⚠️ Warning: This is a draft ⚠️

This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.

If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.

{{task}}A number may be represented as a [[wp:Continued fraction|continued fraction]] (see [http://mathworld.wolfram.com/ContinuedFraction.html Mathworld] for more information) as follows:

:$a_0 + \cfrac\left\{b_1\right\}\left\{a_1 + \cfrac\left\{b_2\right\}\left\{a_2 + \cfrac\left\{b_3\right\}\left\{a_3 + \ddots\right\}\right\}\right\}$

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 $a_0 = 1$ then $a_N = 2$. $b_N$ is always $1$.

:$\sqrt\left\{2\right\} = 1 + \cfrac\left\{1\right\}\left\{2 + \cfrac\left\{1\right\}\left\{2 + \cfrac\left\{1\right\}\left\{2 + \ddots\right\}\right\}\right\}$

For Napier's Constant, use $a_0 = 2$, then $a_N = N$. $b_1 = 1$ then $b_N = N-1$.

:$e = 2 + \cfrac\left\{1\right\}\left\{1 + \cfrac\left\{1\right\}\left\{2 + \cfrac\left\{2\right\}\left\{3 + \cfrac\left\{3\right\}\left\{4 + \ddots\right\}\right\}\right\}\right\}$

For Pi, use $a_0 = 3$ then $a_N = 6$. $b_N = \left(2N-1\right)^2$.

:$\pi = 3 + \cfrac\left\{1\right\}\left\{6 + \cfrac\left\{9\right\}\left\{6 + \cfrac\left\{25\right\}\left\{6 + \ddots\right\}\right\}\right\}$

• [[Continued fraction/Arithmetic]] for tasks that do arithmetic over continued fractions.

## 11l

F calc(f_a, f_b, =n = 1000)
V r = 0.0
L n > 0
r = f_b(n) / (f_a(n) + r)
n--
R f_a(0) + r

print(calc(n -> I n > 0 {2} E 1, n -> 1))
print(calc(n -> I n > 0 {n} E 2, n -> I n > 1 {n - 1} E 1))
print(calc(n -> I n > 0 {6} E 3, n -> (2 * n - 1) ^ 2))


(The source text for these examples can also be found on [https://bitbucket.org/ada_on_rosetta_code/solutions 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;


{{out}}

 1.41421356237310
2.71828182845905
3.14159265358954


## ALGOL 68

{{works with|Algol 68 Genie|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)))



{{out}}


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



## ATS

A fairly direct translation of the [[#C|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 $\pi$ 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

{{works with|BBC BASIC for Windows}}

      *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, ")"))  {{out}}  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; }  {{out}}  1.414213562 2.718281828 3.141592653  ## 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 }  {{out}}  1.41421356237309 2.71828182845905 3.14159265358954  ## C# {{trans|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)); } } } }  {{out}} 1.4142135623731 2.71828182845905 3.14159262280485  ## 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))  {{out}}  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.  {{out}} 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)


{{out}}


> coffee continued_fraction.coffee
1.4142135623730951
1.4142135623730951
---
2.718281828459045
2.7182818284590455
---
3.141592653589793
3.141592653589793
---



## Common Lisp

{{trans|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))


{{out}}

sqrt(2) = 1.4142135623730947d0
napier's = 2.7182818284590464d0
pi = 3.141592653589543d0


## Chapel

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

proc calc(f, n) {
var r = 0.0;

for k in 1..n by -1 {
var v = f.pair(k);
r = v(2) / (v(1) + r);
}

return f.pair(0)(1) + r;
}

record Sqrt2 {
proc pair(n) {
return (if n == 0 then 1 else 2,
1);
}
}

record Napier {
proc pair(n) {
return (if n == 0 then 2 else n,
if n == 1 then 1 else n - 1);
}
}
record Pi {
proc pair(n) {
return (if n == 0 then 3 else 6,
(2*n - 1)**2);
}
}

config const n = 200;
writeln(calc(new Sqrt2(), n));
writeln(calc(new Napier(), n));
writeln(calc(new Pi(), n));


## D

import std.stdio, std.functional, std.traits;

