# Continued fraction

(Redirected from 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))

## Action!

INCLUDE "D2:REAL.ACT" ;from the Action! Tool Kit

DEFINE PTR="CARD"
DEFINE JSR="$20" DEFINE RTS="$60"

PROC CoeffA=*(INT n REAL POINTER res)
[JSR $00$00 ;JSR to address set by SetCoeffA
RTS]

PROC CoeffB=*(INT n REAL POINTER res)
[JSR $00$00 ;JSR to address set by SetCoeffB
RTS]

PROC SetCoeffA(PTR p)

RETURN

PROC SetCoeffB(PTR p)

RETURN

PROC Calc(PTR funA,funB INT count REAL POINTER res)
INT i
REAL a,b,tmp

SetCoeffA(funA)
SetCoeffB(funB)

IntToReal(0,res)
i=count
WHILE i>0
DO
CoeffA(i,a)
CoeffB(i,b)
RealDiv(b,tmp,res)
i==-1
OD
CoeffA(0,a)
RealAssign(tmp,res)
RETURN

PROC sqrtA(INT n REAL POINTER res)
IF n>0 THEN
IntToReal(2,res)
ELSE
IntToReal(1,res)
FI
RETURN

PROC sqrtB(INT n REAL POINTER res)
IntToReal(1,res)
RETURN

PROC napierA(INT n REAL POINTER res)
IF n>0 THEN
IntToReal(n,res)
ELSE
IntToReal(2,res)
FI
RETURN

PROC napierB(INT n REAL POINTER res)
IF n>1 THEN
IntToReal(n-1,res)
ELSE
IntToReal(1,res)
FI
RETURN

PROC piA(INT n REAL POINTER res)
IF n>0 THEN
IntToReal(6,res)
ELSE
IntToReal(3,res)
FI
RETURN

PROC piB(INT n REAL POINTER res)
REAL tmp

IntToReal(2*n-1,tmp)
RealMult(tmp,tmp,res)
RETURN

PROC Main()
REAL res

Put(125) PutE() ;clear the screen

Calc(sqrtA,sqrtB,50,res)
Print(" Sqrt2=") PrintRE(res)

Calc(napierA,napierB,50,res)
Print("Napier=") PrintRE(res)

Calc(piA,piB,500,res)
Print("    Pi=") PrintRE(res)
RETURN
Output:
 Sqrt2=1.41421356
Napier=2.71828182
Pi=3.14159265


(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);
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;

with Ada.Text_IO;

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);
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;

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) + result
END;

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


## Arturo

Translation of: Nim
calc: function [f, n][
[a, b, temp]: 0.0

loop n..1 'i [
[a, b]: call f @[i]
temp: b // a + temp
]
[a, b]: call f @[0]
return a + temp
]

sqrt2: function [n][
(n > 0)? -> [2.0, 1.0] -> [1.0, 1.0]
]

napier: function [n][
a: (n > 0)? -> to :floating n -> 2.0
b: (n > 1)? -> to :floating n-1 -> 1.0
@[a, b]
]

Pi: function [n][
a: (n > 0)? -> 6.0 -> 3.0
b: ((2 * to :floating n)-1) ^ 2
@[a, b]
]

print calc 'sqrt2 20
print calc 'napier 15
print calc 'Pi 10000

Output:
1.414213562373095
2.718281828459046
3.141592653589544

## ATS

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

#include
//
(* ****** ****** *)
//
(*
** 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] *)

## 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 concatenation
f_a:=f . "_a",f_b:=f . "_b"

while i>0 {
; You can see two dynamic function calls here
r:=%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 = 1.414214
e = 2.718282
pi = 3.141593


Output with Autohotkey v2 (currently alpha 56):

sqrt 2 = 1.4142135623730951
e = 2.7182818284590455
pi = 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 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 ${\displaystyle \pi }$ 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+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

## 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  ## 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# Translation of: Java using System; using System.Collections.Generic; namespace ContinuedFraction { class Program { static double Calc(Func<int, int[]> f, int n) { double temp = 0.0; for (int ni = n; ni >= 1; ni--) { int[] p = f(ni); temp = p[1] / (p[0] + temp); } return f(0)[0] + temp; } static void Main(string[] args) { List<Func<int, int[]>> fList = new List<Func<int, int[]>>(); fList.Add(n => new int[] { n > 0 ? 2 : 1, 1 }); fList.Add(n => new int[] { n > 0 ? n : 2, n > 1 ? (n - 1) : 1 }); fList.Add(n => new int[] { n > 0 ? 6 : 3, (int) Math.Pow(2 * n - 1, 2) }); foreach (var f in fList) { Console.WriteLine(Calc(f, 200)); } } } }  Output: 1.4142135623731 2.71828182845905 3.14159262280485 ## 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  ## 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));  ## 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

## 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

## dc

[20k 0 200 si [li lbx r li lax + / li 1 - dsi 0<:]ds:x 0 lax +]sf

[[2q]s2[0<2 1]sa[R1]sb]sr                   # sqrt(2)
[[R2q]s2[d 0=2]sa[R1q]s1[d 1=1 1-]sb]se     # e
[[3q]s3[0=3 6]sa[2*1-d*]sb]sp               # pi

c lex lfx p
lrx lfx p
lpx lfx p
Output:
3.14159262280484694855
1.41421356237309504880
2.71828182845904523536

20 decimal places and 200 iterations.

