Hi,
And regarding how to extend the nls algorithm to a larger dataset it is a
question of indicating in the nls() functions which data.frame to use.
In you example you were using a small set of x's ad y's, so if you want to
use a large set put it in a new data.frame and pass it to nls().
And if y
Your formula could be simplified to
y ~ 1 + a * exp(-b * x)
Solve this equation for x
x = ln[(y - 1)/a]/b
Use this equation to find the intersection point at a given value of y.
For example, when
y = 1.01
x = ln(0.01/a)/b
Jean
Karen Vandepoel wrote on 04/09/2012 0
Hi,
I will try to explain what it is I need to do, how far I am in doing it yet
and where my problem is:
I have a lot of x,y values I need to fit a non linear function through.
Subsequently, I need to find the intersection point of this fitted curve
with y=1.01
The problem is I have a lot of val
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