[R] solving x in a polynomial function

[Rui Barradas]

Thanks.

Rui Barradas

Em 01-03-2013 20:35, arun escreveu:
> 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.
>>
>
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