# Continued fraction

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:
${\displaystyle a_{0}+{\cfrac {b_{1}}{a_{1}+{\cfrac {b_{2}}{a_{2}+{\cfrac {b_{3}}{a_{3}+\ddots }}}}}}}$

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 ${\displaystyle a_{0}=1}$ then ${\displaystyle a_{N}=2}$. ${\displaystyle b_{N}}$ is always ${\displaystyle 1}$.

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

For Napier's Constant, use ${\displaystyle a_{0}=2}$, then ${\displaystyle a_{N}=N}$. ${\displaystyle b_{1}=1}$ then ${\displaystyle b_{N}=N-1}$.

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

For Pi, use ${\displaystyle a_{0}=3}$ then ${\displaystyle a_{N}=6}$. ${\displaystyle b_{N}=(2N-1)^{2}}$.

${\displaystyle \pi =3+{\cfrac {1}{6+{\cfrac {9}{6+{\cfrac {25}{6+\ddots }}}}}}}$

## 11l

F calc(f_a, f_b, =n = 1000)   V r = 0.0   L n > 0      r = f_b(n) / (f_a(n) + r)      n--   R f_a(0) + r print(calc(n -> I n > 0 {2} E 1, n -> 1))print(calc(n -> I n > 0 {n} E 2, n -> I n > 1 {n - 1} E 1))print(calc(n -> I n > 0 {6} E 3, n -> (2 * n - 1) ^ 2))

(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;

### 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

## ALGOL 68

Works with: Algol 68 Genie version 2.8
 PROC cf = (INT steps, PROC (INT) INT a, PROC (INT) INT b) REAL:BEGIN  REAL result;  result := 0;  FOR n FROM steps BY -1 TO 1 DO      result := b(n) / (a(n) + result)  OD;  a(0) + resultEND; PROC asqr2 = (INT n) INT: (n = 0 | 1 | 2);PROC bsqr2 = (INT n) INT: 1; PROC anap = (INT n) INT: (n = 0 | 2 | n);PROC bnap = (INT n) INT: (n = 1 | 1 | n - 1); PROC api = (INT n) INT: (n = 0 | 3 | 6);PROC bpi = (INT n) INT: (n = 1 | 1 | (2 * n - 1) ** 2); INT precision = 10000; print (("Precision: ", precision, newline));print (("Sqr(2):    ", cf(precision, asqr2, bsqr2), newline));print (("Napier:    ", cf(precision, anap, bnap), newline));print (("Pi:        ", cf(precision, api, bpi)))
Output:

Precision:      +10000
Sqr(2):    +1.41421356237310e  +0
Napier:    +2.71828182845905e  +0
Pi:        +3.14159265358954e  +0


## 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} (* ****** ****** *) implementmain0 () =(  println! ("sqrt2  = ", calc(sqrt2,  100));  println! ("napier = ", calc(napier, 100));  println! ("  pi   = ", calc(  pi  , 100));) (* end of [main0] *)

## AutoHotkey

sqrt2_a(n) ; function definition is as simple as that{return n?2.0:1.0} sqrt2_b(n){return 1.0} napier_a(n){return n?n:2.0} napier_b(n){return n>1.0?n-1.0:1.0} pi_a(n){return n?6.0:3.0} pi_b(n){return (2.0*n-1.0)**2.0 ; exponentiation operator} calc(f,expansions){r:=0,i:=expansions; A nasty trick: the names of the two coefficient functions are generated dynamically; a dot surrounded by spaces means string concatenationf_a:=f . "_a",f_b:=f . "_b" while i>0 {; You can see two dynamic function calls herer:=%f_b%(i)/(%f_a%(i)+r)i--} return %f_a%(0)+r} Msgbox, % "sqrt 2 = " . calc("sqrt2", 1000) . "ne = " . calc("napier", 1000) . "npi = " . calc("pi", 1000)

Output with Autohotkey v1 (currently 1.1.16.05):

sqrt 2 = 1.414214e = 2.718282pi = 3.141593

Output with Autohotkey v2 (currently alpha 56):

sqrt 2 = 1.4142135623730951e = 2.7182818284590455pi = 3.1415926533405418

Note the far superiour accuracy of v2.

## Axiom

Axiom provides a ContinuedFraction domain:

get(obj) == convergents(obj).1000 -- utility to extract the 1000th valueget continuedFraction(1, repeating [1], repeating [2]) :: Floatget continuedFraction(2, cons(1,[i for i in 1..]), [i for i in 1..]) :: Floatget 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 ${\displaystyle \pi }$ 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+tempcf(1, repeating [1], repeating [2], 1000) :: Floatcf(2, cons(1,[i for i in 1..]), [i for i in 1..], 1000) :: Floatcf(3, [i^2 for i in 1.. by 2], repeating [6], 1000) :: Float