## EasyLang

numfmt 8 0
func calc_sqrt .
n = 100
sum = n
while n >= 1
a = 1
if n > 1
a = 2
.
b = 1
sum = a + b / sum
n -= 1
.
return sum
.
func calc_e .
n = 100
sum = n
while n >= 1
a = 2
if n > 1
a = n - 1
.
b = 1
if n > 1
b = n - 1
.
sum = a + b / sum
n -= 1
.
return sum
.
func calc_pi .
n = 100
sum = n
while n >= 1
a = 3
if n > 1
a = 6
.
b = 2 * n - 1
b *= b
sum = a + b / sum
n -= 1
.
return sum
.
print calc_sqrt
print calc_e
print calc_pi

## Elixir

defmodule CFrac do
def compute([a | _], []), do: a
def compute([a | as], [b | bs]), do: a + b/compute(as, bs)

def sqrt2 do
a = [1 | Stream.cycle([2]) |> Enum.take(1000)]
b = Stream.cycle([1]) |> Enum.take(1000)
IO.puts compute(a, b)
end

def exp1 do
a = [2 | Stream.iterate(1, &(&1 + 1)) |> Enum.take(1000)]
b = [1 | Stream.iterate(1, &(&1 + 1)) |> Enum.take(999)]
IO.puts compute(a, b)
end

def pi do
a = [3 | Stream.cycle([6]) |> Enum.take(1000)]
b = 1..1000 |> Enum.map(fn k -> (2*k - 1)**2 end)
IO.puts compute(a, b)
end
end

Output:
>elixir -e CFrac.sqrt2()
1.4142135623730951

>elixir -e CFrac.exp1()
2.7182818284590455

>elixir -e CFrac.pi()
3.141592653340542


## 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


## 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., 2018
let 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


### Apéry's constant

See Continued fractions for Zeta(2) and Zeta(3) section II. Zeta(3)

let aπ()=let mutable n=0 in (fun ()->n<-n+1;-decimal(pown n 6))
let bπ()=let mutable n=0M in (fun ()->n<-n+1M; (2M*n-1M)*(17M*n*n-17M*n+5M))
cf2S (aπ()) (bπ()) |>Seq.map(fun n->6M/n) |> Seq.take 10 |> Seq.pairwise |> Seq.iter(fun(n,g)->printfn "%1.20f < p < %- 1.20f" (min n g) (max n g));;

Output:
1.20205479452054794521 < p < 1.20205690119184928874
1.20205690119184928874 < p < 1.20205690315781676650
1.20205690315781676650 < p < 1.20205690315959270361
1.20205690315959270361 < p < 1.20205690315959428400
1.20205690315959428400 < p < 1.20205690315959428540
1.20205690315959428540 < p < 1.20205690315959428540
1.20205690315959428540 < p < 1.20205690315959428540
1.20205690315959428540 < p < 1.20205690315959428540
1.20205690315959428540 < p < 1.20205690315959428540


## 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

## Felix

fun pi (n:int) : (double*double) =>
let a = match n with | 0 => 3.0 | _ => 6.0 endmatch in
let b = pow(2.0 * n.double - 1.0, 2.0) in
(a,b);

fun sqrt_2 (n:int) : (double*double) =>
let a = match n with | 0 => 1.0 | _ => 2.0 endmatch in
let b = 1.0 in
(a,b);

fun napier (n:int) : (double*double) =>
let a = match n with | 0 => 2.0 | _ => n.double endmatch in
let b = match n with | 1 => 1.0 | _ => (n.double - 1.0) endmatch in
(a,b);

fun cf_iter (steps:int) (f:int -> double*double)  = {
var acc = 0.0;
for var n in steps downto 0 do
var a, b = f(n);
acc = if (n > 0) then (b / (a + acc)) else (acc + a);
done
return acc;
}

