Check Machin-like formulas: Difference between revisions

Content added Content deleted

Inline

Revision as of 12:57, 22 May 2013

Machin-like formulas are useful for efficiently computing numerical approximations to Pi. Verify the following Machin-like formulas are correct by calculating the value of tan(right hand side) for each equation using exact arithmetic and showing they equal 1:

{\pi  \over 4}=\arctan {1 \over 2}+\arctan {1 \over 3}

{\pi  \over 4}=2\arctan {1 \over 3}+\arctan {1 \over 7}

{\pi  \over 4}=4\arctan {1 \over 5}-\arctan {1 \over 239}

{\pi  \over 4}=5\arctan {1 \over 7}+2\arctan {3 \over 79}

{\pi  \over 4}=5\arctan {29 \over 278}+7\arctan {3 \over 79}

{\pi  \over 4}=\arctan {1 \over 2}+\arctan {1 \over 5}+\arctan {1 \over 8}

{\pi  \over 4}=4\arctan {1 \over 5}-\arctan {1 \over 70}+\arctan {1 \over 99}

{\pi  \over 4}=5\arctan {1 \over 7}+4\arctan {1 \over 53}+2\arctan {1 \over 4443}

{\pi  \over 4}=6\arctan {1 \over 8}+2\arctan {1 \over 57}+\arctan {1 \over 239}

{\pi  \over 4}=8\arctan {1 \over 10}-\arctan {1 \over 239}-4\arctan {1 \over 515}

{\pi  \over 4}=12\arctan {1 \over 18}+8\arctan {1 \over 57}-5\arctan {1 \over 239}

{\pi  \over 4}=16\arctan {1 \over 21}+3\arctan {1 \over 239}+4\arctan {3 \over 1042}

{\pi  \over 4}=22\arctan {1 \over 28}+2\arctan {1 \over 443}-5\arctan {1 \over 1393}-10\arctan {1 \over 11018}

{\pi  \over 4}=22\arctan {1 \over 38}+17\arctan {7 \over 601}+10\arctan {7 \over 8149}

{\pi  \over 4}=44\arctan {1 \over 57}+7\arctan {1 \over 239}-12\arctan {1 \over 682}+24\arctan {1 \over 12943}

{\pi  \over 4}=88\arctan {1 \over 172}+51\arctan {1 \over 239}+32\arctan {1 \over 682}+44\arctan {1 \over 5357}+68\arctan {1 \over 12943}

and confirm that the following formula is incorrect by showing tan(right hand side) is not 1:

{\pi  \over 4}=88\arctan {1 \over 172}+51\arctan {1 \over 239}+32\arctan {1 \over 682}+44\arctan {1 \over 5357}+68\arctan {1 \over 12944}

These identities are useful in calculating the values:

\tan(a+b)={\tan(a)+\tan(b) \over 1-\tan(a)\tan(b)}

\tan \left(\arctan {a \over b}\right)={a \over b}

\tan(-a)=-\tan(a)

You can store the equations in any convenient data structure, but for extra credit parse them from human-readable text input.

D

This uses the module of the Arithmetic Rational Task.

Translation of: Python

<lang d>import std.stdio, std.regex, std.conv, std.string, std.typecons,

      std.array, std.range;

import arithmetic_rational: Rational;

immutable equationText = "pi/4 = arctan(1/2) + arctan(1/3) pi/4 = 2*arctan(1/3) + arctan(1/7) pi/4 = 4*arctan(1/5) - arctan(1/239) pi/4 = 5*arctan(1/7) + 2*arctan(3/79) pi/4 = 5*arctan(29/278) + 7*arctan(3/79) pi/4 = arctan(1/2) + arctan(1/5) + arctan(1/8) pi/4 = 4*arctan(1/5) - arctan(1/70) + arctan(1/99) pi/4 = 5*arctan(1/7) + 4*arctan(1/53) + 2*arctan(1/4443) pi/4 = 6*arctan(1/8) + 2*arctan(1/57) + arctan(1/239) pi/4 = 8*arctan(1/10) - arctan(1/239) - 4*arctan(1/515) pi/4 = 12*arctan(1/18) + 8*arctan(1/57) - 5*arctan(1/239) pi/4 = 16*arctan(1/21) + 3*arctan(1/239) + 4*arctan(3/1042) pi/4 = 22*arctan(1/28) + 2*arctan(1/443) - 5*arctan(1/1393) - 10*arctan(1/11018) pi/4 = 22*arctan(1/38) + 17*arctan(7/601) + 10*arctan(7/8149) pi/4 = 44*arctan(1/57) + 7*arctan(1/239) - 12*arctan(1/682) + 24*arctan(1/12943) pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12943) pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12944)";

alias Pair = Tuple!(int, Rational);

Pair[][] parseEquations(in string text) {

   auto r = regex(r"\s*(?P<sign>[+-])?\s*(?:(?P<mul>\d+)\s*\*)?\s*" ~
                  r"arctan\((?P<num>\d+)/(?P<denom>\d+)\)", "g");
   Pair[][] machins;
   foreach (const line; text.splitLines()) {
       Pair[] formula;
       foreach (part; std.string.split(line, "=")[1].match(r)) {
           immutable string mul = part["mul"];
           immutable string num = part["num"];
           immutable string denom = part["denom"];
           formula ~= Pair((part["sign"] == "-" ? -1 : 1) *
                           (mul.empty ? 1 : to!int(mul)),
                           Rational(to!int(num),
                                    denom.empty ? 1 : to!int(denom)));
       }
       machins ~= formula;
   }
   return machins;

}

Rational tans(/*in*/ Pair[] xs) /*pure nothrow*/ {

   static Rational tanEval(in int coef, /*in*/ Rational f)
   /*pure nothrow*/ {
       if (coef == 1)
           return f;
       if (coef < 0)
           return -tanEval(-coef, f);
       /*const*/ auto a = tanEval(coef / 2, f);
       /*const*/ auto b = tanEval(coef - coef / 2, f);
       return (a + b) / (1 - a * b);
   }

   if (xs.length == 1)
       return tanEval(xs[0].tupleof);
   /*const*/ auto a = tans(xs[0 .. xs.length / 2]);
   /*const*/ auto b = tans(xs[xs.length / 2 .. $]);
   return (a + b) / (1 - a * b);

}

void main() {

   /*const*/ auto machins = parseEquations(equationText);
   foreach (machin, eqn; zip(machins, equationText.splitLines())) {
       /*const*/ auto ans = tans(machin);
       writefln("%5s: %s", ans == 1 ? "OK" : "ERROR", eqn);
   }