FP calc(FP, F)(in F fun, in int n) pure nothrow if (isCallable!F) {
FP temp = 0;

foreach_reverse (immutable ni; 1 .. n + 1) {
immutable p = fun(ni);
temp = p[1] / (FP(p[0]) + temp);
}
return fun(0)[0] + temp;
}

int[2] fSqrt2(in int n) pure nothrow {
return [n > 0 ? 2 : 1,   1];
}

int[2] fNapier(in int n) pure nothrow {
return [n > 0 ? n : 2,   n > 1 ? (n - 1) : 1];
}

int[2] fPi(in int n) pure nothrow {
return [n > 0 ? 6 : 3,   (2 * n - 1) ^^ 2];
}

alias print = curry!(writefln, "%.19f");

void main() {
calc!real(&fSqrt2, 200).print;
calc!real(&fNapier, 200).print;
calc!real(&fPi, 200).print;
}


{{out}}

1.4142135623730950487
2.7182818284590452354
3.1415926228048469486


## Erlang


-module(continued).
-compile([export_all]).

pi_a (0) -> 3;
pi_a (_N) -> 6.

pi_b (N) ->
(2*N-1)*(2*N-1).

sqrt2_a (0) ->
1;
sqrt2_a (_N) ->
2.

sqrt2_b (_N) ->
1.

nappier_a (0) ->
2;
nappier_a (N) ->
N.

nappier_b (1) ->
1;
nappier_b (N) ->
N-1.

continued_fraction(FA,_FB,0) -> FA(0);
continued_fraction(FA,FB,N) ->
continued_fraction(FA,FB,N-1,FB(N)/FA(N)).

continued_fraction(FA,_FB,0,Acc) -> FA(0) + Acc;
continued_fraction(FA,FB,N,Acc) ->
continued_fraction(FA,FB,N-1,FB(N)/ (FA(N) + Acc)).

test_pi (N) ->
continued_fraction(fun pi_a/1,fun pi_b/1,N).

test_sqrt2 (N) ->
continued_fraction(fun sqrt2_a/1,fun sqrt2_b/1,N).

test_nappier (N) ->
continued_fraction(fun nappier_a/1,fun nappier_b/1,N).



{{out}}


29> continued:test_pi(1000).
3.141592653340542
30> continued:test_sqrt2(1000).
1.4142135623730951
31> continued:test_nappier(1000).
2.7182818284590455



### 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))



{{out}}


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")



{{out}}


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))



{{out}}


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))



{{out}}


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))



{{out}}


2.66666666666667 < e < 3.00000000000000
2.66666666666667 < e < 2.72727272727273
2.71698113207547 < e < 2.72727272727273
2.71698113207547 < e < 2.71844660194175
2.71826333176026 < e < 2.71844660194175
2.71826333176026 < e < 2.71828369389345
2.71828165766640 < e < 2.71828369389345
2.71828165766640 < e < 2.71828184277783
2.71828182735187 < e < 2.71828184277783



## Factor

''cfrac-estimate'' uses [[Arithmetic/Rational|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


