How to resolve the algorithm Continued fraction step by step in the ATS programming language

Published on 12 May 2024 09:40 PM
#Ats

How to resolve the algorithm Continued fraction step by step in the ATS programming language

Table of Contents

Problem Statement

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

{\displaystyle a_{0}=1}

then

a

N

= 2

{\displaystyle a_{N}=2}

.

b

N

{\displaystyle b_{N}}

is always

1

{\displaystyle 1}

. For Napier's Constant, use

a

0

= 2

{\displaystyle a_{0}=2}

, then

a

N

= N

{\displaystyle a_{N}=N}

.

b

1

= 1

{\displaystyle b_{1}=1}

then

b

N

= N − 1

{\displaystyle b_{N}=N-1}

. For Pi, use

a

0

= 3

{\displaystyle a_{0}=3}

then

a

N

= 6

{\displaystyle a_{N}=6}

.

b

N

= ( 2 N − 1

)

2

{\displaystyle b_{N}=(2N-1)^{2}}

.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Continued fraction step by step in the ATS programming language

Source code in the ats programming language

#include
"share/atspre_staload.hats"
//
(* ****** ****** *)
//
(*
** a coefficient function creates double values from in paramters
*)
typedef coeff_f = int -> double
//
(*
** a continued fraction is described by a record of two coefficent
** functions a and b
*)
typedef frac = @{a= coeff_f, b= coeff_f}
//
(* ****** ****** *)

fun calc
(
  f: frac, n: int
) : double = let
//
(*
** recursive definition of the approximation
*)
fun loop
(
  n: int, r: double
) : double =
(
if n = 0
  then f.a(0) + r
  else loop (n - 1, f.b(n) / (f.a(n) + r))
// end of [if]
)
//
in
  loop (n, 0.0)
end // end of [calc]

(* ****** ****** *)

val sqrt2 = @{
  a= lam (n: int): double => if n = 0 then 1.0 else 2.0
,
  b= lam (n: int): double => 1.0
} (* end of [val] *)

val napier = @{
  a= lam (n: int): double => if n = 0 then 2.0 else 1.0 * n
,
  b= lam (n: int): double => if n = 1 then 1.0 else n - 1.0
} (* end of [val] *)

val pi = @{
  a= lam (n: int): double => if n = 0 then 3.0 else 6.0
,
  b= lam (n: int): double => let val x = 2.0 * n - 1 in x * x end
}

(* ****** ****** *)

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

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