# [R] Query on finding root

Ben Bolker bbo|ker @end|ng |rom gm@||@com
Sun Aug 27 23:59:55 CEST 2023

```    This doesn't look like homework to me -- too specific.  The posting
guide <https://www.r-project.org/posting-guide.html> says that the list
is not intended for "Basic statistics and classroom homework" -- again,
this doesn't seem to fall into that category.

tl;dr, I think the difference between the two approaches is just
whether the lower or upper tail is considered (i.e., adding lower.tail =
FALSE to pdavies(), or more simply taking (1-x), makes the two answers
agree).

If you look at the source code for pdavies() you'll see that it's
essentially doing the same uniroot() calculation that you are.

## Q(u)=(c*u^lamda1)/((1-u)^lamda2)
mean <- 28353.7 # mean calculated from data
lambda1 <- .03399381 # estimates c, lambda1 and lambda2 calculated from data
lambda2 <- .1074444
c <- 26104.50
library(Davies)# using package
params <- c(c,lambda1,lambda2)
u <- pdavies(x = mean, params = params, lower.tail = FALSE)
u
fun <- function(u) {
with(as.list(params), (c*u^lambda1)/((1-u)^lambda2)) - mean
}
curve(fun, from = 0.01, to = 1)
uniroot <- uniroot(fun,c(0.01,1))
abline(h = 0)
uniroot\$root

On 2023-08-27 5:40 p.m., Rolf Turner wrote:
>
> On Fri, 25 Aug 2023 22:17:05 +0530
> ASHLIN VARKEY <ashlinvarkey using gmail.com> wrote:
>
>> Sir,
>
> Please note that r-help is a mailing list, not a knight! 😊️
>
>> I want to solve the equation Q(u)=mean, where Q(u) represents the
>> quantile function. Here my Q(u)=(c*u^lamda1)/((1-u)^lamda2), which is
>> the quantile function of Davies (Power-pareto) distribution.  Hence I
>> want to solve , *(c*u^lamda1)/((1-u)^lamda2)=28353.7....(Eq.1)*
>> where lamda1=0.03399381, lamda2=0.1074444 and c=26104.50. When I used
>> the package 'Davies' and solved Eq 1, I got the answer u=0.3952365.
>> But when I use the function  'uniroot' to solve the Eq.1, I got a
>> different answer which is  u=0.6048157.  Why did this difference
>> happen?  Which is the correct method to solve Eq.1. Using the value
>> of *u *from the first method my further calculation was nearer to
>> empirical values.  The R-code I used is herewith. Kindly help me to
>> solve this issue.
>>
>> R-code
>> Q(u)=(c*u^lamda1)/((1-u)^lamda2)
>> mean=28353.7 # mean calculated from data
>> lamda1=.03399381 # estimates c, lamda1 and lamda2 calculated from data
>> lamda2=.1074444
>> c=26104.50
>> library(Davies)# using package
>> params=c(c,lamda1,lamda2)
>> u=pdavies(28353.7,params)
>> u
>> fun=function(u){((26104.50*u^0.03399381)/((1-u)^0.1074444))-28353.7}
>> uniroot= uniroot(fun,c(0.01,1))
>> uniroot
>
> As Prof. Nash has pointed out, this looks like homework.
>
> Some general advice:  graphics can be very revealing, and are easy to
> effect in R.  Relevant method: plot.function(); relevant utility:
> abline().  Look at the help for these.
>
> cheers,
>
> Rolf Turner
>

--
Dr. Benjamin Bolker
Professor, Mathematics & Statistics and Biology, McMaster University
Director, School of Computational Science and Engineering
(Acting) Graduate chair, Mathematics & Statistics
> E-mail is sent at my convenience; I don't expect replies outside of
working hours.

```