      *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  ## 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 ## C++ #include <iomanip>#include <iostream>#include <tuple> typedef std::tuple<double,double> coeff_t; // coefficients typetypedef 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  ## Clojure  (defn cfrac [a b n] (letfn [(cfrac-iter [[x k]] [(+ (a k) (/ (b (inc k)) x)) (dec k)])] (ffirst (take 1 (drop (inc n) (iterate cfrac-iter [1 n])))))) (def sq2 (cfrac #(if (zero? %) 1.0 2.0) (constantly 1.0) 100))(def e (cfrac #(if (zero? %) 2.0 %) #(if (= 1 %) 1.0 (double (dec %))) 100))(def pi (cfrac #(if (zero? %) 3.0 6.0) #(let [x (- (* 2.0 %) 1.0)] (* x x)) 900000))  Output: user=> sq2 e pi 1.4142135623730951 2.7182818284590455 3.141592653589793  ## COBOL Works with: GnuCOBOL  identification division. program-id. show-continued-fractions. environment division. configuration section. repository. function continued-fractions function all intrinsic. procedure division. fractions-main. display "Square root 2 approximately : " continued-fractions("sqrt-2-alpha", "sqrt-2-beta", 100) display "Napier constant approximately : " continued-fractions("napier-alpha", "napier-beta", 40) display "Pi approximately : " continued-fractions("pi-alpha", "pi-beta", 10000) goback. end program show-continued-fractions. *> ************************************************************** identification division. function-id. continued-fractions. data division. working-storage section. 01 alpha-function usage program-pointer. 01 beta-function usage program-pointer. 01 alpha usage float-long. 01 beta usage float-long. 01 running usage float-long. 01 i usage binary-long. linkage section. 01 alpha-name pic x any length. 01 beta-name pic x any length. 01 iterations pic 9 any length. 01 approximation usage float-long. procedure division using alpha-name beta-name iterations returning approximation. set alpha-function to entry alpha-name if alpha-function = null then display "error: no " alpha-name " function" upon syserr goback end-if set beta-function to entry beta-name if beta-function = null then display "error: no " beta-name " function" upon syserr goback end-if move 0 to alpha beta running perform varying i from iterations by -1 until i = 0 call alpha-function using i returning alpha call beta-function using i returning beta compute running = beta / (alpha + running) end-perform call alpha-function using 0 returning alpha compute approximation = alpha + running goback. end function continued-fractions. *> ****************************** identification division. program-id. sqrt-2-alpha. data division. working-storage section. 01 result usage float-long. linkage section. 01 iteration usage binary-long unsigned. procedure division using iteration returning result. if iteration equal 0 then move 1.0 to result else move 2.0 to result end-if goback. end program sqrt-2-alpha. *> ****************************** identification division. program-id. sqrt-2-beta. data division. working-storage section. 01 result usage float-long. linkage section. 01 iteration usage binary-long unsigned. procedure division using iteration returning result. move 1.0 to result goback. end program sqrt-2-beta. *> ****************************** identification division. program-id. napier-alpha. data division. working-storage section. 01 result usage float-long. linkage section. 01 iteration usage binary-long unsigned. procedure division using iteration returning result. if iteration equal 0 then move 2.0 to result else move iteration to result end-if goback. end program napier-alpha. *> ****************************** identification division. program-id. napier-beta. data division. working-storage section. 01 result usage float-long. linkage section. 01 iteration usage binary-long unsigned. procedure division using iteration returning result. if iteration = 1 then move 1.0 to result else compute result = iteration - 1.0 end-if goback. end program napier-beta. *> ****************************** identification division. program-id. pi-alpha. data division. working-storage section. 01 result usage float-long. linkage section. 01 iteration usage binary-long unsigned. procedure division using iteration returning result. if iteration equal 0 then move 3.0 to result else move 6.0 to result end-if goback. end program pi-alpha. *> ****************************** identification division. program-id. pi-beta. data division. working-storage section. 01 result usage float-long. linkage section. 01 iteration usage binary-long unsigned. procedure division using iteration returning result. compute result = (2 * iteration - 1) ** 2 goback. end program pi-beta.  Output: prompt$ cobc -xj continued-fractions.cob
Square root 2 approximately   : 1.414213562373095
Napier constant approximately : 2.718281828459045
Pi approximately              : 3.141592653589543

## 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
---


## 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

## 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));

## 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

## 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).3.14159265334054230> continued:test_sqrt2(1000).1.414213562373095131> continued:test_nappier(1000).2.7182818284590455

