[R] Taking the derivative of a quadratic B-spline
jp.mcdermott at gmail.com
Wed Jul 20 03:50:15 CEST 2005
Would the unique quadratic defined by the three points be the same
curve as the curve predicted by a quadratic B-spline (fit to all of
the data) through those same three points?
On 7/19/05, Duncan Murdoch <murdoch at stats.uwo.ca> wrote:
> On 7/19/2005 3:34 PM, James McDermott wrote:
> > I wish it were that simple (perhaps it is and I am just not seeing
> > it). The output from cobs( ) includes the B-spline coefficients and
> > the knots. These coefficients are not the same as the a, b, and c
> > coefficients in a quadratic polynomial. Rather, they are the
> > coefficients of the quadratic B-spline representation of the fitted
> > curve. I need to evaluate a linear combination of basis functions and
> > it is not clear to me how to accomplish this easily. I was hoping to
> > find an alternative way of getting the derivatives.
> I don't know COBS, but doesn't predict just evaluate the B-spline? The
> point of what I posted is that the particular basis doesn't matter if
> you can evaluate the quadratic at 3 points.
> Duncan Murdoch
> > Jim McDermott
> > On 7/19/05, Duncan Murdoch <murdoch at stats.uwo.ca> wrote:
> >> On 7/19/2005 2:53 PM, James McDermott wrote:
> >> > Hello,
> >> >
> >> > I have been trying to take the derivative of a quadratic B-spline
> >> > obtained by using the COBS library. What I would like to do is
> >> > similar to what one can do by using
> >> >
> >> > fit<-smooth.spline(cdf)
> >> > xx<-seq(-10,10,.1)
> >> > predict(fit, xx, deriv = 1)
> >> >
> >> > The goal is to fit the spline to data that is approximating a
> >> > cumulative distribution function (e.g. in my example, cdf is a
> >> > 2-column matrix with x values in column 1 and the estimate of the cdf
> >> > evaluated at x in column 2) and then take the first derivative over a
> >> > range of values to get density estimates.
> >> >
> >> > The reason I don't want to use smooth.spline is that there is no way
> >> > to impose constraints (e.g. >=0, <=1, and monotonicity) as there is
> >> > with COBS. However, since COBS doesn't have the 'deriv =' option, the
> >> > only way I can think of doing it with COBS is to evaluate the
> >> > derivatives numerically.
> >> Numerical estimates of the derivatives of a quadratic should be easy to
> >> obtain accurately. For example, if the quadratic ax^2 + bx + c is
> >> defined on [-1, 1], then the derivative 2ax + b, has 2a = f(1) - f(0) +
> >> f(-1), and b = (f(1) - f(-1))/2.
> >> You should be able to generalize this to the case where the spline is
> >> quadratic between knots k1 and k2 pretty easily.
> >> Duncan Murdoch
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