Continued fraction

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Task
Continued fraction
You are encouraged to solve this task according to the task description, using any language you may know.
A number may be represented as a continued fraction (see Mathworld for more information) as follows:
a_0 + \cfrac{b_1}{a_1 + \cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + \ddots}}}

The task is to write a program which generates such a number and prints a real representation of it. The code should be tested by calculating and printing the square root of 2, Napier's Constant, and Pi, using the following coefficients:

For the square root of 2, use a0 = 1 then aN = 2. bN is always 1.

\sqrt{2} = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}

For Napier's Constant, use a0 = 2, then aN = N. b1 = 1 then bN = N − 1.

e = 2 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{2}{3 + \cfrac{3}{4 + \ddots}}}}

For Pi, use a0 = 3 then aN = 6. bN = (2N − 1)2.

\pi = 3 + \cfrac{1}{6 + \cfrac{9}{6 + \cfrac{25}{6 + \ddots}}}
See also:

Contents

[edit] Ada

(The source text for these examples can also be found on Bitbucket.)

Generic function for estimating continued fractions:

generic
type Scalar is digits <>;
 
with function A (N : in Natural) return Natural;
with function B (N : in Positive) return Natural;
function Continued_Fraction (Steps : in Natural) return Scalar;
function Continued_Fraction (Steps : in Natural) return Scalar is
function A (N : in Natural) return Scalar is (Scalar (Natural'(A (N))));
function B (N : in Positive) return Scalar is (Scalar (Natural'(B (N))));
 
Fraction : Scalar := 0.0;
begin
for N in reverse Natural range 1 .. Steps loop
Fraction := B (N) / (A (N) + Fraction);
end loop;
return A (0) + Fraction;
end Continued_Fraction;

Test program using the function above to estimate the square root of 2, Napiers constant and pi:

with Ada.Text_IO;
 
with Continued_Fraction;
 
procedure Test_Continued_Fractions is
type Scalar is digits 15;
 
package Square_Root_Of_2 is
function A (N : in Natural) return Natural is (if N = 0 then 1 else 2);
function B (N : in Positive) return Natural is (1);
 
function Estimate is new Continued_Fraction (Scalar, A, B);
end Square_Root_Of_2;
 
package Napiers_Constant is
function A (N : in Natural) return Natural is (if N = 0 then 2 else N);
function B (N : in Positive) return Natural is (if N = 1 then 1 else N-1);
 
function Estimate is new Continued_Fraction (Scalar, A, B);
end Napiers_Constant;
 
package Pi is
function A (N : in Natural) return Natural is (if N = 0 then 3 else 6);
function B (N : in Positive) return Natural is ((2 * N - 1) ** 2);
 
function Estimate is new Continued_Fraction (Scalar, A, B);
end Pi;
 
package Scalar_Text_IO is new Ada.Text_IO.Float_IO (Scalar);
use Ada.Text_IO, Scalar_Text_IO;
begin
Put (Square_Root_Of_2.Estimate (200), Exp => 0); New_Line;
Put (Napiers_Constant.Estimate (200), Exp => 0); New_Line;
Put (Pi.Estimate (10000), Exp => 0); New_Line;
end Test_Continued_Fractions;

[edit] Using only Ada 95 features

This example is exactly the same as the preceding one, but implemented using only Ada 95 features.

generic
type Scalar is digits <>;
 
with function A (N : in Natural) return Natural;
with function B (N : in Positive) return Natural;
function Continued_Fraction_Ada95 (Steps : in Natural) return Scalar;
function Continued_Fraction_Ada95 (Steps : in Natural) return Scalar is
function A (N : in Natural) return Scalar is
begin
return Scalar (Natural'(A (N)));
end A;
 
function B (N : in Positive) return Scalar is
begin
return Scalar (Natural'(B (N)));
end B;
 
Fraction : Scalar := 0.0;
begin
for N in reverse Natural range 1 .. Steps loop
Fraction := B (N) / (A (N) + Fraction);
end loop;
return A (0) + Fraction;
end Continued_Fraction_Ada95;
with Ada.Text_IO;
 
with Continued_Fraction_Ada95;
 
procedure Test_Continued_Fractions_Ada95 is
type Scalar is digits 15;
 
package Square_Root_Of_2 is
function A (N : in Natural) return Natural;
function B (N : in Positive) return Natural;
 
function Estimate is new Continued_Fraction_Ada95 (Scalar, A, B);
end Square_Root_Of_2;
 
package body Square_Root_Of_2 is
function A (N : in Natural) return Natural is
begin
if N = 0 then
return 1;
else
return 2;
end if;
end A;
 
function B (N : in Positive) return Natural is
begin
return 1;
end B;
end Square_Root_Of_2;
 
package Napiers_Constant is
function A (N : in Natural) return Natural;
function B (N : in Positive) return Natural;
 
function Estimate is new Continued_Fraction_Ada95 (Scalar, A, B);
end Napiers_Constant;
 
package body Napiers_Constant is
function A (N : in Natural) return Natural is
begin
if N = 0 then
return 2;
else
return N;
end if;
end A;
 
function B (N : in Positive) return Natural is
begin
if N = 1 then
return 1;
else
return N - 1;
end if;
end B;
end Napiers_Constant;
 
package Pi is
function A (N : in Natural) return Natural;
function B (N : in Positive) return Natural;
 
function Estimate is new Continued_Fraction_Ada95 (Scalar, A, B);
end Pi;
 
package body Pi is
function A (N : in Natural) return Natural is
begin
if N = 0 then
return 3;
else
return 6;
end if;
end A;
 
function B (N : in Positive) return Natural is
begin
return (2 * N - 1) ** 2;
end B;
end Pi;
 
package Scalar_Text_IO is new Ada.Text_IO.Float_IO (Scalar);
use Ada.Text_IO, Scalar_Text_IO;
begin
Put (Square_Root_Of_2.Estimate (200), Exp => 0); New_Line;
Put (Napiers_Constant.Estimate (200), Exp => 0); New_Line;
Put (Pi.Estimate (10000), Exp => 0); New_Line;
end Test_Continued_Fractions_Ada95;
Output:
 1.41421356237310
 2.71828182845905
 3.14159265358954

[edit] ATS

A fairly direct translation of the C version without using advanced features of the type system:

#include
"share/atspre_staload.hats"
//
(* ****** ****** *)
//
(*
** a coefficient function creates double values from in paramters
*)
typedef coeff_f = int -> double
//
(*
** a continued fraction is described by a record of two coefficent
** functions a and b
*)
typedef frac = @{a= coeff_f, b= coeff_f}
//
(* ****** ****** *)
 
fun calc
(
f: frac, n: int
) : double = let
//
(*
** recursive definition of the approximation
*)
fun loop
(
n: int, r: double
) : double =
(
if n = 0
then f.a(0) + r
else loop (n - 1, f.b(n) / (f.a(n) + r))
// end of [if]
)
//
in
loop (n, 0.0)
end // end of [calc]
 
(* ****** ****** *)
 
val sqrt2 = @{
a= lam (n: int): double => if n = 0 then 1.0 else 2.0
,
b= lam (n: int): double => 1.0
} (* end of [val] *)
 
val napier = @{
a= lam (n: int): double => if n = 0 then 2.0 else 1.0 * n
,
b= lam (n: int): double => if n = 1 then 1.0 else n - 1.0
} (* end of [val] *)
 
val pi = @{
a= lam (n: int): double => if n = 0 then 3.0 else 6.0
,
b= lam (n: int): double => let val x = 2.0 * n - 1 in x * x end
}
 
(* ****** ****** *)
 
implement
main0 () =
(
println! ("sqrt2 = ", calc(sqrt2, 100));
println! ("napier = ", calc(napier, 100));
println! (" pi = ", calc( pi , 100));
) (* end of [main0] *)

[edit] Axiom

Axiom provides a ContinuedFraction domain:

get(obj) == convergents(obj).1000 -- utility to extract the 1000th value
get continuedFraction(1, repeating [1], repeating [2]) :: Float
get continuedFraction(2, cons(1,[i for i in 1..]), [i for i in 1..]) :: Float
get continuedFraction(3, [i^2 for i in 1.. by 2], repeating [6]) :: Float
Output:
   (1)  1.4142135623 730950488
Type: Float
 
(2) 2.7182818284 590452354
Type: Float
 
(3) 3.1415926538 39792926
Type: Float

The value for π has an accuracy to only 9 decimal places after 1000 iterations, with an accuracy to 12 decimal places after 10000 iterations.

