Bert Gunter gmail.com> writes:
>
> Use ?uniroot to do it numerically instead of polyroot()?
>
> Cheers,
> Bert
> Bert Gunter
The problem with uniroot() is that we don't know how many intersections/
roots we might be looking for. With polyroot(), we know that there can
be at most 3 roots that
... (should have added)
However one might ask: Isn't this just a bit silly? The density()
function gives kernel density estimates (perhaps interpolated by
?approx -- see ?density) on as fine a grid as one likes, so why use
splines thereafter? And since these are density estimates -- i.e.
fitted ap
Use ?uniroot to do it numerically instead of polyroot()?
Cheers,
Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
-- Clifford Stoll
On Mon, Sep 28, 2015 at 9:17 AM, Ben Bolker wrote:
> Dario Strbenac uni.sydney.edu.au> writes:
Dario Strbenac uni.sydney.edu.au> writes:
>
> Good day,
>
> I have two probability densities, each with a function determined
> by splinefun(densityResult[['x']],
> densityResult[['y']], "natural"), where densityResult is the
> output of the density function in stats.
> How can I determine all
Good day,
I have two probability densities, each with a function determined by
splinefun(densityResult[['x']], densityResult[['y']], "natural"), where
densityResult is the output of the density function in stats. How can I
determine all of the x values at which the densities cross ?
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