> Sent: Thursday, April 05, 2018 at 4:40 PM
> From: "Jeff Newmiller" <jdnew...@dcn.davis.ca.us>
>
> the coef function.
>
For the benefit of other novices, used the following command to read the
documentation:
?coef
Then tried and obtained:
> cvalue100<-coef(graphmodelp~100)
> cvalue100
NULL
Then looked at the model values which of course correspond to original
non-modelled values.
graphmodelp
1 2 3 4 5 6 7 8
91.244636 69.457794 52.873083 40.248368 30.638107 23.322525 17.753714 13.514590
9 10 11
10.287658 7.831233 5.961339
This prompted to think that interpolation is required, but the function
'approx' only seems to perform constant interpolation.
Is the correct thinking to find a function to perform interpolation, then
find/write a function to differentiate the model at a specific value of x, to
find gradient at that point?
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