how can I ensure the following definition to be associative?
Unprotect[Times];
a_ f[x_] + b_ f[y_] ^:= f[a x + b y]
Protect[Times];
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]
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1 Answer
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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.
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$\begingroup$ Thank you for the explanation. But isn't the case
c==1a 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? $\endgroup$ Commented Sep 11, 2018 at 9:28 -
1$\begingroup$ The
UpValuein your attempt is associated withTimes. I replace scriptcapitalN with justNfor brevity:N[x] + b N[y]isPlus[N[x], Times[b, N[y]]].Timesis present at level 1 of the expression, so MMA checksUpValuesfor the symbolTimes. However,N[x] + N[y]isPlus[N[x], N[y]], so MMA has no waying of realizing that it has to apply theUpValuedefined forTimes. So it doesn't even get round to doing that and attempting to substitute default values in the process. $\endgroup$– LLlAMnYPCommented Sep 11, 2018 at 9:33 -
$\begingroup$ Thank you, makes perfectly sense. $\endgroup$ Commented Sep 11, 2018 at 9:47
a_andb_likea_.andb_., otherwise\[ScriptCapitalN][a x + b y]does not match the pattern ofmultiplier_ \[ScriptCapitalN][a x + b y]$\endgroup$Timesvs.Pluswas of course wrong. $\endgroup$['\\\\\\[ScriptCapitalN\\]', 'f'],on line 599 =D $\endgroup$