}</lang>

Output:

   OK: pi/4 = arctan(1/2) + arctan(1/3)
   OK: pi/4 = 2*arctan(1/3) + arctan(1/7)
   OK: pi/4 = 4*arctan(1/5) - arctan(1/239)
   OK: pi/4 = 5*arctan(1/7) + 2*arctan(3/79)
   OK: pi/4 = 5*arctan(29/278) + 7*arctan(3/79)
   OK: pi/4 = arctan(1/2) + arctan(1/5) + arctan(1/8)
   OK: pi/4 = 4*arctan(1/5) - arctan(1/70) + arctan(1/99)
   OK: pi/4 = 5*arctan(1/7) + 4*arctan(1/53) + 2*arctan(1/4443)
   OK: pi/4 = 6*arctan(1/8) + 2*arctan(1/57) + arctan(1/239)
   OK: pi/4 = 8*arctan(1/10) - arctan(1/239) - 4*arctan(1/515)
   OK: pi/4 = 12*arctan(1/18) + 8*arctan(1/57) - 5*arctan(1/239)
   OK: pi/4 = 16*arctan(1/21) + 3*arctan(1/239) + 4*arctan(3/1042)
   OK: pi/4 = 22*arctan(1/28) + 2*arctan(1/443) - 5*arctan(1/1393) - 10*arctan(1/11018)
   OK: pi/4 = 22*arctan(1/38) + 17*arctan(7/601) + 10*arctan(7/8149)
   OK: pi/4 = 44*arctan(1/57) + 7*arctan(1/239) - 12*arctan(1/682) + 24*arctan(1/12943)
   OK: pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12943)
ERROR: pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12944)

Go

Translation of: Python

<lang go>package main

import (

   "fmt"
   "math/big"

)

type mTerm struct {

   a, n, d int64

}

var testCases = [][]mTerm{

   {{1, 1, 2}, {1, 1, 3}},
   {{2, 1, 3}, {1, 1, 7}},
   {{4, 1, 5}, {-1, 1, 239}},
   {{5, 1, 7}, {2, 3, 79}},
   {{1, 1, 2}, {1, 1, 5}, {1, 1, 8}},
   {{4, 1, 5}, {-1, 1, 70}, {1, 1, 99}},
   {{5, 1, 7}, {4, 1, 53}, {2, 1, 4443}},
   {{6, 1, 8}, {2, 1, 57}, {1, 1, 239}},
   {{8, 1, 10}, {-1, 1, 239}, {-4, 1, 515}},
   {{12, 1, 18}, {8, 1, 57}, {-5, 1, 239}},
   {{16, 1, 21}, {3, 1, 239}, {4, 3, 1042}},
   {{22, 1, 28}, {2, 1, 443}, {-5, 1, 1393}, {-10, 1, 11018}},
   {{22, 1, 38}, {17, 7, 601}, {10, 7, 8149}},
   {{44, 1, 57}, {7, 1, 239}, {-12, 1, 682}, {24, 1, 12943}},
   {{88, 1, 172}, {51, 1, 239}, {32, 1, 682}, {44, 1, 5357}, {68, 1, 12943}},
   {{88, 1, 172}, {51, 1, 239}, {32, 1, 682}, {44, 1, 5357}, {68, 1, 12944}},

}

func main() {

   for _, m := range testCases {
       fmt.Printf("tan %v = %v\n", m, tans(m))
   }

}

var one = big.NewRat(1, 1)

func tans(m []mTerm) *big.Rat {

   if len(m) == 1 {
       return tanEval(m[0].a, big.NewRat(m[0].n, m[0].d))
   }
   half := len(m) / 2
   a := tans(m[:half])
   b := tans(m[half:])
   r := new(big.Rat)
   return r.Quo(new(big.Rat).Add(a, b), r.Sub(one, r.Mul(a, b)))

}

func tanEval(coef int64, f *big.Rat) *big.Rat {

   if coef == 1 {
       return f
   }
   if coef < 0 {
       r := tanEval(-coef, f)
       return r.Neg(r)
   }
   ca := coef / 2
   cb := coef - ca
   a := tanEval(ca, f)
   b := tanEval(cb, f)
   r := new(big.Rat)
   return r.Quo(new(big.Rat).Add(a, b), r.Sub(one, r.Mul(a, b)))

}</lang>

Output:

Last line edited to show only most significant digits of fraction which is near, but not exactly equal to 1.

tan [{1 1 2} {1 1 3}] = 1/1
tan [{2 1 3} {1 1 7}] = 1/1
tan [{4 1 5} {-1 1 239}] = 1/1
tan [{5 1 7} {2 3 79}] = 1/1
tan [{1 1 2} {1 1 5} {1 1 8}] = 1/1
tan [{4 1 5} {-1 1 70} {1 1 99}] = 1/1
tan [{5 1 7} {4 1 53} {2 1 4443}] = 1/1
tan [{6 1 8} {2 1 57} {1 1 239}] = 1/1
tan [{8 1 10} {-1 1 239} {-4 1 515}] = 1/1
tan [{12 1 18} {8 1 57} {-5 1 239}] = 1/1
tan [{16 1 21} {3 1 239} {4 3 1042}] = 1/1
tan [{22 1 28} {2 1 443} {-5 1 1393} {-10 1 11018}] = 1/1
tan [{22 1 38} {17 7 601} {10 7 8149}] = 1/1
tan [{44 1 57} {7 1 239} {-12 1 682} {24 1 12943}] = 1/1
tan [{88 1 172} {51 1 239} {32 1 682} {44 1 5357} {68 1 12943}] = 1/1
tan [{88 1 172} {51 1 239} {32 1 682} {44 1 5357} {68 1 12944}] =
100928801... /
100928883...

