📂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

It is undeniably intriguing and useful to know that not only through series but also through products one can calculate the value of pi. The original proof is more complicated and is essentially considered a part of proving the Euler’s representation of the sinc function.

Proof

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

By substituting $\displaystyle x = {{ \pi } \over {2}}$, we get $$ {{2} \over {\pi}} = \prod_{n=1}^{\infty} \left( 1 - { {1} \over { 4 n^2} } \right) $$ Taking reciprocals on both sides yields the desired equation.

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