We could re-implement this, with the same output:

cf(initial, a, b, n) ==
n=1 => initial
temp := 0
for i in (n-1)..1 by -1 repeat
temp := a.i/(b.i+temp)
initial+temp
cf(1, repeating [1], repeating [2], 1000) :: Float
cf(2, cons(1,[i for i in 1..]), [i for i in 1..], 1000) :: Float
cf(3, [i^2 for i in 1.. by 2], repeating [6], 1000) :: Float

[edit] BBC BASIC

      *FLOAT64
@% = &1001010
 
PRINT "SQR(2) = " ; FNcontfrac(1, 1, "2", "1")
PRINT " e = " ; FNcontfrac(2, 1, "N", "N")
PRINT " PI = " ; FNcontfrac(3, 1, "6", "(2*N+1)^2")
END
 
REM a$ and b$ are functions of N
DEF FNcontfrac(a0, b1, a$, b$)
LOCAL N, expr$
REPEAT
N += 1
expr$ += STR$(EVAL(a$)) + "+" + STR$(EVAL(b$)) + "/("
UNTIL LEN(expr$) > (65500 - N)
= a0 + b1 / EVAL (expr$ + "1" + STRING$(N, ")"))
Output:
SQR(2) = 1.414213562373095
     e = 2.718281828459046
    PI = 3.141592653588017

[edit] C

Works with: ANSI C
/* calculate approximations for continued fractions */
#include <stdio.h>
 
/* kind of function that returns a series of coefficients */
typedef double (*coeff_func)(unsigned n);
 
/* calculates the specified number of expansions of the continued fraction
* described by the coefficient series f_a and f_b */

double calc(coeff_func f_a, coeff_func f_b, unsigned expansions)
{
double a, b, r;
a = b = r = 0.0;
 
unsigned i;
for (i = expansions; i > 0; i--) {
a = f_a(i);
b = f_b(i);
r = b / (a + r);
}
a = f_a(0);
 
return a + r;
}
 
/* series for sqrt(2) */
double sqrt2_a(unsigned n)
{
return n ? 2.0 : 1.0;
}
 
double sqrt2_b(unsigned n)
{
return 1.0;
}
 
/* series for the napier constant */
double napier_a(unsigned n)
{
return n ? n : 2.0;
}
 
double napier_b(unsigned n)
{
return n > 1.0 ? n - 1.0 : 1.0;
}
 
/* series for pi */
double pi_a(unsigned n)
{
return n ? 6.0 : 3.0;
}
 
double pi_b(unsigned n)
{
double c = 2.0 * n - 1.0;
 
return c * c;
}
 
int main(void)
{
double sqrt2, napier, pi;
 
sqrt2 = calc(sqrt2_a, sqrt2_b, 1000);
napier = calc(napier_a, napier_b, 1000);
pi = calc(pi_a, pi_b, 1000);
 
printf("%12.10g\n%12.10g\n%12.10g\n", sqrt2, napier, pi);
 
return 0;
}
Output:
 1.414213562
 2.718281828
 3.141592653

[edit] C++

#include <iomanip>
#include <iostream>
#include <tuple>
 
typedef std::tuple<double,double> coeff_t; // coefficients type
typedef coeff_t (*func_t)(int); // callback function type
 
double calc(func_t func, int n)
{
double a, b, temp = 0;
for (; n > 0; --n) {
std::tie(a, b) = func(n);
temp = b / (a + temp);
}
std::tie(a, b) = func(0);
return a + temp;
}
 
coeff_t sqrt2(int n)
{
return coeff_t(n > 0 ? 2 : 1, 1);
}
 
coeff_t napier(int n)
{
return coeff_t(n > 0 ? n : 2, n > 1 ? n - 1 : 1);
}
 
coeff_t pi(int n)
{
return coeff_t(n > 0 ? 6 : 3, (2 * n - 1) * (2 * n - 1));
}
 
int main()
{
std::streamsize old_prec = std::cout.precision(15); // set output digits
std::cout
<< calc(sqrt2, 20) << '\n'
<< calc(napier, 15) << '\n'
<< calc(pi, 10000) << '\n'
<< std::setprecision(old_prec); // reset precision
}
Output:
1.41421356237309
2.71828182845905
3.14159265358954

[edit] CoffeeScript

# Compute a continuous fraction of the form
# a0 + b1 / (a1 + b2 / (a2 + b3 / ...
continuous_fraction = (f) ->
a = f.a
b = f.b
c = 1
for n in [100000..1]
c = b(n) / (a(n) + c)
a(0) + c
 
# A little helper.
p = (a, b) ->
console.log a
console.log b
console.log "---"
 
do ->
fsqrt2 =
a: (n) -> if n is 0 then 1 else 2
b: (n) -> 1
p Math.sqrt(2), continuous_fraction(fsqrt2)
 
fnapier =
a: (n) -> if n is 0 then 2 else n
b: (n) -> if n is 1 then 1 else n - 1
p Math.E, continuous_fraction(fnapier)
 
fpi =
a: (n) ->
return 3 if n is 0
6
b: (n) ->
x = 2*n - 1
x * x
p Math.PI, continuous_fraction(fpi)
Output:
> coffee continued_fraction.coffee 
1.4142135623730951
1.4142135623730951
---
2.718281828459045
2.7182818284590455
---
3.141592653589793
3.141592653589793
---

[edit] Common Lisp

Translation of: C++
(defun estimate-continued-fraction (generator n)
(let ((temp 0))
(loop for n1 from n downto 1
do (multiple-value-bind (a b)
(funcall generator n1)
(setf temp (/ b (+ a temp)))))
(+ (funcall generator 0) temp)))
 
(format t "sqrt(2) = ~a~%" (coerce (estimate-continued-fraction
(lambda (n)
(values (if (> n 0) 2 1) 1)) 20)
'double-float))
(format t "napier's = ~a~%" (coerce (estimate-continued-fraction
(lambda (n)
(values (if (> n 0) n 2)
(if (> n 1) (1- n) 1))) 15)
'double-float))
 
(format t "pi = ~a~%" (coerce (estimate-continued-fraction
(lambda (n)
(values (if (> n 0) 6 3)
(* (1- (* 2 n))
(1- (* 2 n))))) 10000)
'double-float))
Output:
sqrt(2) = 1.4142135623730947d0
napier's = 2.7182818284590464d0
pi = 3.141592653589543d0

[edit] Chapel

Functions don't take other functions as arguments, so I wrapped them in a dummy record each.

proc calc(f, n) {
var r = 0.0;
 
for k in 1..n by -1 {
var v = f.pair(k);
r = v(2) / (v(1) + r);
}
 
return f.pair(0)(1) + r;
}
 
record Sqrt2 {
proc pair(n) {
return (if n == 0 then 1 else 2,
1);
}
}
 
record Napier {
proc pair(n) {
return (if n == 0 then 2 else n,
if n == 1 then 1 else n - 1);
}
}
record Pi {
proc pair(n) {
return (if n == 0 then 3 else 6,
(2*n - 1)**2);
}
}
 
config const n = 200;
writeln(calc(new Sqrt2(), n));
writeln(calc(new Napier(), n));
writeln(calc(new Pi(), n));

[edit] D

import std.stdio, std.functional, std.traits;
 
FP calc(FP, F)(in F fun, in int n) pure nothrow if (isCallable!F) {
FP temp = 0;
 
foreach_reverse (immutable ni; 1 .. n + 1) {
immutable p = fun(ni);
temp = p[1] / (FP(p[0]) + temp);
}
return fun(0)[0] + temp;
}
 
int[2] fSqrt2(in int n) pure nothrow {
return [n > 0 ? 2 : 1, 1];
}
 
int[2] fNapier(in int n) pure nothrow {
return [n > 0 ? n : 2, n > 1 ? (n - 1) : 1];
}
 
int[2] fPi(in int n) pure nothrow {
return [n > 0 ? 6 : 3, (2 * n - 1) ^^ 2];
}
 
alias print = curry!(writefln, "%.19f");
 
void main() {
calc!real(&fSqrt2, 200).print;
calc!real(&fNapier, 200).print;
calc!real(&fPi, 200).print;
}
Output:
1.4142135623730950487
2.7182818284590452354
3.1415926228048469486

[edit] Erlang

 
-module(continued).
-compile([export_all]).
 
pi_a (0) -> 3;
pi_a (_N) -> 6.
 
pi_b (N) ->
(2*N-1)*(2*N-1).
 
sqrt2_a (0) ->
1;
sqrt2_a (_N) ->
2.
 
sqrt2_b (_N) ->
1.
 
nappier_a (0) ->
2;
nappier_a (N) ->
N.
 
nappier_b (1) ->
1;
nappier_b (N) ->
N-1.
 
continued_fraction(FA,_FB,0) -> FA(0);
continued_fraction(FA,FB,N) ->
continued_fraction(FA,FB,N-1,FB(N)/FA(N)).
 
continued_fraction(FA,_FB,0,Acc) -> FA(0) + Acc;
continued_fraction(FA,FB,N,Acc) ->
continued_fraction(FA,FB,N-1,FB(N)/ (FA(N) + Acc)).
 
test_pi (N) ->
continued_fraction(fun pi_a/1,fun pi_b/1,N).
 
test_sqrt2 (N) ->
continued_fraction(fun sqrt2_a/1,fun sqrt2_b/1,N).
 
test_nappier (N) ->
continued_fraction(fun nappier_a/1,fun nappier_b/1,N).
 
Output:
 
29> continued:test_pi(1000).
3.141592653340542
30> continued:test_sqrt2(1000).
1.4142135623730951
31> continued:test_nappier(1000).
2.7182818284590455
 

[edit] Factor

cfrac-estimate uses rational arithmetic and never truncates the intermediate result. When terms is large, cfrac-estimate runs slow because numerator and denominator grow big.

