RE: Re: Function as an argument of the function
- To: mathgroup at smc.vnet.net
- Subject: [mg34696] RE: [mg34670] Re: Function as an argument of the function
- From: "DrBob" <majort at cox-internet.com>
- Date: Sat, 1 Jun 2002 04:29:02 -0400 (EDT)
- Reply-to: <drbob at bigfoot.com>
- Sender: owner-wri-mathgroup at wolfram.com
That solution fails in this case: ClearAll[func] func[y] = 1; Length[Cases[func, Slot[_], Infinity]] == 1 || (func =!= Head[func[x]]) Bobby -----Original Message----- From: Jens-Peer Kuska [mailto:kuska at informatik.uni-leipzig.de] To: mathgroup at smc.vnet.net Subject: [mg34696] [mg34670] Re: Function as an argument of the function Sorry for my last answer, you have to test: Length[Cases[func, Slot[_], Infinity]] == 1 || (func =!= Head[func[x]]) to "find" a function of one variable. Regards Jens Tomek wrote: > > Hi. > I have to write a short function in Mathematica. But I have one > problem with it. One of arguments of my function have to be a > function of one variable. > How can I check if the argument of the function is a function of one > variable (I wish there's FunctionQ)?