On Jan 15, 2009, at 10:04 AM, e-letter wrote:

On 13/01/2009, David Winsemius <dwinsem...@comcast.net> wrote:
It's fairly clear from the documentation that approxfun() will not
extrapolate.

help.search("extrapolate")
library(Hmisc)
?approxExtrap

Some sort of minimization approach:

approxExtrap(x=c(0,5,10,15,20), y=c(16,45,77,101,125),xout=c(-4,0,4))
$x
[1] -4  0  4

$y
[1] -7.2 16.0 39.2

approxExtrap(x=c(0,5,10,15,20),
y=c(16,45,77,101,125),xout=seq(-2.8,-2.6, by=0.01))
$x
 [1] -2.80 -2.79 -2.78 -2.77 -2.76 -2.75 -2.74 -2.73 -2.72 -2.71
-2.70 -2.69 -2.68
[14] -2.67 -2.66 -2.65 -2.64 -2.63 -2.62 -2.61 -2.60

$y
 [1] -0.240 -0.182 -0.124 -0.066 -0.008  0.050  0.108  0.166  0.224
0.282  0.340
[12]  0.398  0.456  0.514  0.572  0.630  0.688  0.746  0.804  0.862
0.920

How accurate do you need the answer?

I tried Hmisc's inverseFunction(), but it returned 0 for an argument
of zero:

invF <- inverseFunction(x=c(0,5,10,15,20), y=c(16,45,77,101,125))
invF(0)

So I then hacked Harrell's inverseFunction by substituting
approxExtrap in every in instance
where approx appeared, creating invFunc2:

then

invF <- invFunc2(x=c(0,5,10,15,20), y=c(16,45,77,101,125))

invF(0)
[1] -2.758621
I have compared your answer to those obtained from gnuplot, scilab and
qtiplot; all report a result of x=-3.28. Why is r different?

Perhaps a coding error on my part (or on your part). Perhaps different methods (none of which you describe)?

I suspect that my method only used the first two points (I just checked by plotting and -2.7 is closer to the paper and pen result I get than is -3.28. Perhaps you made an extrapolation from a linear fit of a dataset that is not co-linear?

> lm(c(0,5) ~ c(16,45))

Call:
lm(formula = c(0, 5) ~ c(16, 45))

Coefficients:
(Intercept)    c(16, 45)
    -2.7586       0.1724

It not that "R is different", it is merely that I used it differently than you used your other tools.

Here's another method ( using all points and again reversing the roles of x and y) :
> lm(c(0,5,10,15,20) ~ c(16,45,77,101,125))

Call:
lm(formula = c(0, 5, 10, 15, 20) ~ c(16, 45, 77, 101, 125))

Coefficients:
            (Intercept)  c(16, 45, 77, 101, 125)
                -3.2332                   0.1818




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