Haskell

<lang haskell>import Data.Ratio import Data.List (foldl')

tanPlus :: Fractional a => a -> a -> a tanPlus a b = (a + b) / (1 - a * b)

tanEval :: (Integral a, Fractional b) => (a, b) -> b tanEval (0,_) = 0 tanEval (coef,f) | coef < 0 = -tanEval (-coef, f) | odd coef = tanPlus f $ tanEval (coef - 1, f) | otherwise = tanPlus a a where a = tanEval (coef `div` 2, f)

tans :: (Integral a, Fractional b) => [(a, b)] -> b tans = foldl' tanPlus 0 . map tanEval

machins = [ [(1, 1%2), (1, 1%3)], [(2, 1%3), (1, 1%7)], [(12, 1%18), (8, 1%57), (-5, 1%239)], [(88, 1%172), (51, 1%239), (32 , 1%682), (44, 1%5357), (68, 1%12943)]]

not_machin = [(88, 1%172), (51, 1%239), (32 , 1%682), (44, 1%5357), (68, 1%12944)]

main = do putStrLn "Machins:" mapM_ (\x -> putStrLn $ show (tans x) ++ " <-- " ++ show x) machins

putStr "\nnot Machin: "; print not_machin print (tans not_machin)</lang>

A crazier way to do the above, exploiting the built-in exponentiation algorithms: <lang haskell>import Data.Ratio

-- Private type. Do not use outside of the tans function newtype Tan a = Tan a deriving (Eq, Show) instance Fractional a => Num (Tan a) where

 _ + _ = undefined
 Tan a * Tan b = Tan $ (a + b) / (1 - a * b)
 negate _ = undefined
 abs _ = undefined
 signum _ = undefined
 fromInteger 1 = Tan 0 -- identity for the (*) above
 fromInteger _ = undefined

instance Fractional a => Fractional (Tan a) where

 fromRational _ = undefined
 recip (Tan f) = Tan (-f) -- inverse for the (*) above

tans :: (Integral a, Fractional b) => [(a, b)] -> b tans xs = x where

 Tan x = product [Tan f ^^ coef | (coef,f) <- xs]

machins = [ [(1, 1%2), (1, 1%3)], [(2, 1%3), (1, 1%7)], [(12, 1%18), (8, 1%57), (-5, 1%239)], [(88, 1%172), (51, 1%239), (32 , 1%682), (44, 1%5357), (68, 1%12943)]]

not_machin = [(88, 1%172), (51, 1%239), (32 , 1%682), (44, 1%5357), (68, 1%12944)]

main = do putStrLn "Machins:" mapM_ (\x -> putStrLn $ show (tans x) ++ " <-- " ++ show x) machins

putStr "\nnot Machin: "; print not_machin print (tans not_machin)</lang>

J

Solution:<lang j> machin =: 1r4p1 = [: +/ ({. * _3 o. %/@:}.)"1@:x:</lang> Example (test cases from task description):<lang j> R =: <@:(0&".);._2 ];._2 noun define 1 1 2 1 1 3

2 1 3 1 1 7

4 1 5 _1 1 239

5 1 7 2 3 79

5 29 278 7 3 79

1 1 2 1 1 5 1 1 8

4 1 5 _1 1 70 1 1 99

5 1 7 4 1 53 2 1 4443

6 1 8 2 1 57 1 1 239

8 1 10 _1 1 239 _4 1 515

12 1 18 8 1 57 _5 1 239

16 1 21 3 1 239 4 3 1042

22 1 28 2 1 443 _5 1 1393 _10 1 11018

22 1 38 17 7 601 10 7 8149

44 1 57 7 1 239 _12 1 682 24 1 12943

88 1 172 51 1 239 32 1 682 44 1 5357 68 1 12943

)

machin&> R 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1</lang> Example (counterexample):<lang j> counterExample=. 12944 (<_1;_1)} >{:R counterExample NB. Same as final test case with 12943 incremented to 12944 88 1 172 51 1 239 32 1 682 44 1 5357 68 1 12944 machin counterExample 0</lang> Notes: The function machin compares the results of each formula to π/4 (expressed as 1r4p1 in J's numeric notation). The first example above shows the results of these comparisons for each formula (with 1 for true and 0 for false). In J, arctan is expressed as _3 o. values and the function x: coerces values to exact representation; thereafter J will maintain exactness throughout its calculations, as long as it can.

Mathematica

<lang>Tan[ArcTan[1/2] + ArcTan[1/3]] == 1 Tan[2 ArcTan[1/3] + ArcTan[1/7]] == 1 Tan[4 ArcTan[1/5] - ArcTan[1/239]] == 1 Tan[5 ArcTan[1/7] + 2 ArcTan[3/79]] == 1 Tan[5 ArcTan[29/278] + 7 ArcTan[3/79]] == 1 Tan[ArcTan[1/2] + ArcTan[1/5] + ArcTan[1/8]] == 1 Tan[4 ArcTan[1/5] - ArcTan[1/70] + ArcTan[1/99]] == 1 Tan[5 ArcTan[1/7] + 4 ArcTan[1/53] + 2 ArcTan[1/4443]] == 1 Tan[6 ArcTan[1/8] + 2 ArcTan[1/57] + ArcTan[1/239]] == 1 Tan[8 ArcTan[1/10] - ArcTan[1/239] - 4 ArcTan[1/515]] == 1 Tan[12 ArcTan[1/18] + 8 ArcTan[1/57] - 5 ArcTan[1/239]] == 1 Tan[16 ArcTan[1/21] + 3 ArcTan[1/239] + 4 ArcTan[3/1042]] == 1 Tan[22 ArcTan[1/28] + 2 ArcTan[1/443] - 5 ArcTan[1/1393] -

  10 ArcTan[1/11018]] == 1

Tan[22 ArcTan[1/38] + 17 ArcTan[7/601] + 10 ArcTan[7/8149]] == 1 Tan[44 ArcTan[1/57] + 7 ArcTan[1/239] - 12 ArcTan[1/682] +

  24 ArcTan[1/12943]] == 1

Tan[88 ArcTan[1/172] + 51 ArcTan[1/239] + 32 ArcTan[1/682] +

  44 ArcTan[1/5357] + 68 ArcTan[1/12943]] == 1

Tan[88 ArcTan[1/172] + 51 ArcTan[1/239] + 32 ArcTan[1/682] +

  44 ArcTan[1/5357] + 68 ArcTan[1/12944]] == 1</lang>

Output:

True

True

True

True

True

True

True

True

True

True

True

True

True

True

True

True

False

Maxima

<lang maxima>trigexpand:true$ is(tan(atan(1/2)+atan(1/3))=1); is(tan(2*atan(1/3)+atan(1/7))=1); is(tan(4*atan(1/5)-atan(1/239))=1); is(tan(5*atan(1/7)+2*atan(3/79))=1); is(tan(5*atan(29/278)+7*atan(3/79))=1); is(tan(atan(1/2)+atan(1/5)+atan(1/8))=1); is(tan(4*atan(1/5)-atan(1/70)+atan(1/99))=1); is(tan(5*atan(1/7)+4*atan(1/53)+2*atan(1/4443))=1); is(tan(6*atan(1/8)+2*atan(1/57)+atan(1/239))=1); is(tan(8*atan(1/10)-atan(1/239)-4*atan(1/515))=1); is(tan(12*atan(1/18)+8*atan(1/57)-5*atan(1/239))=1); is(tan(16*atan(1/21)+3*atan(1/239)+4*atan(3/1042))=1); is(tan(22*atan(1/28)+2*atan(1/443)-5*atan(1/1393)-10*atan(1/11018))=1); is(tan(22*atan(1/38)+17*atan(7/601)+10*atan(7/8149))=1); is(tan(44*atan(1/57)+7*atan(1/239)-12*atan(1/682)+24*atan(1/12943))=1); is(tan(88*atan(1/172)+51*atan(1/239)+32*atan(1/682)+44*atan(1/5357)+68*atan(1/12943))=1); is(tan(88*atan(1/172)+51*atan(1/239)+32*atan(1/682)+44*atan(1/5357)+68*atan(1/12944))=1);</lang>

Output:

(%i2) 
(%o2)                                true
(%i3) 
(%o3)                                true
(%i4) 
(%o4)                                true
(%i5) 
(%o5)                                true
(%i6) 
(%o6)                                true
(%i7) 
(%o7)                                true
(%i8) 
(%o8)                                true
(%i9) 
(%o9)                                true
(%i10) 
(%o10)                               true
(%i11) 
(%o11)                               true
(%i12) 
(%o12)                               true
(%i13) 
(%o13)                               true
(%i14) 
(%o14)                               true
(%i15) 
(%o15)                               true
(%i16) 
(%o16)                               true
(%i17) 
(%o17)                               true
(%i18) 
(%o18)                               false

OCaml

<lang ocaml>open Num;; (* use exact rationals for results *)

let tadd p q = (p +/ q) // ((Int 1) -/ (p */ q)) in

(* tan(n*arctan(a/b)) *) let rec tan_expr (n,a,b) =

 if n = 1 then (Int a)//(Int b) else
 if n = -1 then (Int (-a))//(Int b) else
   let m = n/2 in
   let tm = tan_expr (m,a,b) in
   let m2 = tadd tm tm and k = n-m-m in
   if k = 0 then m2 else tadd (tan_expr (k,a,b)) m2 in

let verify (k, tlist) =

 Printf.printf "Testing: pi/%d = " k;
 let t_str = List.map (fun (x,y,z) -> Printf.sprintf "%d*atan(%d/%d)" x y z) tlist in
 print_endline (String.concat " + " t_str);
 let ans_terms = List.map tan_expr tlist in
 let answer = List.fold_left tadd (Int 0) ans_terms in
 Printf.printf "  tan(RHS) is %s\n" (if answer = (Int 1) then "one" else "not one") in

(* example: prog 4 5 29 278 7 3 79 represents pi/4 = 5*atan(29/278) + 7*atan(3/79) *) let args = Sys.argv in let nargs = Array.length args in let v k = int_of_string args.(k) in let rec triples n =

 if n+2 > nargs-1 then []
 else (v n, v (n+1), v (n+2)) :: triples (n+3) in

if nargs > 4 then let dat = (v 1, triples 2) in verify dat else List.iter verify [

 (4,[(1,1,2);(1,1,3)]);
 (4,[(2,1,3);(1,1,7)]);
 (4,[(4,1,5);(-1,1,239)]);
 (4,[(5,1,7);(2,3,79)]);
 (4,[(5,29,278);(7,3,79)]);
 (4,[(1,1,2);(1,1,5);(1,1,8)]);
 (4,[(4,1,5);(-1,1,70);(1,1,99)]);
 (4,[(5,1,7);(4,1,53);(2,1,4443)]);
 (4,[(6,1,8);(2,1,57);(1,1,239)]);
 (4,[(8,1,10);(-1,1,239);(-4,1,515)]);
 (4,[(12,1,18);(8,1,57);(-5,1,239)]);
 (4,[(16,1,21);(3,1,239);(4,3,1042)]);
 (4,[(22,1,28);(2,1,443);(-5,1,1393);(-10,1,11018)]);
 (4,[(22,1,38);(17,7,601);(10,7,8149)]);
 (4,[(44,1,57);(7,1,239);(-12,1,682);(24,1,12943)]);
 (4,[(88,1,172);(51,1,239);(32,1,682);(44,1,5357);(68,1,12943)]);
 (4,[(88,1,172);(51,1,239);(32,1,682);(44,1,5357);(68,1,12944)])

]</lang>

Compile with

ocamlopt -o verify_machin.opt nums.cmxa verify_machin.ml

or run with

ocaml nums.cma verify_machin.ml

Output:

Testing: pi/4 = 1*atan(1/2) + 1*atan(1/3)
  tan(RHS) is one
Testing: pi/4 = 2*atan(1/3) + 1*atan(1/7)
  tan(RHS) is one
Testing: pi/4 = 4*atan(1/5) + -1*atan(1/239)
  tan(RHS) is one
Testing: pi/4 = 5*atan(1/7) + 2*atan(3/79)
  tan(RHS) is one
Testing: pi/4 = 5*atan(29/278) + 7*atan(3/79)
  tan(RHS) is one
Testing: pi/4 = 1*atan(1/2) + 1*atan(1/5) + 1*atan(1/8)
  tan(RHS) is one
Testing: pi/4 = 4*atan(1/5) + -1*atan(1/70) + 1*atan(1/99)
  tan(RHS) is one
Testing: pi/4 = 5*atan(1/7) + 4*atan(1/53) + 2*atan(1/4443)
  tan(RHS) is one
Testing: pi/4 = 6*atan(1/8) + 2*atan(1/57) + 1*atan(1/239)
  tan(RHS) is one
Testing: pi/4 = 8*atan(1/10) + -1*atan(1/239) + -4*atan(1/515)
  tan(RHS) is one
Testing: pi/4 = 12*atan(1/18) + 8*atan(1/57) + -5*atan(1/239)
  tan(RHS) is one
Testing: pi/4 = 16*atan(1/21) + 3*atan(1/239) + 4*atan(3/1042)
  tan(RHS) is one
Testing: pi/4 = 22*atan(1/28) + 2*atan(1/443) + -5*atan(1/1393) + -10*atan(1/11018)
  tan(RHS) is one
Testing: pi/4 = 22*atan(1/38) + 17*atan(7/601) + 10*atan(7/8149)
  tan(RHS) is one
Testing: pi/4 = 44*atan(1/57) + 7*atan(1/239) + -12*atan(1/682) + 24*atan(1/12943)
  tan(RHS) is one
Testing: pi/4 = 88*atan(1/172) + 51*atan(1/239) + 32*atan(1/682) + 44*atan(1/5357) + 68*atan(1/12943)
  tan(RHS) is one
Testing: pi/4 = 88*atan(1/172) + 51*atan(1/239) + 32*atan(1/682) + 44*atan(1/5357) + 68*atan(1/12944)
  tan(RHS) is not one