## F#

### The Functions

 // I provide four functions:-// cf2S general purpose continued fraction to sequence of float approximations// cN2S Normal continued fractions (a-series always 1)// cfSqRt uses cf2S to calculate sqrt of float// π takes a sequence of b values returning the next until the list is exhausted after which  it injects infinity// Nigel Galloway: December 19th., 2018let cf2S α β=let n0,g1,n1,g2=β(),α(),β(),β()             seq{let (Π:decimal)=g1/n1 in yield n0+Π; yield! Seq.unfold(fun(n,g,Π)->let a,b=α(),β() in let Π=Π*g/n in Some(n0+Π,(b+a/n,b+a/g,Π)))(g2+α()/n1,g2,Π)}let cN2S = cf2S (fun()->1M)let cfSqRt n=(cf2S (fun()->n-1M) (let mutable n=false in fun()->if n then 2M else (n<-true; 1M)))let π n=let mutable π=n in (fun ()->match π with h::t->π<-t; h |_->9999999999999999999999999999M)

 cfSqRt 2M |> Seq.take 10 |> Seq.pairwise |> Seq.iter(fun(n,g)->printfn "%1.14f < √2 < %1.14f" (min n g) (max n g))
Output:
1.40000000000000 < √2 < 1.50000000000000
1.40000000000000 < √2 < 1.41666666666667
1.41379310344828 < √2 < 1.41666666666667
1.41379310344828 < √2 < 1.41428571428571
1.41420118343195 < √2 < 1.41428571428571
1.41420118343195 < √2 < 1.41421568627451
1.41421319796954 < √2 < 1.41421568627451
1.41421319796954 < √2 < 1.41421362489487
1.41421355164605 < √2 < 1.41421362489487

 cfSqRt 0.25M |> Seq.take 30 |> Seq.iter (printfn "%1.14f")
Output:
0.62500000000000
0.53846153846154
0.51250000000000
0.50413223140496
0.50137362637363
0.50045745654163
0.50015243902439
0.50005080784473
0.50001693537461
0.50000564506114
0.50000188167996
0.50000062722587
0.50000020907520
0.50000006969172
0.50000002323057
0.50000000774352
0.50000000258117
0.50000000086039
0.50000000028680
0.50000000009560
0.50000000003187
0.50000000001062
0.50000000000354
0.50000000000118
0.50000000000039
0.50000000000013
0.50000000000004
0.50000000000001
0.50000000000000
0.50000000000000

 let aπ()=let mutable n=0M in (fun ()->n<-n+1M;let b=n+n-1M in b*b)let bπ()=let mutable n=true in (fun ()->match n with true->n<-false;3M |_->6M)cf2S (aπ()) (bπ()) |> Seq.take 10 |> Seq.pairwise |> Seq.iter(fun(n,g)->printfn "%1.14f < π < %1.14f" (min n g) (max n g))
Output:
3.13333333333333 < π < 3.16666666666667
3.13333333333333 < π < 3.14523809523810
3.13968253968254 < π < 3.14523809523810
3.13968253968254 < π < 3.14271284271284
3.14088134088134 < π < 3.14271284271284
3.14088134088134 < π < 3.14207181707182
3.14125482360776 < π < 3.14207181707182
3.14125482360776 < π < 3.14183961892940
3.14140671849650 < π < 3.14183961892940

 let pi = π [3M;7M;15M;1M;292M;1M;1M;1M;2M;1M;3M;1M;14M;2M;1M;1M;2M;2M;2M;2M]cN2S pi |> Seq.take 10 |> Seq.pairwise |> Seq.iter(fun(n,g)->printfn "%1.14f < π < %1.14f" (min n g) (max n g))
Output:
3.14150943396226 < π < 3.14285714285714
3.14150943396226 < π < 3.14159292035398
3.14159265301190 < π < 3.14159292035398
3.14159265301190 < π < 3.14159265392142
3.14159265346744 < π < 3.14159265392142
3.14159265346744 < π < 3.14159265361894
3.14159265358108 < π < 3.14159265361894
3.14159265358108 < π < 3.14159265359140
3.14159265358939 < π < 3.14159265359140

 let ae()=let mutable n=0.5M in (fun ()->match n with 0.5M->n<-0M; 1M |_->n<-n+1M; n)let be()=let mutable n=0.5M in (fun ()->match n with 0.5M->n<-0M; 2M |_->n<-n+1M; n)cf2S (ae()) (be()) |> Seq.take 10 |> Seq.pairwise |> Seq.iter(fun(n,g)->printfn "%1.14f < e < %1.14f" (min n g) (max n g))
Output:
2.66666666666667 < e < 3.00000000000000
2.66666666666667 < e < 2.72727272727273
2.71698113207547 < e < 2.72727272727273
2.71698113207547 < e < 2.71844660194175
2.71826333176026 < e < 2.71844660194175
2.71826333176026 < e < 2.71828369389345
2.71828165766640 < e < 2.71828369389345
2.71828165766640 < e < 2.71828184277783
2.71828182735187 < e < 2.71828184277783


## 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 2SINGLETON: sqrt2M: 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 constantSINGLETON: napierM: 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: piM: 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 ] [email protected]    ! 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

