solving x in a polynomial function

Hi guys,

Perhaps on the right track? However, not sure if it's correct. I fixed
the bug that A.K. indicated above (== vs =), but the values don't seem
right. From the original data, an a of 3 should give b of 2.5.

realroots <- function(model, b){

+         is.zero <- function(x, tol = .Machine$double.eps^0.5) abs(x) < tol
+         if(names(model)[1] == "(Intercept)")
+                 r <- polyroot(c(coef(model)[1] - b, coef(model)[-1]))
+         else
+                 r <- polyroot(c(-b, coef(model)))
+         Re(r[is.zero(Im(r))])
+ }

r <- realroots(po.lm, 3)
predict(po.lm, newdata = data.frame(b = r)) # confirm

    1
1.69

So I think there's a calculation error somehwere.






On 3/1/13, arun <smartpink111 at yahoo.com> wrote:

Hi Rui,

Looks like a bug:
###if(names(model)[1] = "(Intercept)")
Should this be:

if(names(model)[1] == "(Intercept)")

A.K.



----- Original Message -----
From: Rui Barradas <ruipbarradas at sapo.pt>
To: Mike Rennie <mikerennie.r at gmail.com>
Cc: r-help Mailing List <r-help at r-project.org>
Sent: Friday, March 1, 2013 3:18 PM
Subject: Re: [R] solving x in a polynomial function

Hello,

Try the following.


a <- 1:10
b <- c(1, 2, 2.5, 3, 3.5, 4, 6, 7, 7.5, 8)

dat <- data.frame(a = a, b = b)  # for lm(), it's better to use a df
po.lm <- lm(a~b+I(b^2)+I(b^3)+I(b^4), data = dat); summary(po.lm)


realroots <- function(model, b){
     is.zero <- function(x, tol = .Machine$double.eps^0.5) abs(x) < tol
     if(names(model)[1] = "(Intercept)")
         r <- polyroot(c(coef(model)[1] - b, coef(model)[-1]))
     else
         r <- polyroot(c(-b, coef(model)))
     Re(r[is.zero(Im(r))])
}

r <- realroots(po.lm, 5.5)
predict(po.lm, newdata = data.frame(b = r))  # confirm



Hope this helps,

Rui Barradas

Em 01-03-2013 18:47, Mike Rennie escreveu:

Hi there,

Does anyone know how I solve for x from a given y in a polynomial
function? Here's some example code:

##example file

a<-1:10

b<-c(1,2,2.5,3,3.5,4,6,7,7.5,8)

po.lm<-lm(a~b+I(b^2)+I(b^3)+I(b^4)); summary(po.lm)

(please ignore that the model is severely overfit- that's not the point).

Let's say I want to solve for the value b where a = 5.5.

Any thoughts? I did come across the polynom package, but I don't think
that does it- I suspect the answer is simpler than I am making it out
to be. Any help would be welcome.