USING: arrays combinators io kernel locals math math.functions
math.ranges prettyprint sequences ;
IN: rosetta.cfrac
 
! Every continued fraction must implement these two words.
GENERIC: cfrac-a ( n cfrac -- a )
GENERIC: cfrac-b ( n cfrac -- b )
 
! square root of 2
SINGLETON: sqrt2
M: sqrt2 cfrac-a
 ! If n is 1, then a_n is 1, else a_n is 2.
drop { { 1 [ 1 ] } [ drop 2 ] } case ;
M: sqrt2 cfrac-b
 ! Always b_n is 1.
2drop 1 ;
 
! Napier's constant
SINGLETON: napier
M: napier cfrac-a
 ! If n is 1, then a_n is 2, else a_n is n - 1.
drop { { 1 [ 2 ] } [ 1 - ] } case ;
M: napier cfrac-b
 ! If n is 1, then b_n is 1, else b_n is n - 1.
drop { { 1 [ 1 ] } [ 1 - ] } case ;
 
SINGLETON: pi
M: pi cfrac-a
 ! If n is 1, then a_n is 3, else a_n is 6.
drop { { 1 [ 3 ] } [ drop 6 ] } case ;
M: pi cfrac-b
 ! Always b_n is (n * 2 - 1)^2.
drop 2 * 1 - 2 ^ ;
 
:: cfrac-estimate ( cfrac terms -- number )
terms cfrac cfrac-a  ! top = last a_n
terms 1 - 1 [a,b] [ :> n
n cfrac cfrac-b swap /  ! top = b_n / top
n cfrac cfrac-a +  ! top = top + a_n
] each ;
 
:: decimalize ( rational prec -- string )
rational 1 /mod  ! split whole, fractional parts
prec 10^ *  ! multiply fraction by 10 ^ prec
[ >integer unparse ] bi@  ! convert digits to strings
 :> fraction
"."  ! push decimal point
prec fraction length -
dup 0 < [ drop 0 ] when
"0" <repetition> concat  ! push padding zeros
fraction 4array concat ;
 
<PRIVATE
: main ( -- )
" Square root of 2: " write
sqrt2 50 cfrac-estimate 30 decimalize print
"Napier's constant: " write
napier 50 cfrac-estimate 30 decimalize print
" Pi: " write
pi 950 cfrac-estimate 10 decimalize print ;
PRIVATE>
 
MAIN: main
Output:
 Square root of 2: 1.414213562373095048801688724209
Napier's constant: 2.718281828459045235360287471352
               Pi: 3.1415926538

[edit] Forth

Translation of: D
: fsqrt2 1 s>f 0> if 2 s>f else fdup then ;
: fnapier dup dup 1 > if 1- else drop 1 then s>f dup 1 < if drop 2 then s>f ;
: fpi dup 2* 1- dup * s>f 0> if 6 else 3 then s>f ;
( n -- f1 f2)
: cont.fraction ( xt n -- f)
1 swap 1+ 0 s>f \ calculate for 1 .. n
do i over execute frot f+ f/ -1 +loop
0 swap execute fnip f+ \ calcucate for 0
;
Output:
' fsqrt2  200 cont.fraction f. cr 1.4142135623731
 ok
' fnapier 200 cont.fraction f. cr 2.71828182845905
 ok
' fpi     200 cont.fraction f. cr 3.14159268391981
 ok

[edit] Fortran

module continued_fractions
implicit none
 
integer, parameter :: long = selected_real_kind(7,99)
 
type continued_fraction
integer :: a0, b1
procedure(series), pointer, nopass :: a, b
end type
 
interface
pure function series (n)
integer, intent(in) :: n
integer :: series
end function
end interface
 
contains
 
pure function define_cf (a0,a,b1,b) result(x)
integer, intent(in) :: a0
procedure(series) :: a
integer, intent(in), optional :: b1
procedure(series), optional :: b
type(continued_fraction) :: x
x%a0 = a0
x%a => a
if ( present(b1) ) then
x%b1 = b1
else
x%b1 = 1
end if
if ( present(b) ) then
x%b => b
else
x%b => const_1
end if
end function define_cf
 
pure integer function const_1(n)
integer,intent(in) :: n
const_1 = 1
end function
 
pure real(kind=long) function output(x,iterations)
type(continued_fraction), intent(in) :: x
integer, intent(in) :: iterations
integer :: i
output = x%a(iterations)
do i = iterations-1,1,-1
output = x%a(i) + (x%b(i+1) / output)
end do
output = x%a0 + (x%b1 / output)
end function output
 
end module continued_fractions
 
 
program examples
use continued_fractions
 
type(continued_fraction) :: sqr2,napier,pi
 
sqr2 = define_cf(1,a_sqr2)
napier = define_cf(2,a_napier,1,b_napier)
pi = define_cf(3,a=a_pi,b=b_pi)
 
write (*,*) output(sqr2,10000)
write (*,*) output(napier,10000)
write (*,*) output(pi,10000)
 
contains
 
pure integer function a_sqr2(n)
integer,intent(in) :: n
a_sqr2 = 2
end function
 
pure integer function a_napier(n)
integer,intent(in) :: n
a_napier = n
end function
 
pure integer function b_napier(n)
integer,intent(in) :: n
b_napier = n-1
end function
 
pure integer function a_pi(n)
integer,intent(in) :: n
a_pi = 6
end function
 
pure integer function b_pi(n)
integer,intent(in) :: n
b_pi = (2*n-1)*(2*n-1)
end function
 
end program examples
Output:
   1.4142135623730951
   2.7182818284590455
   3.1415926535895435

[edit] Go

package main
 
import "fmt"
 
type cfTerm struct {
a, b int
}
 
// follows subscript convention of mathworld and WP where there is no b(0).
// cf[0].b is unused in this representation.
type cf []cfTerm
 
func cfSqrt2(nTerms int) cf {
f := make(cf, nTerms)
for n := range f {
f[n] = cfTerm{2, 1}
}
f[0].a = 1
return f
}
 
func cfNap(nTerms int) cf {
f := make(cf, nTerms)
for n := range f {
f[n] = cfTerm{n, n - 1}
}
f[0].a = 2
f[1].b = 1
return f
}
 
func cfPi(nTerms int) cf {
f := make(cf, nTerms)
for n := range f {
g := 2*n - 1
f[n] = cfTerm{6, g * g}
}
f[0].a = 3
return f
}
 
func (f cf) real() (r float64) {
for n := len(f) - 1; n > 0; n-- {
r = float64(f[n].b) / (float64(f[n].a) + r)
}
return r + float64(f[0].a)
}
 
func main() {
fmt.Println("sqrt2:", cfSqrt2(20).real())
fmt.Println("nap: ", cfNap(20).real())
fmt.Println("pi: ", cfPi(20).real())
}
Output:
sqrt2: 1.4142135623730965
nap:   2.7182818284590455
pi:    3.141623806667839

[edit] Haskell

import Data.List (unfoldr)
import Data.Char (intToDigit)
 
-- continued fraction represented as a (possibly infinite) list of pairs
sqrt2, napier, myPi :: [(Integer, Integer)]
sqrt2 = zip (1 : [2,2..]) [1,1..]
napier = zip (2 : [1..]) (1 : [1..])
myPi = zip (3 : [6,6..]) (map (^2) [1,3..])
 
-- approximate a continued fraction after certain number of iterations
approxCF :: (Integral a, Fractional b) => Int -> [(a, a)] -> b
approxCF t =
foldr (\(a,b) z -> fromIntegral a + fromIntegral b / z) 1 . take t
 
-- infinite decimal representation of a real number
decString :: RealFrac a => a -> String
decString frac = show i ++ '.' : decString' f where
(i,f) = properFraction frac
decString'
= map intToDigit . unfoldr (Just . properFraction . (10*))
 
main :: IO ()
main = mapM_ (putStrLn . take 200 . decString .
(approxCF 950 :: [(Integer, Integer)] -> Rational))
[sqrt2, napier, myPi]
Output:
1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641572735013846230912297024924836055850737212644121497099935831413222665927505592755799950501152782060571
2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921817413596629043572900334295260595630738132328627943490763233829880753195251019
3.141592653297590947683406834261190738869139611505752231394089152890909495973464508817163306557131591579057202097715021166512662872910519439747609829479577279606075707015622200744006783543589980682386
import Data.Ratio
 
-- ignoring the task-given pi sequence: sucky convergence
-- pie = zip (3:repeat 6) (map (^2) [1,3..])
 
pie = zip (0:[1,3..]) (4:map (^2) [1..])
sqrt2 = zip (1:repeat 2) (repeat 1)
napier = zip (2:[1..]) (1:[1..])
 