   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 =: ([email protected]:[ (}.,'.',{.) ":@:<[email protected]:(* 10x&^)~)"0 100 10 100 dec sqrt2, pi, e1.41421356237309504880168872420969807856967187537694807317667973799073247846212055511094575957753221653.1415924109 2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274 Note that there are two kinds of continued fractions. The kind here where we alternate between a and b values, but in some other tasks b is always 1 (and not included in the list we use to represent the continued fraction). The other kind is evaluated in J using (+%)/ instead of +%/. ## Java Translation of: D Works with: Java version 8 import static java.lang.Math.pow;import java.util.*;import java.util.function.Function; public class Test { static double calc(Function<Integer, Integer[]> f, int n) { double temp = 0; for (int ni = n; ni >= 1; ni--) { Integer[] p = f.apply(ni); temp = p[1] / (double) (p[0] + temp); } return f.apply(0)[0] + temp; } public static void main(String[] args) { List<Function<Integer, Integer[]>> fList = new ArrayList<>(); fList.add(n -> new Integer[]{n > 0 ? 2 : 1, 1}); fList.add(n -> new Integer[]{n > 0 ? n : 2, n > 1 ? (n - 1) : 1}); fList.add(n -> new Integer[]{n > 0 ? 6 : 3, (int) pow(2 * n - 1, 2)}); for (Function<Integer, Integer[]> f : fList) System.out.println(calc(f, 200)); }} 1.4142135623730951 2.7182818284590455 3.141592622804847 ## 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:
$jq -M -n -r -f Continued_fraction.jqValue : Direct : Continued Fraction2|sqrt : 1.4142135623730951 : 1.41421356237309511|exp : 2.718281828459045 : 2.7182818284590455pi : 3.141592653589793 : 3.1415926535892935 (1e-12) ## Julia Works with: Julia version 0.6 function _sqrt(a::Bool, n) if a return n > 0 ? 2.0 : 1.0 else return 1.0 endend function _napier(a::Bool, n) if a return n > 0 ? Float64(n) : 2.0 else return n > 1 ? n - 1.0 : 1.0 endend function _pi(a::Bool, n) if a return n > 0 ? 6.0 : 3.0 else return (2.0 * n - 1.0) ^ 2.0 # exponentiation operator endend function calc(f::Function, expansions::Integer) a, b = true, false r = 0.0 for i in expansions:-1:1 r = f(b, i) / (f(a, i) + r) end return f(a, 0) + rend for (v, f) in (("√2", _sqrt), ("e", _napier), ("π", _pi)) @printf("%3s = %f\n", v, calc(f, 1000))end Output:  √2 = 1.414214 e = 2.718282 π = 3.141593 ## Klong  cf::{[f g i];f::x;g::y;i::z; f(0)+z{i::i-1;g(i+1)%f(i+1)+x}:*0}cf({:[0=x;1;2]};{x;1};1000)cf({:[0=x;2;x]};{:[x>1;x-1;x]};1000)cf({:[0=x;3;6]};{((2*x)-1)^2};1000)  Output: :triad 1.41421356237309504 2.71828182845904523 3.14159265334054205  ## Kotlin Translation of: D // version 1.1.2 typealias Func = (Int) -> IntArray fun calc(f: Func, n: Int): Double { var temp = 0.0 for (i in n downTo 1) { val p = f(i) temp = p[1] / (p[0] + temp) } return f(0)[0] + temp} fun main(args: Array<String>) { val pList = listOf<Pair<String, Func>>( "sqrt(2)" to { n -> intArrayOf(if (n > 0) 2 else 1, 1) }, "e " to { n -> intArrayOf(if (n > 0) n else 2, if (n > 1) n - 1 else 1) }, "pi " to { n -> intArrayOf(if (n > 0) 6 else 3, (2 * n - 1) * (2 * n - 1)) } ) for (pair in pList) println("${pair.first} = ${calc(pair.second, 200)}")} Output: sqrt(2) = 1.4142135623730951 e = 2.7182818284590455 pi = 3.141592622804847  ## Maple  contfrac:=n->evalf(Value(NumberTheory:-ContinuedFraction(n)));contfrac(2^(0.5));contfrac(Pi);contfrac(exp(1));  ## Mathematica / Wolfram Language sqrt2=Function[n,{1,[email protected]{Array[2&,n],Array[1&,n]}}];napier=Function[n,{2,[email protected]{Range[n],Prepend[Range[n-1],1]}}];pi=Function[n,{3,[email protected]{Array[6&,n],Array[(2#-1)^2&,n]}}];approx=Function[l, N[[email protected]@[email protected][{{#2.#[[;;,1]],#2.#[[;;,2]]},#[[1]]}&,{{l[[2,1,1]]l[[1]]+l[[2,1,2]],l[[2,1,1]]},{l[[1]],1}},l[[2,2;;]]],10]];r2=approx/@{[email protected]#,[email protected]#,[email protected]#}&@10000;r2//TableForm Output: 1.414213562 2.718281828 3.141592654  ## Maxima cfeval(x) := block([a, b, n, z], a: x[1], b: x[2], n: length(a), z: 0, for i from n step -1 thru 2 do z: b[i]/(a[i] + z), a[1] + z)$ 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 */ ## 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 ## 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 ## 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 ## 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/($n++ || 1) } }, sub { 1 }, 1_000; Output: √2 ≈ 1.414213562 e ≈ 2.718281828 π ≈ 3.141592653 π/2 ≈ 1.570717797 ## Perl 6 Works with: rakudo version 2015-10-31 sub continued-fraction(:@a, :@b, Int :$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(Nil, |(1 xx *)));printf "e  ≈%.9f\n", continued-fraction(:a(2, |(1 .. *)), :b(Nil, 1, |(1 .. *)));printf "π  ≈%.9f\n", continued-fraction(:a(3, |(6 xx *)), :b(Nil, |((1, 3, 5 ... *) X** 2)));
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:

