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MathGroup Archive 2002

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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)?





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