Re: [R] How to define specially nested functions

2011-04-29 Thread Chee Chen
I will have to recode in Mathematica). Any of your further help is much appreciated! Best regards, -Chee From: Jerome Asselin Sent: Friday, April 29, 2011 12:25 AM To: Chee Chen Cc: R -Help Subject: Re: [R] How to define specially nested functions On Thu, 2011-04-28 at 23:08 -0400, Chee

Re: [R] How to define specially nested functions

2011-04-28 Thread Petr Savicky
On Thu, Apr 28, 2011 at 11:08:23PM -0400, Chee Chen wrote: > Dear All, > I would like to define a function: f(x,y,z) with three arguments x,y,z, such > that: given values for x,y, f(x,y,z) is still a function of z and that I am > still allowed to find the root in terms of z when x,y are given. >

Re: [R] How to define specially nested functions

2011-04-28 Thread Phil Spector
Here's one possibility: funmaker = function(x,y,z)function(z)x + y + (x^2 - z) uniroot(funmaker(1,3,z),c(0,10))$root [1] 5 uniroot(funmaker(5,2,z),c(30,40))$root [1] 32 (The third argument to the function doesn't really do anything.) - Phil Spector

Re: [R] How to define specially nested functions

2011-04-28 Thread Jerome Asselin
On Thu, 2011-04-28 at 23:08 -0400, Chee Chen wrote: > Dear All, > I would like to define a function: f(x,y,z) with three arguments x,y,z, such > that: given values for x,y, f(x,y,z) is still a function of z and that I am > still allowed to find the root in terms of z when x,y are given. > For ex

[R] How to define specially nested functions

2011-04-28 Thread Chee Chen
Dear All, I would like to define a function: f(x,y,z) with three arguments x,y,z, such that: given values for x,y, f(x,y,z) is still a function of z and that I am still allowed to find the root in terms of z when x,y are given. For example: f(x,y,z) = x+y + (x^2-z), given x=1,y=3, f(1,3,z)= 1+