ArcTan Related Formulas

By Jörg Arndt <arndt@jjj.de> This document is online at http://www.jjj.de/ 7-January-1997

Warning: the stuff here is a bit cryptic.

\underline{general}

arctan (\frac{1}{x}) = arctan (\frac{1}{x + d}) + arctan (\frac{1}{x + \frac{x^{2}+1}{d}} )

arctan (\frac{1}{x}) + arctan (\frac{1}{y} ) = arctan ( \frac{1}{\frac{x y−1}{x+y}} )

arctan (\frac{1}{z}) = arctan (\frac{1}{x_{1}} ) + arctan (\frac{1}{x_{2}} ) + ... + arctan (\frac{1}{x_{n}} ) ⟺ z = \frac{Re(∏_{k=1}^{n} (x_{k} + i))}{Im(∏_{k=1}^{n} (x_{k} + i))}

⟺ (z^{2} + 1) = \begin{matrix}∏ & {n \atop k=1} & (x_{k}^{2} + 1) \\ (Im(∏_{k=1}^{n} (x_{k} + i))) & 2 \end{matrix}

arctan (\frac{1}{x}) = arctan (\frac{1}{y} ) + arctan (\frac{1}{z}) ⟺ (x^{2} + 1) = \frac{(y^{2} + 1)(z^{2} + 1)}{(y + z)^{2}} = (y − x)(z − x)

\frac{k π}{4} = arctan \frac{b_{1}}{a_{1}} + arctan \frac{b_{2}}{a_{2}} + ... + arctan \frac{b_{n}}{a_{n}} ⟺ (a_{1} + i b_{1})(a_{2} + i b_{2})...(a_{n} + i b_{n})(1 − i)^{k} ∈ IR

\frac{k π}{4} = arctan (\frac{1}{a_{1}} ) + arctan (\frac{1}{a_{2}} ) + ... + arctan (\frac{1}{a_{n}} ) ⟺ (a_{1} + i)(a_{2} + i)...(a_{n} + i)(1 − i)^{k} ∈ IR

\frac{k π}{4} = m_{1} arctan (\frac{1}{a_{1}} ) + m_{2} arctan (\frac{1}{a_{2}} ) + ... + m_{n} arctan (\frac{1}{a_{n}} )

⟺ (a_{1} + i)^{m_{1}}(a_{2} + i)^{m_{2}} … (a_{n} + i)^{m_{n}}(1 − i)^{k} ∈ IR

a > b, gcd(a, b) = 1, a α + b β = 1 ⟺ arctan (\frac{a}{b}) = arctan (\frac{−β}{α} ) + arctan (\frac{1}{a β − b α})

proof: r := a β − b α, arctan (\frac{1}{z}) = arctan (\frac{1}{a/b}) + arctan (\frac{1}{r})

⟺ z = \frac{(a/b) r − 1}{(a/b) + r} = \frac{a(α b − β a) − b}{a + b(α b − β a)} = \frac{a(α b − β a) − b(1)}{a(1) + b(α b − β a)}

= \frac{a(α b − β a) − b(α a + β b)}{a(α a + β b) + b(α b − β a)} = \frac{−a^{2} β − b^{2} β}{α a^{2} + α b^{2}} = \frac{−β}{α}

\underline{definitions}

q(n) := \frac{3n + n^{3}}{2} F_{n} := F ibonacci(n) = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...

jt(a_{0}, a_{1}, m_{1}, m_{2}, 0) := a0, jt(a_{0}, a_{1}, m_{1}, m_{2}, 1) := a1,

jt(a_{0}, a_{1}, m_{1}, m_{2}, n) := m_{1} jt(a_{0}, a_{1}, m_{1}, m_{2}, n − 1) + m_{2} jt(a_{0}, a_{1}, m_{1}, m_{2}, n − 2)

jt(a_{0}, a_{1}, m_{1}, m_{2}, n) = C_{1} x_{1}^{n} + C_{2} x_{2}^{n} where: x_{1,2}^{2} = m_{1} x_{1,2} + m_{2}, C_{1} + C_{2} = a_{0}, C_{1} x_{1} + C_{2} x_{2} = a_{1}

\underline{semi-crypt}

arctan (\frac{1}{x}) = arctan (\frac{1}{x + 1}) + arctan (\frac{1}{x^{2} + x + 1})

\frac{π}{4} = arctan (\frac{1}{1}) = ∑_{n=1}^{∞} arctan (\frac{1}{n^{2} + n + 1}) =