Perl 6

Translation of: maxima

<lang Perl6>use Test; plan *;

is tan(atan(1/2)+atan(1/3)), 1; is tan(2*atan(1/3)+atan(1/7)), 1; is tan(4*atan(1/5)-atan(1/239)), 1; is tan(5*atan(1/7)+2*atan(3/79)), 1; is tan(5*atan(29/278)+7*atan(3/79)), 1; is tan(atan(1/2)+atan(1/5)+atan(1/8)), 1; is tan(4*atan(1/5)-atan(1/70)+atan(1/99)), 1; is tan(5*atan(1/7)+4*atan(1/53)+2*atan(1/4443)), 1; is tan(6*atan(1/8)+2*atan(1/57)+atan(1/239)), 1; is tan(8*atan(1/10)-atan(1/239)-4*atan(1/515)), 1; is tan(12*atan(1/18)+8*atan(1/57)-5*atan(1/239)), 1; is tan(16*atan(1/21)+3*atan(1/239)+4*atan(3/1042)), 1; is tan(22*atan(1/28)+2*atan(1/443)-5*atan(1/1393)-10*atan(1/11018)), 1; is tan(22*atan(1/38)+17*atan(7/601)+10*atan(7/8149)), 1; is tan(44*atan(1/57)+7*atan(1/239)-12*atan(1/682)+24*atan(1/12943)), 1; is tan(88*atan(1/172)+51*atan(1/239)+32*atan(1/682)+44*atan(1/5357)+68*atan(1/12943)), 1; is tan(88*atan(1/172)+51*atan(1/239)+32*atan(1/682)+44*atan(1/5357)+68*atan(1/12944)), 1; </lang>

Output:

ok 1 - 
ok 2 - 
ok 3 - 
ok 4 - 
ok 5 - 
ok 6 - 
ok 7 - 
ok 8 - 
ok 9 - 
ok 10 - 
ok 11 - 
ok 12 - 
ok 13 - 
ok 14 - 
ok 15 - 
ok 16 - 
not ok 17 - 
#      got: '0.999999188225744'
# expected: '1'

The text input is almost directly parsable as Perl6 code, so we can just write: <lang Perl 6>use Test; plan *;

for lines() {

   is tan(.eval), /INCORRECT/ ?? none(1) !! 1 given
   .subst(/'pi/4 = '/, )\
   .subst(/arctan/, 'atan', :g)\
   .subst(/'[INCORRECT]'/, '# INCORRECT')

}</lang>

This programs reads stdin.

Python

This example parses the original equations to form an intermediate representation then does the checks.
Function tans and tanEval are translations of the Haskel functions of the same names. <lang python>import re from fractions import Fraction from pprint import pprint as pp

equationtext = \

 pi/4 = arctan(1/2) + arctan(1/3) 
 pi/4 = 2*arctan(1/3) + arctan(1/7)
 pi/4 = 4*arctan(1/5) - arctan(1/239)
 pi/4 = 5*arctan(1/7) + 2*arctan(3/79)
 pi/4 = 5*arctan(29/278) + 7*arctan(3/79)
 pi/4 = arctan(1/2) + arctan(1/5) + arctan(1/8) 
 pi/4 = 4*arctan(1/5) - arctan(1/70) + arctan(1/99) 
 pi/4 = 5*arctan(1/7) + 4*arctan(1/53) + 2*arctan(1/4443)
 pi/4 = 6*arctan(1/8) + 2*arctan(1/57) + arctan(1/239)
 pi/4 = 8*arctan(1/10) - arctan(1/239) - 4*arctan(1/515)
 pi/4 = 12*arctan(1/18) + 8*arctan(1/57) - 5*arctan(1/239)
 pi/4 = 16*arctan(1/21) + 3*arctan(1/239) + 4*arctan(3/1042)
 pi/4 = 22*arctan(1/28) + 2*arctan(1/443) - 5*arctan(1/1393) - 10*arctan(1/11018)
 pi/4 = 22*arctan(1/38) + 17*arctan(7/601) + 10*arctan(7/8149)
 pi/4 = 44*arctan(1/57) + 7*arctan(1/239) - 12*arctan(1/682) + 24*arctan(1/12943)
 pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12943)
 pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12944)

def parse_eqn(equationtext=equationtext):

   eqn_re = re.compile(r"""(?mx)
   (?P<lhs> ^ \s* pi/4 \s* = \s*)?             # LHS of equation
   (?:                                         # RHS
       \s* (?P<sign> [+-])? \s* 
       (?: (?P<mult> \d+) \s* \*)? 
       \s* arctan\( (?P<numer> \d+) / (?P<denom> \d+)
   )""")

   found = eqn_re.findall(equationtext)
   machins, part = [], []
   for lhs, sign, mult, numer, denom in eqn_re.findall(equationtext):
       if lhs and part:
           machins.append(part)
           part = []
       part.append( ( (-1 if sign == '-' else 1) * ( int(mult) if mult else 1),
                      Fraction(int(numer), (int(denom) if denom else 1)) ) )
   machins.append(part)
   return machins

def tans(xs):

   xslen = len(xs)
   if xslen == 1:
       return tanEval(*xs[0])
   aa, bb = xs[:xslen//2], xs[xslen//2:]
   a, b = tans(aa), tans(bb)
   return (a + b) / (1 - a * b)

def tanEval(coef, f):

   if coef == 1:
       return f
   if coef < 0:
       return -tanEval(-coef, f)
   ca = coef // 2
   cb = coef - ca
   a, b = tanEval(ca, f), tanEval(cb, f)
   return (a + b) / (1 - a * b)

if __name__ == '__main__':

   machins = parse_eqn()
   #pp(machins, width=160)
   for machin, eqn in zip(machins, equationtext.split('\n')):
       ans = tans(machin)
       print('%5s: %s' % ( ('OK' if ans == 1 else 'ERROR'), eqn))</lang>