-- truncate after n terms
cf2rat n = foldr (\(a,b) f -> (a%1) + ((b%1) / f)) (1%1) . take n
 
-- truncate after error is at most 1/p
cf2rat_p p s = f $ map (\i -> (cf2rat i s, cf2rat (1+i) s)) $ map (2^) [0..]
where f ((x,y):ys) = if abs (x-y) < (1/fromIntegral p) then x else f ys
 
-- returns a decimal string of n digits after the dot; all digits should
-- be correct (doesn't mean it's the best approximation! the decimal
-- string is simply truncated to given digits: pi=3.141 instead of 3.142)
cf2dec n = (ratstr n) . cf2rat_p (10^n) where
ratstr l a = (show t) ++ '.':fracstr l n d where
d = denominator a
(t, n) = quotRem (numerator a) d
fracstr 0 _ _ = []
fracstr l n d = (show t)++ fracstr (l-1) n1 d where (t,n1) = quotRem (10 * n) d
 
main = do
putStrLn $ cf2dec 200 sqrt2
putStrLn $ cf2dec 200 napier
putStrLn $ cf2dec 200 pie

[edit] J

   cfrac=: +`% / NB. Evaluate a list as a continued fraction
 
sqrt2=: cfrac 1 1,200$2 1x
pi=:cfrac 3, , ,&6"0 *:<:+:>:i.100x
e=: cfrac 2 1, , ,~"0 >:i.100x
 
NB. translate from fraction to decimal string
NB. translated from factor
dec =: (-@:[ (}.,'.',{.) ":@:<.@:(* 10x&^)~)"0
 
100 10 100 dec sqrt2, pi, e
1.4142135623730950488016887242096980785696718753769480731766797379907324784621205551109457595775322165
3.1415924109
2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274

[edit] jq

Works with: jq version 1.4

We take one of the points of interest here to be the task of representing the infinite series a0, a1, .... and b0, b1, .... compactly, preferably functionally. For the type of series typically encountered in continued fractions, this is most readily accomplished in jq 1.4 using a filter (a function), here called "next", which, given the triple [i, [a[i], b[i]], will produce the next triple [i+1, a[i+1], b[i+1]].

Another point of interest is avoiding having to specify the number of iterations. The approach adopted here allows one to specify the desired accuracy; in some cases, it is feasible to specify that the computation should continue until the accuracy permitted by the underlying floating point representation is achieved. This is done by specifying delta as 0, as shown in the examples.

We therefore proceed in two steps: continued_fraction( first; next; count ) computes an approximation based on the first "count" terms; and then continued_fraction_delta(first; next; delta) computes the continued fraction until the difference in approximations is less than or equal to delta, which may be 0, as previously noted.

 
# "first" is the first triple,
# e.g. [1,a,b]; count specifies the number of terms to use.
def continued_fraction( first; next; count ):
# input: [i, a, b]]
def cf:
if .[0] == count then 0
else next as $ab
| .[1] + (.[2] / ($ab | cf))
end ;
first | cf;
 
# "first" and "next" are as above;
# if delta is 0 then continue until there is no detectable change.
def continued_fraction_delta(first; next; delta):
def abs: if . < 0 then -. else . end;
def cf:
# state: [n, prev]
.[0] as $n | .[1] as $prev
| continued_fraction(first; next; $n+1) as $this
| if $prev == null then [$n+1, $this] | cf
elif delta <= 0 and ($prev == $this) then $this
elif (($prev - $this)|abs) <= delta then $this
else [$n+1, $this] | cf
end;
[2,null] | cf;
 

Examples:

The convergence for pi is slow so we select delta = 1e-12 in that case.

"Value  :        Direct      : Continued Fraction",
"2|sqrt : \(2|sqrt) : \(continued_fraction_delta( [1,1,1]; [.[0]+1, 2, 1]; 0))",
"1|exp  : \(1|exp)  : \(2 + (1 / (continued_fraction_delta( [1,1,1]; [.[0]+1, .[1]+1, .[2]+1]; 0))))",
"pi  : \(1|atan * 4)  : \(continued_fraction_delta( [1,3,1]; .[0]+1 | [., 6, ((2*. - 1) | (.*.))]; 1e-12)) (1e-12)"
 
Output:
$ jq -M -n -r -f Continued_fraction.jq
Value  : Direct  : Continued Fraction
2|sqrt : 1.4142135623730951 : 1.4142135623730951
1|exp  : 2.718281828459045  : 2.7182818284590455
pi  : 3.141592653589793  : 3.1415926535892935 (1e-12)

[edit] Mathematica

sqrt2=Function[n,{1,Transpose@{Array[2&,n],Array[1&,n]}}];
napier=Function[n,{2,Transpose@{Range[n],Prepend[Range[n-1],1]}}];
pi=Function[n,{3,Transpose@{Array[6&,n],Array[(2#-1)^2&,n]}}];
approx=Function[l,
N[Divide@@First@Fold[{{#2.#[[;;,1]],#2.#[[;;,2]]},#[[1]]}&,{{l[[2,1,1]]l[[1]]+l[[2,1,2]],l[[2,1,1]]},{l[[1]],1}},l[[2,2;;]]],10]];
r2=approx/@{sqrt2@#,napier@#,pi@#}&@10000;r2//TableForm
Output:
1.414213562
2.718281828
3.141592654

[edit] Maxima

cfeval(x) := block([a, b, n, z], a: x[1], b: x[2], n: length(a), z: 0,
for i from n step -1 thru 2 do z: b[i]/(a[i] + z), a[1] + z)$
 
cf_sqrt2(n) := [cons(1, makelist(2, i, 2, n)), cons(0, makelist(1, i, 2, n))]$
 
cf_e(n) := [cons(2, makelist(i, i, 1, n - 1)), append([0, 1], makelist(i, i, 1, n - 2))]$
 
cf_pi(n) := [cons(3, makelist(6, i, 2, n)), cons(0, makelist((2*i - 1)^2, i, 1, n - 1))]$
 
cfeval(cf_sqrt2(20)), numer; /* 1.414213562373097 */
% - sqrt(2), numer; /* 1.3322676295501878*10^-15 */
 
cfeval(cf_e(20)), numer; /* 2.718281828459046 */
% - %e, numer; /* 4.4408920985006262*10^-16 */
 
cfeval(cf_pi(20)), numer; /* 3.141623806667839 */
% - %pi, numer; /* 3.115307804568701*10^-5 */
 
 
/* convergence is much slower for pi */
fpprec: 20$
x: cfeval(cf_pi(10000))$
bfloat(x - %pi); /* 2.4999999900104930006b-13 */

[edit] NetRexx

/* REXX ***************************************************************
* Derived from REXX ... Derived from PL/I with a little "massage"
* SQRT2= 1.41421356237309505 <- PL/I Result
* 1.41421356237309504880168872421 <- NetRexx Result 30 digits
* NAPIER= 2.71828182845904524
* 2.71828182845904523536028747135
* PI= 3.14159262280484695
* 3.14159262280484694855146925223
* 07.09.2012 Walter Pachl
* 08.09.2012 Walter Pachl simplified (with the help of a friend)
**********************************************************************/

options replace format comments java crossref savelog symbols
class CFB public
 
properties static
Numeric Digits 30
Sqrt2 =1
napier=2
pi =3
a =0
b =0
 
method main(args = String[]) public static
Say 'SQRT2='.left(7) calc(sqrt2, 200)
Say 'NAPIER='.left(7) calc(napier, 200)
Say 'PI='.left(7) calc(pi, 200)
Return
 
method get_Coeffs(form,n) public static
select
when form=Sqrt2 Then do
if n > 0 then a = 2; else a = 1
b = 1
end
when form=Napier Then do
if n > 0 then a = n; else a = 2
if n > 1 then b = n - 1; else b = 1
end
when form=pi Then do
if n > 0 then a = 6; else a = 3
b = (2*n - 1)**2
end
end
Return
 
method calc(form,n) public static
temp=0
loop ni = n to 1 by -1
Get_Coeffs(form,ni)
temp = b/(a + temp)
end
Get_Coeffs(form,0)
return (a + temp)

Who could help me make a,b,sqrt2,napier,pi global (public) variables? This would simplify the solution:-)

I got this help and simplified the program.

However, I am told that 'my' value of pi is incorrect. I will investigate!

Apparently the coefficients given in the task description are only good for an approximation. One should, therefore, not SHOW more that 15 digits. See http://de.wikipedia.org/wiki/Kreiszahl

See Rexx for a better computation

[edit] OCaml

let pi = 3, fun n -> ((2*n-1)*(2*n-1), 6)
and nap = 2, fun n -> (max 1 (n-1), n)
and root2 = 1, fun n -> (1, 2) in
 
let eval (i,f) k =
let rec frac n =
let a, b = f n in
float a /. (float b +.
if n >= k then 0.0 else frac (n+1)) in
float i +. frac 1 in
 
Printf.printf "sqrt(2)\t= %.15f\n" (eval root2 1000);
Printf.printf "e\t= %.15f\n" (eval nap 1000);
Printf.printf "pi\t= %.15f\n" (eval pi 1000);

Output (inaccurate due to too few terms):

sqrt(2)	= 1.414213562373095
e	= 2.718281828459046
pi	= 3.141592653340542

[edit] PARI/GP

Partial solution for simple continued fractions.

back(v)=my(t=contfracpnqn(v));t[1,1]/t[2,1]*1.
back(vector(100,i,2-(i==1)))

Output:

%1 = 1.4142135623730950488016887242096980786

[edit] Perl

We'll use closures to implement the infinite lists of coeffficients.

sub continued_fraction {
my ($a, $b, $n) = (@_[0,1], $_[2] // 100);
 
