TaylorSpline: Taylor representation of a B-Spline

Description

Convert B-Spline coefficients into a local Taylor series representation expanded about the midpoint between each pair of distinct knots.

Usage

TaylorSpline(object, ...)
# S3 method for fd
TaylorSpline(object, ...)
# S3 method for fdPar
TaylorSpline(object, ...)
# S3 method for fdSmooth
TaylorSpline(object, ...)

Arguments

object

a spline object possibly of some other class

...

optional arguments

Value

a list with the following components:

knots

a numeric vector of knots(object, interior=FALSE)

midpoints

midpoints of intervals defined by unique(knots)

coef

A matrix of dim = c(nKnots-1, norder) containing the coeffients of a polynomial in (x-midpoints[i]) for interval i, where nKnots = length(unique(knots)).

deriv

A matrix of dim = c(nKnots-1, norder) containing the derivatives of the spline evaluated at midpoints.

% \item{Min, Max}{ % matrices with one row for each segment giving \code{x} at which % local extreme are found, with the corresponding function values. % If a local minimum or maximum is not achieved in that interval, % then that row of Min or Max is NA. % % If the minimum is achieved at multiple locations in an interval, % the smallest such \code{x} value is returned. Similarly, if the % maximum is achieved at multiple locations, the largest such % \code{x} is returned. % }

Details

1. Is object a spline object with a B-spline basis? If no, throw an error.

2. Find knots and midpoints.

3. Obtain coef(object).

4. Determine the number of dimensions of coef(object) and create empty coef and deriv arrays to match. Then fill the arrays.

Examples

Run this code

# NOT RUN {
##
## The simplest b-spline basis:  order 1, degree 0, zero interior knots:
##       a single step function
##
bspl1.1 <- create.bspline.basis(norder=1, breaks=0:1)
# ... jump to pi to check the code
fd.bspl1.1pi <- fd(pi, basisobj=bspl1.1)
bspl1.1pi <- TaylorSpline(fd.bspl1.1pi)
# }
# NOT RUN {
##
## Cubic spline:  4  basis functions
##
bspl4 <- create.bspline.basis(nbasis=4)
plot(bspl4)
parab4.5 <- fd(c(3, -1, -1, 3)/3, bspl4)
# = 4*(x-.5)
TaylorSpline(parab4.5)

##
## A more realistic example
##
data(titanium)
#  Cubic spline with 5 interior knots (6 segments)
# do not run this example on CRAN to save test time
#if(!CRAN()){
#titan10 <- with(titanium, curfit.free.knot(x, y))
#(titan10T <- TaylorSpline(titan10) )
#}

# }