Machin-like formulas   are useful for efficiently computing numerical approximations for 



π


{\displaystyle \pi }





Task

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:

 





π
4


=
arctan
⁡


1
2


+
arctan
⁡


1
3




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

 
 





π
4


=
2
arctan
⁡


1
3


+
arctan
⁡


1
7




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


 





π
4


=
4
arctan
⁡


1
5


−
arctan
⁡


1
239




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


 





π
4


=
5
arctan
⁡


1
7


+
2
arctan
⁡


3
79




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


 





π
4


=
5
arctan
⁡


29
278


+
7
arctan
⁡


3
79




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


 





π
4


=
arctan
⁡


1
2


+
arctan
⁡


1
5


+
arctan
⁡


1
8




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

 
 





π
4


=
4
arctan
⁡


1
5


−
arctan
⁡


1
70


+
arctan
⁡


1
99




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

 
 





π
4


=
5
arctan
⁡


1
7


+
4
arctan
⁡


1
53


+
2
arctan
⁡


1
4443




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


 





π
4


=
6
arctan
⁡


1
8


+
2
arctan
⁡


1
57


+
arctan
⁡


1
239




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


 





π
4


=
8
arctan
⁡


1
10


−
arctan
⁡


1
239


−
4
arctan
⁡


1
515




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


 





π
4


=
12
arctan
⁡


1
18


+
8
arctan
⁡


1
57


−
5
arctan
⁡


1
239




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


 





π
4


=
16
arctan
⁡


1
21


+
3
arctan
⁡


1
239


+
4
arctan
⁡


3
1042




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


 





π
4


=
22
arctan
⁡


1
28


+
2
arctan
⁡


1
443


−
5
arctan
⁡


1
1393


−
10
arctan
⁡


1
11018




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


 





π
4


=
22
arctan
⁡


1
38


+
17
arctan
⁡


7
601


+
10
arctan
⁡


7
8149




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


 





π
4


=
44
arctan
⁡


1
57


+
7
arctan
⁡


1
239


−
12
arctan
⁡


1
682


+
24
arctan
⁡


1
12943




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


 





π
4


=
88
arctan
⁡


1
172


+
51
arctan
⁡


1
239


+
32
arctan
⁡


1
682


+
44
arctan
⁡


1
5357


+
68
arctan
⁡


1
12943




{\displaystyle {\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:

 





π
4


=
88
arctan
⁡


1
172


+
51
arctan
⁡


1
239


+
32
arctan
⁡


1
682


+
44
arctan
⁡


1
5357


+
68
arctan
⁡


1
12944




{\displaystyle {\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
)


1
−
tan
⁡
(
a
)
tan
⁡
(
b
)





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


 



tan
⁡

(
arctan
⁡


a
b


)

=


a
b




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


 



tan
⁡
(
−
a
)
=
−
tan
⁡
(
a
)


{\displaystyle \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.

Note: to formally prove the formula correct, it would have to be shown that 






−
3
p
i

4




{\displaystyle {-3pi \over 4}}

 < right hand side < 






5
p
i

4




{\displaystyle {5pi \over 4}}

 due to 



tan
⁡
(
)


{\displaystyle \tan()}

 periodicity.



import "/big" for BigRat
import "/fmt" for Fmt
 
/** represents a term of the form: c * atan(n / d) */
class Term {
    construct new(c, n, d) {
        _c = c
        _n = n
        _d = d
    }
 
    c { _c }
    n { _n }
    d { _d }
 
    toString {
        var a = "atan(%(n)/%(d))"
        return ((_c ==  1) ? " + " :
                (_c == -1) ? " - " :
                (_c <   0) ? " - %(-c)*" : " + %(c)*") + a
    }
}
 
var tanEval // recursive function
tanEval = Fn.new { |c, f|
    if (c == 1)  return f
    if (c < 0) return -tanEval.call(-c, f)
    var ca = (c/2).truncate
    var cb = c - ca
    var a = tanEval.call(ca, f)
    var b = tanEval.call(cb, f)
    return (a + b) / (BigRat.one - (a * b))
}
 
var tanSum // recursive function
tanSum = Fn.new { |terms|
    if (terms.count == 1) return tanEval.call(terms[0].c, BigRat.new(terms[0].n, terms[0].d))
    var half = (terms.count/2).floor
    var a = tanSum.call(terms.take(half).toList)
    var b = tanSum.call(terms.skip(half).toList)
    return (a + b) / (BigRat.one - (a * b))
}
 
var T = Term // type alias
 
var termsList = [
    [T.new(1, 1, 2), T.new(1, 1, 3)],
    [T.new(2, 1, 3), T.new(1, 1, 7)],
    [T.new(4, 1, 5), T.new(-1, 1, 239)],
    [T.new(5, 1, 7), T.new(2, 3, 79)],
    [T.new(5, 29, 278), T.new(7, 3, 79)],
    [T.new(1, 1, 2), T.new(1, 1, 5), T.new(1, 1, 8)],
    [T.new(4, 1, 5), T.new(-1, 1, 70), T.new(1, 1, 99)],
    [T.new(5, 1, 7), T.new(4, 1, 53), T.new(2, 1, 4443)],
    [T.new(6, 1, 8), T.new(2, 1, 57), T.new(1, 1, 239)],
    [T.new(8, 1, 10), T.new(-1, 1, 239), T.new(-4, 1, 515)],
    [T.new(12, 1, 18), T.new(8, 1, 57), T.new(-5, 1, 239)],
    [T.new(16, 1, 21), T.new(3, 1, 239), T.new(4, 3, 1042)],
    [T.new(22, 1, 28), T.new(2, 1, 443), T.new(-5, 1, 1393), T.new(-10, 1, 11018)],
    [T.new(22, 1, 38), T.new(17, 7, 601), T.new(10, 7, 8149)],
    [T.new(44, 1, 57), T.new(7, 1, 239), T.new(-12, 1, 682), T.new(24, 1, 12943)],
    [T.new(88, 1, 172), T.new(51, 1, 239), T.new(32, 1, 682), T.new(44, 1, 5357), T.new(68, 1, 12943)],
    [T.new(88, 1, 172), T.new(51, 1, 239), T.new(32, 1, 682), T.new(44, 1, 5357), T.new(68, 1, 12944)]
]
 
for (terms in termsList) {
    var f = Fmt.swrite("$-5s: 1 == tan(", tanSum.call(terms) == BigRat.one)
    System.write(f)
    System.write(terms[0].toString.skip(3).join())
    for (i in 1...terms.count) System.write(terms[i])
    System.print(")")
}