println$cf_iter 200 sqrt_2; // => 1.41421 println$ cf_iter 200 napier; // => 2.71818
println$cf_iter 1000 pi; // => 3.14159  ## 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  ## 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  ## FreeBASIC #define MAX 70000 function sqrt2_a( n as uinteger ) as uinteger return iif(n,2,1) end function function sqrt2_b( n as uinteger ) as uinteger return 1 end function function napi_a( n as uinteger ) as uinteger return iif(n,n,2) end function function napi_b( n as uinteger ) as uinteger return iif(n>1,n-1,1) end function function pi_a( n as uinteger ) as uinteger return iif(n,6,3) end function function pi_b( n as uinteger ) as uinteger return (2*n-1)^2 end function function calc_contfrac( an as function (as uinteger) as uinteger, bn as function (as uinteger) as uinteger, byval iter as uinteger ) as double dim as double r dim as integer i for i = iter to 1 step -1 r = bn(i)/(an(i)+r) next i return an(0)+r end function print calc_contfrac( @sqrt2_a, @sqrt2_b, MAX ) print calc_contfrac( @napi_a, @napi_b, MAX ) print calc_contfrac( @pi_a, @pi_b, MAX )  ## Fōrmulæ Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition. Programs in Fōrmulæ are created/edited online in its website. In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation. Solution The following function definition creates a continued fraction: The function accepts the following parameters: Parameter Description a₀ Value for a₀ λa Lambda expression to define aᵢ λb Lambda expression to define bᵢ depth Depth to calculate the continued fraction Since Fōrmulæ arithmetic simplifies numeric results as they are generated, the result is not a continued fraction by default. If we want to create the structure, we can introduce the parameters as string or text expressions (or lambda expressions that produce them). Because string or text expressions are not reduced when they are operands of additions and divisions, the structure is preserved, such as follows: Case 1. ${\displaystyle {\sqrt {2}}}$ In this case • a₀ is 1 • λa is n ↦ 2 • λb is n ↦ 1 Let us show the results as a table, for several levels of depth (1 to 10). The columns are: • The depth • The "textual" call, in order to generate the structure • The normal (numeric) call. Since arithmetic operations are exact by default, it is usually a rational number. • The value of the normal (numeric) call, forced to be shown as a decimal number, by using the Math.Numeric expression (the N(x) expression) Case 2. ${\displaystyle e}$ In this case • a₀ is 2 • λa is n ↦ n • λb is n ↦ 1 if n = 1, n - 1 elsewhere Case 3. ${\displaystyle \pi }$ In this case • a₀ is 3 • λa is n ↦ 6 • λb is n ↦ 2(n - 1)² ## 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  ## Groovy Translation of: Java import java.util.function.Function import static java.lang.Math.pow 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 } static void main(String[] args) { List<Function<Integer, Integer[]>> fList = new ArrayList<>() fList.add({ n -> [n > 0 ? 2 : 1, 1] }) fList.add({ n -> [n > 0 ? n : 2, n > 1 ? (n - 1) : 1] }) fList.add({ n -> [n > 0 ? 6 : 3, (int) pow(2 * n - 1, 2)] }) for (Function<Integer, Integer[]> f : fList) System.out.println(calc(f, 200)) } }  Output: 1.4142135623730951 2.7182818284590455 3.141592622804847 ## 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 ..]) ((^ 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 ((%), denominator, numerator)
import Data.Bool (bool)

-- 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)) . (2 ^)) [0 ..] where f ((x, y):ys) | abs (x - y) < (1 / fromIntegral p) = x | otherwise = 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 :: IO () main = mapM_ putStrLn [cf2dec 200 sqrt2, cf2dec 200 napier, cf2dec 200 pie]  ## Icon $define EVAL_DEPTH 100

# A generalized continued fraction, represented by two functions. Each
# function maps from an index to a floating-point value.
record continued_fraction (a, b)

procedure main ()
writes (" sqrt 2.0 = ")
write (evaluate_continued_fraction (continued_fraction (sqrt2_a, sqrt2_b),
EVAL_DEPTH))
writes ("        e = ")
write (evaluate_continued_fraction (continued_fraction (e_a, e_b),
EVAL_DEPTH))
writes ("       pi = ")
write (evaluate_continued_fraction (continued_fraction (pi_a, pi_b),
EVAL_DEPTH))
end

procedure evaluate_continued_fraction (frac, depth)
local i, retval
retval := frac.a (depth)
every i := depth to 1 by -1 do {
retval := frac.a (i - 1) + (frac.b (i) / retval)
}
return retval
end

procedure sqrt2_a (i)
return (if i = 0 then 1.0 else 2.0)
end

procedure sqrt2_b (i)
return 1.0
end

procedure e_a (i)
return (if i = 0 then 2.0 else real (i))
end

procedure e_b (i)
return (if i = 1 then 1.0 else real (i - 1))
end

procedure pi_a (i)
return (if i = 0 then 3.0 else 6.0)
end

procedure pi_b (i)
return real (((2 * i) - 1)^2)
end

Output:
$icon continued-fraction-task.icn sqrt 2.0 = 1.414213562 e = 2.718281828 pi = 3.141592411  ## 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


