> On Apr 5, 2018, at 2:00 PM, g l wrote:
>
>> Sent: Thursday, April 05, 2018 at 4:40 PM
>> From: "Jeff Newmiller"
>>
>> the coef function.
>>
>
> For the benefit of other novices, used the following command to read the
> documentation:
>
> ?coef
>
> Then tried and obtained:
>
>> cvalue10
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 wrote:
>> Sent: Thurs
> Sent: Thursday, April 05, 2018 at 4:40 PM
> From: "Jeff Newmiller"
>
> 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
This smells like homework, which the Posting Guide indicates is off topic.
I am not aware of "the function" that will solve this, but if you know what a
gradient is analytically then you should be able to put together a solution
very similar to the code you already have with the addition of usin
Readers,
Data set:
t,c
0,100
40,78
80,59
120,38
160,25
200,21
240,16
280,12
320,10
360,9
400,7
graphdata<-read.csv('~/tmp/data.csv')
graphmodeld<-lm(log(graphdata[,2])~graphdata[,1])
graphmodelp<-exp(predict(graphmodeld))
plot(graphdata[,2]~graphdata[,1])
lines(graphdata[,1],graphmodelp)
Please
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