙{g3, h3} ? $\endgroup$ Commented Mar 12, 2021 at 2:46
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    $\begingroup$ @BillyRubina SetAttributes[CircleDot, {Flat, OneIdentity}]; CircleDot[{x_, y_}, {a_, b_}] := {x ** \[Theta][y] ** a, y ** b}; {g1, h1}\[CircleDot]{g2, h2}\[CircleDot]{g3, h3} $\endgroup$ Commented Mar 12, 2021 at 3:44

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