Check Machin-like formulas: Difference between revisions
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[[wp:Machin-like_formula|Machin-like formulas]] are useful for efficiently computing numerical approximations to Pi. |
[[wp:Machin-like_formula|Machin-like formulas]] are useful for efficiently computing numerical approximations to Pi. |
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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: |
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: |
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Revision as of 18:57, 9 December 2012
You are encouraged to solve this task according to the task description, using any language you may know.
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:
and confirm that the following formula is incorrect by showing tan(right hand side) is not 1:
These identities are useful in calculating the values:
You can store the equations in any convenient data structure, but for extra credit parse them from human-readable text input.
Go
<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 extended precision; thereafter J will maintain extended precision throughout its calculations, as long as it can.
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
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.
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