how can I ensure the following definition to be associative?

a_ f[x_] + b_ f[y_] ^:= f[a x + b y]

gives me

a f[x] + b f[y] + c f[z]

f[a x + b y] + c f[z]

I'd like it to be

f[a x + b y + c z]

  • $\begingroup$ You probably need to set default values for the patterns a_ and b_ like a_. and b_., otherwise \[ScriptCapitalN][a x + b y] does not match the pattern of multiplier_ \[ScriptCapitalN][a x + b y] $\endgroup$ – LLlAMnYP Sep 11 '18 at 8:58
  • $\begingroup$ My previous comment about Times vs. Plus was of course wrong. $\endgroup$ – LLlAMnYP Sep 11 '18 at 8:59
  • $\begingroup$ Well it works with the default value. But I don't really understand why I need it. $\endgroup$ – Display Name Sep 11 '18 at 9:01
  • $\begingroup$ And if c=1 it does not work anymore. $\endgroup$ – Display Name Sep 11 '18 at 9:07
  • 1
    $\begingroup$ @LLlAMnYP That's easy to achieve. Just go to thw NinjaKot script and add ['\\\\\\[ScriptCapitalN\\]', 'f'], on line 599 =D $\endgroup$ – Henrik Schumacher Sep 11 '18 at 9:43
f[a x + b y] + c f[z]

is not the same as

1 * f[a x + b y] + c f[z]

so the pattern

a_ f[x] + b_ f[y]

does not match it. But if you use a_. then it's replaced with the default value (1, for Times) if the multiplier is omitted. So then it will match.

However, you observe that if c == 1, then it doesn't work again. This is understandable, since if c == 1, the expression

f[a x + b y] + f[z]

does not contain Times at level 1 anymore. I suggest using more granular rules:

a_ f[x_] ^:= f[a x]

f[x_] + f[y_] ^:= f[x + y]

This achieves two things: the upvalue is attached to f and not to system symbols, and it solves the problem you mentioned in the comments.

  • $\begingroup$ Thank you for the explanation. But isn't the case c==1 a variant of the first case, now with the two multipliers omitted? And shouldn't then the two multipliers be replaced with their default values? Or did I get it completely wrong what you tried to point out? Edit Or can only be one default value be replaced at a time? $\endgroup$ – Display Name Sep 11 '18 at 9:28
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    $\begingroup$ The UpValue in your attempt is associated with Times. I replace scriptcapitalN with just N for brevity: N[x] + b N[y] is Plus[N[x], Times[b, N[y]]]. Times is present at level 1 of the expression, so MMA checks UpValues for the symbol Times. However, N[x] + N[y] is Plus[N[x], N[y]], so MMA has no waying of realizing that it has to apply the UpValue defined for Times. So it doesn't even get round to doing that and attempting to substitute default values in the process. $\endgroup$ – LLlAMnYP Sep 11 '18 at 9:33
  • $\begingroup$ Thank you, makes perfectly sense. $\endgroup$ – Display Name Sep 11 '18 at 9:47

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