Brouncker's Formula

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Theorem

$\dfrac 4 \pi = 1 + \cfrac {1^2} {2 + \cfrac {3^2} {2 + \cfrac {5^2} {2 + \cfrac {7^2} {2 + \cfrac {9^2} {2 + \cdots} } } } }$


Proof

From Continued Fraction for Real Arctangent Function, we have:

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

From Arctangent of One, we have:

$\map \arctan 1 = \dfrac \pi 4$

Therefore:

\(\ds \frac \pi 4\) \(=\) \(\ds \map \arctan 1\)
\(\ds \) \(=\) \(\ds \cfrac 1 {1 + \cfrac 1 {3 - 1 + \cfrac {3^2} {5 - 3 + \cfrac {5^2} {7 - 5 + \cfrac {7^2} {\ddots } } } } }\)
\(\ds \) \(=\) \(\ds \cfrac 1 {1 + \cfrac 1 {2+ \cfrac {3^2} {2 + \cfrac {5^2} {2 + \cfrac {7^2} {\ddots } } } } }\)

Taking the reciprocal of both sides, we obtain:

$\dfrac 4 \pi = 1 + \cfrac {1^2} {2 + \cfrac {3^2} {2 + \cfrac {5^2} {2 + \cfrac {7^2} {2 + \cfrac {9^2} {2 + \cdots} } } } }$


$\blacksquare$

Also presented as

Brouncker's Formula can also be presented as:

$\dfrac \pi 4 = \dfrac 1 {1 + \cfrac {1^2} {2 + \cfrac {3^2} {2 + \cfrac {5^2} {2 + \cfrac {7^2} {2 + \cfrac {9^2} {2 + \cdots} } } } } }$


Source of Name

This entry was named for William Brouncker.


Sources

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