${\displaystyle \mathrm {CF} _{3}(x)=a_{0}+{\cfrac {b_{1}}{a_{1}+{\cfrac {b_{2}}{a_{2}+{\cfrac {b_{3}}{a_{3}+x}}}}}}}$

Or, more consistently:

${\displaystyle \mathrm {CF} _{3}(x)=a_{0}+{\cfrac {b_{0}}{a_{1}+{\cfrac {b_{1}}{a_{2}+{\cfrac {b_{2}}{a_{3}+{\cfrac {b_{3}}{x}}}}}}}}}$

Viewed as such, ${\displaystyle \mathrm {CF} _{n}(x)}$ could be written recursively:

${\displaystyle \mathrm {CF} _{n}(x)=\mathrm {CF} _{n-1}(a_{n}+{\frac {b_{n}}{x}})}$

Or in other words:

${\displaystyle \mathrm {CF} _{n}=\mathrm {CF} _{n-1}\circ f_{n}=\mathrm {CF} _{n-2}\circ f_{n-1}\circ f_{n}=\ldots =f_{0}\circ f_{1}\ldots \circ f_{n}}$

where ${\displaystyle f_{n}(x)=a_{n}+{\frac {b_{n}}{x}}}$

Perl6 has a builtin composition operator. We can use it with the triangular reduction metaoperator, and evaluate each resulting function at infinity (any value would do actually, but infinite makes it consistent with this particular task).

sub continued-fraction(@a, @b) {    map { .(Inf) }, [\o] map { @a[$_] + @b[$_] / * }, ^Inf} 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];
Output:
√2 ≈ 1.414213552
e  ≈ 2.718281827
π  ≈ 3.141592411

## Phix

Translation of: ALGOL_68
function continued_fraction(integer steps, integer rid_a, integer rid_b)atom res = 0  for n=steps to 1 by -1 do     res := call_func(rid_b,{n}) / (call_func(rid_a,{n}) + res)  end for  return call_func(rid_a,{0}) + resend function function sqr2_a(integer n) return iff(n=0?1:2) end functionfunction sqr2_b(integer n) return 1 end function function nap_a(integer n) return iff(n=0?2:n) end functionfunction nap_b(integer n) return iff(n=1?1:n-1) end function function pi_a(integer n) return iff(n=0?3:6) end functionfunction pi_b(integer n) return iff(n=1?1:power(2*n-1,2)) end function constant precision = 10000 printf(1,"Precision: %d\n", {precision})printf(1,"Sqr(2):    %.10g\n", {continued_fraction(precision, routine_id("sqr2_a"), routine_id("sqr2_b"))})printf(1,"Napier:    %.10g\n", {continued_fraction(precision, routine_id("nap_a"), routine_id("nap_b"))})printf(1,"Pi:        %.10g\n", {continued_fraction(precision, routine_id("pi_a"), routine_id("pi_b"))})
Output:
Precision: 10000
Sqr(2):    1.414213562
Napier:    2.718281828
Pi:        3.141592654


## 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


## 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 fractionscontinued_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 2sqrt_2_ab(0, 1, 1).sqrt_2_ab(_, 2, 1). % definitions for napiernapier_ab(0, 2, _).napier_ab(1, 1, 1).napier_ab(N, N, V) :-	V is N - 1. % definitions for pipi_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 .


## Python

Works with: Python version 2.6+ and 3.x
from fractions import Fractionimport itertoolstry: zip = itertools.izipexcept: pass # The Continued Fractiondef 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 stringdef 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 sqrt2def 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 Constantdef 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 Pidef 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

### 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 = 50print calc(fsqrt2, 200)print calc(fnapier, 200)print calc(fpi, 200)
Output:
1.4142135623730950488016887242096980785696718753770
2.7182818284590452353602874713526624977572470937000
3.1415926839198062649342019294083175420335002640134

## Racket

### 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:

 1.41421356237309512.71828182845904553.1415926839198063

### 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)

## REXX

### 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 for the decimal digits that can be used, although 100k digits would be a bit unwieldy to display.

A generalized       function was added to calculate a few low integers   (and also   1/2).

