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

> 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".

> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote:
> 
>> 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".
Besides symbolic differentiation, there is also the option of numeric
differentiation. Here's an amateurish attempt:

myNumDeriv <- function(x){ (exp( predict (graphmodeld,
newdata=data.frame(t=x+.0001))) -
                                            exp( predict (graphmodeld,
newdata=data.frame(t=x) )))/
                                          .0001 }
myNumDeriv(c(100, 250, 350))



David Winsemius
Alameda, CA, USA

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

> On Apr 6, 2018, at 8:03 AM, David Winsemius <dwinsemius at
comcast.net> wrote:
> 
> 
>> On Apr 6, 2018, at 3:43 AM, g l <gnulinux at gmx.com> wrote:
>> 
>>> 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".
> 
> Besides symbolic differentiation, there is also the option of numeric
differentiation. Here's an amateurish attempt:
> 
> myNumDeriv <- function(x){ (exp( predict (graphmodeld,
newdata=data.frame(t=x+.0001))) -
>                                            exp( predict (graphmodeld,
newdata=data.frame(t=x) )))/
>                                          .0001 }
> myNumDeriv(c(100, 250, 350))
I realized that this would not work in the context of your construction. I had
earlier made a more symbolic version using R formulae:

 graphdata<-read.csv(text='t,c
 0,100
 40,78
 80,59
 120,38
 160,25
 200,21
 240,16
 280,12
 320,10
 360,9
 400,7')
 graphmodeld<-lm(log(c)~t, graphdata)
 graphmodelp<-exp(predict(graphmodeld))
 plot(c~t, graphdata)
 lines(graphdata[,1],graphmodelp)
 myNumDeriv(c(100, 250, 350), graphmodeld )
#----------------------------------------------
          1           2           3 
-0.31464102 -0.11310753 -0.05718414 

> 
> 
> 
> David Winsemius
> Alameda, CA, USA
> 
> 'Any technology distinguishable from magic is insufficiently
advanced.'   -Gehm's Corollary to Clarke's Third Law
> 
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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