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. -- Michael Rennie, Research Scientist Fisheries and Oceans Canada, Freshwater Institute Winnipeg, Manitoba, CANADA
help.search("polynomial")
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Mike Rennie <mikerennie.r at gmail.com> wrote:
>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.
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.
>