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.jq
Value  :        Direct      : Continued Fraction
2|sqrt : 1.4142135623730951 : 1.4142135623730951
1|exp  : 2.718281828459045  : 2.7182818284590455
pi     : 3.141592653589793  : 3.1415926535892935 (1e-12)


## Julia

Works with: Julia version 1.8.5

High performant lazy evaluation on demand with Julias iterators.

using .Iterators: countfrom, flatten, repeated, zip
using .MathConstants: ℯ
using Printf

function cf(a₀, a, b = repeated(1))
m = BigInt[a₀ 1; 1 0]
for (aᵢ, bᵢ) ∈ zip(a, b)
m *= [aᵢ 1; bᵢ 0]
isapprox(m[1]/m[2], m[3]/m[4]; atol = 1e-12) && break
end
m[1]/m[2]
end

out((k, v)) = @printf "%2s: %.12f ≈ %.12f\n" k v eval(k)

foreach(out, (
:(√2) => cf(1, repeated(2)),
:ℯ    => cf(2, countfrom(), flatten((1, countfrom()))),
:π    => cf(3, repeated(6), (k^2 for k ∈ countfrom(1, 2)))))

Output:
√2: 1.414213562373 ≈ 1.414213562373
ℯ: 2.718281828459 ≈ 2.718281828459
π: 3.141592653590 ≈ 3.141592653590

## 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


## Lambdatalk

{def gcf
{def gcf.rec
{lambda {:f :n :r}
{if {< :n 1}
then {+ {car {:f 0}} :r}
else {gcf.rec :f
{- :n 1}
{let { {:r :r}
{:ab {:f :n}}
} {/ {cdr :ab}
{+ {car :ab} :r}} }}}}}
{lambda {:f :n}
{gcf.rec :f :n 0}}}

{def phi
{lambda {:n}
{cons 1 1}}}

{gcf phi 50}
-> 1.618033988749895

{def sqrt2
{lambda {:n}
{cons {if {> :n 0} then 2 else 1} 1}}}

{gcf sqrt2 25}
-> 1.4142135623730951

{def napier
{lambda {:n}
{cons {if {> :n 0} then :n else 2} {if {> :n 1} then {- :n 1} else 1} }}}

{gcf napier 20}
-> 2.7182818284590455

{def fpi
{lambda {:n}
{cons {if {> :n 0} then 6 else 3} {pow {- {* 2 :n} 1} 2} }}}

{gcf fpi 500}
-> 3.1415926 516017554
// only 8 exact decimals for 500 iterations
// A very very slow convergence.
// Here is a quicker version without any obvious pattern

{def pi
{lambda {:n}
{cons {A.get :n {A.new 3 7 15 1 292 1 1 1 2 1 3 1 14 2 1 1}} 1}}}

{gcf pi 15}
-> 3.1415926 53589793

// Much quicker, 15 exact decimals after 15 iterations


## Lua

Translation of: C
function calc(fa, fb, expansions)
local a = 0.0
local b = 0.0
local r = 0.0
local i = expansions
while i > 0 do
a = fa(i)
b = fb(i)
r = b / (a + r)
i = i - 1
end
a = fa(0)
return a + r
end

function sqrt2a(n)
if n ~= 0 then
return 2.0
else
return 1.0
end
end

function sqrt2b(n)
return 1.0
end

function napiera(n)
if n ~= 0 then
return n
else
return 2.0
end
end

function napierb(n)
if n > 1.0 then
return n - 1.0
else
return 1.0
end
end

function pia(n)
if n ~= 0 then
return 6.0
else
return 3.0
end
end

function pib(n)
local c = 2.0 * n - 1.0
return c * c
end

function main()
local sqrt2  = calc(sqrt2a,  sqrt2b,  1000)
local napier = calc(napiera, napierb, 1000)
local pi     = calc(pia,     pib,     1000)
print(sqrt2)
print(napier)
print(pi)
end

main()

Output:
1.4142135623731
2.718281828459
3.1415926533405

## Maple

contfrac:=n->evalf(Value(NumberTheory:-ContinuedFraction(n)));
contfrac(2^(0.5));
contfrac(Pi);
contfrac(exp(1));

