Re: Two related question. Question 1

*To*: mathgroup at smc.vnet.net*Subject*: [mg57529] Re: [mg57498] Two related question. Question 1*From*: "David Park" <djmp at earthlink.net>*Date*: Tue, 31 May 2005 04:59:49 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Vlad, f = #1^2 + #2 &; c = #^2 &; cf = Composition[c, f];(a^2 + b)^2 cf[a, b] (a^2 + b)^2 Since f and cf are functions with two arguments, you need to use two arguments in the Derivative statement. Derivative[1, 0][f][a, b] 2 a Derivative[0, 1][f][a, b] 1 Derivative[1, 0][cf][a, b] 4*a*(a^2 + b) David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ From: Kazimir [mailto:kazimir04 at yahoo.co.uk] To: mathgroup at smc.vnet.net 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