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MathGroup Archive 2002

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Re: Function as an argument of the function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg34732] Re: Function as an argument of the function
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Tue, 4 Jun 2002 03:41:42 -0400 (EDT)
  • Organization: Universitaet Leipzig
  • References: <ada17i$kc8$1@smc.vnet.net>
  • Reply-to: kuska at informatik.uni-leipzig.de
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,

it fails not ! it returns False as it should, because
func[y]=1

is *not* a function definition. It defines a constant, but not
a constant function.
A function should map a parameter space like all real number onto
an image space and the symbol y is not a parameter space.

Regards
  Jens

DrBob wrote:
> 
> 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: [mg34732]  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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