## Mathematica / Wolfram Language

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


## 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

## Nim

proc calc(f: proc(n: int): tuple[a, b: float], n: int): float =
var a, b, temp = 0.0
for i in countdown(n, 1):
(a, b) = f(i)
temp = b / (a + temp)
(a, b) = f(0)
a + temp

proc sqrt2(n: int): tuple[a, b: float] =
if n > 0:
(2.0, 1.0)
else:
(1.0, 1.0)

proc napier(n: int): tuple[a, b: float] =
let a = if n > 0: float(n) else: 2.0
let b = if n > 1: float(n - 1) else: 1.0
(a, b)

proc pi(n: int): tuple[a, b: float] =
let a = if n > 0: 6.0 else: 3.0
let b = (2 * float(n) - 1) * (2 * float(n) - 1)
(a, b)

echo calc(sqrt2, 20)
echo calc(napier, 15)
echo calc(pi, 10000)

Output:
1.414213562373095
2.718281828459046
3.141592653589544

## 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

## Pascal

This console application is written in Delphi, which allows the results to be displayed to 17 correct decimal places (Free Pascal seems to allow only 16). As in the jq solution, we aim to work forwards and stop as soon the desired precision has been reached, rather than guess a suitable number of terms and work backwards. In this program, the continued fraction is converted to an infinite sum, each term after the first being the difference between consecutive convergents. The convergence for pi is very slow (as others have noted) so as well as the c.f. in the task description an alternative is given from the Wikipedia article "Continued fraction".

program ContFrac_console;

{$APPTYPE CONSOLE} uses SysUtils; type TCoeffFunction = function( n : integer) : extended; // Calculate continued fraction as a sum, working forwards. // Stop on reaching a term with absolute value less than epsilon, // or on reaching the maximum number of terms. procedure CalcContFrac( a, b : TCoeffFunction; epsilon : extended; maxNrTerms : integer = 1000); // optional, with default var n : integer; sum, term, u, v : extended; whyStopped : string; begin sum := a(0); term := b(1)/a(1); v := a(1); n := 1; repeat sum := sum + term; inc(n); u := v; v := a(n) + b(n)/u; term := -term * b(n)/(u*v); until (Abs(term) < epsilon) or (n >= maxNrTerms); if n >= maxNrTerms then whyStopped := 'too many terms' else whyStopped := 'converged'; WriteLn( SysUtils.Format( '%21.17f after %d terms (%s)', [sum, n, whyStopped])); end; //---------------- a and b for sqrt(2) ---------------- function a_sqrt2( n : integer) : extended; begin if n = 0 then result := 1 else result := 2; end; function b_sqrt2( n : integer) : extended; begin result := 1; end; //---------------- a snd b for e ---------------- function a_e( n : integer) : extended; begin if n = 0 then result := 2 else result := n; end; function b_e( n : integer) : extended; begin if n = 1 then result := 1 else result := n - 1; end; //-------- Rosetta Code a and b for pi -------- function a_pi( n : integer) : extended; begin if n = 0 then result := 3 else result := 6; end; function b_pi( n : integer) : extended; var temp : extended; begin temp := 2*n - 1; result := temp*temp; end; //-------- More efficient a and b for pi -------- function a_pi_alt( n : integer) : extended; begin if n = 0 then result := 0 else result := 2*n - 1; end; function b_pi_alt( n : integer) : extended; var temp : extended; begin if n = 1 then result := 4 else begin temp := n - 1; result := temp*temp; end; end; //---------------- Main routine ---------------- // Unlike Free Pascal, Delphi does not require // an @ sign before the function names. begin WriteLn( 'sqrt(2)'); CalcContFrac( a_sqrt2, b_sqrt2, 1E-20); WriteLn( 'e'); CalcContFrac( a_e, b_e, 1E-20); WriteLn( 'pi'); CalcContFrac( a_pi, b_pi, 1E-20); WriteLn( 'pi (alternative formula)'); CalcContFrac( a_pi_alt, b_pi_alt, 1E-20); end.  Output: sqrt(2) 1.41421356237309505 after 27 terms (converged) e 2.71828182845904524 after 20 terms (converged) pi 3.14159265383979293 after 1000 terms (too many terms) pi (alternative formula) 3.14159265358979324 after 29 terms (converged)  ## Perl Use closures to implement the infinite lists of coeffficients. use strict; use warnings; no warnings 'recursion'; use experimental 'signatures'; sub continued_fraction ($a, $b,$n = 100) {
$a->() + ($n and $b->() / continued_fraction($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++ or 2     } },  do { my $n; sub {$n++ or 1       } };
printf "π   ≈ %.9f\n", continued_fraction do { my $n; sub {$n++ ? 6 : 3  } },  do { my $n; sub { (2*$n++ + 1)**2 } }, 1000;
printf "π/2 ≈ %.9f\n", continued_fraction do { my $n; sub { 1/($n++ or 1) } },              sub { 1                 }, 1000;