{{out}}

 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  =={{header|Fōrmulæ}}== In [http://wiki.formulae.org/Continued_fraction this] page you can see the solution of this task. Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text ([http://wiki.formulae.org/Editing_F%C5%8Drmul%C3%A6_expressions more info]). 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 transportation effects more than visualization and edition. The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code. ## Forth {{trans|D}} 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 ;  {{out}}  ' 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  {{out}}  1.4142135623730951 2.7182818284590455 3.1415926535895435  ## 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()) }  {{out}}  sqrt2: 1.4142135623730965 nap: 2.7182818284590455 pi: 3.141623806667839  ## Haskell import Data.List (unfoldr) import Data.Char (intToDigit) -- continued fraction represented as a (possibly infinite) list of pairs sqrt2, napier, myPi :: [(Integer, Integer)] sqrt2 = zip (1 : [2,2..]) [1,1..] napier = zip (2 : [1..]) (1 : [1..]) myPi = zip (3 : [6,6..]) (map (^2) [1,3..]) -- approximate a continued fraction after certain number of iterations approxCF :: (Integral a, Fractional b) => Int -> [(a, a)] -> b approxCF t = foldr (\(a,b) z -> fromIntegral a + fromIntegral b / z) 1 . take t -- infinite decimal representation of a real number decString :: RealFrac a => a -> String decString frac = show i ++ '.' : decString' f where (i,f) = properFraction frac decString' = map intToDigit . unfoldr (Just . properFraction . (10*)) main :: IO () main = mapM_ (putStrLn . take 200 . decString . (approxCF 950 :: [(Integer, Integer)] -> Rational)) [sqrt2, napier, myPi]  {{out}}  1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641572735013846230912297024924836055850737212644121497099935831413222665927505592755799950501152782060571 2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921817413596629043572900334295260595630738132328627943490763233829880753195251019 3.141592653297590947683406834261190738869139611505752231394089152890909495973464508817163306557131591579057202097715021166512662872910519439747609829479577279606075707015622200744006783543589980682386  import Data.Ratio -- ignoring the task-given pi sequence: sucky convergence -- pie = zip (3:repeat 6) (map (^2) [1,3..]) pie = zip (0:[1,3..]) (4:map (^2) [1..]) sqrt2 = zip (1:repeat 2) (repeat 1) napier = zip (2:[1..]) (1:[1..]) -- truncate after n terms cf2rat n = foldr (\(a,b) f -> (a%1) + ((b%1) / f)) (1%1) . take n -- truncate after error is at most 1/p cf2rat_p p s = f$ map (\i -> (cf2rat i s, cf2rat (1+i) s)) $map (2^) [0..] where f ((x,y):ys) = if abs (x-y) < (1/fromIntegral p) then x else f ys -- returns a decimal string of n digits after the dot; all digits should -- be correct (doesn't mean it's the best approximation! the decimal -- string is simply truncated to given digits: pi=3.141 instead of 3.142) cf2dec n = (ratstr n) . cf2rat_p (10^n) where ratstr l a = (show t) ++ '.':fracstr l n d where d = denominator a (t, n) = quotRem (numerator a) d fracstr 0 _ _ = [] fracstr l n d = (show t)++ fracstr (l-1) n1 d where (t,n1) = quotRem (10 * n) d main = do putStrLn$ cf2dec 200 sqrt2
