[英]uniroot in R when there are two unknowns
Consider a function f
of two arguments x
and a
. 考虑两个参数x
和a
的函数f
。 First I take integration of f
with respect to x
, which becomes a function g
of a
. 第一予取的集成f
相对于x
,这成为一个功能g
的a
。 Second, I want to find the root of the resulting function g
of a
. 其次,我想找到a
的结果函数g
的根。 Can I do this using uniroot
and integrate
in R
? 我可以使用uniroot
并integrate
到R
吗? If so, how? 如果是这样,怎么办? If not, is there a way to do this at all? 如果没有,有没有办法做到这一点? Thanks. 谢谢。
b <- 2
truncfn <- function(x) pmin(b, pmax(x, -b))
# thetashape and thetascale are constants
# x and a are arguments
f <- function(x, thetashape, thetascale, a){
term1 <- -1/thetascale
term2 <- (1-thetashape)/thetascale
term3 <- x/(thetascale-thetashape*x)
term1 + term2*term3 - a
}
# First, integrate f with respect to x
g <- integrate(truncfn(f), lower=0, upper=Inf)
# Second, find root of g
uniroot(g, ...)
You can define a function (I call it truncfn2
) that calls truncfn
on the result of the call to f
, and then g
integrates truncfn2
. 您可以定义一个函数(我称其为truncfn2
),该truncfn
在对f
的调用结果上调用truncfn
,然后g
集成truncfn2
。 Finally uniroot
searches for the root of g
: 最后, uniroot
搜索g
的根:
b <- 2
truncfn <- function(x) pmin(b, pmax(x, -b))
# thetashape and thetascale are constants
# x and a are arguments
f <- function(x, thetashape, thetascale, a){
term1 <- -1/thetascale
term2 <- (1-thetashape)/thetascale
term3 <- x/(thetascale-thetashape*x)
term1 + term2*term3 - a
}
truncfn2 <- function(x, thetashape, thetascale, a) truncfn(f(x, thetashape, thetascale, a))
g <- function(a) integrate(truncfn2, thetascale=1, thetashape=0.6, a=a, lower=0, upper=10)$value
uniroot(g, lower=-10, upper=10)
# $root
# [1] -1.867932
#
# $f.root
# [1] 1.134733e-07
#
# $iter
# [1] 7
#
# $init.it
# [1] NA
#
# $estim.prec
# [1] 6.103516e-05
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