Output:
√2  ≈ 1.414213562
e   ≈ 2.718281828
π   ≈ 3.141592653
π/2 ≈ 1.570717797

## Phix

with javascript_semantics
constant precision = 10000

function continued_fraction(integer f, steps=precision)
atom a, b, res = 0
for n=steps to 1 by -1 do
{a, b} = f(n)
res := b / (a + res)
end for
{a} = f(0)
return a + res
end function

function sqr2(integer n) return {iff(n=0?1:2),1} end function

function nap(integer n) return {iff(n=0?2:n),iff(n=1?1:n-1)} end function

function pi(integer n) return {iff(n=0?3:6),power(2*n-1,2)} end function

printf(1,"Precision: %d\n", {precision})
printf(1,"Sqr(2):    %.10g\n", {continued_fraction(sqr2)})
printf(1,"Napier:    %.10g\n", {continued_fraction(nap)})
printf(1,"Pi:        %.10g\n", {continued_fraction(pi)})

Output:
Precision: 10000
Sqr(2):    1.414213562
Napier:    2.718281828
Pi:        3.141592654


## Picat

For Pi a test is added with a higher precision (200 -> 2000) to get a better result.

### Recursion

Translation of: Prolog
go =>

% square root 2
continued_fraction(200, sqrt_2_ab, V1),
printf("sqrt(2) = %w (diff: %0.15f)\n", V1, V1-sqrt(2)),

% napier
continued_fraction(200, napier_ab, V2),
printf("e       = %w (diff: %0.15f)\n", V2, V2-math.e),

% pi
continued_fraction(200, pi_ab, V3),
printf("pi      = %w (diff: %0.15f)\n", V3, V3-math.pi),
% get a better precision
continued_fraction(20000, pi_ab, V3b),
printf("pi      = %w (diff: %0.15f)\n", V3b, V3b-math.pi),
nl.

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 = A + Temp.

continued_fraction(N, Compute_ab, Tmp, V) =>
call(Compute_ab, N, A, B),
Tmp1 = B / (A + Tmp),
N1 = N - 1,
continued_fraction(N1, Compute_ab, Tmp1, V).

% 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:
sqrt(2) = 1.414213562373095 (diff: 0.000000000000000)
e       = 2.718281828459046 (diff: 0.000000000000000)
pi      = 3.141592622804847 (diff: -0.000000030784946)
pi      = 3.141592653589762 (diff: -0.000000000000031)


### Iterative

Translation of: Python

(from Python's Fast Iterative version)

continued_fraction_it(Fun, N) = Ret =>
Temp = 0.0,
foreach(I in N..-1..1)
[A,B] = apply(Fun,I),
Temp := B / (A + Temp)
end,
F = apply(Fun,0),
Ret = F[1] + Temp.

fsqrt2(N)  = [cond(N > 0, 2, 1),1].
fnapier(N) = [cond(N > 0, N,2), cond(N>1,N-1,1)].
fpi(N)     = [cond(N>0,6,3), (2*N-1) ** 2].

Which has exactly the same output as the recursive version.

## PicoLisp

(scl 49)
(de fsqrt2 (N A)
(default A 1)
(cond
((> A (inc N)) 2)
(T
(+
(if (=1 A) 1.0 2.0)
(*/ (* 1.0 1.0) (fsqrt2 N (inc A))) ) ) ) )
(de pi (N A)
(default A 1)
(cond
((> A (inc N)) 6.0)
(T
(+
(if (=1 A) 3.0 6.0)
(*/
(* (** (dec (* 2 A)) 2) 1.0)
1.0
(pi N (inc A)) ) ) ) ) )
(de napier (N A)
(default A 0)
(cond
((> A N) (* A 1.0))
(T
(+
(if (=0 A) 2.0 (* A 1.0))
(*/
(if (> 1 A) 1.0 (* A 1.0))
1.0
(napier N (inc A)) ) ) ) ) )
(prinl (format (fsqrt2 200) *Scl))
(prinl (format (napier 200) *Scl))
(prinl (format (pi 200) *Scl))
Output:
1.4142135623730950488016887242096980785696718753770
2.7182818284590452353602874713526624977572470937000
3.1415926839198062649342019294083175420335002640134


## 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 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 .


## 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


### 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

## Quackery

  [ $"bigrat.qky" loadfile ] now! [ 1 min [ table [ 1 1 ] [ 2 1 ] ] do ] is sqrt2 ( n --> n/d ) [ dup 2 min [ table [ drop 2 1 ] [ 1 ] [ dup 1 - ] ] do ] is napier ( n --> n/d ) [ dup 1 min [ table [ drop 3 1 ] [ 2 * 1 - dup * 6 swap ] ] do ] is pi ( n --> n/d ) [ ]'[ temp put 0 1 rot times [ i 1+ temp share do v+ 1/v ] 0 temp take do v+ ] is cf ( n --> n/d ) 1000 cf sqrt2 10 point$ echo$cr 1000 cf napier 10 point$ echo$cr 1000 cf pi 10 point$ echo$cr Output: 1.4142135624 2.7136688544 3.1413776152 ## 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: 1.4142135623730951 2.7182818284590455 3.1415926839198063  ### 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)  ## Raku (formerly 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 ${\displaystyle f_{n}(x)=a_{n}+{\frac {b_{n}}{x}}}$