= arctan (\frac{1}{3}) + arctan (\frac{1}{7}) + arctan (\frac{1}{13}) + arctan (\frac{1}{21}) + arctan (\frac{1}{31}) + arctan (\frac{1}{43}) + arctan (\frac{1}{57}) + ...

arctan (\frac{1}{2x + 1}) = arctan (\frac{1}{2x + 3}) + arctan (\frac{1}{2x^{2} + 4x + 2})

\frac{π}{4} = ∑_{n=1}^{∞} arctan (\frac{1}{2n^{2}} ) =

= arctan (\frac{1}{2}) + arctan (\frac{1}{8}) + arctan (\frac{1}{18}) + arctan (\frac{1}{50}) + arctan (\frac{1}{98}) + arctan (\frac{1}{128}) + …

\frac{π}{4} = arctan (\frac{1}{2}) + arctan (\frac{1}{3}) = ∑_{n=0}^{∞} (arctan (\frac{1}{5 + 11x + 5x^{2}} ) + arctan (\frac{1}{5 + 9x + 5x^{2}} ))

= ∑_{n=0}^{∞} (arctan ( \frac{1}{\frac{(5x+2)^{2}+1}{5} + (5x + 2)}) + arctan ( \frac{1}{\frac{(5x+3)^{2}+1}{5} + (5x + 3)}))

continued fraction related ...

partial quotients P_{n}/Q_{n} = of \sqrt{1 + x^{2}} :

P_{n} = \frac{(x + \sqrt{1 + x^{2}})^{n} + (x − \sqrt{1 + x^{2}})^{n}}{2}

Q_{n} = \frac{(x + \sqrt{1 + x^{2}})^{n} − (x − \sqrt{1 + x^{2}})^{n}}{2 \sqrt{1 + x^{2}}}

arctan (\frac{1}{sinh((2k − 1)ϕ)}) = 2 arctan (\frac{1}{sinh(2kϕ)/(cosh(ϕ))}) − arctan (\frac{1}{sinh((2k + 1)ϕ)})

integer denominators if ϕ = arcsinh(x) integer x

arctan (\frac{1}{x}) = 2 ∑_{n=1}^{∞}(−1)^{n}arctan (\frac{1}{jt(0, 1, 2x, 1, 2n)}) =

= 2 (arctan (\frac{1}{2x}) − arctan (\frac{1}{4x + 8x^{3}} ) + arctan (\frac{1}{6x + 32x^{3} + 32x^{5}} ) − + …) =

= 2 ∑_{n=1}^{∞}(−1)^{n}arctan (\frac{1}{sinh(2n arcosh(x))/\sqrt{x^{2} + 1}})

\frac{π}{8} = ∑_{n=1}^{∞}(−1)^{n}arctan (\frac{1}{jt(0, 1, 2, 1, 2n)}) =

= arctan (\frac{1}{2}) − arctan (\frac{1}{12}) + arctan (\frac{1}{70}) − arctan (\frac{1}{408}) + arctan (\frac{1}{2378}) − + …

\frac{π}{8} = ∑_{n=1}^{∞} arctan (\frac{1}{jt(1, 1, 2, 1, 2n)}) =

= ∑_{n=1}^{∞}(−1)^{n} arctan (\frac{1}{(1 − \sqrt{2})^{2n} + (1 + \sqrt{2})^{2n}} ) =

= arctan (\frac{1}{3}) + arctan (\frac{1}{17}) + arctan (\frac{1}{99}) + arctan (\frac{1}{577}) + arctan (\frac{1}{3363}) + … f(n) := jt(0, 1, 2, 1, 2n) jt(0, 1, 2, 1, 2n + 1)^{2} + jt(1, 1, 2, 1, 2n + 1) eg. f(0) = 2 · 5^{2} + 7 = 57, f(1) = 12 · 29^{2} + 41 = 10133, ...

