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Exercises for Wallis's Product 
Last revised: September 16, 2004
Please hand these in, on this sheet of paper, at the class after the last minilecture on the topic is presented.
1.The modern way to approach the "sum of powers of integers" problem that Wallis solved is to use proof by induction.  Here is the easiest case: exponent 1
S(n) = 1 + 2 + 3 + ... + n = n(n+1)/2
Proof: it works for n = 1, since S(1) = 1.
Assume the formula is correct for some positive n, so that S(n) = n(n+1)/2.
Then S(n+1) = S(n) + n+1 = 
This means, by induction, that the formula is true for any positive n.

a. For exponent 2, 12 + 22 + 32 + ... + n2 = 
Check this for n = 1, 2, and 3, then prove it by induction.

2. Start with Wallis's integral  and make the substitution x = sin u to get
. Integrate twice by parts to obtain Wallis's recursion formula