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.



 
using AbstractAlgebra # implements arbitrary precision rationals
 
tanplus(x,y) = (x + y) / (1 - x * y)
 
function taneval(coef, frac)
    if coef == 0
        return 0
    elseif coef < 0
        return -taneval(-coef, frac)
    elseif isodd(coef)
        return tanplus(frac, taneval(coef - 1, frac))
    else
        x = taneval(div(coef, 2), frac)
        return tanplus(x, x)
    end
end
 
taneval(tup::Tuple) = taneval(tup[1], tup[2])
 
tans(v::Vector{Tuple{BigInt, Rational{BigInt}}}) = foldl(tanplus, map(taneval, v), init=0)
 
const testmats = Dict{Vector{Tuple{BigInt, Rational{BigInt}}}, Bool}([
    ([(1, 1//2), (1, 1//3)], true), ([(2, 1//3), (1, 1//7)], true),
    ([(12, 1//18), (8, 1//57), (-5, 1//239)], true),
    ([(88, 1//172), (51, 1//239), (32, 1//682), (44, 1//5357), (68, 1//12943)], true),
    ([(88, 1//172), (51, 1//239), (32, 1//682), (44, 1//5357), (68, 1//12944)], false)])
 
 
function runtestmats()
    println("Testing matrices:")
    for (k, m) in testmats
        ans = tans(k)
        println((ans == 1) == m ? "Verified as $m: " : "Not Verified as $m: ", "tan $k = $ans")
    end
end
 
runtestmats()
 