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MathGroup Archive 2005

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Re: Two related question. Question 1

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57541] Re: Two related question. Question 1
  • From: dh <dh at metrohm.ch>
  • Date: Wed, 1 Jun 2005 06:01:27 -0400 (EDT)
  • References: <d7dp2r$qam$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi Kazimir,
a function is not the same as a variable. A function has input slots 
that you must take care of. There is a samll pitfall here, "Function" 
has the attribute HoldAll, therefore you must use Evaluate to do some 
calculation inside a function definition. Therefore, if
f = #1^2 + #2 &
the the square is obtained by:

c= Evaluate[f[#1,#2]^2]&

this gives:(#1^2 + #2^2)^2 &
Sincerely, Daniel


Kazimir wrote:
> I have two related question. Let me introduce a pure function
> 
> f = #1^2 + #2 &
> 
> Now. I want to make an operation over the function, for example to
> find its square and to call the result (the expected function f = (#1^2
> + #2)^2 & ) c:
> 
> c=f^2
> 
> However, I do not obtain this, as 
> 
> c[a,b]
> 
> does not evaluate to (a+b)^2. Can anybody advise me how to obtain
> such a function without long substitutions. I would like to obtain
> something which is made for derivatives :
> 
> In[11]:=
> Derivative[1][f]
> 
> Out[11]=
> 2 #1&
> 
> In[12]:=
> Derivative[2][f]
> 
> Out[12]=
> 2&
> 
> Regards 
> 
> Vlad
> 


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