putStrLn $cf2dec 200 napier putStrLn$ cf2dec 200 pie


## 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 {{trans|D}} {{works with|Java|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 | 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)"



{{Out}}

$jq -M -n -r -f Continued_fraction.jq Value : Direct : Continued Fraction 2|sqrt : 1.4142135623730951 : 1.4142135623730951 1|exp : 2.718281828459045 : 2.7182818284590455 pi : 3.141592653589793 : 3.1415926535892935 (1e-12)  ## Julia {{works with|Julia|0.6}} function _sqrt(a::Bool, n) if a return n > 0 ? 2.0 : 1.0 else return 1.0 end end function _napier(a::Bool, n) if a return n > 0 ? Float64(n) : 2.0 else return n > 1 ? n - 1.0 : 1.0 end end function _pi(a::Bool, n) if a return n > 0 ? 6.0 : 3.0 else return (2.0 * n - 1.0) ^ 2.0 # exponentiation operator end end function calc(f::Function, expansions::Integer) a, b = true, false r = 0.0 for i in expansions:-1:1 r = f(b, i) / (f(a, i) + r) end return f(a, 0) + r end for (v, f) in (("√2", _sqrt), ("e", _napier), ("π", _pi)) @printf("%3s = %f\n", v, calc(f, 1000)) end  {{out}}  √2 = 1.414214 e = 2.718282 π = 3.141593  ## Klong  cf::{[f g i];f::x;g::y;i::z; f(0)+z{i::i-1;g(i+1)%f(i+1)+x}:*0} cf({:[0=x;1;2]};{x;1};1000) cf({:[0=x;2;x]};{:[x>1;x-1;x]};1000) cf({:[0=x;3;6]};{((2*x)-1)^2};1000)  {{out}}  :triad 1.41421356237309504 2.71828182845904523 3.14159265334054205  ## Kotlin {{trans|D}} // version 1.1.2 typealias Func = (Int) -> IntArray fun calc(f: Func, n: Int): Double { var temp = 0.0 for (i in n downTo 1) { val p = f(i) temp = p[1] / (p[0] + temp) } return f(0)[0] + temp } fun main(args: Array<String>) { val pList = listOf<Pair<String, Func>>( "sqrt(2)" to { n -> intArrayOf(if (n > 0) 2 else 1, 1) }, "e " to { n -> intArrayOf(if (n > 0) n else 2, if (n > 1) n - 1 else 1) }, "pi " to { n -> intArrayOf(if (n > 0) 6 else 3, (2 * n - 1) * (2 * n - 1)) } ) for (pair in pList) println("${pair.first} = ${calc(pair.second, 200)}") }  {{out}}  sqrt(2) = 1.4142135623730951 e = 2.7182818284590455 pi = 3.141592622804847  ## Maple  contfrac:=n->evalf(Value(NumberTheory:-ContinuedFraction(n))); contfrac(2^(0.5)); contfrac(Pi); contfrac(exp(1));  =={{header|Mathematica}} / {{header|Wolfram Language}}== sqrt2=Function[n,{1,Transpose@{Array[2&,n],Array[1&,n]}}]; napier=Function[n,{2,Transpose@{Range[n],Prepend[Range[n-1],1]}}]; pi=Function[n,{3,Transpose@{Array[6&,n],Array[(2#-1)^2&,n]}}]; approx=Function[l, N[Divide@@First@Fold[{{#2.#[[;;,1]],#2.#[[;;,2]]},#[[1]]}&,{{l[[2,1,1]]l[[1]]+l[[2,1,2]],l[[2,1,1]]},{l[[1]],1}},l[[2,2;;]]],10]]; r2=approx/@{sqrt2@#,napier@#,pi@#}&@10000;r2//TableForm  {{out}}  1.414213562 2.718281828 3.141592654  ## Maxima cfeval(x) := block([a, b, n, z], a: x[1], b: x[2], n: length(a), z: 0, for i from n step -1 thru 2 do z: b[i]/(a[i] + z), a[1] + z)$

