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?
Jim On 7/19/05, Duncan Murdoch <[EMAIL PROTECTED]> 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 <[EMAIL PROTECTED]> 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 > >> > > ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
