[R] Obtain gradient at multiple values for exponetial decay model

[David Winsemius]

> On Apr 5, 2018, at 2:00 PM, g l <gnulinux using gmx.com> wrote:
> 
>> Sent: Thursday, April 05, 2018 at 4:40 PM
>> From: "Jeff Newmiller" <jdnewmil using 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

Should have been:

  coef(graphmodelp)


> 
> 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

Read up on ?predict and what it delivers when only a model is offered as input.

> 
> 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?

Not correct. You already have `predict`. It is capale of using the `newdata` values to do interpolation with the values of the coefficients in the model. See: 

?predict

The original question asked for a derivative (i.e. a "gradient"), but so far it's not clear that you understand the mathematical definiton of that term. We also remain unclear whether this is homework.


> 
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David Winsemius
Alameda, CA, USA

'Any technology distinguishable from magic is insufficiently advanced.'   -Gehm's Corollary to Clarke's Third Law