TaylorSpline: Taylor representation of a B-Spline
In fda: Functional Data Analysis

Description Usage Arguments Details Value Author(s) See Also Examples

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

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

`object`	a spline object possibly of some other class
`...`	optional arguments

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.

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`.

Spencer Graves

fd create.bspline.basis

##
## 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)


##
## 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) )
#}