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

/*REXX program  calculates and displays  values of  various  continued fractions.       */parse arg terms digs .if terms=='' | terms==","  then terms=500if  digs=='' |  digs==","  then  digs=100numeric digits digs                              /*use  100  decimal digits for display.*/b.=1                                             /*omitted ß terms are assumed to be  1.*//*══════════════════════════════════════════════════════════════════════════════════════*/a.=2;                                                           call tell '√2',      cf(1)/*══════════════════════════════════════════════════════════════════════════════════════*/a.=1;  do N=2  by  2  to terms; a.N=2; end;                     call tell '√3',      cf(1)     /*also:  2∙sin(π/3) *//*══════════════════════════════════════════════════════════════════════════════════════*/a.=2                  /*              ___ */      do N=2  to 17   /*generalized  √ N  */      b.=N-1;                          NN=right(N, 2);          call tell 'gen √'NN, cf(1)      end   /*N*//*══════════════════════════════════════════════════════════════════════════════════════*/a.=2;   b.=-1/2;                                                call tell 'gen √ ½', cf(1)/*══════════════════════════════════════════════════════════════════════════════════════*/  do j=1 for terms; a.j=j;  if j>1   then b.j=a.p; p=j; end;    call tell 'e',       cf(2)/*══════════════════════════════════════════════════════════════════════════════════════*/a.=1;                                                           call tell 'φ, phi',  cf(1)/*══════════════════════════════════════════════════════════════════════════════════════*/a.=1;    do j=1 for terms;  if j//2  then a.j=j;        end;    call tell 'tan(1)',  cf(1)/*══════════════════════════════════════════════════════════════════════════════════════*/         do j=1 for terms;                a.j=2*j+1;    end;    call tell 'coth(1)', cf(1)/*══════════════════════════════════════════════════════════════════════════════════════*/         do j=1 for terms;                a.j=4*j+2;    end;    call tell 'coth(½)', cf(2)    /*also:  [e+1]÷[e-1] *//*══════════════════════════════════════════════════════════════════════════════════════*/                     terms=100000a.=6;    do j=1  for terms;  b.j=(2*j-1)**2;            end;    call tell 'π, pi',   cf(3)exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/cf:      procedure expose a. b. terms;  parse arg C;     !=0;    numeric digits 9+digits()                                          do k=terms  by -1  for terms;  d=a.k+!;  !=b.k/d                                          end   /*k*/         return !+C/*──────────────────────────────────────────────────────────────────────────────────────*/tell:    parse arg ?,v;   $=left(format(v)/1,1+digits()); w=50 /*50 bytes of terms*/ aT=; do k=1; _=space(aT a.k); if length(_)>w then leave; aT=_; end /*k*/ bT=; do k=1; _=space(bT b.k); if length(_)>w then leave; bT=_; end /*k*/ say right(?,8) "="$     '  α terms='aT  ...         if b.1\==1  then say right("",12+digits())     '  ß terms='bT  ...         a=;   b.=1;  return       /*only 50 bytes of  α & ß terms  ↑   are displayed.  */

output

      √2 = 1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641573   α 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 ...
√3 = 1.732050807568877293527446341505872366942805253810380628055806979451933016908800037081146186757248576   α terms=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 = 1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641573   α 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 ...
gen √ 3 = 1.732050807568877293527446341505872366942805253810380628055806979451933016908800037081146186757248576   α 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 ...
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 ...
gen √ 4 = 2                                                                                                       α 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 ...
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 ...
gen √ 5 = 2.236067977499789696409173668731276235440618359611525724270897245410520925637804899414414408378782275   α 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 ...
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 ...
gen √ 6 = 2.449489742783178098197284074705891391965947480656670128432692567250960377457315026539859433104640235   α 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 ...
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 ...
gen √ 7 = 2.645751311064590590501615753639260425710259183082450180368334459201068823230283627760392886474543611   α 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 ...
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 ...
gen √ 8 = 2.828427124746190097603377448419396157139343750753896146353359475981464956924214077700775068655283145   α 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 ...
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 ...
gen √ 9 = 3                                                                                                       α 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 ...
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 ...
gen √10 = 3.162277660168379331998893544432718533719555139325216826857504852792594438639238221344248108379300295   α 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 ...
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 ...
gen √11 = 3.316624790355399849114932736670686683927088545589353597058682146116484642609043846708843399128290651   α 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 ...
ß terms=10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 ...
gen √12 = 3.464101615137754587054892683011744733885610507620761256111613958903866033817600074162292373514497151   α 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 ...
ß terms=11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 ...
gen √13 = 3.605551275463989293119221267470495946251296573845246212710453056227166948293010445204619082018490718   α 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 ...
ß terms=12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 ...
gen √14 = 3.741657386773941385583748732316549301756019807778726946303745467320035156306939027976809895194379572   α 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 ...
ß terms=13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 ...
gen √15 = 3.872983346207416885179265399782399610832921705291590826587573766113483091936979033519287376858673518   α 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 ...
ß terms=14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 ...
gen √16 = 4                                                                                                       α 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 ...
ß terms=15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 ...
gen √17 = 4.123105625617660549821409855974077025147199225373620434398633573094954346337621593587863650810684297   α 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 ...
ß terms=16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 ...
gen √ ½ = 0.707106781186547524400844362104849039284835937688474036588339868995366239231053519425193767163820786   α 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 ...
ß terms=-0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 ...
e = 2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427   α terms=1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ...
φ, phi = 1.618033988749894848204586834365638117720309179805762862135448622705260462818902449707207204189391137   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 ...
tan(1) = 1.557407724654902230506974807458360173087250772381520038383946605698861397151727289555099965202242984   α terms=1 1 3 1 5 1 7 1 9 1 11 1 13 1 15 1 17 1 19 1 21 1 ...
coth(1) = 1.313035285499331303636161246930847832912013941240452655543152967567084270461874382674679241480856303   α terms=3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 ...
coth(½) = 2.163953413738652848770004010218023117093738602150792272533574119296087634783339486574409418809750115   α terms=6 10 14 18 22 26 30 34 38 42 46 50 54 58 62 66 70 ...
π, pi = 3.141592653589792988470143264530440384041017830472772036746332303472711537960073664096818977224037083   α 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 ...


