Pi as Sum of Odd Reciprocals Alternating in Sign in Pairs

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Theorem

$\dfrac {\pi \sqrt 2} 4 = 1 + \dfrac 1 3 - \dfrac 1 5 - \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} - \dfrac 1 {13} - \dfrac 1 {15} \cdots$


Proof

Let $f: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map f x = \dfrac {x^1} 1 + \dfrac {x^3} 3 - \dfrac {x^5} 5 - \dfrac {x^7} 7 + \dfrac {x^9} 9 + \dfrac {x^{11} } {11} - \dfrac {x^{13} } {13} - \dfrac {x^{15} } {15} \cdots$

We have:

\(\displaystyle \map f x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\frac {x^{4 n + 1} } {4 n + 1} + \frac {x^{4 n + 3} } {4 n + 3} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {f'} x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {x^{4 n} + x^{4 n + 2} }\) Power Rule for Derivatives
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {1 + x^2} \paren {x^{4 n} }\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {1 + x ^ 2} \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {x^{4 n} }\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {1 + x ^ 2} {1 + x ^ 4}\) Sum of Infinite Geometric Progression
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map f 1\) \(=\) \(\displaystyle \int_0^1 \dfrac {1 + x^2} {1 + x^4} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \intlimits {\dfrac 1 {\sqrt 2} \map \arctan {\dfrac 1 {\sqrt 2} \paren {x - \dfrac 1 x} } } 0 1\) Primitive of One plus x Squared over One plus Fourth Power of x
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {\sqrt 2} \map \arctan {\dfrac 1 {\sqrt 2} \paren {1 - \dfrac 1 1} } - \dfrac 1 { \sqrt 2} \map \arctan {\dfrac 1 {\sqrt 2} \paren {0 - \dfrac 1 0} }\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {\sqrt 2} \paren {\map \arctan 0 - \map \arctan {-\infty} }\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {\sqrt 2} \paren {0 - \paren {-\dfrac \pi 2} }\) Definition of Real Arctangent
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\pi \sqrt 2} 4\)

$\blacksquare$


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