## Related questions with answers

The formula for $\pi$ 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.

$\frac{\pi}{2}=\left(\frac{2 \cdot 2}{1 \cdot 3}\right)\left(\frac{4 \cdot 4}{3 \cdot 5}\right)\left(\frac{6 \cdot 6}{5 \cdot 7}\right) \cdots\left(\frac{(2 n) \cdot(2 n)}{(2 n-1) \cdot(2 n+1)}\right) \cdots$

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 $\pi$. Let

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

From Wallis's Formulas, $I(n)=\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\left(\frac{5}{6}\right) \cdots\left(\frac{n-1}{n}\right)\left(\frac{\pi}{2}\right), n$ is even $(n \geq 2)$

or

$I(n)=\left(\frac{2}{3}\right)\left(\frac{4}{5}\right)\left(\frac{6}{7}\right) \cdots\left(\frac{n-1}{n}\right), n$ is odd $(n \geq 3)$

Show that $I(n+1) \leq I(n)$ for $n \geq 2$.

Solution

VerifiedIn this exercise, we have to prove that for $n\geq 2$,

$I(n)\geq I(n+1)$ where

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

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Thomas' Calculus

14th Edition•ISBN: 9780134438986 (11 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir#### Calculus: Early Transcendentals

9th Edition•ISBN: 9781337613927 (3 more)Daniel K. Clegg, James Stewart, Saleem Watson## More related questions

1/4

1/7