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

### 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)

### 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)

## Ring

 # Project : Continued fraction see "SQR(2) = " + contfrac(1, 1, "2", "1") + nlsee "        e = " + contfrac(2, 1, "n", "n") + nlsee "       PI = " + contfrac(3, 1, "6", "(2*n+1)^2") + nl func contfrac(a0, b1, a, b)        expr = ""        n = 0        while len(expr) < (700 - n)                 n = n + 1                 eval("temp1=" + a)                 eval("temp2=" + b)                 expr = expr + string(temp1) + char(43) + string(temp2) + "/("        end         str = copy(")",n)        eval("temp3=" + expr + "1" + str)        return a0 + b1 / temp3

Output:

SQR(2) = 1.414213562373095
e = 2.718281828459046
PI = 3.141592653588017


## Ruby

require 'bigdecimal' # square root of 2sqrt2 = Object.newdef sqrt2.a(n); n == 1 ? 1 : 2; enddef sqrt2.b(n); 1; end # Napier's constantnapier = Object.newdef napier.a(n); n == 1 ? 2 : n - 1; enddef napier.b(n); n == 1 ? 1 : n - 1; end pi = Object.newdef pi.a(n); n == 1 ? 3 : 6; enddef 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  endend puts estimate(sqrt2, 50).to_s('F')puts estimate(napier, 50).to_s('F')puts estimate(pi, 10).to_s('F')
Output:
$ruby cfrac.rb 1.41421356237309504880168872420969807856967187537695 2.71828182845904523536028747135266249775724709369996 3.1415926536 ## Rust  use std::iter; // Calculating a continued fraction is quite easy with iterators, however// writing a proper iterator adapter is less so. We settle for a macro which// for most purposes works well enough.//// One limitation with this iterator based approach is that we cannot reverse// input iterators since they are not usually DoubleEnded. To circumvent this// we can collect the elements and then reverse them, however this isn't ideal// as we now have to store elements equal to the number of iterations.//// Another is that iterators cannot be resused once consumed, so it is often// required to make many clones of iterators.macro_rules! continued_fraction { ($a:expr, $b:expr ;$iterations:expr) => (        ($a).zip($b)            .take($iterations) .collect::<Vec<_>>().iter() .rev() .fold(0 as f64, |acc: f64, &(x, y)| { x as f64 + (y as f64 / acc) }) ); ($a:expr, $b:expr) => (continued_fraction!($a, $b ; 1000));} fn main() { // Sqrt(2) let sqrt2a = (1..2).chain(iter::repeat(2)); let sqrt2b = iter::repeat(1); println!("{}", continued_fraction!(sqrt2a, sqrt2b)); // Napier's Constant let napiera = (2..3).chain(1..); let napierb = (1..2).chain(1..); println!("{}", continued_fraction!(napiera, napierb)); // Pi let pia = (3..4).chain(iter::repeat(6)); let pib = (1i64..).map(|x| (2 * x - 1).pow(2)); println!("{}", continued_fraction!(pia, pib));}  Output: 1.4142135623730951 2.7182818284590455 3.141592653339042  ## 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 ## Scheme The following code relies on a library implementing SRFI 41 (lazy streams). Most Scheme interpreters include an implementation. #!r6rs(import (rnrs base (6)) (srfi :41 streams)) (define nats (stream-cons 0 (stream-map (lambda (x) (+ x 1)) nats))) (define (build-stream fn) (stream-map fn nats)) (define (stream-cycle s . S) (cond ((stream-null? (car S)) stream-null) (else (stream-cons (stream-car s) (apply stream-cycle (append S (list (stream-cdr s)))))))) (define (cf-floor cf) (stream-car cf))(define (cf-num cf) (stream-car (stream-cdr cf)))(define (cf-denom cf) (stream-cdr (stream-cdr cf))) (define (cf-integer? x) (stream-null? (stream-cdr x))) (define (cf->real x) (let refine ((x x) (n 65536)) (cond ((= n 0) +inf.0) ((cf-integer? x) (cf-floor x)) (else (+ (cf-floor x) (/ (cf-num x) (refine (cf-denom x) (- n 1)))))))) (define (real->cf x) (let-values (((integer-part fractional-part) (div-and-mod x 1))) (if (= fractional-part 0.0) (stream (exact integer-part)) (stream-cons (exact integer-part) (stream-cons 1 (real->cf (/ fractional-part))))))) (define sqrt2 (stream-cons 1 (stream-constant 1 2))) (define napier (stream-append (stream 2 1) (stream-cycle (stream-cdr nats) (stream-cdr nats)))) (define pi (stream-cons 3 (stream-cycle (build-stream (lambda (n) (expt (- (* 2 (+ n 1)) 1) 2))) (stream-constant 6)))) Test: > (cf->real sqrt2)1.4142135623730951> (cf->real napier)2.7182818284590455> (cf->real pi)3.141592653589794 ## Sidef func continued_fraction(a, b, f, n = 1000, r = 1) { f(func (r) { r < n ? (a(r) / (b(r) + __FUNC__(r+1))) : 0 }(r))} var params = Hash( "φ" => [ { 1 }, { 1 }, { 1 + _ } ], "√2" => [ { 1 }, { 2 }, { 1 + _ } ], "e" => [ { _ }, { _ }, { 1 + 1/_ } ], "π" => [ { (2*_ - 1)**2 }, { 6 }, { 3 + _ } ], "τ" => [ { _**2 }, { 2*_ + 1 }, { 8 / (1 + _) } ],) for k in (params.keys.sort) { printf("%2s ≈ %s\n", k, continued_fraction(params{k}...))} Output:  e ≈ 2.7182818284590452353602874713526624977572470937 π ≈ 3.14159265383979292596359650286939597045138933078 τ ≈ 6.28318530717958647692528676655900576839433879875 φ ≈ 1.61803398874989484820458683436563811772030917981 √2 ≈ 1.41421356237309504880168872420969807856967187538  ## 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 pairsproc 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 $a [expr {(2*[incr n]-1)**2}]] set a 6 }} # Continued fraction calculatorproc cf {generator {termCount 50}} { # Get the chunk of terms we want to work with set terms [list [coroutine cf.c$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} # Demonstrationputs [cf r2]puts [cf e]puts [cf pi 250]; # Converges more slowly
Output:
1.4142135623730951
2.7182818284590455
3.1415926373965735