Raku 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

## 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=500
if  digs=='' |  digs==","  then  digs=100
numeric 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=100000
a.=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") + nl
see "        e = " + contfrac(2, 1, "n", "n") + nl
see "       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


## RPL

This task demonstrates how both global and local variables, arithmetic expressions and stack can be used together to build a compact and versatile piece of code.

Works with: Halcyon Calc version 4.2.7
RPL code Comment
≪
4 ROLL ROT → an bn
≪ 'N' STO 0 WHILE N 2 ≥ REPEAT
an + INV bn * EVAL
'N' 1 STO- END
an + / + EVAL
'N' PURGE
≫ ≫ ‘→CFRAC’ STO

→CFRAC ( a0 an b1 bn N -- x )
Unstack an and bn
Loop from N to 2
Calculate Nth fraction
Decrement counter
Calculate last fraction with b1 in stack, then add a0
Discard N variable - not mandatory but hygienic


Input:
1 2 1 1 100 →CFRAC
2 'N' 1 'N-1' 100 →CFRAC
3 6 1 '(2*N-1)^2' 1000 →CFRAC
1 1 1 1 100 →CFRAC
1 '2-MOD(N,2)' 1 1 100 →CFRAC

Output:
5: 1.41421356237
4: 2.71828182846
3: 3.14159265334
2: 1.61803398875
1: 1.73205080757


## 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:
$ruby cfrac.rb 1.41421356237309504880168872420969807856967187537695 2.71828182845904523536028747135266249775724709369996 3.1415926536 ## Rust use std::iter; // Calculating a continued fraction is quite easy with iterators, however // writing a proper iterator adapter is less so. We settle for a macro which // for most purposes works well enough. // // One limitation with this iterator based approach is that we cannot reverse // input iterators since they are not usually DoubleEnded. To circumvent this // we can collect the elements and then reverse them, however this isn't ideal // as we now have to store elements equal to the number of iterations. // // Another is that iterators cannot be resused once consumed, so it is often // required to make many clones of iterators. macro_rules! continued_fraction { ($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  ## Swift Translation of: Rust extension BinaryInteger { @inlinable public func power(_ n: Self) -> Self { return stride(from: 0, to: n, by: 1).lazy.map({_ in self }).reduce(1, *) } } public struct CycledSequence<WrappedSequence: Sequence> { private var seq: WrappedSequence private var iter: WrappedSequence.Iterator init(seq: WrappedSequence) { self.seq = seq self.iter = seq.makeIterator() } } extension CycledSequence: Sequence, IteratorProtocol { public mutating func next() -> WrappedSequence.Element? { if let ele = iter.next() { return ele } else { iter = seq.makeIterator() return iter.next() } } } extension Sequence { public func cycled() -> CycledSequence<Self> { return CycledSequence(seq: self) } } public struct ChainedSequence<Element> { private var sequences: [AnySequence<Element>] private var iter: AnyIterator<Element> private var curSeq = 0 init(chain: ChainedSequence) { self.sequences = chain.sequences self.iter = chain.iter self.curSeq = chain.curSeq } init<Seq: Sequence>(_ seq: Seq) where Seq.Element == Element { sequences = [AnySequence(seq)] iter = sequences[curSeq].makeIterator() } func chained<Seq: Sequence>(with seq: Seq) -> ChainedSequence where Seq.Element == Element { var res = ChainedSequence(chain: self) res.sequences.append(AnySequence(seq)) return res } } extension ChainedSequence: Sequence, IteratorProtocol { public mutating func next() -> Element? { if let el = iter.next() { return el } curSeq += 1 guard curSeq != sequences.endIndex else { return nil } iter = sequences[curSeq].makeIterator() return iter.next() } } extension Sequence { public func chained<Seq: Sequence>(with other: Seq) -> ChainedSequence<Element> where Seq.Element == Element { return ChainedSequence(self).chained(with: other) } } func continuedFraction<T: Sequence, V: Sequence>( _ seq1: T, _ seq2: V, iterations: Int = 1000 ) -> Double where T.Element: BinaryInteger, T.Element == V.Element { return zip(seq1, seq2).prefix(iterations).reversed().reduce(0.0, { Double($1.0) + (Double($1.1) /$0) })
}

let sqrtA = [1].chained(with: [2].cycled())
let sqrtB = [1].cycled()

print("√2 ≈ \(continuedFraction(sqrtA, sqrtB))")

let napierA = [2].chained(with: 1...)
let napierB = [1].chained(with: 1...)

