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
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
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 example: f(x,y,z) = x+y + (x^2-z), given x=1,y=3, f(1,3,z)= 1+3+1-z is a
function of
z, and then I can use R to find the root z=5.
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