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The formula for π\pi as an infinite product was derived by English mathematician John Wallis in 1655. This product, called the Wallis Product, appeared in his book Arithmetica Infinitorum.

π2=(2213)(4435)(6657)((2n)(2n)(2n1)(2n+1))\frac{\pi}{2}=\left(\frac{2 \cdot 2}{1 \cdot 3}\right)\left(\frac{4 \cdot 4}{3 \cdot 5}\right)\left(\frac{6 \cdot 6}{5 \cdot 7}\right) \cdots\left(\frac{(2 n) \cdot(2 n)}{(2 n-1) \cdot(2 n+1)}\right) \cdots

In 2015, physicists Carl Hagen and Tamar Friedmann (also a mathematician) stumbled upon a connection between quantum mechanics and the Wallis Product when they applied the variational principle to higher energy states of the hydrogen atom. This principle was previously used only on the ground energy state. The Wallis Product appeared naturally in the midst of their calculations involving gamma functions.

Quantum mechanics is the study of matter and light on the atomic and subatomic scale.

Consider Wallis's method of finding a formula for π\pi. Let

I(n)=0π/2sinnxdxI(n)=\int_0^{\pi / 2} \sin ^n x d x

From Wallis's Formulas, I(n)=(12)(34)(56)(n1n)(π2),nI(n)=\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\left(\frac{5}{6}\right) \cdots\left(\frac{n-1}{n}\right)\left(\frac{\pi}{2}\right), n is even (n2)(n \geq 2)


I(n)=(23)(45)(67)(n1n),nI(n)=\left(\frac{2}{3}\right)\left(\frac{4}{5}\right)\left(\frac{6}{7}\right) \cdots\left(\frac{n-1}{n}\right), n is odd (n3)(n \geq 3)

Find I(n)I(n) for n=2,3,4n=2,3,4, and 5 . What do you observe?


Answered 10 months ago
Answered 10 months ago
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In this exercise, we have to use Wallis's formula for I(n)I(n) where

I(n)=0π2sinnx dxI(n)=\int_{0}^{\frac{\pi}{2}} \sin^n x \ dx

and calculate I(n)I(n) for n=2,3,4,5n=2,3,4,5.

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