Chapter 7 Conceptual Project:

Infinite Wisdom

In this project, we will derive a famous infinite product named after its discoverer, the English mathematician John Wallis (1616–1703). Wallis introduced the symbol $\mathrm{\infty }$ for infinity, and in turn he used ${\displaystyle \frac{1}{\infty }}$ to denote an infinitesimal quantity. He contributed to the development of infinitesimal calculus (it wasn't until the 19^th century that infinitesimals were replaced by limits in the works of Bolzano, Cauchy, and Weierstrass).

1. For a nonnegative integer n, let

   $I_n={\displaystyle {\int }_0^{{\displaystyle \frac{\pi }{2}}}{\sin }^{n}x\mathit{d}x}={\displaystyle {\int }_0^{{\displaystyle \frac{\pi }{2}}}{\cos }^{n}x\mathit{d}x}$.

   Find $I_0$, $I_1$, $I_2$, and $I_3$.

2. Show that if $n\ge 2$,

   $I_n={\displaystyle \frac{n-1}{n}}{I}_{n-2}$.

   (Hint: See Exercise 81 of Section 7.1.)

3. Use Questions 1 and 2 to find $I_4$, $I_5$, $I_6$, and $I_7$.

4. Show that in general,

   $I_{2n}={\displaystyle \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}\cdot \dots \cdot \frac{1}{2}\cdot \frac{\pi }{2}}$,

   while

   $I_{2n+1}={\displaystyle \frac{2n}{2n+1}\cdot \frac{2n-2}{2n-1}\cdot \frac{2n-4}{2n-3}\cdot \dots \cdot \frac{2}{3}}$.

   (Hint: Observe a pattern or use induction.)

5. Use Question 4 to show that

   ${\displaystyle \frac{I_{2n}}{I_{2n+1}}={\displaystyle \frac{3^2{5}^2\cdot \dots \cdot {\left(2n-1\right)}^2\left(2n+1\right)}{2^2{4}^2\cdot \dots \cdot {\left(2n\right)}^2}}\cdot {\displaystyle \frac{\pi }{2}}}$

   holds for all n.

6. Show that

   ${\displaystyle \frac{I_{2n-1}}{I_{2n+1}}=1+\frac{1}{2n}}$.

7. Prove the inequalities

   $I_{2n-1}\ge I_{2n}\ge I_{2n+1}$.

   (Hint: Use the definition of $I_n$ from Question 1 and compare the integrands.)

8. Use Questions 6 and 7 to show that

   $1\le {\displaystyle \frac{I_{2n}}{I_{2n+1}}}\le 1+{\displaystyle \frac{1}{2n}}$,

   and use this observation to prove that

   $\underset{n\to \infty }{\lim }{\displaystyle \frac{I_{2n}}{I_{2n+1}}}=1$.

9. Use your answers to the previous questions to derive Wallis' product, as follows.

   ${\displaystyle \frac{\pi }{2}}=\underset{n\to \infty }{\lim }{\displaystyle \frac{2^2{4}^2\cdot \dots \cdot {\left(2n\right)}^2}{3^2{5}^2\cdot \dots \cdot {\left(2n-1\right)}^2\left(2n+1\right)}}$