[size=150]The Wallis product is used to calculate the value of [math]\frac{\pi}{2}[/math] and is defined as:[br][br][math]\frac{\pi}{2}=\prod\left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right)[/math]; from n=1 to [math]\infty[/math][br][br]To find an approximation of \(\pi\), we can compute the product by taking a finite number of terms. The more terms we use, the closer the approximation will be to the actual value of [math]\pi[/math].[br][br]Let's calculate an approximation of [math]\pi[/math] using the Wallis product with 10 terms:[br][br][math]\frac{\pi}{2}=\left(\frac{2\cdot1}{2\cdot1-1}\cdot\frac{2\cdot1}{2\cdot1+1}\right)\cdot\left(\frac{2\cdot2}{2\cdot2-1}\cdot\frac{2\cdot2}{2\cdot3+1}\right)\cdot.....\cdot\left(\frac{2\cdot10}{2\cdot10-1}\cdot\frac{2\cdot10}{2\cdot10+1}\right)[/math][br][br][math]\frac{\pi}{2}=\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdot\frac{8}{7}\cdot\frac{8}{9}\cdot\frac{10}{9}\cdot\frac{10}{11}[/math][br][br][math]\frac{\pi}{2}=\frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdot8\cdot8\cdot10\cdot10}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot7\cdot9\cdot9\cdot11}[/math][math]\frac{\pi}{2}=\frac{320}{33}[/math][br][br][math]\frac{\pi}{2}=\frac{2^5\cdot3^2\cdot4^2\cdot5^2}{11\cdot9\cdot7\cdot5\cdot3}[/math][br][br][math]\frac{\pi}{2}=\frac{2^5\cdot5^2}{11\cdot3}[/math][br][br][math]\frac{\pi}{2}=\frac{320}{33}[/math][br][br]Finally, to find an approximation of [math]\pi[/math], we multiply the above result by 2:[br][br][math]\pi[/math]=[math]2\cdot\frac{320}{33}\approx6.096[/math][br][br]As we used only 10 terms in the Wallis product, this is a relatively rough approximation. The more terms we include in the product, the more accurate the approximation of [math]\pi[/math] will be.[/size]