R help - Apr 2018 - Obtain gradient at multiple values for exponential decay model

> Sent: Thursday, April 05, 2018 at 4:40 PM
> From: "Jeff Newmiller" <jdnewmil at 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?

Try

coef( graphmodeld )

And you don't need to approximate if you use the newdata argument to the
predict function.

I think reading the "Introduction to R" that comes with R would help.
-- 
Sent from my phone. Please excuse my brevity.

On April 5, 2018 2:00:45 PM PDT, g l <gnulinux at gmx.com>
wrote:
>> Sent: Thursday, April 05, 2018 at 4:40 PM
>> From: "Jeff Newmiller" <jdnewmil at 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?

> On Apr 5, 2018, at 2:00 PM, g l <gnulinux at gmx.com> wrote:
> 
>> Sent: Thursday, April 05, 2018 at 4:40 PM
>> From: "Jeff Newmiller" <jdnewmil at 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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> PLEASE do read the posting guide
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> and provide commented, minimal, self-contained, reproducible code.
David Winsemius
Alameda, CA, USA

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

> Sent: Friday, April 06, 2018 at 4:53 AM
> From: "Jeff Newmiller" <jdnewmil at dcn.davis.ca.us>
> To: "g l" <gnulinux at gmx.com>
> coef( graphmodeld )
> 
coef(graphmodelp)
Error: $ operator is invalid for atomic vectors

A quick search engine query revealed primarily references to the dollar sign ($)
operator which does not seem relevant to this question.

> Sent: Friday, April 06, 2018 at 5:55 AM
> From: "David Winsemius" <dwinsemius at comcast.net>
> 
> 
> 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 ? details did not mention interpolation explicity; thanks.
 
> 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.
> 
The motivation of this post was simple differentiation of a tangent point
(dy/dx) manually, then wondering how to re-think in modern-day computing terms.
Hence the original question about asking the appropriate functions/syntax to
read further ("curiosity"), not the answer (indeed,
"homework"). :)

Personal curiosity should be considered "homework".