\frac{π}{8} = ∑_{n=1}^{∞} 3^{n} arctan (\frac{1}{jt(0, 1, 2, 1, 2n + 1)}) − ∑_{n=1}^{∞} 3^{n} arctan (\frac{1}{f(n)})

= arctan (\frac{1}{5}) + 3 (arctan (\frac{1}{29}) + 3 (arctan (\frac{1}{169}) + 3 (arctan (\frac{1}{5741}) + …))) −

− (arctan (\frac{1}{57}) + 3 (arctan (\frac{1}{10133}) + 3 (arctan (\frac{1}{1999509}) + …)))

arctan (\frac{1}{P_{n}} ) = 4 arctan (\frac{1}{Q_{n+2}} ) + arctan (\frac{1}{P_{n+4}} ) − 2 arctan (\frac{1}{q(Q_{n+2})})(n odd)

\frac{π}{4} = arctan (\frac{1}{P_{1}} ) =

= 4 (arctan (\frac{1}{5}) + arctan (\frac{1}{169}) + arctan (\frac{1}{5741}) + …) −

−2 (arctan (\frac{1}{q(5)}) + arctan (\frac{1}{q(169)}) + arctan (\frac{1}{q(5741)}) + …)

with f(n) := jt(0, 1, 2, 1, n + 3) jt(0, 1, 2, 1, n + 2)^{2} + jt(1, 1, 2, 1, n + 2)

arctan (\frac{1}{jt(1, 1, 2, 1, n)}) = 5 arctan (\frac{1}{jt(1, 1, 2, 1, n + 2)}) + 2 arctan (\frac{1}{jt(1, 1, 2, 1, n + 3)})−

−arctan (\frac{1}{q(jt(0, 1, 2, 1, n + 2))}) + 2 arctan (\frac{1}{f(n)}) (n odd)

\frac{π}{4} = 2 arctan (\frac{1}{3}) + arctan (\frac{1}{7}) = 2 arctan (\frac{1}{3}) + 2 arctan (\frac{1}{17}) + arctan (\frac{1}{41}) =

= (12 arctan (\frac{1}{17}) + 5 arctan (\frac{1}{41})) − 2 arctan (\frac{1}{70}) − 4 arctan (\frac{1}{307}) =

= (70 arctan (\frac{1}{99}) + 29 arctan (\frac{1}{239})) − 2 arctan (\frac{1}{70}) − 4 arctan (\frac{1}{307}) − 12 arctan (\frac{1}{12238}) − 24 arctan (\frac{1}{58911}) =

= (408 arctan (\frac{1}{577}) + 169 arctan (\frac{1}{1393}))−(2 arctan (\frac{1}{70})+4 arctan (\frac{1}{307})+12 arctan (\frac{1}{12238})+24 arctan (\frac{1}{58911})+70 arctan (\frac{1}{...})+140 arctan (\frac{1}{...})) =

\frac{π}{4} = 2(\sqrt{2} − 1) − ∑_{n=1}^{∞} (jt(0, 1, 2, 1, 2n) arctan (\frac{1}{q(jt(0, 1, 2, 1, 2n + 1))}) + 2 jt(0, 1, 2, 1, 2n) arctan (\frac{1}{f(2n + 1)}))

\frac{π}{4} = 3 ∑_{n=1}^{∞} arctan (\frac{1}{jt(1, 1, 3, 1, 2n)}) − ∑_{n=1}^{∞} arctan (\frac{1}{jt(1, 1, 3, 1, 2n)}) =

= 3 (arctan (\frac{1}{4}) + arctan (\frac{1}{43}) + arctan (\frac{1}{469}) + arctan (\frac{1}{5116}) + …) −

− (arctan (\frac{1}{q(4)}) + arctan (\frac{1}{q(43)}) + arctan (\frac{1}{q(469)}) + arctan (\frac{1}{q(5116)}) + …) =

= 3 (arctan (\frac{1}{4}) + arctan (\frac{1}{43}) + arctan (\frac{1}{469}) + arctan (\frac{1}{5116}) + …) −

− (arctan (\frac{1}{38}) + arctan (\frac{1}{39818}) + arctan (\frac{1}{51581558}) + …)

arctan (\frac{1}{x}) = 2 arctan (\frac{1}{2x}) − arctan (\frac{1}{3x + 4x^{3}} )

\frac{π}{4} = 1 − ∑_{n=0}^{∞} 2^{n} arctan (\frac{1}{3 2^{n} + 4 2^{3n}} ) = 1 − ∑_{n=0}^{∞} 2^{n} arctan (\frac{1}{q(2^{n+1})})

