[R] Nonlinear regression

[Timothy Axberg]

Thank you! I didn't realize I wasn't linearizing it correctly. This was
very helpful.

-Timothy

On Tue, Dec 19, 2017 at 10:29 PM, Berwin A Turlach <berwin.turlach at gmail.com
> wrote:

> G'day Timothy,
>
> On Tue, 19 Dec 2017 18:28:00 -0600
> Timothy Axberg <axbergtimothy at gmail.com> wrote:
>
> > Should I repost the question with reply-all?
>
> Nope, we got all from Jeff's post. :)
>
> > On Tue, Dec 19, 2017 at 6:13 PM, Jeff Newmiller
> > <jdnewmil at dcn.davis.ca.us> wrote:
> >
> > > You also need to reply-all so the mailing list stays in the loop.
> > > --
> > > Sent from my phone. Please excuse my brevity.
> > >
> > > On December 19, 2017 4:00:29 PM PST, Timothy Axberg <
> > > axbergtimothy at gmail.com> wrote:
> > > >Sorry about that. Here is the code typed directly on the email.
> > > >
> > > >qe = (Qmax * Kl * ce) / (1 + Kl * ce)
> [...]
> > > >##The linearized data
> > > >celin <- 1/ce
> > > >qelin <- 1/qe
>
> Plotting qelin against celin, I can see why you call this the
> linearized data.  But fitting a linear model to these data obviously
> does not give you good starting values for nls().
>
> Given your model equation, I would linearize the model to:
>
>    qe = Qmax*KL * ce + KL * ce*qe
>
> and fit a non-intercept linear model to predict qe by ce and
> ce*qe.  From this model I would then determine starting values for
> nls().  It seems to work with your data set:
>
> R> ce <- c(15.17, 42.15, 69.12, 237.7, 419.77)
> R> qe <- c(17.65, 30.07, 65.36, 81.7, 90.2)
> R> fit2 <- lm(qe ~ ce + I(ce*qe) - 1)
> R> summary(fit2)
> R> Kl <- - coef(fit2)[2]
> R> Qmax <- coef(fit2)[1]/Kl
> R> plot(ce, qe)
> R> c <- seq(min(ce), max(ce))
> R> q <- (Qmax*Kl*c)/(1+(Kl*c))
> R> lines(c, q)
> R> fit2 <- nls(qe ~ ((Qmax*Kl*ce)/(1+(Kl*ce))), start = list(Qmax =
> Qmax,Kl =Kl))
> R> summary(fit2)
>
> Formula: qe ~ ((Qmax * Kl * ce)/(1 + (Kl * ce)))
>
> Parameters:
>       Estimate Std. Error t value Pr(>|t|)
> Qmax 106.42602   12.82808   8.296  0.00367 **
> Kl     0.01456    0.00543   2.681  0.07496 .
> ---
> Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
>
> Residual standard error: 9.249 on 3 degrees of freedom
>
> Number of iterations to convergence: 2
> Achieved convergence tolerance: 6.355e-06
>
> HTH.
>
> Cheers,
>
>         Berwin
>
>

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