📂Functions

Wallis Product

Theorem

$$ \prod_{n=1}^{\infty} {{4n^2} \over {4n^2 - 1}} = \lim_{n \to \infty} {{2 \cdot 2 } \over { 1 \cdot 3 } } \cdot {{4 \cdot 4 } \over { 3 \cdot 5 } } \cdot \cdots \cdot {{2n \cdot 2n } \over { (2n-1) \cdot (2n+1) } } = {{ \pi } \over {2}} $$

Explanation

Needless to say, the fact that pi can be obtained not only through a series but also through a product is fascinating and a useful fact. The original proof is more difficult than this, and can in fact be regarded as being contained within the process of proving the Euler representation of the sinc function.

Proof

Euler representation of the sinc function: $${{\sin x} \over {x}} = \prod_{n=1}^{\infty} \left( 1 - {{x^2} \over { \pi^2 n^2}} \right)$$

Substituting $\displaystyle x = {{ \pi } \over {2}}$ gives $$ {{2} \over {\pi}} = \prod_{n=1}^{\infty} \left( 1 - { {1} \over { 4 n^2} } \right) $$ Taking the reciprocal of both sides yields the desired equation.

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