$a->() + ($n && $b->() / continued_fraction($a, $b, $n-1));
}
 
printf "√2 ≈ %.9f\n", continued_fraction do { my $n; sub { $n++ ? 2 : 1 } }, sub { 1 };
printf "e ≈ %.9f\n", continued_fraction do { my $n; sub { $n++ || 2 } }, do { my $n; sub { $n++ || 1 } };
printf "π ≈ %.9f\n", continued_fraction do { my $n; sub { $n++ ? 6 : 3 } }, do { my $n; sub { (2*$n++ + 1)**2 } }, 1_000;
printf "π/2 ≈ %.9f\n", continued_fraction do { my $n; sub { 1/($n++ || 1) } }, sub { 1 }, 1_000;
Output:
√2  ≈ 1.414213562
e   ≈ 2.718281828
π   ≈ 3.141592653
π/2 ≈ 1.570717797

[edit] Perl 6

sub continued-fraction(:@a, :@b, Int :$n = 100)
{
my $x = @a[$n - 1];
$x = @a[$_ - 1] + @b[$_] / $x for reverse 1 ..^ $n;
$x;
}
 
printf "√2 ≈ %.9f\n", continued-fraction(:a(1, 2 xx *), :b(*, 1 xx *));
printf "e ≈ %.9f\n", continued-fraction(:a(2, 1 .. *), :b(*, 1, 1 .. *));
printf "π ≈ %.9f\n", continued-fraction(:a(3, 6 xx *), :b(*, [\+] 1, (8, 16 ... *)), :n(1000));
Output:
√2 ≈ 1.414213562
e  ≈ 2.718281828
π  ≈ 3.141592654

A more original and a bit more abstract method would consist in viewing a continued fraction on rank n as a function of a variable x:

\mathrm{CF}_3(x) = a_0 + \cfrac{b_1}{a_1 + \cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + x}}}

Or, more consistently:

\mathrm{CF}_3(x) = a_0 + \cfrac{b_0}{a_1 + \cfrac{b_1}{a_2 + \cfrac{b_2}{a_3 + \cfrac{b_3}{x}}}}

Viewed as such, CFn(x) could be written recursively:

\mathrm{CF}_n(x) = \mathrm{CF}_{n-1}(a_n + \frac{b_n}{x})

Or in other words:

\mathrm{CF}_n= \mathrm{CF}_{n-1}\circ f_n = \mathrm{CF}_{n-2}\circ f_{n-1}\circ f_n=\ldots=f_0\circ f_1 \ldots \circ f_n

where f_n(x) = a_n + \frac{b_n}{x}

Perl6 allows us to define a custom composition operator. We can then use it with the triangular reduction metaoperator, and map each resulting function with an infinite value for x (any value would do actually, but infinite make it consistent with this particular task).

sub infix:<>(&f, &g) { -> $x { &f(&g($x)) } }
sub continued-fraction(@a, @b, $x = Inf) {
map { .($x) },
[\⚬] map { @a[$_] + @b[$_] / * }, 0 .. *
}
 
printf "√2 ≈ %.9f\n", continued-fraction((1, 2 xx *), (1 xx *))[10];
printf "e ≈ %.9f\n", continued-fraction((2, 1 .. *), (1, 1 .. *))[10];
printf "π ≈ %.9f\n", continued-fraction((3, 6 xx *), ((1, 3, 5 ... *) X** 2))[100];

The main advantage is that the definition of the function does not need to know for which rank n it is computed. This is arguably closer to the mathematical definition.

[edit] PL/I

/* Version for SQRT(2) */
test: proc options (main);
declare n fixed;
 
denom: procedure (n) recursive returns (float (18));
declare n fixed;
n = n + 1;
if n > 100 then return (2);
return (2 + 1/denom(n));
end denom;
 
put (1 + 1/denom(2));
 
end test;
Output:
 1.41421356237309505E+0000 

Version for NAPIER:

test: proc options (main);
declare n fixed;
 
denom: procedure (n) recursive returns (float (18));
declare n fixed;
n = n + 1;
if n > 100 then return (n);
return (n + n/denom(n));
end denom;
 
put (2 + 1/denom(0));
 
end test;
 2.71828182845904524E+0000 

Version for SQRT2, NAPIER, PI

/* Derived from continued fraction in Wiki Ada program */
 
continued_fractions: /* 6 Sept. 2012 */
procedure options (main);
declare (Sqrt2 initial (1), napier initial (2), pi initial (3)) fixed (1);
 
Get_Coeffs: procedure (form, n, coefA, coefB);
declare form fixed (1), n fixed, (coefA, coefB) float (18);
 
select (form);
when (Sqrt2) do;
if n > 0 then coefA = 2; else coefA = 1;
coefB = 1;
end;
when (Napier) do;
if n > 0 then coefA = n; else coefA = 2;
if n > 1 then coefB = n - 1; else coefB = 1;
end;
when (Pi) do;
if n > 0 then coefA = 6; else coefA = 3;
coefB = (2*n - 1)**2;
end;
end;
end Get_Coeffs;
 
Calc: procedure (form, n) returns (float (18));
declare form fixed (1), n fixed;
declare (A, B) float (18);
declare Temp float (18) initial (0);
declare ni fixed;
 
do ni = n to 1 by -1;
call Get_Coeffs (form, ni, A, B);
Temp = B/(A + Temp);
end;
call Get_Coeffs (form, 0, A, B);
return (A + Temp);
end Calc;
 
put edit ('SQRT2=', calc(sqrt2, 200)) (a(10), f(20,17));
put skip edit ('NAPIER=', calc(napier, 200)) (a(10), f(20,17));
put skip edit ('PI=', calc(pi, 99999)) (a(10), f(20,17));
 
end continued_fractions;
Output:
SQRT2=     1.41421356237309505
NAPIER=    2.71828182845904524
PI=        3.14159265358979349

[edit] Prolog

continued_fraction :-
% square root 2
continued_fraction(200, sqrt_2_ab, V1),
format('sqrt(2) = ~w~n', [V1]),
 
% napier
continued_fraction(200, napier_ab, V2),
format('e = ~w~n', [V2]),
 
% pi
continued_fraction(200, pi_ab, V3),
format('pi = ~w~n', [V3]).
 
 
% code for continued fractions
continued_fraction(N, Compute_ab, V) :-
continued_fraction(N, Compute_ab, 0, V).
 
continued_fraction(0, Compute_ab, Temp, V) :-
call(Compute_ab, 0, A, _),
V is A + Temp.
 
continued_fraction(N, Compute_ab, Tmp, V) :-
call(Compute_ab, N, A, B),
Tmp1 is B / (A + Tmp),
N1 is N - 1,
continued_fraction(N1, Compute_ab, Tmp1, V).
 
% specific codes for examples
% definitions for square root of 2
sqrt_2_ab(0, 1, 1).
sqrt_2_ab(_, 2, 1).
 
% definitions for napier
napier_ab(0, 2, _).
napier_ab(1, 1, 1).
napier_ab(N, N, V) :-
V is N - 1.
 
% definitions for pi
pi_ab(0, 3, _).
pi_ab(N, 6, V) :-
V is (2 * N - 1)*(2 * N - 1).
Output:
 ?- continued_fraction.
sqrt(2) = 1.4142135623730951
e       = 2.7182818284590455
pi      = 3.141592622804847
true .

[edit] Python

Works with: Python version 2.6+ and 3.x
from fractions import Fraction
import itertools
try: zip = itertools.izip
except: pass
 
# The Continued Fraction
def CF(a, b, t):
terms = list(itertools.islice(zip(a, b), t))
z = Fraction(1,1)
for a, b in reversed(terms):
z = a + b / z
return z
 
# Approximates a fraction to a string
def pRes(x, d):
q, x = divmod(x, 1)
res = str(q)
res += "."
for i in range(d):
x *= 10
q, x = divmod(x, 1)
res += str(q)
return res
 
# Test the Continued Fraction for sqrt2
def sqrt2_a():
yield 1
for x in itertools.repeat(2):
yield x
 
def sqrt2_b():
for x in itertools.repeat(1):
yield x
 
cf = CF(sqrt2_a(), sqrt2_b(), 950)
print(pRes(cf, 200))
#1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206057147
 
 
# Test the Continued Fraction for Napier's Constant
def Napier_a():
yield 2
for x in itertools.count(1):
yield x
 
def Napier_b():
yield 1
for x in itertools.count(1):
yield x
 
cf = CF(Napier_a(), Napier_b(), 950)
print(pRes(cf, 200))
#2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642742746639193200305992181741359662904357290033429526059563073813232862794349076323382988075319525101901
 
# Test the Continued Fraction for Pi
def Pi_a():
yield 3
for x in itertools.repeat(6):
yield x
 
def Pi_b():
for x in itertools.count(1,2):
yield x*x
 
cf = CF(Pi_a(), Pi_b(), 950)
print(pRes(cf, 10))
#3.1415926532

[edit] Fast iterative version

Translation of: D
from decimal import Decimal, getcontext
 
def calc(fun, n):
temp = Decimal("0.0")
 
for ni in xrange(n+1, 0, -1):
(a, b) = fun(ni)
temp = Decimal(b) / (a + temp)
 
return fun(0)[0] + temp
 
def fsqrt2(n):
return (2 if n > 0 else 1, 1)
 
def fnapier(n):
return (n if n > 0 else 2, (n - 1) if n > 1 else 1)
 
def fpi(n):
return (6 if n > 0 else 3, (2 * n - 1) ** 2)
 
getcontext().prec = 50
print calc(fsqrt2, 200)
print calc(fnapier, 200)
print calc(fpi, 200)
Output:
1.4142135623730950488016887242096980785696718753770
2.7182818284590452353602874713526624977572470937000
3.1415926839198062649342019294083175420335002640134

