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A problem related to symbolic function.

3 vues (au cours des 30 derniers jours)
Angus Wong
Angus Wong le 2 Juil 2019
Commenté : Angus Wong le 2 Juil 2019
I know that if a want to check if a variable is a function handle, then I can use isa(fun,'function_handle').
What should I type in the 2nd input on the isa function to determine if a variable is a symbolic function? I mean I want the output to be 1 if the variable is a symbolic function, and 0 if it isn't. I am expecting something like: isa(fun,'symbolic_function'). (However, it does not work on my computer.)
* fun is the name of the variable to be determined.

Réponse acceptée

Alex Mcaulley
Alex Mcaulley le 2 Juil 2019
Modifié(e) : Alex Mcaulley le 2 Juil 2019
Try this:
isa(fun,'symfun')

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