## VBA

Translation of: Phix
Public Const precision = 10000Private Function continued_fraction(steps As Integer, rid_a As String, rid_b As String) As Double    Dim res As Double    res = 0    For n = steps To 1 Step -1       res = Application.Run(rid_b, n) / (Application.Run(rid_a, n) + res)    Next n    continued_fraction = Application.Run(rid_a, 0) + resEnd Function Function sqr2_a(n As Integer) As Integer    sqr2_a = IIf(n = 0, 1, 2)End Function Function sqr2_b(n As Integer) As Integer    sqr2_b = 1End Function Function nap_a(n As Integer) As Integer    nap_a = IIf(n = 0, 2, n)End Function Function nap_b(n As Integer) As Integer    nap_b = IIf(n = 1, 1, n - 1)End Function Function pi_a(n As Integer) As Integer    pi_a = IIf(n = 0, 3, 6)End Function Function pi_b(n As Integer) As Long    pi_b = IIf(n = 1, 1, (2 * n - 1) ^ 2)End Function Public Sub main()    Debug.Print "Precision:", precision    Debug.Print "Sqr(2):", continued_fraction(precision, "sqr2_a", "sqr2_b")    Debug.Print "Napier:", continued_fraction(precision, "nap_a", "nap_b")    Debug.Print "Pi:", continued_fraction(precision, "pi_a", "pi_b")End Sub
Output:
Precision:     10000
Sqr(2):        1,4142135623731
Napier:        2,71828182845905
Pi:            3,14159265358954 

## 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


## 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


## ZX Spectrum Basic

Translation of: BBC_BASIC
10 LET a0=1: LET b1=1: LET a$="2": LET b$="1": PRINT "SQR(2) = ";: GO SUB 100020 LET a0=2: LET b1=1: LET a$="N": LET b$="N": PRINT "e = ";: GO SUB 100030 LET a0=3: LET b1=1: LET a$="6": LET b$="(2*N+1)^2": PRINT "PI = ";: GO SUB 1000100 STOP 1000 LET n=0: LET e$="": LET p$=""1010 LET n=n+11020 LET e$=e$+STR$VAL a$+"+"+STR$VAL b$+"/("1030 IF LEN e$<(4000-n) THEN GO TO 10101035 FOR i=1 TO n: LET p$=p$+")": NEXT i1040 PRINT a0+b1/VAL (e$+"1"+p\$)1050 RETURN