[edit] Racket

[edit] Using Doubles

This version uses standard double precision floating point numbers:

 
#lang racket
(define (calc cf n)
(match/values (cf 0)
[(a0 b0)
(+ a0
(for/fold ([t 0.0]) ([i (in-range (+ n 1) 0 -1)])
(match/values (cf i)
[(a b) (/ b (+ a t))])))]))
 
(define (cf-sqrt i) (values (if (> i 0) 2 1) 1))
(define (cf-napier i) (values (if (> i 0) i 2) (if (> i 1) (- i 1) 1)))
(define (cf-pi i) (values (if (> i 0) 6 3) (sqr (- (* 2 i) 1))))
 
(calc cf-sqrt 200)
(calc cf-napier 200)
(calc cf-pi 200)
 

Output:

 
1.4142135623730951
2.7182818284590455
3.1415926839198063
 

[edit] Version - Using Doubles

This versions uses big floats (arbitrary precision floating point):

 
#lang racket
(require math)
(bf-precision 2048) ; in bits
 
(define (calc cf n)
(match/values (cf 0)
[(a0 b0)
(bf+ (bf a0)
(for/fold ([t (bf 0)]) ([i (in-range (+ n 1) 0 -1)])
(match/values (cf i)
[(a b) (bf/ (bf b) (bf+ (bf a) t))])))]))
 
(define (cf-sqrt i) (values (if (> i 0) 2 1) 1))
(define (cf-napier i) (values (if (> i 0) i 2) (if (> i 1) (- i 1) 1)))
(define (cf-pi i) (values (if (> i 0) 6 3) (sqr (- (* 2 i) 1))))
 
(calc cf-sqrt 200)
(calc cf-napier 200)
(calc cf-pi 200)
 

Output:

 
(bf #e1.4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727350138462309122970249248360558507372126441214970999358960036439214262599769155193770031712304888324413327207659690547583107739957489062466508437105234564161085482146113860092820802430986649987683947729823677905101453725898480737256099166805538057375451207262441039818826744940289448489312217214883459060818483750848688583833366310472320771259749181255428309841375829513581694269249380272698662595131575038315461736928338289219865139248048189188905788104310928762952913687232022557677738108337499350045588767581063729)
(bf #e2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642742746639193200305992181741359662904357290033429526059563073813232862794349076323382988075319525101901157383418793070215408914993488416750924476146066808226480016847741185374234544243710753907774499206955170276183860626133138458300075204493382656029760673711320070932870912744374704723624212700454495421842219077173525899689811474120614457405772696521446961165559468253835854362096088934714907384964847142748311021268578658461064714894910680584249490719358138073078291397044213736982988247857479512745588762993966446075)
(bf #e3.14159268391980626493420192940831754203350026401337226640663040854412059241988978103217808449508253393479795573626200366332733859609651462659489470805432281782785922056335606047700127154963266242144951481397480765182268219697420028007903565511884267297358842935537138583640066772149177226656227031792115896439889412205871076985598822285367358003457939603015797225018209619662200081521930463480571130673429337524564941105654923909951299948539893933654293161126559643573974163405197696633200469475250152247413175932572922175467223988860975105100904322239324381097207835036465269418118204894206705789759765527734394105147)
 

[edit] REXX

[edit] Version 1

The CF subroutine (for continued fractions) isn't limited to positive integers. Any form of REXX numbers (negative, exponentiated, decimal fractions) can be used; [note the use of negative fractions for the ß terms when computing √½].

There isn't any practical limit on the precision that can be used, although 100k digits would be a bit unwieldly to display.

A generalized function was added to calculate a few low integers (and also ½). In addition, ½π was calculated (as described in the talk page under Gold Credit).

More code is used for nicely formatting the output than the continued fraction calculation.

/*REXX program calculates and displays values of some specific continued*/
/*───────────── fractions (along with their α and ß terms). */
/*───────────── Continued fractions: also known as anthyphairetic ratio.*/
T=500 /*use 500 terms for calculations.*/
showDig=100; numeric digits 2*showDig /*use 100 digits for the display.*/
a=; @=; b= /*omitted ß terms are assumed = 1*/
/*══════════════════════════════════════════════════════════════════════*/
a=1 rep(2); call tell '√2'
/*══════════════════════════════════════════════════════════════════════*/
a=1 rep(1 2); call tell '√3' /*also: 2∙sin(π/3) */
/*══════════════════════════════════════ ___ ════════════════════════*/
/*generalized √ N */
do N=2 to 11; a=1 rep(2); b=rep(N-1); call tell 'gen √'N; end
N=1/2; a=1 rep(2); b=rep(N-1); call tell 'gen √½'
/*══════════════════════════════════════════════════════════════════════*/
do j=1 for T; a=a j; end; b=1 a; a=2 a; call tell 'e'
/*══════════════════════════════════════════════════════════════════════*/
do j=1 for T by 2; a=a j; b=b j+1; end; call tell '1÷[√e-1]'
/*══════════════════════════════════════════════════════════════════════*/
do j=1 for T; a=a j; end; b=a; a=0 a; call tell '1÷[e-1]'
/*══════════════════════════════════════════════════════════════════════*/
a=1 rep(1); call tell 'φ, phi'
/*══════════════════════════════════════════════════════════════════════*/
a=1; do j=1 for T by 2; a=a j 1; end; call tell 'tan(1)'
/*══════════════════════════════════════════════════════════════════════*/
a=1; do j=1 for T; a=a 2*j+1; end; call tell 'coth(1)'
/*══════════════════════════════════════════════════════════════════════*/
a=2; do j=1 for T; a=a 4*j+2; end; call tell 'coth(½)' /*also: [e+1] ÷ [e-1] */
/*══════════════════════════════════════════════════════════════════════*/
T=10000
a=1 rep(2)
do j=1 for T by 2; b=b j**2; end; call tell '4÷π'
/*══════════════════════════════════════════════════════════════════════*/
T=10000
a=1; do j=1 for T; a=a 1/j; @=@ '1/'j; end; call tell '½π, ½pi'
/*══════════════════════════════════════════════════════════════════════*/
T=10000
a=0 1 rep(2)
do j=1 for T by 2; b=b j**2; end; b=4 b; call tell 'π, pi'
/*══════════════════════════════════════════════════════════════════════*/
T=10000
a=0; do j=1 for T; a=a j*2-1; b=b j**2; end; b=4 b; call tell 'π, pi'
/*══════════════════════════════════════════════════════════════════════*/
T=100000
a=3 rep(6)
do j=1 for T by 2; b=b j**2; end; call tell 'π, pi'
exit /*stick a fork in it, we're done.*/
 
/*────────────────────────────────CF subroutine─────────────────────────*/
cf: procedure; parse arg C x,y;  !=0; numeric digits digits()+5
do k=words(x) to 1 by -1; a=word(x,k); b=word(word(y,k) 1,1)
d=a+!; if d=0 then call divZero /*in case divisor is bogus.*/
 !=b/d /*here's a binary mosh pit.*/
end /*k*/
return !+C
/*────────────────────────────────DIVZERO subroutine────────────────────*/
divZero: say; say '***error!***'; say 'division by zero.'; say; exit 13
/*────────────────────────────────GETT subroutine───────────────────────*/
getT: parse arg stuff,width,ma,mb,_
do m=1; mm=m+ma; mn=max(1,m-mb); w=word(stuff,m)
w=right(w,max(length(word(As,mm)),length(word(Bs,mn)),length(w)))
if length(_ w)>width then leave /*stop getting terms?*/
_=_ w /*whole, don't chop. */
end /*m*/ /*done building terms*/
return strip(_) /*strip leading blank*/
/*────────────────────────────────REP subroutine────────────────────────*/
rep: parse arg rep; return space(copies(' 'rep, T%words(rep)))
/*────────────────────────────────RF subroutine─────────────────────────*/
rf: parse arg xxx,z
do m=1 for T; w=word(xxx,m)  ; if w=='1/1' | w=1 then w=1
if w=='1/2' | w=1/2 then w='½'; if w=-.5 then w='-½'
if w=='1/4' | w=1/4 then w='¼'; if w=-.25 then w='-¼'
z=z w
end
return z /*done re-formatting.*/
/*────────────────────────────────TELL subroutine───────────────────────*/
tell: parse arg ?; v=cf(a,b); numeric digits showdig; As=rf(@ a); Bs=rf(b)
say right(?,8) '=' left(v/1,showdig) ' α terms= ' getT(As,72 ,0,1)
if b\=='' then say right('',8+2+showdig+1) ' ß terms= ' getT(Bs,72-2,1,0)
a=; @=; b=; return

