# [R] Find Crossover Points of Two Spline Functions

Bert Gunter bgunter.4567 at gmail.com
Mon Sep 28 18:47:21 CEST 2015

```... (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 approximations -- anyway, why search for "exact" solutions? Of
course I don't know the detailed context, but this whole question
sounds somewhat bogus.

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 9:36 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote:
> 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 <bbolker at gmail.com> wrote:
>> Dario Strbenac <dstr7320 <at> 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 of the x values at which the densities cross ?
>>>
>>
>>   My initial thought was this is non-trivial, because the two densities could
>> cross (or nearly-but-not-quite cross) at an unlimited number of points.
>> I thought it would essentially boils down to "how do I find all
>> the roots of an arbitrary (continuous, smooth) function?
>>
>>   However, after thinking about it for a few more seconds I realize
>> that at least the functions are piecewise cubic.  I still don't see
>> a *convenient* way to do it ... if the knots were all coincident between
>> the two densities (maybe you could constrain them to be so?) then you
>> just have a difference of cubics within each segment, and you can use
>> polyroot() to find the roots (and throw out any that are complex or
>> don't fall within the segment).
>>   If the knots are not coincident it's more of a pain but you should
>> still be able to do it by considering overlapping segments ...
>>
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