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 >