Output:

   OK:   pi/4 = arctan(1/2) + arctan(1/3) 
   OK:   pi/4 = 2*arctan(1/3) + arctan(1/7)
   OK:   pi/4 = 4*arctan(1/5) - arctan(1/239)
   OK:   pi/4 = 5*arctan(1/7) + 2*arctan(3/79)
   OK:   pi/4 = 5*arctan(29/278) + 7*arctan(3/79)
   OK:   pi/4 = arctan(1/2) + arctan(1/5) + arctan(1/8) 
   OK:   pi/4 = 4*arctan(1/5) - arctan(1/70) + arctan(1/99) 
   OK:   pi/4 = 5*arctan(1/7) + 4*arctan(1/53) + 2*arctan(1/4443)
   OK:   pi/4 = 6*arctan(1/8) + 2*arctan(1/57) + arctan(1/239)
   OK:   pi/4 = 8*arctan(1/10) - arctan(1/239) - 4*arctan(1/515)
   OK:   pi/4 = 12*arctan(1/18) + 8*arctan(1/57) - 5*arctan(1/239)
   OK:   pi/4 = 16*arctan(1/21) + 3*arctan(1/239) + 4*arctan(3/1042)
   OK:   pi/4 = 22*arctan(1/28) + 2*arctan(1/443) - 5*arctan(1/1393) - 10*arctan(1/11018)
   OK:   pi/4 = 22*arctan(1/38) + 17*arctan(7/601) + 10*arctan(7/8149)
   OK:   pi/4 = 44*arctan(1/57) + 7*arctan(1/239) - 12*arctan(1/682) + 24*arctan(1/12943)
   OK:   pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12943)
ERROR:   pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12944)

Note: the Kodos tool was used in developing the regular expression.

Seed7

<lang seed7>$ include "seed7_05.s7i";

 include "bigint.s7i";
 include "bigrat.s7i";

const type: mTerms is array array bigInteger;

const array mTerms: testCases is [] (

   [] ([] ( 1_, 1_,   2_), [] ( 1_, 1_,   3_)),
   [] ([] ( 2_, 1_,   3_), [] ( 1_, 1_,   7_)),
   [] ([] ( 4_, 1_,   5_), [] (-1_, 1_, 239_)),
   [] ([] ( 5_, 1_,   7_), [] ( 2_, 3_,  79_)),
   [] ([] ( 1_, 1_,   2_), [] ( 1_, 1_,   5_), [] (  1_, 1_,    8_)),
   [] ([] ( 4_, 1_,   5_), [] (-1_, 1_,  70_), [] (  1_, 1_,   99_)),
   [] ([] ( 5_, 1_,   7_), [] ( 4_, 1_,  53_), [] (  2_, 1_, 4443_)),
   [] ([] ( 6_, 1_,   8_), [] ( 2_, 1_,  57_), [] (  1_, 1_,  239_)),
   [] ([] ( 8_, 1_,  10_), [] (-1_, 1_, 239_), [] ( -4_, 1_,  515_)),
   [] ([] (12_, 1_,  18_), [] ( 8_, 1_,  57_), [] ( -5_, 1_,  239_)),
   [] ([] (16_, 1_,  21_), [] ( 3_, 1_, 239_), [] (  4_, 3_, 1042_)),
   [] ([] (22_, 1_,  28_), [] ( 2_, 1_, 443_), [] ( -5_, 1_, 1393_), [] (-10_, 1_, 11018_)),
   [] ([] (22_, 1_,  38_), [] (17_, 7_, 601_), [] ( 10_, 7_, 8149_)),
   [] ([] (44_, 1_,  57_), [] ( 7_, 1_, 239_), [] (-12_, 1_,  682_), [] ( 24_, 1_, 12943_)),
   [] ([] (88_, 1_, 172_), [] (51_, 1_, 239_), [] ( 32_, 1_,  682_), [] ( 44_, 1_,  5357_), [] (68_, 1_, 12943_)),
   [] ([] (88_, 1_, 172_), [] (51_, 1_, 239_), [] ( 32_, 1_,  682_), [] ( 44_, 1_,  5357_), [] (68_, 1_, 12944_))
 );

const func bigRational: tanEval (in bigInteger: coef, in bigRational: f) is func

 result
   var bigRational: tanEval is bigRational.value;
 local
   var bigRational: a is bigRational.value;
   var bigRational: b is bigRational.value;
 begin
   if coef = 1_ then
     tanEval := f;
   elsif coef < 0_ then
     tanEval := -tanEval(-coef, f);
   else
     a := tanEval(coef div 2_, f);
     b := tanEval(coef - coef div 2_, f);
     tanEval := (a + b) / (1_/1_ - a * b);
   end if;
 end func;

const func bigRational: tans (in mTerms: terms) is func

 result
   var bigRational: tans is bigRational.value;
 local
   var bigRational: a is bigRational.value;
   var bigRational: b is bigRational.value;
 begin
   if length(terms) = 1 then
     tans := tanEval(terms[1][1], terms[1][2] / terms[1][3]);
   else
     a := tans(terms[.. length(terms) div 2]);
     b := tans(terms[succ(length(terms) div 2) ..]);
     tans := (a + b) / (1_/1_ - a * b);
   end if;
 end func;

const proc: main is func

 local
   var integer: index is 0;
   var array bigInteger: term is 0 times 0_;
 begin
   for key index range testCases do
     write(tans(testCases[index]) = 1_/1_ <& ": pi/4 = ");
     for term range testCases[index] do
       write([0] ("+", "-")[ord(term[1] < 0_)] <& abs(term[1]) <& "*arctan(" <& term[2] <& "/" <& term[3] <& ")");
     end for;
     writeln;
   end for;
 end func;</lang>

Output:

