numbers (version 0.9-2)

cf2num: Generalized Continous Fractions

Description

Evaluate a generalized continuous fraction as an alternating sum.

Usage

cf2num(b0, b, a = 1, scaled = FALSE, tol = 1e-12)
num2cf(x, nterms=20)

Value

Returns a numerical value, an approximation of the continued fraction.

Arguments

b0

absolute term, integer part of the continuous fraction.

b

numeric vector of length greater than 2.

a

numeric vector of length 1 or the same length as a.

scaled

logical; shall the convergents be scaled.

tol

relative tolerance.

x

real number.

nterms

number of terms.

Details

cf2num calculates the numerical value of (simple or generalized) continued fractions of the form $$ b_0 + \frac{a1}{b1+} \frac{a2}{b2+} \frac{a3}{b3+...} $$ by converting its convergents into an alternating sum. The argument \(a\) is by default set to \(a = (1, 1, ...)\), that is the continued fraction is treated in its simple form.

The convergents may grow and become too big, especially for large and growing coefficients a. Then NA will probably be returned. In this case use scaled = TRUE and the convergents will be scaled in every iteration, thus avoiding FP overflow .

num2cf applies the direct calculation of the continued fraction. Attention: The input is not checked -- this allows for applying 'mpfr' numbers for longer, correct continued fractions, see the examples.

The relation between the number of terms nterms and the precision of the 'mpfr' numbers should be about nterm = 0.30..35 precBits.

References

Press, Teukolsky, Vetterling, Flannery. NUMERICAL RECIPES: The Art of Scientific Computing. 3rd Edition, Cambridge University Press 2007.

See Also

contfrac

Examples

#-- Continued fraction of sqrt(2) is [1; 2, 2, 2, ...] -----------------
b0 <- 1; b <- rep(2, 20)                # sqrt(2)
cf2num(b0, b)                           # 1.414213562, error = eps()

#-- Approximate an analytic function -----------------------------------
# tan(x) = 0 + x / (1 + (-x^2) / 3 + (-x^2) / (5 + ...))
x  <- 0.5                           # tan(0.5) = 0.546302489844
n  <- 10                            # CF of length 20
b0 <- 0                             # b0 = 0
b  <- seq(1, by=2, length=n)        # b = c(1, 3, 5, ...)
a  <- c(x, rep(-x^2, times=n-1))    # a = c(x, -x^2, -x^2, ...)

cf2num(b0, b, a)                # 0.546302489844, error: eps()

if (FALSE) {
#-- Continued fraction of 1/(exp(0.5)-1) -------------------------------
library(Rmpfr)
e0 = 1/(exp(1/mpfr(2, precBits=128)) - 1)  #=> 1.54149408253679828413...
x0 <- num2cf(e0, nterms=20)
## [1]  1  1  1  5  1  1  9  1  1 13  1  1 17  1  1 21  1  1 25  1  1

# Determine e0 from its continued fraction
b0 <- x0[1]; b <- x0[2:19]
cf2num(b0, b)                       # 1.541494082536798, error = eps()

#-- pi as arctan(1.0) --------------------------------------------------
n = 24
b0 = 0
b = seq(1, by=2, length=n)          # b = c(1, 3, 5, 7, ...)
a = c(1, seq(1,by=1, length=n-1)^2) # a = c(1, 1, 4, 9, ...)
4 * cf2num(b0, b, a) - pi           # error: 0

#-- Leibniz-Wallis continued fraction for pi ---------------------------
# 4/pi = 1 + 1^2/(2 + 3^2 / (2 + 5^2 / 2 + ...))
pi4 = 4 / pi                        # 1.27323954474

n  = 20
b0 = 1
b  = rep(2, times=n)
a  = seq(1, by = 2, length=n)^2
cf2num(b0, b, a)                    # NA, i.e. convergents overflow
cf2num(b0, b, a, scaled = TRUE)
# 1.273272 with estimated precision 0.004032851 
# [1] 1.273272                      # actual error: 3.3e-05
  }