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