TRUE: pi/4 = +1*arctan(1/2)+1*arctan(1/3)
TRUE: pi/4 = +2*arctan(1/3)+1*arctan(1/7)
TRUE: pi/4 = +4*arctan(1/5)-1*arctan(1/239)
TRUE: pi/4 = +5*arctan(1/7)+2*arctan(3/79)
TRUE: pi/4 = +1*arctan(1/2)+1*arctan(1/5)+1*arctan(1/8)
TRUE: pi/4 = +4*arctan(1/5)-1*arctan(1/70)+1*arctan(1/99)
TRUE: pi/4 = +5*arctan(1/7)+4*arctan(1/53)+2*arctan(1/4443)
TRUE: pi/4 = +6*arctan(1/8)+2*arctan(1/57)+1*arctan(1/239)
TRUE: pi/4 = +8*arctan(1/10)-1*arctan(1/239)-4*arctan(1/515)
TRUE: pi/4 = +12*arctan(1/18)+8*arctan(1/57)-5*arctan(1/239)
TRUE: pi/4 = +16*arctan(1/21)+3*arctan(1/239)+4*arctan(3/1042)
TRUE: pi/4 = +22*arctan(1/28)+2*arctan(1/443)-5*arctan(1/1393)-10*arctan(1/11018)
TRUE: pi/4 = +22*arctan(1/38)+17*arctan(7/601)+10*arctan(7/8149)
TRUE: pi/4 = +44*arctan(1/57)+7*arctan(1/239)-12*arctan(1/682)+24*arctan(1/12943)
TRUE: pi/4 = +88*arctan(1/172)+51*arctan(1/239)+32*arctan(1/682)+44*arctan(1/5357)+68*arctan(1/12943)
FALSE: pi/4 = +88*arctan(1/172)+51*arctan(1/239)+32*arctan(1/682)+44*arctan(1/5357)+68*arctan(1/12944)

Tcl

<lang tcl>package require Tcl 8.5

Compute tan(atan(p)+atan(q)) using rationals

proc tadd {p q} {

   lassign $p pp pq
   lassign $q qp qq
   set topp [expr {$pp*$qq + $qp*$pq}]
   set topq [expr {$pq*$qq}]
   set prodp [expr {$pp*$qp}]
   set prodq [expr {$pq*$qq}]
   set lowp [expr {$prodq - $prodp}]
   set resultp [set gcd1 [expr {$topp * $prodq}]]
   set resultq [set gcd2 [expr {$topq * $lowp}]]
   # Critical! Normalize using the GCD
   while {$gcd2 != 0} {

lassign [list $gcd2 [expr {$gcd1 % $gcd2}]] gcd1 gcd2

   }
   list [expr {$resultp / abs($gcd1)}] [expr {$resultq / abs($gcd1)}]

} proc termTan {n a b} {

   if {$n < 0} {

set n [expr {-$n}] set a [expr {-$a}]

   }
   if {$n == 1} {

return [list $a $b]

   }
   set k [expr {$n - [set m [expr {$n / 2}]]*2}]
   set t2 [termTan $m $a $b]
   set m2 [tadd $t2 $t2]
   if {$k == 0} {

return $m2

   }
   return [tadd [termTan $k $a $b] $m2]

} proc machinTan {terms} {

   set sum {0 1}
   foreach term $terms {

set sum [tadd $sum [termTan {*}$term]]

   }
   return $sum

}

Assumes that the formula is in the very specific form below!

proc parseFormula {formula} {

   set RE {(-?\s*\d*\s*\*?)\s*arctan\s*\(\s*(-?\s*\d+)\s*/\s*(-?\s*\d+)\s*\)}
   set nospace {" " "" "*" ""}
   foreach {all n a b} [regexp -inline -all $RE $formula] {

if {![regexp {\d} $n]} {append n 1} lappend result [list [string map $nospace $n] [string map $nospace $a] [string map $nospace $b]]

   }
   return $result

}

foreach formula {

   "pi/4 = arctan(1/2) + arctan(1/3)"
   "pi/4 = 2*arctan(1/3) + arctan(1/7)"
   "pi/4 = 4*arctan(1/5) - arctan(1/239)"
   "pi/4 = 5*arctan(1/7) + 2*arctan(3/79)"
   "pi/4 = 5*arctan(29/278) + 7*arctan(3/79)"
   "pi/4 = arctan(1/2) + arctan(1/5) + arctan(1/8)"
   "pi/4 = 4*arctan(1/5) - arctan(1/70) + arctan(1/99)"
   "pi/4 = 5*arctan(1/7) + 4*arctan(1/53) + 2*arctan(1/4443)"
   "pi/4 = 6*arctan(1/8) + 2*arctan(1/57) + arctan(1/239)"
   "pi/4 = 8*arctan(1/10) - arctan(1/239) - 4*arctan(1/515)"
   "pi/4 = 12*arctan(1/18) + 8*arctan(1/57) - 5*arctan(1/239)"
   "pi/4 = 16*arctan(1/21) + 3*arctan(1/239) + 4*arctan(3/1042)"
   "pi/4 = 22*arctan(1/28) + 2*arctan(1/443) - 5*arctan(1/1393) - 10*arctan(1/11018)"
   "pi/4 = 22*arctan(1/38) + 17*arctan(7/601) + 10*arctan(7/8149)"
   "pi/4 = 44*arctan(1/57) + 7*arctan(1/239) - 12*arctan(1/682) + 24*arctan(1/12943)"
   "pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12943)"
   "pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12944)"

} {

   if {[tcl::mathop::== {*}[machinTan [parseFormula $formula]]]} {

puts "Yes! '$formula' is true"

   } else {

puts "No! '$formula' not true"

}</lang>

Output:

Yes! 'pi/4 = arctan(1/2) + arctan(1/3)' is true
Yes! 'pi/4 = 2*arctan(1/3) + arctan(1/7)' is true
Yes! 'pi/4 = 4*arctan(1/5) - arctan(1/239)' is true
Yes! 'pi/4 = 5*arctan(1/7) + 2*arctan(3/79)' is true
Yes! 'pi/4 = 5*arctan(29/278) + 7*arctan(3/79)' is true
Yes! 'pi/4 = arctan(1/2) + arctan(1/5) + arctan(1/8)' is true
Yes! 'pi/4 = 4*arctan(1/5) - arctan(1/70) + arctan(1/99)' is true
Yes! 'pi/4 = 5*arctan(1/7) + 4*arctan(1/53) + 2*arctan(1/4443)' is true
Yes! 'pi/4 = 6*arctan(1/8) + 2*arctan(1/57) + arctan(1/239)' is true
Yes! 'pi/4 = 8*arctan(1/10) - arctan(1/239) - 4*arctan(1/515)' is true
Yes! 'pi/4 = 12*arctan(1/18) + 8*arctan(1/57) - 5*arctan(1/239)' is true
Yes! 'pi/4 = 16*arctan(1/21) + 3*arctan(1/239) + 4*arctan(3/1042)' is true
Yes! 'pi/4 = 22*arctan(1/28) + 2*arctan(1/443) - 5*arctan(1/1393) - 10*arctan(1/11018)' is true
Yes! 'pi/4 = 22*arctan(1/38) + 17*arctan(7/601) + 10*arctan(7/8149)' is true
Yes! 'pi/4 = 44*arctan(1/57) + 7*arctan(1/239) - 12*arctan(1/682) + 24*arctan(1/12943)' is true
Yes! 'pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12943)' is true
No! 'pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12944)' not true