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|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)


{{out}}

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


## Perl

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

sub continued_fraction {
my ($a,$b, $n) = (@_[0,1],$_[2] // 100);

$a->() + ($n && $b->() / continued_fraction($a, $b,$n-1));
}

printf "√2  ≈ %.9f\n", continued_fraction do { my $n; sub {$n++ ? 2 : 1 } }, sub { 1 };
printf "e   ≈ %.9f\n", continued_fraction do { my $n; sub {$n++ || 2 } }, do { my $n; sub {$n++ || 1 } };
printf "π   ≈ %.9f\n", continued_fraction do { my $n; sub {$n++ ? 6 : 3 } }, do { my $n; sub { (2*$n++ + 1)**2 } }, 1_000;
printf "π/2 ≈ %.9f\n", continued_fraction do { my $n; sub { 1/($n++ || 1) } }, sub { 1 }, 1_000;


{{out}}

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


## Perl 6

{{Works with|rakudo|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;
$x; } printf "√2 ≈%.9f\n", continued-fraction(:a(1, |(2 xx *)), :b(Nil, |(1 xx *))); printf "e ≈%.9f\n", continued-fraction(:a(2, |(1 .. *)), :b(Nil, 1, |(1 .. *))); printf "π ≈%.9f\n", continued-fraction(:a(3, |(6 xx *)), :b(Nil, |((1, 3, 5 ... *) X** 2)));  {{out}} √2 ≈ 1.414213562 e ≈ 2.718281828 π ≈ 3.141592654  A more original and a bit more abstract method would consist in viewing a continued fraction on rank n as a function of a variable x: :$\mathrm\left\{CF\right\}_3\left(x\right) = a_0 + \cfrac\left\{b_1\right\}\left\{a_1 + \cfrac\left\{b_2\right\}\left\{a_2 + \cfrac\left\{b_3\right\}\left\{a_3 + x\right\}\right\}\right\}$ Or, more consistently: :$\mathrm\left\{CF\right\}$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, $\mathrm\left\{CF\right\}$n(x) could be written recursively: :$\mathrm\left\{CF\right\}$n(x) = \mathrm{CF}{n-1}(a_n + \frac{b_n}{x}) Or in other words: :$\mathrm\left\{CF\right\}$n= \mathrm{CF}{n-1}\circ f_n = \mathrm{CF}{n-2}\circ f{n-1}\circ f_n=\ldots=f_0\circ f_1 \ldots \circ f_n where $f_n\left(x\right) = a_n + \frac\left\{b_n\right\}\left\{x\right\}$ Perl6 has a builtin composition operator. We can use it with the triangular reduction metaoperator, and evaluate each resulting function at infinity (any value would do actually, but infinite makes it consistent with this particular task). sub continued-fraction(@a, @b) { map { .(Inf) }, [\o] map { @a[$_] + @b[$_] / * }, ^Inf } printf "√2 ≈ %.9f\n", continued-fraction((1, |(2 xx *)), (1 xx *))[10]; printf "e ≈ %.9f\n", continued-fraction((2, |(1 .. *)), (1, |(1 .. *)))[10]; printf "π ≈ %.9f\n", continued-fraction((3, |(6 xx *)), ((1, 3, 5 ... *) X** 2))[100];  {{out}} √2 ≈ 1.414213552 e ≈ 2.718281827 π ≈ 3.141592411  ## Phix {{trans|ALGOL_68}} function continued_fraction(integer steps, integer rid_a, integer rid_b) atom res = 0 for n=steps to 1 by -1 do res := call_func(rid_b,{n}) / (call_func(rid_a,{n}) + res) end for return call_func(rid_a,{0}) + res end function function sqr2_a(integer n) return iff(n=0?1:2) end function function sqr2_b(integer n) return 1 end function function nap_a(integer n) return iff(n=0?2:n) end function function nap_b(integer n) return iff(n=1?1:n-1) end function function pi_a(integer n) return iff(n=0?3:6) end function function pi_b(integer n) return iff(n=1?1:power(2*n-1,2)) end function constant precision = 10000 printf(1,"Precision: %d\n", {precision}) printf(1,"Sqr(2): %.10g\n", {continued_fraction(precision, routine_id("sqr2_a"), routine_id("sqr2_b"))}) printf(1,"Napier: %.10g\n", {continued_fraction(precision, routine_id("nap_a"), routine_id("nap_b"))}) printf(1,"Pi: %.10g\n", {continued_fraction(precision, routine_id("pi_a"), routine_id("pi_b"))})  {{Out}}  Precision: 10000 Sqr(2): 1.414213562 Napier: 2.718281828 Pi: 3.141592654  ## 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))  {{out}}  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;  {{out}}  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;  {{out}}  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).  {{out}}  ?- continued_fraction. sqrt(2) = 1.4142135623730951 e = 2.7182818284590455 pi = 3.141592622804847 true .  ## Python {{works with|Python|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 {{trans|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)  {{out}} 1.4142135623730950488016887242096980785696718753770 2.7182818284590452353602874713526624977572470937000 3.1415926839198062649342019294083175420335002640134  ## Racket ### Using Doubles This version uses standard double precision floating point numbers:  #lang racket (define (calc cf n) (match/values (cf 0) [(a0 b0) (+ a0 (for/fold ([t 0.0]) ([i (in-range (+ n 1) 0 -1)]) (match/values (cf i) [(a b) (/ b (+ a t))])))])) (define (cf-sqrt i) (values (if (> i 0) 2 1) 1)) (define (cf-napier i) (values (if (> i 0) i 2) (if (> i 1) (- i 1) 1))) (define (cf-pi i) (values (if (> i 0) 6 3) (sqr (- (* 2 i) 1)))) (calc cf-sqrt 200) (calc cf-napier 200) (calc cf-pi 200)  Output:  1.4142135623730951 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)  ## REXX ### version 1 The '''cf''' subroutine (for '''C'''ontinued '''F'''ractions) 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 '''{{overline| 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 '''{{overline| }}''' 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|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  ## 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')  {{out}} $ 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));
}



{{out}}


1.4142135623730951
2.7182818284590455
3.141592653339042



## Scala

{{works with|Scala|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()
}
}


{{out}}

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()
}
}


{{out}}

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}...))
}


{{out}}


e ≈ 2.7182818284590452353602874713526624977572470937
π ≈ 3.14159265383979292596359650286939597045138933078
τ ≈ 6.28318530717958647692528676655900576839433879875
φ ≈ 1.61803398874989484820458683436563811772030917981
√2 ≈ 1.41421356237309504880168872420969807856967187538



## Tcl

{{works with|tclsh|8.6}} {{trans|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  {{out}} 1.4142135623730951 2.7182818284590455 3.1415926373965735  ## VBA {{trans|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  {{out}} Precision: 10000 Sqr(2): 1,4142135623731 Napier: 2,71828182845905 Pi: 3,14159265358954  ## Visual Basic .NET {{trans|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  {{out}} 1.4142135623731 2.71828182845905 3.14159262280485  ## 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); ]  {{out}}  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 {{out}}  1.41421356237309514547e+00 1.33227e-15 -2.49251e-10  ## ZX Spectrum Basic {{trans|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
`