output

      √2 = 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157   α terms=  1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
      √3 = 1.73205080756887729352744634150587236694280525381038062805580697945193301690880003708114618675724857   α terms=  1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1
  gen √2 = 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157   α terms=  1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
                                                                                                                  ß terms=    1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  gen √3 = 1.73205080756887729352744634150587236694280525381038062805580697945193301690880003708114618675724857   α terms=  1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
                                                                                                                  ß terms=    2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
  gen √4 = 2                                                                                                      α terms=  1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
                                                                                                                  ß terms=    3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
  gen √5 = 2.23606797749978969640917366873127623544061835961152572427089724541052092563780489941441440837878227   α terms=  1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
                                                                                                                  ß terms=    4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4
  gen √6 = 2.44948974278317809819728407470589139196594748065667012843269256725096037745731502653985943310464023   α terms=  1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
                                                                                                                  ß terms=    5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
  gen √7 = 2.64575131106459059050161575363926042571025918308245018036833445920106882323028362776039288647454361   α terms=  1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
                                                                                                                  ß terms=    6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6
  gen √8 = 2.82842712474619009760337744841939615713934375075389614635335947598146495692421407770077506865528314   α terms=  1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
                                                                                                                  ß terms=    7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7
  gen √9 = 3                                                                                                      α terms=  1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
                                                                                                                  ß terms=    8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8
 gen √10 = 3.16227766016837933199889354443271853371955513932521682685750485279259443863923822134424810837930029   α terms=  1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
                                                                                                                  ß terms=    9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
 gen √11 = 3.31662479035539984911493273667068668392708854558935359705868214611648464260904384670884339912829065   α terms=  1  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2
                                                                                                                  ß terms=    10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10
  gen √½ = 0.70710678118654752440084436210484903928483593768847403658833986899536623923105351942519376716382078   α terms=  1  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2
                                                                                                                  ß terms=    -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½ -½
       e = 2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642   α terms=  2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
                                                                                                                  ß terms=    1 1 2 3 4 5 6 7 8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
1÷[√e-1] = 1.54149408253679828413110344447251463834045923684188210947413695663754263914331480707182572408500774   α terms=  1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
                                                                                                                  ß terms=    2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48
 1÷[e-1] = 0.58197670686932642438500200510901155854686930107539613626678705964804381739166974328720470940487505   α terms=  0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
                                                                                                                  ß terms=    1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
  φ, phi = 1.61803398874989484820458683436563811772030917980576286213544862270526046281890244970720720418939113   α terms=  1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
  tan(1) = 1.55740772465490223050697480745836017308725077238152003838394660569886139715172728955509996520224298   α terms=  1 1 1 3 1 5 1 7 1 9 1 11 1 13 1 15 1 17 1 19 1 21 1 23 1 25 1 27 1 29 1
 coth(1) = 1.31303528549933130363616124693084783291201394124045265554315296756708427046187438267467924148085630   α terms=  1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
 coth(½) = 2.16395341373865284877000401021802311709373860215079227253357411929608763478333948657440941880975011   α terms=  2 6 10 14 18 22 26 30 34 38 42 46 50 54 58 62 66 70 74 78 82 86 90 94
     4÷π = 1.27283479368898552763028955877501991659132774562422849516943825241995907438793488819711543252100078   α terms=  1 2 2  2  2  2   2   2   2   2   2   2   2   2   2   2   2    2    2
                                                                                                                  ß terms=    1 9 25 49 81 121 169 225 289 361 441 529 625 729 841 961 1089 1225
 ½π, ½pi = 1.57071779679495409624134340842172803914665374867756685353559415360226497865248110814032818773478787   α terms=  1 ½ 1/3 ¼ 1/5 1/6 1/7 1/8 1/9 1/10 1/11 1/12 1/13 1/14 1/15 1/16 1/17
   π, pi = 3.14259165433954305090112773725220456615353825631695587367530386050342717161955770321769607013860474   α terms=  0 1 2 2  2  2  2   2   2   2   2   2   2   2   2   2   2   2    2    2
                                                                                                                  ß terms=    4 1 9 25 49 81 121 169 225 289 361 441 529 625 729 841 961 1089 1225
   π, pi = 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706   α terms=  0 1 3 5 7  9 11 13 15 17 19  21  23  25  27  29  31  33  35  37  39  41
                                                                                                                  ß terms=    4 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400
   π, pi = 3.14159265358979298847014326453044038404101783047277203674633230347271153796007366409681897722403708   α terms=  3 6 6  6  6  6   6   6   6   6   6   6   6   6   6   6   6    6    6
                                                                                                                  ß terms=    1 9 25 49 81 121 169 225 289 361 441 529 625 729 841 961 1089 1225

Note: even with 200 digit accuracy and 100,000 terms, the last calculation of π (pi) is only accurate to 15 digits.

[edit] Version 2 derived from PL/I

/* REXX **************************************************************
* Derived from PL/I with a little "massage"
* SQRT2= 1.41421356237309505 <- PL/I Result
* 1.41421356237309504880168872421 <- REXX Result 30 digits
* NAPIER= 2.71828182845904524
* 2.71828182845904523536028747135
* PI= 3.14159262280484695
* 3.14159262280484694855146925223
* 06.09.2012 Walter Pachl
**********************************************************************/

Numeric Digits 30
Parse Value '1 2 3 0 0' with Sqrt2 napier pi a b
Say left('SQRT2=' ,10) calc(sqrt2, 200)
Say left('NAPIER=',10) calc(napier, 200)
Say left('PI=' ,10) calc(pi, 200)
Exit
 
Get_Coeffs: procedure Expose a b Sqrt2 napier pi
Parse Arg form, n
select
when form=Sqrt2 Then do
if n > 0 then a = 2; else a = 1
b = 1
end
when form=Napier Then do
if n > 0 then a = n; else a = 2
if n > 1 then b = n - 1; else b = 1
end
when form=pi Then do
if n > 0 then a = 6; else a = 3
b = (2*n - 1)**2
end
end
Return
 
Calc: procedure Expose a b Sqrt2 napier pi
Parse Arg form,n
Temp=0
do ni = n to 1 by -1
Call Get_Coeffs form, ni
Temp = B/(A + Temp)
end
call Get_Coeffs form, 0
return (A + Temp)

[edit] Version 3 better approximation

/* REXX *************************************************************
* The task description specifies a continued fraction for pi
* that gives a reasonable approximation.
* Literature shows a better CF that yields pi with a precision of
* 200 digits.
* http://de.wikipedia.org/wiki/Kreiszahl
* 1
* pi = 3 + ------------------------
* 1
* 7 + --------------------
* 1
* 15 + ---------------
* 1
* 1 + -----------
*
* 292 + ...
*
* This program uses that CF and shows the first 50 digits
* PI =3.1415926535897932384626433832795028841971693993751...
* PIX=3.1415926535897932384626433832795028841971693993751...
* 201 correct digits
* 18.09.2012 Walter Pachl
**********************************************************************/

pi='3.1415926535897932384626433832795028841971'||,
'693993751058209749445923078164062862089986280348'||,
'253421170679821480865132823066470938446095505822'||,
'317253594081284811174502841027019385211055596446'||,
'229489549303819644288109756659334461284756482337'||,
'867831652712019091456485669234603486104543266482'||,
'133936072602491412737245870066063155881748815209'||,
'209628292540917153643678925903600113305305488204'||,
'665213841469519415116094330572703657595919530921'||,
'861173819326117931051185480744623799627495673518'||,
'857527248912279381830119491298336733624'
Numeric Digits 1000
al='7 15 1 292 1 1 1 2 1 3 1 14 2 1 1 2 2 2 2 1 84 2',
'1 1 15 3 13 1 4 2 6 6 99 1 2 2 6 3 5 1 1 6 8 1 7 1 2',
'3 7 1 2 1 1 12 1 1 1 3 1 1 8 1 1 2 1 6 1 1 5 2 2 3 1',
'2 4 4 16 1 161 45 1 22 1 2 2 1 4 1 2 24 1 2 1 3 1 2',
'1 1 10 2 5 4 1 2 2 8 1 5 2 2 26 1 4 1 1 8 2 42 2 1 7',
'3 3 1 1 7 2 4 9 7 2 3 1 57 1 18 1 9 19 1 2 18 1 3 7',
'30 1 1 1 3 3 3 1 2 8 1 1 2 1 15 1 2 13 1 2 1 4 1 12',
'1 1 3 3 28 1 10 3 2 20 1 1 1 1 4 1 1 1 5 3 2 1 6 1 4'
a.=3
Do i=1 By 1 while al<>''
Parse Var al a.i al
End
pix=calc(194)
Do e=1 To length(pi)
If substr(pix,e,1)<>substr(pi,e,1) Then Leave
End
Numeric Digits 50
Say 'PI ='||(pi+0)||'...'
Say 'PIX='||(pix+0)||'...'
Say (e-1) 'correct digits'
Exit
 
Get_Coeffs: procedure Expose a b a.
Parse Arg n
a=a.n
b=1
Return
 
Calc: procedure Expose a b a.
Parse Arg n
Temp=0
do ni = n to 1 by -1
Call Get_Coeffs ni
Temp = B/(A + Temp)
end
call Get_Coeffs 0
return (A + Temp)

[edit] Ruby

require 'bigdecimal'
 