XPL0

<lang XPL0>code ChOut=8, Text=12; \intrinsic routines int Number(18); \numbers from equations def LF=$0A; \ASCII line feed (end-of-line character)

func Parse(S); \Convert numbers in string S to binary in Number array char S; int I, Neg;

       proc GetNum;    \Get number from string S
       int  N;
       [while S(0)<^0 ! S(0)>^9 do S:= S+1;
       N:= S(0)-^0;  S:= S+1;
       while S(0)>=^0 & S(0)<=^9 do
               [N:= N*10 + S(0) - ^0;  S:= S+1];
       Number(I):= N;  I:= I+1;
       ];

[while S(0)#^= do S:= S+1; \skip to "=" I:= 0; loop [Neg:= false; \assume positive term

       loop    [S:= S+1;       \next char
               case S(0) of
                 LF:   [Number(I):= 0;  return S+1];   \mark end of array
                 ^-:   Neg:= true;                     \term is negative
                 ^a:   [Number(I):= 1;  I:= I+1; quit] \no coefficient so use 1
               other if S(0)>=^0 & S(0)<=^9 then       \if digit
                       [S:= S-1;  GetNum;  quit];      \backup and get number
               ];
       GetNum;                                         \numerator
       if Neg then Number(I-1):= -Number(I-1);         \tan(-a) = -tan(a)
       GetNum;                                         \denominator
       ];

];

func GCD(U, V); \Return the greatest common divisor of U and V int U, V; int T; [while V do \Euclid's method

   [T:= U;  U:= V;  V:= rem(T/V)];

return abs(U); ];

proc Verify; \Verify that tangent of equation = 1 (i.e: E = F) int E, F, I, J;

   proc Machin(A, B, C, D);
   int  A, B, C, D;
   int  Div;
   \tan(a+b) = (tan(a) + tan(b)) / (1 - tan(a)*tan(b))
   \tan(arctan(A/B) + arctan(C/D))
   \   = (tan(arctan(A/B)) + tan(arctan(C/D))) / (1 - tan(arctan(A/B))*tan(arctan(C/D)))
   \   = (A/B + C/D) / (1 - A/B*C/D)
   \   = (A*D/B*D + B*C/B*D) / (B*D/B*D - A*C/B*D)
   \   = (A*D + B*C) / (B*D - A*C)
   [E:= A*D + B*C;  F:= B*D - A*C;
   Div:= GCD(E, F);    \keep integers from getting too big
   E:= E/Div;  F:= F/Div;
   ];

[E:= 0; F:= 1; I:= 0; while Number(I) do

   [for J:= 1 to Number(I) do
       Machin(E, F, Number(I+1), Number(I+2));
   I:= I+3;
   ];

Text(0, if E=F then "Yes " else "No "); ];

char S, SS; int I; [S:= "pi/4 = arctan(1/2) + arctan(1/3) pi/4 = 2*arctan(1/3) + arctan(1/7) pi/4 = 4*arctan(1/5) - arctan(1/239) pi/4 = 5*arctan(1/7) + 2*arctan(3/79) pi/4 = 5*arctan(29/278) + 7*arctan(3/79) pi/4 = arctan(1/2) + arctan(1/5) + arctan(1/8) pi/4 = 4*arctan(1/5) - arctan(1/70) + arctan(1/99) pi/4 = 5*arctan(1/7) + 4*arctan(1/53) + 2*arctan(1/4443) pi/4 = 6*arctan(1/8) + 2*arctan(1/57) + arctan(1/239) pi/4 = 8*arctan(1/10) - arctan(1/239) - 4*arctan(1/515) pi/4 = 12*arctan(1/18) + 8*arctan(1/57) - 5*arctan(1/239) pi/4 = 16*arctan(1/21) + 3*arctan(1/239) + 4*arctan(3/1042) pi/4 = 22*arctan(1/28) + 2*arctan(1/443) - 5*arctan(1/1393) - 10*arctan(1/11018) pi/4 = 22*arctan(1/38) + 17*arctan(7/601) + 10*arctan(7/8149) pi/4 = 44*arctan(1/57) + 7*arctan(1/239) - 12*arctan(1/682) + 24*arctan(1/12943) pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12943) pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12944)

";                             \Python version of equations (thanks!)

for I:= 1 to 17 do

       [SS:= S;                \save start of string line
       S:= Parse(S);           \returns start of next line
       Verify;                 \correct Machin equation? Yes or No
       repeat ChOut(0, SS(0)); SS:= SS+1 until SS(0)=LF;  ChOut(0, LF); \show equation
       ];

]</lang>

Output:

Yes  pi/4 = arctan(1/2) + arctan(1/3) 
Yes  pi/4 = 2*arctan(1/3) + arctan(1/7)
Yes  pi/4 = 4*arctan(1/5) - arctan(1/239)
Yes  pi/4 = 5*arctan(1/7) + 2*arctan(3/79)
Yes  pi/4 = 5*arctan(29/278) + 7*arctan(3/79)
Yes  pi/4 = arctan(1/2) + arctan(1/5) + arctan(1/8) 
Yes  pi/4 = 4*arctan(1/5) - arctan(1/70) + arctan(1/99) 
Yes  pi/4 = 5*arctan(1/7) + 4*arctan(1/53) + 2*arctan(1/4443)
Yes  pi/4 = 6*arctan(1/8) + 2*arctan(1/57) + arctan(1/239)
Yes  pi/4 = 8*arctan(1/10) - arctan(1/239) - 4*arctan(1/515)
Yes  pi/4 = 12*arctan(1/18) + 8*arctan(1/57) - 5*arctan(1/239)
Yes  pi/4 = 16*arctan(1/21) + 3*arctan(1/239) + 4*arctan(3/1042)
Yes  pi/4 = 22*arctan(1/28) + 2*arctan(1/443) - 5*arctan(1/1393) - 10*arctan(1/11018)
Yes  pi/4 = 22*arctan(1/38) + 17*arctan(7/601) + 10*arctan(7/8149)
Yes  pi/4 = 44*arctan(1/57) + 7*arctan(1/239) - 12*arctan(1/682) + 24*arctan(1/12943)
Yes  pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12943)
No   pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12944)