Continued Fraction for Real Arctangent Function

Theorem

Let $-1 \le x \le 1$

Then:

$\arctan x = \cfrac x {1 + \cfrac {x^2} {3 - x^2 + \cfrac {\paren {3 x}^2} {5 - 3 x^2 + \cfrac {\paren {5 x}^2} {7 - 5 x^2 + \cfrac {\paren {7 x}^2} {\ddots } } } } }$

Proof

$\ds \arctan x$

$=$

$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {2 n + 1}$

Power Series Expansion for Real Arctangent Function for $-1 \le x \le 1$

$\ds $

$=$

$\ds x - \dfrac {x^3} 3 + \dfrac {x^5} 5 - \dfrac {x^7} 7 + \cdots$

$\ds $

$=$

$\ds x + x \paren {\dfrac {-x^2} 3} + x \paren {\dfrac {-x^2} 3} \paren {\dfrac {-3 x^2} 5} + x \paren {\dfrac {-x^2} 3} \paren {\dfrac {-3 x^2} 5} \paren {\dfrac {-5 x^2} 7} + \cdots$

$\ds $

$=$

$\ds a_0 + a_0 a_1 + a_0 a_1 a_2 + a_0 a_1 a_2 a_3 + \cdots$

From Euler's Continued Fraction Formula, we have:

$a_0 + a_0 a_1 + a_0 a_1 a_2 + a_0 a_1 a_2 a_3 + \cdots = \cfrac {a_0} {1 - \cfrac {a_1} {1 + a_1 - \cfrac {a_2} {1 + a_2 - \cfrac {a_3} {1 + a_3 - \cfrac {\ddots} {\ddots } } } } }$

Therefore:

$\ds x + x \paren {\dfrac {-x^2} 3} + x \paren {\dfrac {-x^2} 3} \paren {\dfrac {-3 x^2} 5} + x \paren {\dfrac {-x^2} 3} \paren {\dfrac {-3 x^2} 5} \paren {\dfrac {-5 x^2} 7} + \cdots$

$=$

$\ds \cfrac x {1 - \cfrac {\paren {\dfrac {-x^2} 3} } {1 + \paren {\dfrac {-x^2} 3} - \cfrac {\paren {\dfrac {-3 x^2} 5} } {1 + \paren {\dfrac {-3 x^2} 5} - \cfrac {\paren {\dfrac {-5 x^2} 7} } {1 + \paren {\dfrac {-5 x^2} 7} - \cfrac {\ddots} {\ddots } } } } }$

$\ds $

$=$

$\ds \cfrac x {1 - \cfrac {\paren {\dfrac {-x^2} 3} \times 3 } {3 \times \paren {1 + \paren {\dfrac {-x^2} 3} } - \cfrac {\paren {\dfrac {-3 x^2} 5} \times 3 \times 5 } {5 \times \paren {1 + \paren {\dfrac {-3 x^2} 5} } - \cfrac {\paren {\dfrac {-5 x^2} 7} \times 5 \times 7 } {7 \times \paren {1 + \paren {\dfrac {-5 x^2} 7} } - \cfrac {\ddots} {\ddots } } } } }$

multiplying by $1$

$\ds $

$=$

$\ds \cfrac x {1 + \cfrac {x^2} {3 - x^2 + \cfrac {\paren {3 x}^2} {5 - 3 x^2 + \cfrac {\paren {5 x}^2} {7 - 5 x^2 + \cfrac {\paren {7 x}^2} {\ddots } } } } }$

$\blacksquare$

Also see

Sources

1750: Leonhard Paul Euler: De fractionibus continuis Observationes: Pages $\text {32 - 81}$