fda (version 2.4.7)

exponentiate.fd: Powers of a functional data ('fd') object

Description

Exponentiate a functional data object where feasible.

Usage

# S3 method for fd
^(e1, e2)
exponentiate.fd(e1, e2, tolint=.Machine$double.eps^0.75,
  basisobj=e1$basis,
  tolfd=sqrt(.Machine$double.eps)*
          sqrt(sum(e1$coefs^2)+.Machine$double.eps)^abs(e2),
  maxbasis=NULL, npoints=NULL)

Arguments

e1

object of class 'fd'.

e2

a numeric vector of length 1.

basisobj

reference basis

tolint

if abs(e2-round(e2))<tolint, we assume e2 is an integer. This simplifies the algorithm.

tolfd

the maximum error allowed in the difference between the direct computation eval.fd(e1)^e2 and the computed representation.

maxbasis

The maximum number of basis functions in growing referencebasis to achieve a fit within tolfd. Default = 2*nbasis12+1 where nbasis12 = nbasis of e1^floor(e2).

npoints

The number of points at which to compute eval.fd(e1)^e2 and the computed representation to evaluate the adequacy of the representation. Default = 2*maxbasis-1. For a max Fourier basis, this samples the highest frequency at all its extrema and zeros.

Value

A function data object approximating the desired power.

Details

If e1 has a B-spline basis, this uses the B-spline algorithm.

Otherwise it throws an error unless it finds one of the following special cases:

• e2 = 0 Return an fd object with a constant basis that is everywhere 1

• e2 is a positive integer to within tolint Multiply e1 by itself e2 times

• e2 is positive and e1 has a Fourier basis e120 <- e1^floor(e2)

  outBasis <- e120$basis

  rng <- outBasis$rangeval

  Time <- seq(rng[1], rng[2], npoints)

  e1.2 <- predict(e1, Time)^e2

  fd1.2 <- Data2fd(Time, e1.2, outBasis)

  d1.2 <- (e1.2 - predict(fd1.2, Time))

  if(all(abs(d1.2)<tolfd))return(fd1.2)

  Else if(outBasis$nbasis<maxbasis) increase the size of outBasis and try again.

  Else write a warining with the max(abs(d1.2)) and return fd1.2.

See Also

arithmetic.fd basisfd, basisfd.product

Examples

# NOT RUN {
##
## sin^2
##

basis3 <- create.fourier.basis(nbasis=3)
plot(basis3)
# max = sqrt(2), so
# integral of the square of each basis function (from 0 to 1) is 1
integrate(function(x)sin(2*pi*x)^2, 0, 1) # = 0.5

# sin(theta)
fdsin <- fd(c(0,sqrt(0.5),0), basis3)
plot(fdsin)

fdsin2 <- fdsin^2

# check
fdsinsin <- fdsin*fdsin
# sin^2(pi*time) = 0.5*(1-cos(2*pi*theta) basic trig identity
plot(fdsinsin) # good

# }
# NOT RUN {
all.equal(fdsin2, fdsinsin)
# }
# NOT RUN {
##
## sqrt(sin2)
##
plot(fdsin2)
fdsin. <- sqrt(fdsin2)
plot(fdsin, main='fdsin and sqrt(fdsin^2)')
lines(fdsin., col='red')
# fdsin is positive and negative
# fdsin. = sqrt(fdsin^2)
# has trouble, because it wants to be smooth
# but theoretically has a discontinuous first derivative at 0.5

fdsin.5.2 <- fdsin.^2
resin <- fdsin2-fdsin.5.2
plot(resin)
# }
# NOT RUN {
max(abs(resin$coefs))<0.01
# }
# NOT RUN {
##
## x^2, x = straight line f(x)=x
##
bspl1 <- create.bspline.basis(norder=2)
x <- fd(0:1, bspl1)
plot(x)

x2 <- x^2
plot(x2)

er.x <- x-sqrt(x2)
er.x$coefs
max(er.x$coefs)
# }
# NOT RUN {
max(abs(er.x$coefs))<10*.Machine$double.eps
# }
# NOT RUN {
# }