Try the fastest way to create flashcards
Question

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)

or

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?

Solution

Verified
Answered 10 months ago
Answered 10 months ago
Step 1
1 of 7

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.

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Thomas' Calculus 14th Edition by Christopher E Heil, Joel R. Hass, Maurice D. Weir

Thomas' Calculus

14th EditionISBN: 9780134438986 (11 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir
10,142 solutions
Calculus: Early Transcendentals 8th Edition by James Stewart

Calculus: Early Transcendentals

8th EditionISBN: 9781285741550 (4 more)James Stewart
11,085 solutions
Calculus AP Edition 11th Edition by Bruce H. Edwards, Ron Larson

Calculus AP Edition

11th EditionISBN: 9781337286886 (1 more)Bruce H. Edwards, Ron Larson
12,440 solutions
Calculus: Early Transcendentals 9th Edition by Daniel K. Clegg, James Stewart, Saleem Watson

Calculus: Early Transcendentals

9th EditionISBN: 9781337613927 (1 more)Daniel K. Clegg, James Stewart, Saleem Watson
11,049 solutions

More related questions

1/4

1/7