fitting curve to data

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Just 'cos it's bent doesn't mean you need nls.
Sorry, my bad! :o(
With your data, lm fits (suspicously!) well...
Nothing suspicious....just benchmarking some functions with different sized inputs!

Thanks for the help!
Nathan
y<-c(0.000,0.004,0.008,0.016,0.024,0.032,0.044,0.064,0.072,0.088,0.108,0.140
,0.156,0.180,0.208,0.236,0.264,0.296,0.320,0.360,0.408,0.444,0.472,0.524
,0.576)
x<-c(100,200,300,400,500,600,700,800,900,1000,1100,1200,1300,1400,1500
,1600,1700,1800,1900,2000,2100,2200,2300,2400,2500)
plot(x,y)
lines(x, predict(lm(y~I(x^2))))

"Nathan S. Watson-Haigh" <nathan.watson-haigh at csiro.au> 01/12/09
2:19 AM >>>
I have the following data:

y
 [1] 0.000 0.004 0.008 0.016 0.024 0.032 0.044 0.064 0.072 0.088 0.108
0.140
[13] 0.156 0.180 0.208 0.236 0.264 0.296 0.320 0.360 0.408 0.444 0.472
0.524
[25] 0.576
x
 [1]  100  200  300  400  500  600  700  800  900 1000 1100 1200 1300
1400 1500
[16] 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500

I'd like to plot the points and calculate a curved line of best fit. I
know I need to use nls(), but
I'm unsure how to begin....any pointers?

Cheers,
Nathan

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- --
- --------------------------------------------------------
Dr. Nathan S. Watson-Haigh
OCE Post Doctoral Fellow
CSIRO Livestock Industries
Queensland Bioscience Precinct
St Lucia, QLD 4067
Australia

Tel: +61 (0)7 3214 2922
Fax: +61 (0)7 3214 2900
Web: http://www.csiro.au/people/Nathan.Watson-Haigh.html
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