# square root of 2
sqrt2 = Object.new
def sqrt2.a(n); n == 1 ? 1 : 2; end
def sqrt2.b(n); 1; end
 
# Napier's constant
napier = Object.new
def napier.a(n); n == 1 ? 2 : n - 1; end
def napier.b(n); n == 1 ? 1 : n - 1; end
 
pi = Object.new
def pi.a(n); n == 1 ? 3 : 6; end
def pi.b(n); (2*n - 1)**2; end
 
# Estimates the value of a continued fraction _cfrac_, to _prec_
# decimal digits of precision. Returns a BigDecimal. _cfrac_ must
# respond to _cfrac.a(n)_ and _cfrac.b(n)_ for integer _n_ >= 1.
def estimate(cfrac, prec)
last_result = nil
terms = prec
 
loop do
# Estimate continued fraction for _n_ from 1 to _terms_.
result = cfrac.a(terms)
(terms - 1).downto(1) do |n|
a = BigDecimal cfrac.a(n)
b = BigDecimal cfrac.b(n)
digits = [b.div(result, 1).exponent + prec, 1].max
result = a + b.div(result, digits)
end
result = result.round(prec)
 
if result == last_result
return result
else
# Double _terms_ and try again.
last_result = result
terms *= 2
end
end
end
 
puts estimate(sqrt2, 50).to_s('F')
puts estimate(napier, 50).to_s('F')
puts estimate(pi, 10).to_s('F')
Output:
$ ruby cfrac.rb                                                              
1.41421356237309504880168872420969807856967187537695
2.71828182845904523536028747135266249775724709369996
3.1415926536

[edit] Scala

Works with: Scala version 2.9.1

Note that Scala-BigDecimal provides a precision of 34 digits. Therefore we take a limitation of 32 digits to avoiding rounding problems.

object CF extends App {
import Stream._
val sqrt2 = 1 #:: from(2,0) zip from(1,0)
val napier = 2 #:: from(1) zip (1 #:: from(1))
val pi = 3 #:: from(6,0) zip (from(1,2) map {x=>x*x})
 
// reference values, source: wikipedia
val refPi = "3.14159265358979323846264338327950288419716939937510"
val refNapier = "2.71828182845904523536028747135266249775724709369995"
val refSQRT2 = "1.41421356237309504880168872420969807856967187537694"
 
def calc(cf: Stream[(Int, Int)], numberOfIters: Int=200): BigDecimal = {
(cf take numberOfIters toList).foldRight[BigDecimal](1)((a, z) => a._1+a._2/z)
}
 
def approx(cfV: BigDecimal, cfRefV: String): String = {
val p: Pair[Char,Char] => Boolean = pair =>(pair._1==pair._2)
((cfV.toString+" "*34).substring(0,34) zip cfRefV.toString.substring(0,34))
.takeWhile(p).foldRight[String]("")((a:Pair[Char,Char],z)=>a._1+z)
}
 
List(("sqrt2",sqrt2,50,refSQRT2),("napier",napier,50,refNapier),("pi",pi,3000,refPi)) foreach {t=>
val (name,cf,iters,refV) = t
val cfV = calc(cf,iters)
println(name+":")
println("ref value: "+refV.substring(0,34))
println("cf value: "+(cfV.toString+" "*34).substring(0,34))
println("precision: "+approx(cfV,refV))
println()
}
}
Output:
sqrt2:
ref value: 1.41421356237309504880168872420969
cf value:  1.41421356237309504880168872420969
precision: 1.41421356237309504880168872420969

napier:
ref value: 2.71828182845904523536028747135266
cf value:  2.71828182845904523536028747135266
precision: 2.71828182845904523536028747135266

pi:
ref value: 3.14159265358979323846264338327950
cf value:  3.14159265358052780404906362935452
precision: 3.14159265358

For higher accuracy of pi we have to take more iterations. Unfortunately the foldRight function in calc isn't tail recursiv - therefore a stack overflow exception will be thrown for higher numbers of iteration, thus we have to implement an iterative way for calculation:

object CFI extends App {
import Stream._
val sqrt2 = 1 #:: from(2,0) zip from(1,0)
val napier = 2 #:: from(1) zip (1 #:: from(1))
val pi = 3 #:: from(6,0) zip (from(1,2) map {x=>x*x})
 
// reference values, source: wikipedia
val refPi = "3.14159265358979323846264338327950288419716939937510"
val refNapier = "2.71828182845904523536028747135266249775724709369995"
val refSQRT2 = "1.41421356237309504880168872420969807856967187537694"
 
def calc_i(cf: Stream[(Int, Int)], numberOfIters: Int=50): BigDecimal = {
val cfl = cf take numberOfIters toList
var z: BigDecimal = 1.0
for (i <- 0 to cfl.size-1 reverse)
z=cfl(i)._1+cfl(i)._2/z
z
}
 
def approx(cfV: BigDecimal, cfRefV: String): String = {
val p: Pair[Char,Char] => Boolean = pair =>(pair._1==pair._2)
((cfV.toString+" "*34).substring(0,34) zip cfRefV.toString.substring(0,34))
.takeWhile(p).foldRight[String]("")((a:Pair[Char,Char],z)=>a._1+z)
}
 
List(("sqrt2",sqrt2,50,refSQRT2),("napier",napier,50,refNapier),("pi",pi,50000,refPi)) foreach {t=>
val (name,cf,iters,refV) = t
val cfV = calc_i(cf,iters)
println(name+":")
println("ref value: "+refV.substring(0,34))
println("cf value: "+(cfV.toString+" "*34).substring(0,34))
println("precision: "+approx(cfV,refV))
println()
}
}
Output:
sqrt2:
ref value: 1.41421356237309504880168872420969
cf value:  1.41421356237309504880168872420969
precision: 1.41421356237309504880168872420969

napier:
ref value: 2.71828182845904523536028747135266
cf value:  2.71828182845904523536028747135266
precision: 2.71828182845904523536028747135266

pi:
ref value: 3.14159265358979323846264338327950
cf value:  3.14159265358983426214354599901745
precision: 3.141592653589

[edit] Tcl

Works with: tclsh version 8.6
Translation of: Python

Note that Tcl does not provide arbitrary precision floating point numbers by default, so all result computations are done with IEEE doubles.

package require Tcl 8.6
 
# Term generators; yield list of pairs
proc r2 {} {
yield {1 1}
while 1 {yield {2 1}}
}
proc e {} {
yield {2 1}
while 1 {yield [list [incr n] $n]}
}
proc pi {} {
set n 0; set a 3
while 1 {
yield [list $a [expr {(2*[incr n]-1)**2}]]
set a 6
}
}
 
# Continued fraction calculator
proc cf {generator {termCount 50}} {
# Get the chunk of terms we want to work with
set terms [list [coroutine cf.c $generator]]
while {[llength $terms] < $termCount} {
lappend terms [cf.c]
}
rename cf.c {}
 
# Merge the terms to compute the result
set val 0.0
foreach pair [lreverse $terms] {
lassign $pair a b
set val [expr {$a + $b/$val}]
}
return $val
}
 
# Demonstration
puts [cf r2]
puts [cf e]
puts [cf pi 250]; # Converges more slowly
Output:
1.4142135623730951
2.7182818284590455
3.1415926373965735

[edit] XPL0

The number of iterations (N) needed to get the 13 digits of accuracy was determined by experiment.

include c:\cxpl\codes;
int N;
real A, B, F;
[Format(1, 15);
A:= 2.0; B:= 1.0; N:= 16;
IntOut(0, N); CrLf(0);
F:= 0.0;
while N>=1 do [F:= B/(A+F); N:= N-1];
RlOut(0, 1.0+F); CrLf(0);
RlOut(0, sqrt(2.0)); CrLf(0);
 
N:= 13;
IntOut(0, N); CrLf(0);
F:= 0.0;
while N>=2 do [F:= float(N-1)/(float(N)+F); N:= N-1];
RlOut(0, 2.0 + 1.0/(1.0+F)); CrLf(0);
RlOut(0, Exp(1.0)); CrLf(0);
 
N:= 10000;
IntOut(0, N); CrLf(0);
F:= 0.0;
while N>=1 do [F:= float(sq(2*N-1))/(6.0+F); N:= N-1];
RlOut(0, 3.0+F); CrLf(0);
RlOut(0, ACos(-1.0)); CrLf(0);
]
Output:
16
1.414213562372820
1.414213562373100
13
2.718281828459380
2.718281828459050
10000
3.141592653589540
3.141592653589790

[edit] zkl

fcn cf(fa,fb,a0){fcn(fa,fb,a0,n){
a0 + [n..1,-1].reduce('wrap(p,n){fb(n)/(fa(n)+p)},0.0) }
.fp(fa,fb,a0) }

cf creates a function that calculates the continued fraction from the bottom up. The new function takes a single parameter, n, which is used to calculate the nth term.

sqrt2:=cf((2.0).noop,(1.0).noop,1.0);
sqrt2(200) : "%.20e".fmt(_).println();
nap:=cf((0.0).create,fcn(n){(n==1) and 1.0 or (n-1).toFloat()},2.0);
println(nap(15) - (1.0).e);
pi:=cf((6.0).noop,fcn(n){n=2*n-1; (n*n).toFloat()},3.0);
println(pi(1000) - (1.0).pi);

(1.0).create(n) --> n, (1.0).noop(n) --> 1.0

Output:
1.41421356237309514547e+00
1.33227e-15
-2.49251e-10
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