= 1 − (arctan (\frac{1}{7}) + 2 arctan (\frac{1}{38}) + 4 arctan (\frac{1}{268}) + 8 arctan (\frac{1}{2072}) + 16 arctan (\frac{1}{16432}) + …)

fibonaccoidal

\frac{π}{4} = ∑_{n=1}^{∞} arctan (\frac{1}{F_{2n+1}} ) =

= arctan (\frac{1}{2}) + arctan (\frac{1}{5}) + arctan (\frac{1}{13}) + arctan (\frac{1}{34}) + arctan (\frac{1}{89}) + arctan (\frac{1}{233}) + …

arctan (\frac{1}{F_{2n}} ) = 3 arctan (\frac{1}{F_{2n+2}} ) − arctan (\frac{1}{F_{2n+4}} ) − arctan (\frac{1}{q(F_{2n+2})})

⇒

\frac{π}{4} = \frac{3\sqrt{5} − 5}{2} − ∑_{n=1}^{∞} F_{2n} arctan (\frac{1}{q(F_{2n+2})})

= \frac{π}{4} = \frac{3\sqrt{5} − 5}{2} − (1 arctan (\frac{1}{18}) + 3 arctan (\frac{1}{268}) + 8 arctan (\frac{1}{4662}) + 21 arctan (\frac{1}{83270}) + …)

arctan (\frac{1}{F_{2n}} ) = ∑_{k=0}^{N} arctan (\frac{1}{F_{2n+2k+1}} ) + arctan (\frac{1}{2n + 2N + 1})

\frac{π}{4} = ∑_{n=1}^{∞} n [arctan (\frac{1}{F_{2n+2}} ) − arctan (\frac{1}{q(F_{2n+2})})] =

= ∑_{n=1}^{∞} n [ ∑_{m=1}^{∞} m [arctan (\frac{1}{F_{2n+2m+2}} ) − arctan (\frac{1}{q(F_{2n+2m+2})})] − arctan (\frac{1}{q(F_{2n+2})})] = etc.

\underline{cryptoidal}

arctan \frac{2}{x + \sqrt{x^{2} − 4}} = ∑_{n=1}^{∞} arctan \frac{1}{x jt(0, 1, x, −1, n)^{2}}

for x=2:

arctan \frac{2}{2 + \sqrt{2^{2} − 4}} = arctan 1 = \frac{π}{4} = ∑_{n=1}^{∞} arctan (\frac{1}{2 n^{2}} )

for x=4:

arctan \frac{2}{4 + \sqrt{4^{2} − 4}} = arctan (\frac{1}{2 + \sqrt{3}}) = \frac{π}{12}

= ∑_{n=1}^{∞} arctan \frac{1}{4 jt(0, 1, 4, −1, n)^{2}}

= arctan (\frac{1}{4}) + arctan (\frac{1}{64}) + arctan (\frac{1}{900}) + arctan (\frac{1}{12544}) + arctan (\frac{1}{174724}) + arctan (\frac{1}{2433600}) + ...

arctan (\frac{1}{2x^{2}} ) = 2 arctan (\frac{1}{4x^{2} − 2x + 1}) − arctan (\frac{1}{4x^{2} + 2x + 1})

⇒ \frac{π}{4} = ∑_{n=1}^{∞} [arctan (\frac{1}{4n^{2} − 2n + 1}) + arctan (\frac{1}{4n^{2} + 2n + 1})]

arctan (\frac{1}{2x^{2}} ) = 2 arctan (\frac{1}{4x^{2} + 2x + 1}) − arctan (\frac{1}{2x^{3} + x}) + arctan (\frac{1}{4x^{3} + 3x})

arctan (\frac{1}{2x^{2}} ) = 2 arctan (\frac{1}{4x^{2} − 2x + 1}) + arctan (\frac{1}{2x^{3} + x}) − arctan (\frac{1}{4x^{3} + 3x})

⇒ \frac{π}{4} = ∑_{n=1}^{∞} [arctan (\frac{1}{4n^{2} + 2n + 1}) + arctan (\frac{1}{2n^{3} + n}) − arctan (\frac{1}{4n^{3} + 3n})]

\frac{π}{6} = ∑_{n=0}^{∞} arctan (\frac{1}{2 jt(1, 3, 4, −1, n)^{2}} )

= arctan (\frac{1}{2}) + arctan (\frac{1}{18}) + arctan (\frac{1}{242}) + arctan (\frac{1}{3362}) + arctan (\frac{1}{46818}) + arctan (\frac{1}{652082}) + ...

\frac{π}{4} = ∑_{n=1}^{∞} arctan (\frac{1}{2n^{2}} ) = (by changing the order of summation)

\frac{π}{4} = ∑_{n=0}^{∞} \frac{(−1)^{n}}{(2n + 1)2^{2n+1}} ζ(4n + 2)