print("e ≈ \(continuedFraction(napierA, napierB))")

let piA = [3].chained(with: [6].cycled())
let piB = (1...).lazy.map({ (2 * $0 - 1).power(2) }) print("π ≈ \(continuedFraction(piA, piB))")  Output: √2 ≈ 1.4142135623730951 e ≈ 2.7182818284590455 π ≈ 3.141592653339042 Translation of: Java import Foundation func calculate(n: Int, operation: (Int) -> [Int])-> Double { var tmp: Double = 0 for ni in stride(from: n, to:0, by: -1) { var p = operation(ni) tmp = Double(p[1])/(Double(p[0]) + tmp); } return Double(operation(0)[0]) + tmp; } func sqrt (n: Int) -> [Int] { return [n > 0 ? 2 : 1, 1] } func napier (n: Int) -> [Int] { var res = [n > 0 ? n : 2, n > 1 ? (n - 1) : 1] return res } func pi(n: Int) -> [Int] { var res = [n > 0 ? 6 : 3, Int(pow(Double(2 * n - 1), 2))] return res } print (calculate(n: 200, operation: sqrt)); print (calculate(n: 200, operation: napier)); print (calculate(n: 200, operation: pi));  ## 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

## VBA

Translation of: Phix
Public Const precision = 10000
Private 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) + res
End 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 = 1
End 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 

## Visual Basic .NET

Translation of: C#
Module Module1
Function Calc(f As Func(Of Integer, Integer()), n As Integer) As Double
Dim temp = 0.0
For ni = n To 1 Step -1
Dim p = f(ni)
temp = p(1) / (p(0) + temp)
Next
Return f(0)(0) + temp
End Function

Sub Main()
Dim fList = {
Function(n As Integer) New Integer() {If(n > 0, 2, 1), 1},
Function(n As Integer) New Integer() {If(n > 0, n, 2), If(n > 1, n - 1, 1)},
Function(n As Integer) New Integer() {If(n > 0, 6, 3), Math.Pow(2 * n - 1, 2)}
}

For Each f In fList
Console.WriteLine(Calc(f, 200))
Next
End Sub

End Module

Output:
1.4142135623731
2.71828182845905
3.14159262280485

## Uiua

# Evaluates some interesting continued fractions.

Cfrac! ← +⊢^!0∧(÷:⊙+:°⊟)⊙0≡^!^.+1⇌⇡
Fsqrt₂ ← [⊃(+1>0)⋅1]
Fe ← [⊃(⨬⋅2∘>0.|⨬⋅1-1>1.)]
Fpi ← [⊃(×3+1>0|ⁿ2-1×2)]
&p$"√2 = _"Cfrac!Fsqrt₂ 200 &p$"e  = _"Cfrac!Fe 200
&p$"π = _"Cfrac!Fpi 200 Output: √2 = 1.4142135623730951 e = 2.7182818284590455 π = 3.141592622804847  ## Wren Translation of: D var calc = Fn.new { |f, n| var t = 0 for (i in n..1) { var p = f.call(i) t = p[1] / (p[0] + t) } return f.call(0)[0] + t } var pList = [ ["sqrt(2)", Fn.new { |n| [(n > 0) ? 2 : 1, 1] }], ["e ", Fn.new { |n| [(n > 0) ? n : 2, (n > 1) ? n - 1 : 1] }], ["pi ", Fn.new { |n| [(n > 0) ? 6 : 3, (2*n - 1) * (2*n - 1)] }] ] for (p in pList) System.print("%(p[0]) = %(calc.call(p[1], 200))")  Output: sqrt(2) = 1.4142135623731 e = 2.718281828459 pi = 3.1415926228048  ## 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$="1": PRINT "SQR(2) = ";: GO SUB 1000 20 LET a0=2: LET b1=1: LET a$="N": LET b$="N": PRINT "e = ";: GO SUB 1000 30 LET a0=3: LET b1=1: LET a$="6": LET b$="(2*N+1)^2": PRINT "PI = ";: GO SUB 1000 100 STOP 1000 LET n=0: LET e$="": LET p$="" 1010 LET n=n+1 1020 LET e$=e$+STR$ VAL a$+"+"+STR$ VAL b$+"/(" 1030 IF LEN e$<(4000-n) THEN GO TO 1010
1035 FOR i=1 TO n: LET p$=p$+")": NEXT i
1040 PRINT a0+b1/VAL (e$+"1"+p$)
1050 RETURN
`