# Proving Factorial Equations involving Double Factorials

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In many problems in mathematical physics, particularly in connection with Legendre polynomials (Chapter 12), we encounter products of the odd positive integers and products of the even positive integers. For convenience, these are given special labels as double factorials:

1*3*5***(2n + 1) = (2n + 1)!!

2*4*6***(2n) = (2n)!!

Show that these are related to the regular factorial functions by

(2n)!! = (2^n)n! and (2n + 1)!! = (2n + 1)!/[(2^n)(n!)].

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see attachment for complete solution.

For example, to see that (2n)!! = (2^n)n!, note that we can factor out a two from every factor in (2n)!!:

(2n)!! = 2*4*6***(2n)

= (2*1)*(2*2)*(2*3)***(2*n)

= (2*2*2***2)(1*2*3***n)

= (2^n)n!.

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