Dear R-Users,
Just out of curiosity:
Which of the 2 methods is the better one?
The results seem to differ slightly.
fun = function(u){((26104.50*u^0.03399381)/((1-u)^0.1074444)) - 28353.7}
uniroot(fun, c(0,1))
# 0.6048184
curve(fun(x), 0, 1)
abline(v=0.3952365, col="red")
abline(v=0.6048184, col="red")
abline(h=0, col="blue")
fun = function(u){ (0.03399381*log(u) - 0.1074444*log(1-u)) -
log(28353.7/26104.50) }
fun = function(u){ (0.03399381*log(u) - 0.1074444*log1p(-u)) -
log(28353.7/26104.50) }
uniroot(fun, c(0,1))
# 0.6047968
curve(fun(x), 0, 1)
abline(v=0.3952365, col="red")
abline(v=0.6047968, col="red")
abline(h=0, col="blue")
Sincerely,
Leonard
Query on finding root
3 messages · Leonard Mada, Ben Bolker
I would probably use the built in qdavies() function... On Mon, Aug 28, 2023, 7:48 AM Leonard Mada via R-help <r-help at r-project.org> wrote:
Dear R-Users,
Just out of curiosity:
Which of the 2 methods is the better one?
The results seem to differ slightly.
fun = function(u){((26104.50*u^0.03399381)/((1-u)^0.1074444)) - 28353.7}
uniroot(fun, c(0,1))
# 0.6048184
curve(fun(x), 0, 1)
abline(v=0.3952365, col="red")
abline(v=0.6048184, col="red")
abline(h=0, col="blue")
fun = function(u){ (0.03399381*log(u) - 0.1074444*log(1-u)) -
log(28353.7/26104.50) }
fun = function(u){ (0.03399381*log(u) - 0.1074444*log1p(-u)) -
log(28353.7/26104.50) }
uniroot(fun, c(0,1))
# 0.6047968
curve(fun(x), 0, 1)
abline(v=0.3952365, col="red")
abline(v=0.6047968, col="red")
abline(h=0, col="blue")
Sincerely,
Leonard
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
(I mean pdavies)
On Mon, Aug 28, 2023, 7:52 AM Ben Bolker <bbolker at gmail.com> wrote:
I would probably use the built in qdavies() function... On Mon, Aug 28, 2023, 7:48 AM Leonard Mada via R-help < r-help at r-project.org> wrote:
Dear R-Users,
Just out of curiosity:
Which of the 2 methods is the better one?
The results seem to differ slightly.
fun = function(u){((26104.50*u^0.03399381)/((1-u)^0.1074444)) - 28353.7}
uniroot(fun, c(0,1))
# 0.6048184
curve(fun(x), 0, 1)
abline(v=0.3952365, col="red")
abline(v=0.6048184, col="red")
abline(h=0, col="blue")
fun = function(u){ (0.03399381*log(u) - 0.1074444*log(1-u)) -
log(28353.7/26104.50) }
fun = function(u){ (0.03399381*log(u) - 0.1074444*log1p(-u)) -
log(28353.7/26104.50) }
uniroot(fun, c(0,1))
# 0.6047968
curve(fun(x), 0, 1)
abline(v=0.3952365, col="red")
abline(v=0.6047968, col="red")
abline(h=0, col="blue")
Sincerely,
Leonard
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.